Mechanics - User contributions [en]

E1. Work-Energy Theorem: differential form

2025-05-21T08:01:43Z

Bros:

<div class="noautonum">__TOC__</div>
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</math>

The '''Work-Energy Theorem''' plays a prominent role in physics, as it bridges all branches of science (electromagnetism, thermodynamics, etc.): all of them deal with energy.

There is a certain parallel between the Vector Theorems and the '''Work-Energy Theorem''': they all allow us to understand the evolution of a magnitude that depends on the mechanical state of the system. In the case of the former, we are dealing with the instantaneous evolution of vector magnitudes (linear momentum, angular momentum), while the latter deals with the finite change (in a finite time interval) of a scalar magnitude ('''mechanical energy''').

This unit presents the differential form of this theorem (the '''power balance'''). The integrated form (which is the Work-Energy Theorem itself) is presented in [[E2. Teorema de la energía: versión integrada|'''unit E2''']].
 
------
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==E1.1 Power balance in a system of particles==

:[[File:E1-1-eng.png|thumb|right|300px|link=]]
<center>'''Figure E1.1''' Forces on a particle of a system with constant matter</center>

Let us consider a system of constant matter, and let’s apply Newton's second law to each particle (or each mass differential) of this system in a general reference frame R (which may be non-Galilean) ('''Figure E1.1'''):

<math>\F{\rightarrow \Ps} + \Fcal{}{\text{ar}\rightarrow\Ps} + \Fcal{}{\text{Cor}\rightarrow\Ps} =\ms_\Ps\acc{P}{R}</math>

<math>\F{\rightarrow \Ps} </math> is the resultant interaction force on '''P''', and it comes from internal interactions (with particles of the same system) and external ones:

<math>\F{\rightarrow \Ps}=\F{\rightarrow \Ps}^\text{int}+\F{\rightarrow \Ps}^\text{ext}</math>

If both sides of the equation are multiplied by <math>\vel{P}{R}</math>, all forces (interaction and inertia forces) orthogonal to <math>\vel{P}{R}</math> are filtered out. Specifically, since <math>\acc{P}{R}=\accs{P}{R} + \accn{P}{R} </math>, where <math>\accn{P}{R} \perp \vel{P}{R}</math>, and since the Coriolis force is always orthogonal to <math>\vel{P}{R} (\Fcal{}{\text{Cor}\rightarrow\Ps} = -2\ms_\Ps \velang{R}{Gal} \times \vel{P}{R})</math>:

<math>\vel{P}{R} \cdot \left( \F{\rightarrow \Ps}^\text{int}+\F{\rightarrow \Ps}^\text{ext} + \Fcal{}{\text{ar}\rightarrow\Ps} \right) = \ms_\Ps\accs{P}{R} \cdot \vel{P}{R} = \frac{\ds}{\ds \ts} \left(\frac{1}{2}\ms_\Ps \vvec_\Rs^2(\Ps) \right) \equiv \frac{\ds\Ts_\Rs(\Ps)}{\ds\ts}=\dot{\Ts}_\Rs(\Ps)</math>

The scalar function <math>\Ts_\Rs(\Ps)</math> is the '''kinetic energy''' of the particle, and is a function of the [[D1. Foundational laws of Newtonian dynamics#D1.3 Newton’s Principle of Determinacy|'''mechanical state''']] of '''P'''. The scalar product of a force and the velocity of its application point is the '''force power''': <math>\vel{P}{R} \cdot \F{\rightarrow \Ps} \equiv \dot{\Ws}_\Rs^{\F{\rightarrow \Ps}} </math>. Both scalar quantities <math>(\Ts_\Rs(\Ps), \dot{\Ws}_\Rs^{\F{\rightarrow \Ps}})</math> depend on the reference frame. For the same set of interaction forces acting on '''P''', the reference frame determines which are filtered out and which are not.

The equation can be rewritten as <math>\dot{\Ws}_\Rs^{\text{ext}\rightarrow \Ps}+\dot{\Ws}_\Rs^{\text{int}\rightarrow \Ps}+\dot{\Ws}_\Rs^{\text{ar}\rightarrow \Ps} =\dot{\Ts}_\Rs(\Ps)</math>, and constitutes the '''power balance''' for particle '''P''' in the R reference frame. Summing for all particles in the system:

<math>\dot{\Ws}_\Rs^{\text{ext}}+\dot{\Ws}^{\text{int}}+\dot{\Ws}_\Rs^{\text{ar}} =\dot{\Ts}_\Rs^\text{syst}</math>

The calculation of kinetic energy is covered in [[E2. Teorema de la energía: versión integrada|'''unit E2''']]. This unit focuses on the calculation of powers. [[E1. Work-Energy Theorem: differential form#E1.2 Potencia de una pareja de acción-reacción|'''section E1-2''']] shows that the power of internal interactions does not depend on the reference frame (which is why the subscript R does not appear in the <math>\dot{\Ws}^{\text{int}}</math> term of the previous equation).

====✏️ EXAMPLE E1-1.1: free particle====
------
:
{|
:[[File:ExE1-1-1-1-eng.jpg|thumb|left|180px|link=]]
|A particle '''P''', with mass m, is released from the ring of a Ferris wheel when it is at its highest point. We want to calculate the power, at this instant, of all forces acting on the particle from three different reference frames: the ground, the ring, and the Ferris wheel cabin.
|}

:Ground reference frame: Since this is a Galilean reference frame, all forces on '''P''' are interaction forces.
{|
:[[File:ExE1-1-2- eng.png|thumb|right|230px|link=]]

|<math>\hspace{0.5cm}\sum\dot{\Ws}_\Ts=\dot{\Ws}_\Ts^\text{weight}= \left(\leftarrow \Rs\Omega_0\right) \cdot \left(\downarrow \ms\gs\right)=0</math>
|}

:Ring reference frame: It is a non-Galilean reference, and the transportation force on '''P''' must be taken into account.
{|
:[[File:ExE1-1-3-eng.png|thumb|right|250px|link=]]
|<math>\hspace{0.5cm}</math> AB:ground, REL:ring 

<math>\hspace{0.5cm}\vel{P}{REL}=\vel{P}{AB}-\vel{P}{ar}= \left(\leftarrow \Rs\Omega_0\right)-\left(\leftarrow \Rs\Omega_0\right)=\overline{0}</math> 

<math>\hspace{0.5cm}</math>

Since the velocity is zero, the total power is zero.
|}

:Cabin reference frame: It is also a non-Galilean reference, and the transportation force on '''P''' must be taken into account.

{|
:[[File:ExE1-1-4-eng.png|thumb|right|250px|link=]]
|<math>\hspace{0.5cm}</math>AB:ground, REL:cabin

<math>\hspace{0.5cm}\vel{P}{REL}=\vel{P}{AB}-\vel{P}{ar}= \left(\leftarrow \Rs\Omega_0\right)-\left(\uparrow \Rs\Omega_0\right)</math>

<math>\hspace{0.5cm}\dot{\Ws}_\text{cabin}^\text{weight}=\left[ \left(\leftarrow \Rs\Omega_0\right)+\left(\downarrow \Rs\Omega_0\right) \right] \cdot \left(\downarrow\ms\gs\right) =\ms\gs\Rs\Omega_0</math>

<math>\hspace{0.5cm}\dot{\Ws}_\text{cabin}^\text{ar}=\left[ \left(\leftarrow \Rs\Omega_0\right)+\left(\downarrow \Rs\Omega_0\right) \right] \cdot \left(\rightarrow\ms\Rs^2\Omega_0^2\right) =-\ms\gs\Rs^2 \Omega_0^3</math>

<math>\hspace{0.5cm}\sum\dot{\Ws}_\text{cabin}=\ms\Rs\Omega_0(\gs-\Rs\Omega_0^2)</math>

|}


 
====✏️ EXAMPLE E1-1.2: pendulum====
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:
{|
:[[File:ExE1-2-1-eng.png|thumb|left|180px|link=]]
|A particle '''P''' of mass m hangs from an inextensible thread of length L attached to the ceiling. The assembly oscillates in a vertical plane fixed to the ground. We want to calculate the power of all the forces acting on the particle in two different reference frames: the ground and a platform rotating relative to the ground with a constant angular velocity <math>\Omega_0</math>. 
|}
 

 

Ground reference frame: Since it is a Galilean reference frame, '''P''' is only subjected to interaction forces.
{|
:[[File:ExE1-2-2-neut.png|thumb|left|300px|link=]]

|<math>\dot{\Ws}_\Es^\Ts=(\nearrow\Ls\dot{\theta})\cdot(\nwarrow\Ts)=0</math>

<math>\dot{\Ws}_\Es^\text{weight}=(\nearrow\Ls\dot{\theta}\cdot(\downarrow \ms\gs)=-\ms\gs\Ls\dot{\theta}\sin\theta</math>
|}

Platform reference frame: It is a non Galilean reference frame, and the transportation force on '''P''' has to be taken into account.

{|
:[[File:ExE1-2-3-eng.png|thumb|left|400px|link=]]

|AB:ground, REL:platform

<math>\vel{P}{REL}=\vel{P}{AB}-\vel{P}{ar}=(\nearrow\Ls\dot{\theta})-(\otimes \Ls \Omega_0 \sin \theta)</math>

<math>\dot{\Ws}_\Es^\Ts\left[(\nearrow\Ls\dot{\theta})+(\odot\Ls\Omega_0\sin\theta)\right]\cdot(\nwarrow\Ts)=0</math>

<math>\dot{\Ws}_\Es^\text{weight}\left[(\nearrow\Ls\dot{\theta})+(\odot\Ls\Omega_0\sin\theta)\right]\cdot(\downarrow \ms\gs)=-\ms\gs\Ls\dot{\theta}\sin\theta</math>

<math>\dot{\Ws}_\Es^\text{ar}\left[(\nearrow\Ls\dot{\theta})+(\odot\Ls\Omega_0\sin\theta)\right]\cdot(\rightarrow \ms\Ls\Omega_0^2\sin\theta)=\ms\Ls^2\Omega_0^2\dot{\theta}\cos\theta</math>
|}



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==E1.2 Power of an action-reaction pair==

The total power associated with internal interactions in a constant-mass system is not zero in principle (since the two forces of each action-reaction pair are applied to particles that may have different velocities relative to the reference frame in which the calculation is made), and it is independent of the reference frame in which it is evaluated. It only depends on the separation velocity of the two particles. The power balance for the system is therefore written as: <math>\dot{\Ts}_\Rs^\text{syst}=\dot{\Ws}_\Rs^\text{ext} + \dot{\Ws}^\text{int} + \dot{\Ws}_\Rs^\text{ar} </math> .

<div>
=====💭 PROOF ➕=====

{|
:[[File:E1-2-neut.png|thumb|left|200px|link=]]
|<math>\dot{\Ws}^\text{ac-reac}=\overline{\Fs}_{\Ps \rightarrow \Qs}\cdot \vel{Q}{R} + \overline{\Fs}_{\Qs \rightarrow \Ps}\cdot \vel{P}{R}= \overline{\Fs}_{\Ps \rightarrow \Qs}\cdot \left[ \vel{Q}{R}-\vel{P}{R}\right]</math>

By the [[D1. Foundational laws of Newtonian dynamics#D1.6 Newton’s third law (action-reaction principle)|'''action-reaction principle''']], <math>\overline{\Fs}_{\Ps \rightarrow \Qs}=-\overline{\Fs}_{\Qs \rightarrow \Ps}</math>. Therefore: <math>\dot{\Ws}^\text{ac-reac}=\overline{\Fs}_{\Ps \rightarrow \Qs}\cdot \left[ \vel{Q}{R}-\vel{P}{R}\right]</math>

As <math>\overline{\Fs}_{\Ps \rightarrow \Qs}</math> has the <math>\PQvec</math> direction, only the components of the velocities in that direction have to be considered:

<math>\dot{\Ws}^\text{ac-reac}=\overline{\Fs}_{\Ps \rightarrow \Qs}\cdot \biggl( \vel{Q}{R} \Bigr]_{||\Ps\Qs}-\ \vel{P}{R} \Bigr]_{||\Ps\Qs}\biggl)</math>
|}

The difference <math>\biggl( \vel{Q}{R} \Bigr]_{\Ps\Qs}-\ \vel{P}{R} \Bigr]_{\Ps\Qs}\biggl)</math> is the separation velocity <math>\dot{\rho}</math> between '''P''' and '''Q''', and it does not dpend on the reference frame since it is the derivative of a scalar (the distance <math>\dot{\rho}</math> between the two particles). For the previous drawing, <math>\dot{\rho}> 0</math>. Since this is an attractive interaction, <math>\dot{\Ws}^\text{ac-reac} < 0</math>.

With a drawing, it is easy to see that:
* attractive interaction and particle approach <math>(\dot{\rho}<0)</math> : <math>\dot{\Ws}^\text{ac-reac} > 0</math> ;
* repulsive interaction and particle separation <math>(\dot{\rho}>0)</math> : <math>\dot{\Ws}^\text{ac-reac} > 0</math>
* repulsive interaction and particle approach <math>(\dot{\rho}<0)</math> : <math>\dot{\Ws}^\text{ac-reac} < 0</math>


</div>

There are two particular cases of interaction forces whose power always has the same sign:

* Friction action-reaction pair between '''P''' and '''Q''': since the friction force always opposes the relative velocity between the two points, its power is always negative:

<math>\hspace{1cm} \dot{\Ws}^\text{friction}=\dot{\Ws}^{\text{fric}\rightarrow\Ps}_\text{RTQ}+\dot{\Ws}^{\text{fric}\rightarrow\Qs}_\text{RTQ}=\overline{\Fs}^{\text{fric}\rightarrow\Ps} \cdot \vel{P}{RTQ} = - \abs{\overline{\Fs}^{\text{fric}\rightarrow\Ps}} \frac{\vel{P}{RTQ}}{\abs{\vel{P}{RTQ}}} \cdot \vel{P}{RTQ}<0;</math>

* Constraint action-reaction pair between '''P''' and '''Q''': from the [[D3. Interactions between rigid bodies#D3.4 Direct constraint interactions|'''analytical characterization equation''']] of the constraint forces between two points '''P''' and '''Q''' belonging to two rigid bodies <math>\mathrm{S}_\Ps</math> and <math>\mathrm{S}_\Qs</math>, it is easy to see that the associated power is zero:

<math>\hspace{1cm} \dot{\Ws}^\text{constraint}=\dot{\Ws}^{\text{constraint}\rightarrow\Ps}_{\mathrm{S}_\Ps}+\dot{\Ws}^{\text{constraint}\rightarrow\Qs}_{\mathrm{S}_\Ps}=\overline{\Fs}^{\text{constraint}\rightarrow\Ps} \cdot \overline{\mathbf{v}}_{\mathrm{S}_\Qs}(\Ps) = - 0;</math>

====✏️ EXAMPLE E1-2.1: power of an action-reaction friction pair====
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:
{|
:[[File:ExE1-2-1-2-eng.png|thumb|left|180px|link=]]
|A homogeneous wheel, with radius r and mass m, moves down an inclined plane without sliding until it joins a conveyor belt moving at a constant velocity <math>\vs_0</math> relative to the ground. At the moment it makes contact with the belt, the center of the wheel has a velocity <math>\vs_0</math> relative to the ground in the same direction as the belt. We want to calculate the power of the friction force between the belt and the wheel at this instant.
|}

At that momento, the <math>\mathbf{ICR}^\text{wheel}_\Es</math> is point <math>\Js_\text{wheel}</math> in contact with the belt. This point slides backwards with respect to the belt, therefore the friction force on the wheel is forwards:

<math>\left.\begin{array}{l}
\text { AB: ground } \\
\text { REL: belt }
\end{array}\right\} \quad \bar{\vs}_{\text {REL }}\left(\Js_{\text {wheel }}\right)=\bar{\vs}_{\text{AB}}\left(\Js_{\text {wheel }}\right)-\bar{\vs}_{\mathrm{ar}}\left(\Js_{\text {wheel }}\right)=-\bar{\vs}_{\mathrm{ar}}\left(\Js_{\text {wheel }}\right)=-\left(\rightarrow \mathrm{v}_0\right)=\left(\leftarrow \mathrm{v}_0\right) .</math>

[[File:ExE1-2-1-3-eng.png|thumb|center|500px|link=]]

Calculation of the friction power: <math>\dot{\Ws}^\text{friction}=\dot{\Ws}_\Rs^{\text{friction}\rightarrow \text{wheel}}+ \dot{\Ws}_\Rs^{\text{friction}\rightarrow \text{belt}}</math>, where R may be any reference frame. To quickly calculate this power, it is advisable to choose a reference frame R where one of the two points subjected to the action-reaction pair has zero speed, so that one of the two terms is directly zero:

<math>\dot{\Ws}^\text{friction}=\dot{\Ws}_\text{belt}^{\text{fric}\rightarrow \text{wheel}}+ \dot{\Ws}_\text{belt}^{\text{fric }\rightarrow \text{belt}}=\overline{\Fs}^{\text{fric}\rightarrow \text{wheel}} \cdot \bar{\vs}_{\text {belt}}\left(\Js_{\text {wheel }}\right)=(\rightarrow \mu\ms\gs)\cdot(\leftarrow \vs_0)=-\mu\ms\gs\vs_0.</math>

If the calculation is done in any other reference frame, the result is the same:

<math>\dot{\Ws}^\text{friction}=\dot{\Ws}_\text{E}^{\text{fric}\rightarrow \text{wheel}}+ \dot{\Ws}_\text{E}^{\text{fric} \rightarrow \text{belt}}=(\rightarrow \mu\ms\gs)\cdot 0 + (\leftarrow \mu\ms\gs)\cdot(\rightarrow \vs_0)=-\mu\ms\gs\vs_0.</math>

<math>\dot{\Ws}^\text{friction}=\dot{\Ws}_\text{vehicle}^{\text{fric}\rightarrow \text{wheel}}+ \dot{\Ws}_\text{vehicle}^{\text{fric} \rightarrow \text{belt}}=(\rightarrow \mu\ms\gs)\cdot (\rightarrow\vs_0) + (\leftarrow \mu\ms\gs) \cdot(\rightarrow 3\vs_0)=-\mu\ms\gs\vs_0.</math>

However, if only the power of one of the two forces is calculated (for example, the friction on the wheel), the result depends on the reference frame and can have any sign:

<math>\dot{\Ws}_\text{belt}^{\text{fric }\rightarrow \text{wheel}}=(\rightarrow \mu\ms\gs)\cdot(\leftarrow \vs_0)=-\mu\ms\gs\vs_0 ,</math>

<math>\dot{\Ws}_\text{E}^{\text{fric} \rightarrow \text{wheel}}=(\rightarrow \mu\ms\gs)\cdot 0 = 0 ,</math>

<math>\dot{\Ws}_\text{vehicle}^{\text{fric} \rightarrow \text{wheel}}=(\rightarrow \mu\ms\gs)\cdot (\rightarrow\vs_0)=\mu\ms\gs\vs_0.</math>


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==E1.3 Power of a system of forces on a rigid body==

{|
|
:[[File:E1-3-eng.png|thumb|left|180px|link=]]
<center>'''Figure E1.3''': Kinematic torsor of a rigid body in a reference frame R, and forces on different points (dm) of that rigid body</center>

|When calculating the power of a system of forces on a rigid body S, the term associated with internal interactions is zero because those interactions are all constraint (cohesion) forces and, as seen in [[E1. Work-Energy Theorem: differential form#E1.2 Power of an action-reaction pair|'''section E1.2''']], their total power is zero.

The total power of a system of forces on S can be calculated from the torsor of the system at any point '''O''' of the rigid body, <math>(\overline{\Fs},\overline{\Ms}(\Os))</math>. Taking into account rigid body kinematics ('''Figure E1.3'''):

<math>\dot{\Ws}_\Rs^{\rightarrow \Ss} =\sum_\Ps\overline{\Fs}^{\rightarrow \Ps} \cdot \vel{P}{R}= \sum_\Ps \left[ \overline{\Fs}^{\rightarrow \Ps} \cdot \vel{O}{R} + \velang{S}{R} \times \OPvec \right] </math>

|}
<math>\dot{\Ws}_\Rs^{\rightarrow \Ss} =\left(\sum_\Ps\overline{\Fs}^{\rightarrow \Ps} \right)\cdot \vel{O}{R} + \sum_\Ps\overline{\Fs}^{\rightarrow \Ps} \cdot \left(\velang{S}{R} \times \OPvec \right) = \left( \sum_\Ps\overline{\Fs}^{\rightarrow \Ps} \right) \cdot \vel{O}{R} + \left[ \sum_\Ps \left( \OPvec \times \overline{\Fs}^{\rightarrow \Ps} \right)\right] \cdot \velang{S}{R} = \overline{\Fs}\cdot \vel{O}{R} + \overline{\Ms}(\Os)\cdot \velang{S}{R} </math>

This calculation also shows that the power of a moment (or a torque <math>\Gamma</math>) on a rigid body in a reference frame R is the product of the moment by the angular velocity of the rigid body: <math>\dot{\Ws}_\Rs^{\Gamma\rightarrow \mathrm{S}} = \overline{\Gamma} \cdot \velang{S}{R}</math>.

The power balance for a rigid body is: <math>\dot{\Ws}_\Rs^{ \rightarrow \mathrm{S}} = \dot{\Ws}_\Rs^{ \mathrm{ext}} + \dot{\Ws}_\Rs^{ \mathrm{ar}}.</math>

====✏️ EXAMPLE E1-3.1: roller sliding on a slope====
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:
{|
|
:[[File:ExE1-3-1-1-eng.png|thumb|left|180px|link=]]
|A homogeneous roller, with mass m and radius r, slides down a slope. The friction coefficient between the ground and the roller is <math>\mu</math>. We want to calculate the power of every force acting on the roller in the ground reference frame.
|}

:It is a 2D problem. The kinematic and dynamic descriptions are:
{|
|
[[File:ExE1-3-1-2-eng.png|thumb|right|250px|link=]]

|<math> \hspace{0.5cm}\vel{J}{E}=\vel{C}{T} + \velang{}{E} \times \CJvec= </math>

<math>\hspace{1.7cm}=\left( \rightarrow \vs_\Es \right) + \left( \otimes \Omega_\Es\right) \times \left( \downarrow \rs \right)</math>

<math> \hspace{0.5cm}\vel{J}{E} = \left[\rightarrow (\vs_\Es -\rs\Omega_\Es)\right]</math>

<math>\hspace{0.5cm} \dot{\Ws}_\Es^{\mathrm{ext}} = \dot{\Ws}_\Es^{\mathrm{weight}} + \dot{\Ws}_\Es^{\mathrm{friction}} + \dot{\Ws}_\Es^{\mathrm{constraint}}= \Bigr[\left(\downarrow \ms\gs\cos\beta\right)+\left(\rightarrow \ms\gs\sin\beta\right)\Bigr] \cdot (\rightarrow \vs_\Es) + \Bigr[(\leftarrow \mu \Ns) + (\uparrow \Ns ) \Bigr] \cdot \Bigr[\rightarrow (\vs_\Es -\rs\Omega_\Es)\Bigr]</math>

|}
:Since the roller center has no motion perpendicular to the plane, the AMT leads to <math>\Ns =\ms\gs\cos\beta</math>. Therefore:

{|
|
[[File:ExE1-3-1-3-eng.png|thumb|right|140px|link=]]
|<math>\hspace{0.5cm} \dot{\Ws}_\Ts^{\mathrm{ext}} = \ms\gs\vs_\Ts \sin \beta - \ms\gs(\vs_\Ts -\rs\Omega_\Ts)\cos\beta=\ms\gs \Bigr[ \vs_\Ts (\sin\beta-\cos\beta)+\rs\Omega_\Ts \cos\beta \Bigr].</math>

<math>\hspace{0.5cm}</math>

If instead of considering the friction at '''J''' we consider the equivalent torsor at point '''C''', we must consider the force applied in '''C''' and the moment about '''C''', which is not zero:

<math>\hspace{0.5cm} \dot{\Ws}_\Es^{ \mathrm{friction}} = (\leftarrow \mu\Ns)\cdot(\rightarrow\vs_\Es)+(\otimes \mu\Ns\rs)\cdot(\otimes \Omega_\Es) =\mu\ms\gs\cos\beta(\rs\Omega_\Es-\vs_\Es)</math>,

<math>\hspace{0.5cm} \dot{\Ws}_\Es^{\mathrm{ext}} = \dot{\Ws}_\Es^{ \mathrm{weight}} + \dot{\Ws}_\Es^{ \mathrm{friction}}+ \dot{\Ws}_\Es^{ \mathrm{constraint}}= \ms\gs\vs_\Es\sin\beta+\mu\ms\gs\cos\beta (\rs\Omega_\Es-\vs_\Es)</math>.

<math>\hspace{0.5cm}</math>

The result is the same.
|}


====✏️ EXAMPLE E1-3.2: block sliding on a truck====
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:
{|
|
:[[File:ExE1-3-2-1-eng.png|thumb|left|220px|link=]]
|A truck brakes with acceleration '''a''' relative to the ground at the instant in which its velocity (also relative to the ground) is v. At the same instant, the homogeneous block, of mass m, slides with velocity v' relative to the truck. We want to perform a power balance for the block in two different reference frames: the ground and the truck.
|}

:Ground reference frame: It is a 2D problem. The kinematic and dynamic descriptions are:
{|
|
[[File:ExE1-3-2-2-eng.png|thumb|right|200px|link=]]

| The constraint between the truck and the block allows only one translational DoF between the two. Therefore, in the 2D approximation, the associated torsor has two independent components. If characterized at the midpoint of the contact surface, these components are a normal force and a moment perpendicular to the plane. The AMT applied to the block leads to <math>\Ns=\ms\gs</math>.

|}
:The powers associated with all interactions on the block in the ground reference are:

:<math>\dot{\Ws}_\Es^{\rightarrow \mathrm{block}}=\dot{\Ws}_\Es^{\mathrm{ext}}=\dot{\Ws}_\Es^{\mathrm{weoght}}+\dot{\Ws}_\Es^{\mathrm{friction}}+\dot{\Ws}_\Es^{\mathrm{constraint}}= \Bigr[(\downarrow \ms\gs)+(\leftarrow \mu\ms\gs)+(\uparrow \Ns) \Bigr] \cdot \Bigr[\rightarrow(\vs+\vs')\Bigr] + (\odot \Ms_\es)\cdot 0 = -\mu\ms\gs(\vs+\vs')</math>

:Since the block has a translational motion relative to the ground, the calculation of its total kinetic energy is straightforward, and the power balance can be easily stated:

:<math>\Ts_\Es^\mathrm{block}=\frac{1}{2}\int_\mathrm{block}\ds\ms \vs_\Es^2(\mathrm{block})=\frac{1}{2}\ms \vs_\Es^2(\mathrm{block})\quad \Rightarrow \quad \dot{\Ts}_\Es^\mathrm{bloque}=\ms\vs_\Es( \mathrm{block})\dot{\vs}_\Es( \mathrm{block}) </math>

:Since the only horizontal force on the block is friction:

<math>\dot{\vs}_\Es(\mathrm{block})=-\mu\gs\quad \Rightarrow \quad \dot{\Ts}_\Es^\mathrm{block}=-\mu\ms\gs(\vs+\vs').</math>

:The power balance is consistent:

:<math>\left.\begin{array}{l}
\dot{\mathrm{W}}_\Es^{\rightarrow \text { block }}=\dot{\mathrm{W}}_{\mathrm{T}}^{\text {ext }}=-\mu \mathrm{mg}\left(\mathrm{v}+\mathrm{v}^{\prime}\right) \\
\dot{\mathrm{T}}_{\mathrm{E}}^{\text {block }}=-\mu \mathrm{mg}\left(\mathrm{v}+\mathrm{v}^{\prime}\right)
\end{array}\right\} \Rightarrow \dot{\mathrm{W}}_{\mathrm{E}}^{\rightarrow \text { block }}=\dot{\mathrm{T}}_{\mathrm{E}}^{\text {block }}</math>

:Truck reference frame: As the truck is accelerated relative to the ground, the transportation forces on all the mass differentials (dm) of the block must be taken into account:

{|
|
[[File:ExE1-3-2-3-eng.png|thumb|right|250px|link=]]
|<math>\cal{\Fs}^{\rightarrow \ds\ms}_{\mathrm{tr}}=-\ds\ms\overline{\as}_\mathrm{tr}(\ds\ms)=-\ds\ms (\leftarrow \as)=(\rightarrow \ds\ms \cdot \as)</math>.

The resulting torsor of this system of forces at the centre of inertia '''G''' contains just a horizontal force.

<math>\dot{\Ws}_\mathrm{truck}^{\rightarrow \mathrm{block}}=\dot{\Ws}_\mathrm{truck}^{\mathrm{ext}} + \dot{\Ws}_\mathrm{truck}^{\mathrm{ar}} =\dot{\Ws}_\mathrm{truck}^{\mathrm{weight}} + \dot{\Ws}_\mathrm{truck}^{\mathrm{friction}} + \dot{\Ws}_\mathrm{truck}^{\mathrm{constraint}} + \dot{\Ws}_\mathrm{truck}^{\mathrm{ar}} , </math>

<math>\dot{\Ws}_\mathrm{truck}^{\rightarrow \mathrm{block}}= \Bigr[(\downarrow \ms\gs)+(\leftarrow \mu\ms\gs)+(\uparrow \Ns) + (\rightarrow \ms\as)\Bigr] \cdot(\rightarrow \vs')+(\odot\Ms_\es) \cdot0=\ms(\as-\mu\gs)\vs'.</math>
|}

:The consistency of the power balance is easily checked:

:<math>\Ts^\mathrm{block}_\mathrm{camión}=\frac{1}{2}\ms\vs^2_\mathrm{truck}(\mathrm{block}) \quad \Rightarrow \quad \dot{\Ts}^\mathrm{block}_\mathrm{truck} = \ms\vs_\mathrm{truck}(\mathrm{block})\dot{\vs}_\mathrm{truck}(\mathrm{block})</math>

:<math>\left.\begin{array}{l}
\mathrm{AB}:\mathrm{ground} \\
\mathrm{REL}:\mathrm{truck}
\end{array}\right\} \quad \dot{\vs}_\mathrm{truck}(\mathrm{block})= \dot{\vs}_\mathrm{T}(\mathrm{block})-\dot{\vs}_\mathrm{ar}(\mathrm{block}) = (-\mu\gs)-(-\as)=\as-\mu\gs \quad \Rightarrow \quad \dot{\Ts}^\mathrm{block}_\mathrm{truck}=\ms\vs'(\as-\mu\gs)</math>

:<math>\left.\begin{array}{l}
\dot{\Ws}_\mathrm{truck}^{\rightarrow \mathrm{block}}=\dot{\Ws}_\mathrm{truck}^{\mathrm{ext}} + \dot{\Ws}_\mathrm{truck}^{\mathrm{ar}}= \ms(\as-\mu\gs)\vs'\\
\dot{\Ts}^\mathrm{block}_\mathrm{truck}=\ms\vs'(\as-\mu\gs)
\end{array}\right\} \Rightarrow\dot{\Ws}_\mathrm{truck}^{\rightarrow \mathrm{block}}= \dot{\Ts}^\mathrm{block}_\mathrm{truck}</math>


====✏️ EXAMPLE E1-3.3: loaded truck braking====
------
:
{|
|
:[[File:ExE1-3-3-1-eng.png|thumb|left|220px|link=]]
|A truck brakes with acceleration '''a''' relative to the ground at the instant in which its speed (also relative to the ground) is v. The braking torques on the front and rear axles are equal with value , and the wheels, of radius r and negligible mass, do not slip on the ground. At the same instant, the homogeneous block, of mass m, slides with velocity v' relative to the truck. We want to perform a power balance for the system (block+truck) in two different reference frames: the ground and to the truck.

|}

:Since the power of the internal interactions is not zero in principle (it is not a single rigid body), it is convenient to draw a DGI of the system so as not to forget any:
{|
|
[[File:ExE1-3-3-2-eng.jpg|thumb|right|250px|link=]]
|
|
s-p.c.n.s.: single-point contact no sliding

m-p.c.w.s.: multiple-point contact with sliding

|}
:The system on which the power balance is performed contains all elements (but the ground). Since the power of the internal constraints is always zero, the only interactions that need to be taken into account are weight, wheel-to-ground constraints, braking torques, and friction between the block and the truck:

[[File:ExE1-3-3-3-eng.jpg|thumb|center|300px|link=]]

:Ground reference frame:
:The ground's constraint forces on the wheels are applied at points of zero speed, and therefore generate no power. Weight is a vertical force, and the motion of the truck and block is horizontal, so its power is also zero.

:<math>\dot{\Ws}_\Es^{\rightarrow \mathrm{syst}}=\dot{\Ws}_\Es^{\mathrm{ext}}+\dot{\Ws}^{\mathrm{int}}=\dot{\Ws}_\Es^{\mathrm{weight}}+\dot{\Ws}_\Es^{\mathrm{constraint}}+\dot{\Ws}^{\mathrm{friction}}+\dot{\Ws}^{\Gamma}=\dot{\Ws}^{\mathrm{friction}}+\dot{\Ws}^{\Gamma}.</math>

:The power of the internal interactions can be calculated in any reference frame. Thus:

:<math>\dot{\Ws}^{\mathrm{friction}}= \dot{\Ws}_\mathrm{truck}^{\mathrm{fric}\rightarrow \mathrm{bloque}} +\dot{\Ws}_\mathrm{truck}^{\mathrm{fric}\rightarrow \mathrm{truck}} = \dot{\Ws}_\mathrm{truck}^{\mathrm{fric}\rightarrow \mathrm{block}}= (\leftarrow \mu\ms\gs)\cdot (\rightarrow \vs')=-\mu\ms\gs\vs'</math>

:<math>\dot{\Ws}^{\Gamma}= \dot{\Ws}_\mathrm{truck}^{\Gamma\rightarrow \mathrm{wheels}} +\dot{\Ws}_\mathrm{truck}^{\Gamma\rightarrow \mathrm{truck}} = \dot{\Ws}_\mathrm{truck}^{\Gamma\rightarrow \mathrm{wheels}}= 2 \left(\odot \Gamma \right)\cdot \left(\otimes \frac{\vs}{\rs}\right)=-2\Gamma\frac{\vs}{\rs}.</math>

:On the other hand: <math>\Ts_\Es^\mathrm{sist}=\frac{1}{2}\ms\vs_\Es^2(\mathrm{bloque})+\frac{1}{2}\ms\vs_\Es^2(\mathrm{camión})\quad \Rightarrow \quad \dot{\Ts}_\Es^\mathrm{sist}=-\ms(\vs+\vs')\mu\gs-\Ms\vs\as. </math>

:Finally: <math>\dot{\Ws}_\Es^{\rightarrow \mathrm{sist}}=\dot{\Ts}_\Es^\mathrm{sist} \quad \Rightarrow \quad -\mu\ms\gs\vs'-2\Gamma\frac{\vs}{\rs}=-\ms(\vs+\vs')\mu\gs-\Ms\vs\as.</math>

:The braking torque can be obtained from that equation: <math>\Gamma=\frac{1}{2}(\mu\ms\gs+\Ms\as)\rs.</math>

:Truck reference frame:

:In this reference frame, transportation forces must be taken into account. On the other hand, the ground's constraint forces on the wheels are applied at points that no longer have zero velocity, and therefore generate power. As before, the power of weight is zero.

:<math>\dot{\Ws}_\mathrm{truck}^{\rightarrow \mathrm{syst}}=\dot{\Ws}_\mathrm{truck}^{\mathrm{ext}}+\dot{\Ws}^{\mathrm{int}}+\dot{\Ws}_\mathrm{truck}^{\mathrm{tr}}=\dot{\Ws}_\mathrm{truck}^{\mathrm{constraint}}+\dot{\Ws}^{\mathrm{friction}}+\dot{\Ws}^{\Gamma}+\dot{\Ws}_\mathrm{truck}^{\mathrm{tr}}.</math>

:The transportation forces on the truck do not generate power because the truck does not move relative to this reference frame. The transportation forces on the block were calculated in [[E1. TWork-Energy Theorem: differential form#✏️ EXAMPLE E1-3.2: block sliding on a truck|'''example E1-3.2''']]. The friction and braking torques are the same as before since they do not depend on the reference frame.

:<math>\dot{\Ws}_\mathrm{truck}^{\rightarrow \mathrm{syst}}=\dot{\Ws}_\mathrm{truck}^{\mathrm{constraint}}+\dot{\Ws}^{\mathrm{fricción}}+\dot{\Ws}^{\Gamma}+\dot{\Ws}_\mathrm{truck}^{\mathrm{tr} \rightarrow \mathrm{block}}= \dot{\Ws}_\mathrm{truck}^{\mathrm{constraint}} -\mu\ms\gs\vs'-2\Gamma\frac{\vs}{\rs} +\ms\as\vs'.</math>

:The kinematic and dynamic descriptions of the wheel are:

[[File:ExE1-3-3-4-eng.png|thumb|center|430px|link=]]

:<math>\dot{\Ws}_\mathrm{truck}^{\mathrm{constraint}}=2\Bigr[(\leftarrow \Ts)+(\uparrow \Ns)\Bigr]\cdot(\leftarrow\vs)=2\Ts\vs.</math>

The tangential constraint force can be calculated from the AMT applied to the wheel centre C. Since it has no mass: <math>\sum\overline{\Ms}_\mathrm{ext}(\Cs)=\overline{0} \quad \Rightarrow \quad (\odot \Gamma)+(\otimes \Ts\rs)=0\quad \Rightarrow \quad \Ts=\Gamma/\rs.</math>

:Therefore: <math>\dot{\Ws}_\mathrm{truck}^{\rightarrow \mathrm{syst}} =2\frac{\Gamma}{\rs}\vs-\mu\ms\gs\vs'-2\Gamma\frac{\vs}{\rs}+\ms\as\vs'=\ms(\as-\mu\gs)\vs'.</math>

:On the other hand, as seen in [[E1. TWork-Energy Theorem: differential form#✏️ EXAMPLE E1-3.2: block sliding on a truck|'''example E1-3.2''']]:

:<math>\Ts_\mathrm{truck}^\mathrm{sist}= \Ts_\mathrm{truck}^\mathrm{bloque}=\frac{1}{2}\ms\vs_\mathrm{truck}^2(\mathrm{block}) \quad \Rightarrow \quad \dot{\Ts}_\mathrm{truck}^\mathrm{syst}= \dot{\Ts}_\mathrm{truck}^\mathrm{block}=\ms\vs'(\as-\mu\gs).</math>

:The power balance is consistent:

:<math>\left.\begin{array}{l}
\dot{\Ws}_\mathrm{truck}^{\rightarrow \mathrm{syst}}= \ms(\as-\mu\gs)\vs'\\
\dot{\Ts}^\mathrm{syst}_\mathrm{truck}=\ms(\as-\mu\gs)\vs'
\end{array}\right\} \Rightarrow\dot{\Ws}_\mathrm{truck}^{\rightarrow \mathrm{syst}}= \dot{\Ts}^\mathrm{syst}_\mathrm{truck}</math>

-----
-----

==E1.4 Power balance in a multibody system: direct and indirect calculation==

The '''direct calculation''' (calculation from its definition) of the power associated with an interaction (a force on a point P or a moment on a rigid body S) requires knowledge of the interaction (force or moment):

<math>\dot{\Ws}_\Rs^{\Fs\rightarrow \Ps}=\overline{\Fs}_{\rightarrow\Ps} \cdot \vel{P}{R} \quad , \quad \dot{\Ws}_\Rs^{\Ms\rightarrow \Ss}=\overline{\Ms}_{\rightarrow\Ss} \cdot \velang{S}{R} </math>

When the interaction cannot be formulated, one must resort to vector theorems to determine it.

The '''indirect calculation''' is an alternative that does not require resort to these theorems. It is based on the evaluation of a power balance for a suitable system and reference frame, such that the power to be determined is the only unknown term in the equation (the only one that cannot be calculated directly). When the balance includes more than one unknown term, several balances (for different systems and reference frames) can be combined until a sufficient number of equations is obtained to allow the determination of all the unknown powers.

====✏️ EXAMPLE E1-4.1: braked pulley====
------
:
{|
|
:[[File:ExE1-4-1-1-eng.png|thumb|left|150px|link=]]
|A block of mass m hangs from an inextensible thread attached to a point on the periphery of a pulley of radius r. A brake acts between the support (fixed to the ground) and the pulley, ensuring that the block descends at a constant speed relative to the ground. We want to calculate he brake's power.

|}

:Direct calculation: <math>\dot{\Ws}^{\mathrm{brake}}= \overline{\Gamma}^{\rightarrow \mathrm{pulley}}\cdot\velang{pulley}{E}=\left(\odot\Gamma\right)\cdot\left( \otimes \frac{\vs_0}{\rs}\right)=-\Gamma \frac{\vs_0}{\rs}.</math>

:The braking torque is not given, and the AMT must be applied to calculate it.

:<math>\left.\begin{array}{l}
\text{SYSTEM: pulley + block}\\
\text{AMT about } \Os
\end{array}\right\} \sum\overline{\Ms}_\mathrm{ext}(\Os)=\dot{\overline{\Hs}}_\mathrm{RTO}(\Os)</math>

:The angular momentum is associated with both pieces. Since '''O''' is not fixed to the block, baricentric decomposition is required to calculate the angular momentum of the block:

:<math>\overline{\Hs}_\mathrm{RTO=E} (\Os) =\overline{\Hs}_\Es^\mathrm{pulley}(\Os) + \overline{\Hs}_\Es^\mathrm{block}(\Os)= \overline{\Hs}_\Es^\mathrm{pulley}(\Os) + \overline{\Hs}_\Es^\mathrm{block}(\Gs_\mathrm{block})+ \overline{\Hs}_\Es^{\oplus\mathrm{block}}(\Os) = </math>

:<math>\hspace{2.3cm}=\Is_\mathrm{O}^\mathrm{polea} \velang{polea}{T}+\Is_{\Gs_\mathrm{block}}^\mathrm{block}+\OGvec_\mathrm{block}\times \ms\overline{\vs}_\Ts(\Gs_\mathrm{block})=\left(\otimes \Is_\mathrm{O}\frac{\vs_0}{\rs}\right)+(\otimes \ms\vs_0\rs).</math>

:[[File:ExE1-4-1-2-eng.png|thumb|right|180px|link=]]
:It is an angular momentum whose direction and value are constant. Therefore:

:<math>\dot{\overline{\Hs}}_\mathrm{RTO}(\Os)=0 \quad \Rightarrow \quad \sum{\overline{\Ms}}_\mathrm{ext}(\Os)=(\odot\Gamma)+(\otimes \ms\gs\rs) =0\quad \Rightarrow \quad \Gamma= \ms\gs\rs</math>.

:Finally: <math>\dot{\Ws}^{\mathrm{freno}}= -\Gamma \frac{\vs_0}{\rs} = -\ms\gs\vs_0.</math>

:Indirect calculation:

:<math>\left.\begin{array}{l}
\text{SYSTEM: ppulley + support + block}\\
\text{REF: ground(Gal)}
\end{array}\right\} \quad \dot{\Ts}^\mathrm{syst}_\Es=\dot{\Ws}^\mathrm{ext}_\Es+ \dot{\Ws}^\mathrm{int}</math>

:<math>\left.\begin{array}{l}
\dot{\Ts}^\mathrm{syst}_\Es=0\\
\dot{\Ws}^\mathrm{ext}_\Es=\left(\downarrow \ms\gs\right)\cdot\left(\downarrow \vs_0\right)= \ms\gs\vs_0\\
\dot{\Ws}^\mathrm{int}=\dot{\Ws}^{\mathrm{brake}}
\end{array}\right\} \quad \Rightarrow \quad \dot{\Ws}^{\mathrm{brake}} =-\ms\gs\vs_0</math>



====✏️ EXAMPLE E1-4.2: monkey and bananas====
------
:
{|
|
:[[File:ExE1-4-2-1-eng.png|thumb|left|180px|link=]]
|A bunch of bananas of mass <math>(2/3)\ms</math> hangs from an inextensible rope, which rests without sliding on a pulley of radius r and negligible mass, fixed to the ground by a hinge. A monkey of mass m is attached to the other end of the rope. The pulley rotates at constant angular velocity <math>\Omega_0</math> with respect to the ground under the action of a motor. The monkey moves relative to the rope without rubbing its hands against it, and in such a way that the distance from its centre of mass to the ground remains constant. We want to calculate the motor's power.
|}

:Direct calculation: <math>\dot{\Ws}^{\mathrm{brake}}= \overline{\Gamma}^{\rightarrow \mathrm{pulley}}\cdot\velang{pulley}{T}=\left(\odot\Gamma\right)\cdot\left( \odot \Omega_0\right)=\Gamma \Omega_0.</math>

:The engine torque is not given, and AMT is required to calculate it.

:<math>\left.\begin{array}{l}
\text{SYST: pulley + monkey + bananas}\\
\text{AMT about } \Os
\end{array}\right\} \sum\overline{\Ms}_\mathrm{ext}(\Os)=\dot{\overline{\Hs}}_\mathrm{RTO}(\Os)</math>

:The angular momentum is associated with both the monkey and the bananas. As a first approximation, the monkey's mass can be considered concentrated in its trunk and head, so it doesn't rotate and its velocity relative to the ground is zero. Taking into account that point '''O''' does not move relative to the monkey but it does relative to the bananas:

:<math>\overline{\Hs}_\mathrm{RTO=E} (\Os) =\overline{\Hs}_\Es^\mathrm{monkey}(\Os) + \overline{\Hs}_\Es^\mathrm{bananas}(\Os)= \overline{\Hs}_\Es^\mathrm{monkey}(\Os) + \overline{\Hs}_\Es^\mathrm{bananas}(\Gs_\mathrm{ban})+ \OGvec_\mathrm{ban} \times \frac{2}{3} \ms \overline{\vs}_\Es(\Gs_\mathrm{ban})= </math>

:<math>\hspace{2.3cm}=\OGvec_\mathrm{ban}\times \frac{2}{3} \ms \overline{\vs}_\Ts(\Gs_\mathrm{ban}) = \Bigr[(\rightarrow \rs)+(\downarrow \xs)\Bigr] \times\frac{2}{3} \ms \left(\downarrow \rs\Omega_0\right)=\left(\otimes \frac{2}{3}\ms\rs^2\Omega_0\right).</math>

:It is an angular momentum with constant value and constant direction. Therefore:

:<math>\dot{\overline{\Hs}}_\mathrm{RTO}(\Os)=0 \quad \Rightarrow \quad \sum{\overline{\Ms}}_\mathrm{ext}(\Os)=(\odot\Gamma)+(\otimes \ms\gs\rs) + \left(\otimes \frac{2}{3}\ms\gs\rs\right)=0\quad \Rightarrow \quad \Gamma= \frac{5}{3}\ms\gs\rs</math>.

:Finally: <math>\dot{\Ws}^{\mathrm{motor}}= \Gamma \Omega_0 = \frac{5}{3}\ms\gs\rs\Omega_0.</math>

:Indirect calculation:

:It should be noted that this is a problem involving two types of actuators: the motor and the monkey’s muscles. There are several biomechanical formulations for muscle forces, but they are beyond the scope of this course. Since the monkey moves, its muscles develop a non-zero power.

:<math>\left.\begin{array}{l}
\text{SYSTEM: pulley + monkey + bananas}\\
\text{REF: ground(Gal)}
\end{array}\right\} \quad \dot{\Ts}^\mathrm{syst}_\Es=\dot{\Ws}^\mathrm{ext}_\Es+ \dot{\Ws}^\mathrm{int}</math>

:<math>\left.\begin{array}{l}
\dot{\Ts}^\mathrm{syst}_\Es=0\\
\dot{\Ws}^\mathrm{ext}_\Es=\left(\downarrow \frac{2}{3}\ms\gs\right)\cdot\left(\downarrow \rs\Omega_0\right)= \frac{2}{3}\ms\gs\rs\Omega_0\\
\dot{\Ws}^\mathrm{int}=\dot{\Ws}^{\mathrm{musc}}+\dot{\Ws}^{\mathrm{motor}}
\end{array}\right\} \quad \Rightarrow \quad \dot{\Ws}^{\mathrm{musc}} + \dot{\Ws}^{\mathrm{motor}}=-\frac{2}{3}\ms\gs\rs\Omega_0</math>

:This balance provides an equation with two unknowns. A second balance can be established for a system where one of the two is missing: a system without the monkey or a system without the motor. If a power balance is performed for the system (monkey + piece of rope), the external forces to consider are the weight and tension of the rope. If we carry on the calculation at the ground reference, this (unknown) tension will result in non-zero power since it is applied at an upward velocity point.
:[[File:ExE1-4-2-2-eng.png|thumb|right|330px|link=]]

:To avoid the appearance of a term associated with the rope tension, we can move to the reference frame of the piece of rope, which, having a constant upward velocity, is Galilean. In this reference frame, the monkey descends at a constant velocity:

:<math>\left.\begin{array}{l}
\text{REL: piece of rope}\\
\text{AB: ground}
\end{array}\right\}</math>

:<math>\overline{\vs}_\mathrm{REL}(\mathrm{monkey})=\overline{\vs}_\mathrm{ar}(\mathrm{monkey})=-\overline{\vs}_\mathrm{ar}(\mathrm{monkey})=(\downarrow \rs\Omega_0)</math>

:<math>\left.\begin{array}{l}
\text{SISTEMA: monkey+ piece of rope}\\
\text{REF: piece of rope(Gal)}
\end{array}\right\} \quad \dot{\Ts}^\mathrm{syst}_\mathrm{rope} =\dot{\Ws}^\mathrm{ext}_\mathrm{rope}+ \dot{\Ws}^\mathrm{int}</math>

:<math>\left.\begin{array}{l}
\dot{\Ts}^\mathrm{syst}_\mathrm{rope}=0\\
\dot{\Ws}^\mathrm{ext}_\mathrm{rope}=\left(\downarrow \ms\gs\right)\cdot\left(\downarrow \rs\Omega_0\right)= \ms\gs\rs\Omega_0\\
\dot{\Ws}^\mathrm{int}=\dot{\Ws}^{\mathrm{musc}}
\end{array}\right\} \quad \Rightarrow \quad \dot{\Ws}^{\mathrm{musc}} =-\ms\gs\rs\Omega_0</math>

:Combining this result with the previous balance:

:<math>\dot{\Ws}^\mathrm{motor}=-\frac{2}{3}\ms\gs\rs\Omega_0-\dot{\Ws}^\mathrm{musc}=-\frac{5}{3}\ms\gs\rs\Omega_0</math>



© Universitat Politècnica de Catalunya. [[Mecánica:Derechos de autor |All rights reserved]]
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[[D8. Conservation of dynamic magnitudes|<<< D8. Conservation of dynamic magnitudes]]

[[E2. Teorema de la energía: versión integrada|E2. Teorema de la energía: versión integrada>>>]]
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E1. Work-Energy Theorem: differential form

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Bros: /* ✏️ EXAMPLE E1-1.2: pendulum */

<div class="noautonum">__TOC__</div>
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{#3}
\end{Bmatrix}}
\newcommand{\vecdosd}[2]{
\begin{Bmatrix}
{#1}\\
{#2}
\end{Bmatrix}}
\newcommand{\vel}[2]{\vvec_{\textrm{#2}} (\textbf{#1})}
\newcommand{\acc}[2]{\vecbf{a}_{\textrm{#2}} (\textbf{#1})}
\newcommand{\accs}[2]{\vecbf{a}_{\textrm{#2}}^{\textrm{s}} (\textbf{#1})}
\newcommand{\accn}[2]{\vecbf{a}_{\textrm{#2}}^{\textrm{n}} (\textbf{#1})}
\newcommand{\velo}[1]{\vvec_{\textrm{#1}}}
\newcommand{\accso}[1]{\vecbf{a}_{\textrm{#1}}^{\textrm{s}}}
\newcommand{\accno}[1]{\vecbf{a}_{\textrm{#1}}^{\textrm{n}}}
\newcommand{\re}[2]{\Re_{\textrm{#2}}(\textbf{#1})}
\newcommand{\psio}{\dot{\psi}_0}
\newcommand{\Pll}{\textbf{P}_\textrm{lliure}}
\newcommand{\agal}[1]{\vecbf{a}_{\textrm{Gal}} (#1)}
\newcommand{\angal}[1]{\vecbf{a}_{\textrm{NGal}} (#1)}
\newcommand{\vgal}[1]{\vecbf{v}_{\textrm{Gal}} (#1)}
\newcommand{\fvec}[2]{\overline{\mathbf{F}}_{\textrm{#1}\rightarrow\textrm{#2}}}
\newcommand{\mvec}[2]{\overline{\mathbf{M}}_{\textrm{#1}\rightarrow\textrm{#2}}}
\newcommand{\F}[1]{\vec{\Fs}_{#1}}
\newcommand{\Fcal}[2]{\vec{\cal{F}}^{\text{#1}}_{#2}}
</math>

The '''Work-Energy Theorem''' plays a prominent role in physics, as it bridges all branches of science (electromagnetism, thermodynamics, etc.): all of them deal with energy.

There is a certain parallel between the Vector Theorems and the '''Work-Energy Theorem''': they all allow us to understand the evolution of a magnitude that depends on the mechanical state of the system. In the case of the former, we are dealing with the instantaneous evolution of vector magnitudes (linear momentum, angular momentum), while the latter deals with the finite change (in a finite time interval) of a scalar magnitude ('''mechanical energy''').

This unit presents the differential form of this theorem (the '''power balance'''). The integrated form (which is the Work-Energy Theorem itself) is presented in [[E2. Teorema de la energía: versión integrada|'''unit E2''']].
 
------
------
==E1.1 Power balance in a system of particles==

:[[File:E1-1-eng.png|thumb|right|300px|link=]]
<center>'''Figure E1.1''' Forces on a particle of a system with constant matter</center>

Let us consider a system of constant matter, and let’s apply Newton's second law to each particle (or each mass differential) of this system in a general reference frame R (which may be non-Galilean) ('''Figure E1.1'''):

<math>\F{\rightarrow \Ps} + \Fcal{}{\text{ar}\rightarrow\Ps} + \Fcal{}{\text{Cor}\rightarrow\Ps} =\ms_\Ps\acc{P}{R}</math>

<math>\F{\rightarrow \Ps} </math> is the resultant interaction force on '''P''', and it comes from internal interactions (with particles of the same system) and external ones:

<math>\F{\rightarrow \Ps}=\F{\rightarrow \Ps}^\text{int}+\F{\rightarrow \Ps}^\text{ext}</math>

If both sides of the equation are multiplied by <math>\vel{P}{R}</math>, all forces (interaction and inertia forces) orthogonal to <math>\vel{P}{R}</math> are filtered out. Specifically, since <math>\acc{P}{R}=\accs{P}{R} + \accn{P}{R} </math>, where <math>\accn{P}{R} \perp \vel{P}{R}</math>, and since the Coriolis force is always orthogonal to <math>\vel{P}{R} (\Fcal{}{\text{Cor}\rightarrow\Ps} = -2\ms_\Ps \velang{R}{Gal} \times \vel{P}{R})</math>:

<math>\vel{P}{R} \cdot \left( \F{\rightarrow \Ps}^\text{int}+\F{\rightarrow \Ps}^\text{ext} + \Fcal{}{\text{ar}\rightarrow\Ps} \right) = \ms_\Ps\accs{P}{R} \cdot \vel{P}{R} = \frac{\ds}{\ds \ts} \left(\frac{1}{2}\ms_\Ps \vvec_\Rs^2(\Ps) \right) \equiv \frac{\ds\Ts_\Rs(\Ps)}{\ds\ts}=\dot{\Ts}_\Rs(\Ps)</math>

The scalar function <math>\Ts_\Rs(\Ps)</math> is the '''kinetic energy''' of the particle, and is a function of the [[D1. Foundational laws of Newtonian dynamics#D1.3 Newton’s Principle of Determinacy|'''mechanical state''']] of '''P'''. The scalar product of a force and the velocity of its application point is the '''force power''': <math>\vel{P}{R} \cdot \F{\rightarrow \Ps} \equiv \dot{\Ws}_\Rs^{\F{\rightarrow \Ps}} </math>. Both scalar quantities <math>(\Ts_\Rs(\Ps), \dot{\Ws}_\Rs^{\F{\rightarrow \Ps}})</math> depend on the reference frame. For the same set of interaction forces acting on '''P''', the reference frame determines which are filtered out and which are not.

The equation can be rewritten as <math>\dot{\Ws}_\Rs^{\text{ext}\rightarrow \Ps}+\dot{\Ws}_\Rs^{\text{int}\rightarrow \Ps}+\dot{\Ws}_\Rs^{\text{ar}\rightarrow \Ps} =\dot{\Ts}_\Rs(\Ps)</math>, and constitutes the '''power balance''' for particle '''P''' in the R reference frame. Summing for all particles in the system:

<math>\dot{\Ws}_\Rs^{\text{ext}}+\dot{\Ws}^{\text{int}}+\dot{\Ws}_\Rs^{\text{ar}} =\dot{\Ts}_\Rs^\text{syst}</math>

The calculation of kinetic energy is covered in [[E2. Teorema de la energía: versión integrada|'''unit E2''']]. This unit focuses on the calculation of powers. [[E1. Work-Energy Theorem: differential form#E1.2 Potencia de una pareja de acción-reacción|'''section E1-2''']] shows that the power of internal interactions does not depend on the reference frame (which is why the subscript R does not appear in the <math>\dot{\Ws}^{\text{int}}</math> term of the previous equation).

====✏️ EXAMPLE E1-1.1: free particle====
------
:
{|
:[[File:ExE1-1-1-1-eng.jpg|thumb|left|180px|link=]]
|A particle '''P''', with mass m, is released from the ring of a Ferris wheel when it is at its highest point. We want to calculate the power, at this instant, of all forces acting on the particle from three different reference frames: the ground, the ring, and the Ferris wheel cabin.
|}

:Ground reference frame: Since this is a Galilean reference frame, all forces on '''P''' are interaction forces.
{|
:[[File:ExE1-1-2- eng.png|thumb|right|230px|link=]]

|<math>\hspace{0.5cm}\sum\dot{\Ws}_\Ts=\dot{\Ws}_\Ts^\text{weight}= \left(\leftarrow \Rs\Omega_0\right) \cdot \left(\downarrow \ms\gs\right)=0</math>
|}

:Ring reference frame: It is a non-Galilean reference, and the transportation force on '''P''' must be taken into account.
{|
:[[File:ExE1-1-3-eng.png|thumb|right|250px|link=]]
|<math>\hspace{0.5cm}</math> AB:ground, REL:ring 

<math>\hspace{0.5cm}\vel{P}{REL}=\vel{P}{AB}-\vel{P}{ar}= \left(\leftarrow \Rs\Omega_0\right)-\left(\leftarrow \Rs\Omega_0\right)=\overline{0}</math> 

<math>\hspace{0.5cm}</math>

Since the velocity is zero, the total power is zero.
|}

:Cabin reference frame: It is also a non-Galilean reference, and the transportation force on '''P''' must be taken into account.

{|
:[[File:ExE1-1-4-eng.png|thumb|right|250px|link=]]
|<math>\hspace{0.5cm}</math>AB:ground, REL:cabin

<math>\hspace{0.5cm}\vel{P}{REL}=\vel{P}{AB}-\vel{P}{ar}= \left(\leftarrow \Rs\Omega_0\right)-\left(\uparrow \Rs\Omega_0\right)</math>

<math>\hspace{0.5cm}\dot{\Ws}_\text{cabin}^\text{weight}=\left[ \left(\leftarrow \Rs\Omega_0\right)+\left(\downarrow \Rs\Omega_0\right) \right] \cdot \left(\downarrow\ms\gs\right) =\ms\gs\Rs\Omega_0</math>

<math>\hspace{0.5cm}\dot{\Ws}_\text{cabin}^\text{ar}=\left[ \left(\leftarrow \Rs\Omega_0\right)+\left(\downarrow \Rs\Omega_0\right) \right] \cdot \left(\rightarrow\ms\Rs^2\Omega_0^2\right) =-\ms\gs\Rs^2 \Omega_0^3</math>

<math>\hspace{0.5cm}\sum\dot{\Ws}_\text{cabin}=\ms\Rs\Omega_0(\gs-\Rs\Omega_0^2)</math>

|}


 
====✏️ EXAMPLE E1-1.2: pendulum====
------
:
{|
:[[File:ExE1-2-1-eng.png|thumb|left|180px|link=]]
|A particle '''P''' of mass m hangs from an inextensible thread of length L attached to the ceiling. The assembly oscillates in a vertical plane fixed to the ground. We want to calculate the power of all the forces acting on the particle in two different reference frames: the ground and a platform rotating relative to the ground with a constant angular velocity <math>\Omega_0</math>. 
|}
 

 

Ground reference frame: Since it is a Galilean reference frame, '''P''' is only subjected to interaction forces.
{|
:[[File:ExE1-2-2-neut.png|thumb|left|300px|link=]]

|<math>\dot{\Ws}_\Es^\Ts=(\nearrow\Ls\dot{\theta})\cdot(\nwarrow\Ts)=0</math>

<math>\dot{\Ws}_\Es^\text{weight}=(\nearrow\Ls\dot{\theta}\cdot(\downarrow \ms\gs)=-\ms\gs\Ls\dot{\theta}\sin\theta</math>
|}

Platform reference frame: It is a non Galilean reference frame, and the transportation force on '''P''' has to be taken into account.

{|
:[[File:ExE1-2-3-eng.png|thumb|left|400px|link=]]

|AB:ground, REL:platform

<math>\vel{P}{REL}=\vel{P}{AB}-\vel{P}{ar}=(\nearrow\Ls\dot{\theta})-(\otimes \Ls \Omega_0 \sin \theta)</math>

<math>\dot{\Ws}_\Es^\Ts\left[(\nearrow\Ls\dot{\theta})+(\odot\Ls\Omega_0\sin\theta)\right]\cdot(\nwarrow\Ts)=0</math>

<math>\dot{\Ws}_\Es^\text{weight}\left[(\nearrow\Ls\dot{\theta})+(\odot\Ls\Omega_0\sin\theta)\right]\cdot(\downarrow \ms\gs)=-\ms\gs\Ls\dot{\theta}\sin\theta</math>

<math>\dot{\Ws}_\Es^\text{ar}\left[(\nearrow\Ls\dot{\theta})+(\odot\Ls\Omega_0\sin\theta)\right]\cdot(\rightarrow \ms\Ls\Omega_0^2\sin\theta)=\ms\Ls^2\Omega_0^2\dot{\theta}\cos\theta</math>
|}



------
------

==E1.2 Power of an action-reaction pair==

The total power associated with internal interactions in a constant-mass system is not zero in principle (since the two forces of each action-reaction pair are applied to particles that may have different velocities relative to the reference frame in which the calculation is made), and it is independent of the reference frame in which it is evaluated. It only depends on the separation velocity of the two particles. The power balance for the system is therefore written as: <math>\dot{\Ts}_\Rs^\text{syst}=\dot{\Ws}_\Rs^\text{ext} + \dot{\Ws}^\text{int} + \dot{\Ws}_\Rs^\text{ar} </math> .

<div>
=====💭 PROOF ➕=====

{|
:[[File:E1-2-neut.png|thumb|left|200px|link=]]
|<math>\dot{\Ws}^\text{ac-reac}=\overline{\Fs}_{\Ps \rightarrow \Qs}\cdot \vel{Q}{R} + \overline{\Fs}_{\Qs \rightarrow \Ps}\cdot \vel{P}{R}= \overline{\Fs}_{\Ps \rightarrow \Qs}\cdot \left[ \vel{Q}{R}-\vel{P}{R}\right]</math>

By the [[D1. Foundational laws of Newtonian dynamics#D1.6 Newton’s third law (action-reaction principle)|'''action-reaction principle''']], <math>\overline{\Fs}_{\Ps \rightarrow \Qs}=-\overline{\Fs}_{\Qs \rightarrow \Ps}</math>. Therefore: <math>\dot{\Ws}^\text{ac-reac}=\overline{\Fs}_{\Ps \rightarrow \Qs}\cdot \left[ \vel{Q}{R}-\vel{P}{R}\right]</math>

As <math>\overline{\Fs}_{\Ps \rightarrow \Qs}</math> has the <math>\PQvec</math> direction, only the components of the velocities in that direction have to be considered:

<math>\dot{\Ws}^\text{ac-reac}=\overline{\Fs}_{\Ps \rightarrow \Qs}\cdot \biggl( \vel{Q}{R} \Bigr]_{||\Ps\Qs}-\ \vel{P}{R} \Bigr]_{||\Ps\Qs}\biggl)</math>
|}

The difference <math>\biggl( \vel{Q}{R} \Bigr]_{\Ps\Qs}-\ \vel{P}{R} \Bigr]_{\Ps\Qs}\biggl)</math> is the separation velocity <math>\dot{\rho}</math> between '''P''' and '''Q''', and it does not dpend on the reference frame since it is the derivative of a scalar (the distance <math>\dot{\rho}</math> between the two particles). For the previous drawing, <math>\dot{\rho}> 0</math>. Since this is an attractive interaction, <math>\dot{\Ws}^\text{ac-reac} < 0</math>.

With a drawing, it is easy to see that:
* attractive interaction and particle approach <math>(\dot{\rho}<0)</math> : <math>\dot{\Ws}^\text{ac-reac} > 0</math> ;
* repulsive interaction and particle separation <math>(\dot{\rho}>0)</math> : <math>\dot{\Ws}^\text{ac-reac} > 0</math>
* repulsive interaction and particle approach <math>(\dot{\rho}<0)</math> : <math>\dot{\Ws}^\text{ac-reac} < 0</math>


</div>

There are two particular cases of interaction forces whose power always has the same sign:

* Friction action-reaction pair between '''P''' and '''Q''': since the friction force always opposes the relative velocity between the two points, its power is always negative:

<math>\hspace{1cm} \dot{\Ws}^\text{friction}=\dot{\Ws}^{\text{fric}\rightarrow\Ps}_\text{RTQ}+\dot{\Ws}^{\text{fric}\rightarrow\Qs}_\text{RTQ}=\overline{\Fs}^{\text{fric}\rightarrow\Ps} \cdot \vel{P}{RTQ} = - \abs{\overline{\Fs}^{\text{fric}\rightarrow\Ps}} \frac{\vel{P}{RTQ}}{\abs{\vel{P}{RTQ}}} \cdot \vel{P}{RTQ}<0;</math>

* Constraint action-reaction pair between '''P''' and '''Q''': from the [[D3. Interactions between rigid bodies#D3.4 Direct constraint interactions|'''analytical characterization equation''']] of the constraint forces between two points '''P''' and '''Q''' belonging to two rigid bodies <math>\mathrm{S}_\Ps</math> and <math>\mathrm{S}_\Qs</math>, it is easy to see that the associated power is zero:

<math>\hspace{1cm} \dot{\Ws}^\text{constraint}=\dot{\Ws}^{\text{constraint}\rightarrow\Ps}_{\mathrm{S}_\Ps}+\dot{\Ws}^{\text{constraint}\rightarrow\Qs}_{\mathrm{S}_\Ps}=\overline{\Fs}^{\text{constraint}\rightarrow\Ps} \cdot \overline{\mathbf{v}}_{\mathrm{S}_\Qs}(\Ps) = - 0;</math>

====✏️ EXAMPLE E1-2.1: power of an action-reaction friction pair====
------
:
{|
:[[File:ExE1-2-1-2-eng.png|thumb|left|180px|link=]]
|A homogeneous wheel, with radius r and mass m, moves down an inclined plane without sliding until it joins a conveyor belt moving at a constant velocity <math>\vs_0</math> relative to the ground. At the moment it makes contact with the belt, the center of the wheel has a velocity <math>\vs_0</math> relative to the ground in the same direction as the belt. We want to calculate the power of the friction force between the belt and the wheel at this instant.
|}

At that momento, the <math>\mathbf{ICR}^\text{wheel}_\Es</math> is point <math>\Js_\text{wheel}</math> in contact with the belt. This point slides backwards with respect to the belt, therefore the friction force on the wheel is forwards:

<math>\left.\begin{array}{l}
\text { AB: ground } \\
\text { REL: belt }
\end{array}\right\} \quad \bar{\vs}_{\text {REL }}\left(\Js_{\text {wheel }}\right)=\bar{\vs}_{\text{AB}}\left(\Js_{\text {wheel }}\right)-\bar{\vs}_{\mathrm{ar}}\left(\Js_{\text {wheel }}\right)=-\bar{\vs}_{\mathrm{ar}}\left(\Js_{\text {wheel }}\right)=-\left(\rightarrow \mathrm{v}_0\right)=\left(\leftarrow \mathrm{v}_0\right) .</math>

[[File:ExE1-2-1-3-eng.png|thumb|center|500px|link=]]

Calculation of the friction power: <math>\dot{\Ws}^\text{friction}=\dot{\Ws}_\Rs^{\text{friction}\rightarrow \text{wheel}}+ \dot{\Ws}_\Rs^{\text{friction}\rightarrow \text{belt}}</math>, where R may be any reference frame. To quickly calculate this power, it is advisable to choose a reference frame R where one of the two points subjected to the action-reaction pair has zero speed, so that one of the two terms is directly zero:

<math>\dot{\Ws}^\text{friction}=\dot{\Ws}_\text{belt}^{\text{fric}\rightarrow \text{wheel}}+ \dot{\Ws}_\text{belt}^{\text{fric }\rightarrow \text{belt}}=\overline{\Fs}^{\text{fric}\rightarrow \text{wheel}} \cdot \bar{\vs}_{\text {belt}}\left(\Js_{\text {wheel }}\right)=(\rightarrow \mu\ms\gs)\cdot(\leftarrow \vs_0)=-\mu\ms\gs\vs_0.</math>

If the calculation is done in any other reference frame, the result is the same:

<math>\dot{\Ws}^\text{friction}=\dot{\Ws}_\text{E}^{\text{fric}\rightarrow \text{wheel}}+ \dot{\Ws}_\text{E}^{\text{fric} \rightarrow \text{belt}}=(\rightarrow \mu\ms\gs)\cdot 0 + (\leftarrow \mu\ms\gs)\cdot(\rightarrow \vs_0)=-\mu\ms\gs\vs_0.</math>

<math>\dot{\Ws}^\text{friction}=\dot{\Ws}_\text{vehicle}^{\text{fric}\rightarrow \text{wheel}}+ \dot{\Ws}_\text{vehicle}^{\text{fric} \rightarrow \text{belt}}=(\rightarrow \mu\ms\gs)\cdot (\rightarrow\vs_0) + (\leftarrow \mu\ms\gs) \cdot(\rightarrow 3\vs_0)=-\mu\ms\gs\vs_0.</math>

However, if only the power of one of the two forces is calculated (for example, the friction on the wheel), the result depends on the reference frame and can have any sign:

<math>\dot{\Ws}_\text{belt}^{\text{fric }\rightarrow \text{wheel}}=(\rightarrow \mu\ms\gs)\cdot(\leftarrow \vs_0)=-\mu\ms\gs\vs_0 ,</math>

<math>\dot{\Ws}_\text{E}^{\text{fric} \rightarrow \text{wheel}}=(\rightarrow \mu\ms\gs)\cdot 0 = 0 ,</math>

<math>\dot{\Ws}_\text{vehicle}^{\text{fric} \rightarrow \text{wheel}}=(\rightarrow \mu\ms\gs)\cdot (\rightarrow\vs_0)=\mu\ms\gs\vs_0.</math>


-----
-----

==E1.3 Power of a system of forces on a rigid body==

{|
|
:[[File:E1-3-eng.png|thumb|left|180px|link=]]
<center>'''Figure E1.3''': Kinematic torsor of a rigid body in a reference frame R, and forces on different points (dm) of that rigid body</center>

|When calculating the power of a system of forces on a rigid body S, the term associated with internal interactions is zero because those interactions are all constraint (cohesion) forces and, as seen in [[E1. Work-Energy Theorem: differential form#E1.2 Power of an action-reaction pair|'''section E1.2''']], their total power is zero.

The total power of a system of forces on S can be calculated from the torsor of the system at any point '''O''' of the rigid body, <math>(\overline{\Fs},\overline{\Ms}(\Os))</math>. Taking into account rigid body kinematics ('''Figure E1.3'''):

<math>\dot{\Ws}_\Rs^{\rightarrow \Ss} =\sum_\Ps\overline{\Fs}^{\rightarrow \Ps} \cdot \vel{P}{R}= \sum_\Ps \left[ \overline{\Fs}^{\rightarrow \Ps} \cdot \vel{O}{R} + \velang{S}{R} \times \OPvec \right] </math>

|}
<math>\dot{\Ws}_\Rs^{\rightarrow \Ss} =\left(\sum_\Ps\overline{\Fs}^{\rightarrow \Ps} \right)\cdot \vel{O}{R} + \sum_\Ps\overline{\Fs}^{\rightarrow \Ps} \cdot \left(\velang{S}{R} \times \OPvec \right) = \left( \sum_\Ps\overline{\Fs}^{\rightarrow \Ps} \right) \cdot \vel{O}{R} + \left[ \sum_\Ps \left( \OPvec \times \overline{\Fs}^{\rightarrow \Ps} \right)\right] \cdot \velang{S}{R} = \overline{\Fs}\cdot \vel{O}{R} + \overline{\Ms}(\Os)\cdot \velang{S}{R} </math>

This calculation also shows that the power of a moment (or a torque <math>\Gamma</math>) on a rigid body in a reference frame R is the product of the moment by the angular velocity of the rigid body: <math>\dot{\Ws}_\Rs^{\Gamma\rightarrow \mathrm{S}} = \overline{\Gamma} \cdot \velang{S}{R}</math>.

The power balance for a rigid body is: <math>\dot{\Ws}_\Rs^{ \rightarrow \mathrm{S}} = \dot{\Ws}_\Rs^{ \mathrm{ext}} + \dot{\Ws}_\Rs^{ \mathrm{ar}}.</math>

====✏️ EXAMPLE E1-3.1: roller sliding on a slope====
------
:
{|
|
:[[File:ExE1-3-1-1-eng.png|thumb|left|180px|link=]]
|A homogeneous roller, with mass m and radius r, slides down a slope. The friction coefficient between the ground and the roller is <math>\mu</math>. We want to calculate the power of every force acting on the roller in the ground reference frame.
|}

:It is a 2D problem. The kinematic and dynamic descriptions are:
{|
|
[[File:ExE1-3-1-2-eng.png|thumb|right|250px|link=]]

|<math> \hspace{0.5cm}\vel{J}{E}=\vel{C}{T} + \velang{}{E} \times \CJvec= </math>

<math>\hspace{1.7cm}=\left( \rightarrow \vs_\Es \right) + \left( \otimes \Omega_\Es\right) \times \left( \downarrow \rs \right)</math>

<math> \hspace{0.5cm}\vel{J}{E} = \left[\rightarrow (\vs_\Es -\rs\Omega_\Es)\right]</math>

<math>\hspace{0.5cm} \dot{\Ws}_\Es^{\mathrm{ext}} = \dot{\Ws}_\Es^{\mathrm{weight}} + \dot{\Ws}_\Es^{\mathrm{friction}} + \dot{\Ws}_\Es^{\mathrm{constraint}}= \Bigr[\left(\downarrow \ms\gs\cos\beta\right)+\left(\rightarrow \ms\gs\sin\beta\right)\Bigr] \cdot (\rightarrow \vs_\Es) + \Bigr[(\leftarrow \mu \Ns) + (\uparrow \Ns ) \Bigr] \cdot \Bigr[\rightarrow (\vs_\Es -\rs\Omega_\Es)\Bigr]</math>

|}
:Since the roller center has no motion perpendicular to the plane, the AMT leads to <math>\Ns =\ms\gs\cos\beta</math>. Therefore:

{|
|
[[File:ExE1-3-1-3-eng.png|thumb|right|140px|link=]]
|<math>\hspace{0.5cm} \dot{\Ws}_\Ts^{\mathrm{ext}} = \ms\gs\vs_\Ts \sin \beta - \ms\gs(\vs_\Ts -\rs\Omega_\Ts)\cos\beta=\ms\gs \Bigr[ \vs_\Ts (\sin\beta-\cos\beta)+\rs\Omega_\Ts \cos\beta \Bigr].</math>

<math>\hspace{0.5cm}</math>

If instead of considering the friction at '''J''' we consider the equivalent torsor at point '''C''', we must consider the force applied in '''C''' and the moment about '''C''', which is not zero:

<math>\hspace{0.5cm} \dot{\Ws}_\Es^{ \mathrm{friction}} = (\leftarrow \mu\Ns)\cdot(\rightarrow\vs_\Es)+(\otimes \mu\Ns\rs)\cdot(\otimes \Omega_\Es) =\mu\ms\gs\cos\beta(\rs\Omega_\Es-\vs_\Es)</math>,

<math>\hspace{0.5cm} \dot{\Ws}_\Es^{\mathrm{ext}} = \dot{\Ws}_\Es^{ \mathrm{weight}} + \dot{\Ws}_\Es^{ \mathrm{friction}}+ \dot{\Ws}_\Es^{ \mathrm{constraint}}= \ms\gs\vs_\Es\sin\beta+\mu\ms\gs\cos\beta (\rs\Omega_\Es-\vs_\Es)</math>.

<math>\hspace{0.5cm}</math>

The result is the same.
|}


====✏️ EXAMPLE E1-3.2: block sliding on a truck====
------
:
{|
|
:[[File:ExE1-3-2-1-eng.png|thumb|left|220px|link=]]
|A truck brakes with acceleration '''a''' relative to the ground at the instant in which its velocity (also relative to the ground) is v. At the same instant, the homogeneous block, of mass m, slides with velocity v' relative to the truck. We want to perform a power balance for the block in two different reference frames: the ground and the truck.
|}

:Ground reference frame: It is a 2D problem. The kinematic and dynamic descriptions are:
{|
|
[[File:ExE1-3-2-2-eng.png|thumb|right|200px|link=]]

| The constraint between the truck and the block allows only one translational DoF between the two. Therefore, in the 2D approximation, the associated torsor has two independent components. If characterized at the midpoint of the contact surface, these components are a normal force and a moment perpendicular to the plane. The AMT applied to the block leads to <math>\Ns=\ms\gs</math>.

|}
:The powers associated with all interactions on the block in the ground reference are:

:<math>\dot{\Ws}_\Es^{\rightarrow \mathrm{bloque}}=\dot{\Ws}_\Ts^{\mathrm{ext}}=\dot{\Ws}_\Ts^{\mathrm{peso}}+\dot{\Ws}_\Ts^{\mathrm{fricción}}+\dot{\Ws}_\Ts^{\mathrm{enlace}}= \Bigr[(\downarrow \ms\gs)+(\leftarrow \mu\ms\gs)+(\uparrow \Ns) \Bigr] \cdot \Bigr[\rightarrow(\vs+\vs')\Bigr] + (\odot \Ms_\es)\cdot 0 = -\mu\ms\gs(\vs+\vs')</math>

:Ya que el bloque se traslada respecto al suelo, su energía cinética total es inmediata, y se puede plantear fácilmente el balance de potencias:

:<math>\Ts_\Ts^\mathrm{bloque}=\frac{1}{2}\int_\mathrm{bloque}\ds\ms \vs_\Ts^2(\mathrm{bloque})=\frac{1}{2}\ms \vs_\Ts^2(\mathrm{bloc})\quad \Rightarrow \quad \dot{\Ts}_\Ts^\mathrm{bloque}=\ms\vs_\Ts( \mathrm{bloque})\dot{\vs}_\Ts( \mathrm{bloque}) </math>

:Como la única fuerza horizontal sobre el bloque es la fricción:

<math>\dot{\vs}_\Ts(\mathrm{bloque})=-\mu\gs\quad \Rightarrow \quad \dot{\Ts}_\Ts^\mathrm{bloque}=-\mu\ms\gs(\vs+\vs').</math>

:El balance de potencias es consistente:

:<math>\left.\begin{array}{l}
\dot{\mathrm{W}}_\Ts^{\rightarrow \text { bloque }}=\dot{\mathrm{W}}_{\mathrm{T}}^{\text {ext }}=-\mu \mathrm{mg}\left(\mathrm{v}+\mathrm{v}^{\prime}\right) \\
\dot{\mathrm{T}}_{\mathrm{T}}^{\text {bloque }}=-\mu \mathrm{mg}\left(\mathrm{v}+\mathrm{v}^{\prime}\right)
\end{array}\right\} \Rightarrow \dot{\mathrm{W}}_{\mathrm{T}}^{\rightarrow \text { bloque }}=\dot{\mathrm{T}}_{\mathrm{T}}^{\text {bloque }}</math>

:Referencia camión: Al estar el camión acelerado respecto al suelo, hay que tener en cuenta las fuerzas de arrastre sobre todos los diferenciales de masa (dm) del bloque:

{|
|

[[Archivo:ExE1-3-2-3-cat-esp.jpg|thumb|right|250px|link=]]
|<math>\cal{\Fs}^{\rightarrow \ds\ms}_{\mathrm{ar}}=-\ds\ms\overline{\as}_\mathrm{ar}(\ds\ms)=-\ds\ms (\leftarrow \as)=(\rightarrow \ds\ms \cdot \as)</math>.

El torsor resultante de este sistema de fuerzas en el centro de inercia '''G''' se reduce a una fuerza horizontal.

<math>\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{bloque}}=\dot{\Ws}_\mathrm{camión}^{\mathrm{ext}} + \dot{\Ws}_\mathrm{camión}^{\mathrm{ar}} =\dot{\Ws}_\mathrm{camión}^{\mathrm{peso}} + \dot{\Ws}_\mathrm{camión}^{\mathrm{fricción}} + \dot{\Ws}_\mathrm{camión}^{\mathrm{enlace}} + \dot{\Ws}_\mathrm{camión}^{\mathrm{ar}} , </math>

<math>\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{bloque}}= \Bigr[(\downarrow \ms\gs)+(\leftarrow \mu\ms\gs)+(\uparrow \Ns) + (\rightarrow \ms\as)\Bigr] \cdot(\rightarrow \vs')+(\odot\Ms_\es) \cdot0=\ms(\as-\mu\gs)\vs'.</math>
|}

:Se puede comprobar que el balance de potencias es consistente:

:<math>\Ts^\mathrm{bloque}_\mathrm{camión}=\frac{1}{2}\ms\vs^2_\mathrm{camión}(\mathrm{bloque}) \quad \Rightarrow \quad \dot{\Ts}^\mathrm{bloque}_\mathrm{camión} = \ms\vs_\mathrm{camión}(\mathrm{bloque})\dot{\vs}_\mathrm{camión}(\mathrm{bloque})</math>

:<math>\left.\begin{array}{l}
\mathrm{AB}:\mathrm{suelo} \\
\mathrm{REL}:\mathrm{camión}
\end{array}\right\} \quad \dot{\vs}_\mathrm{camión}(\mathrm{bloque})= \dot{\vs}_\mathrm{T}(\mathrm{bloque})-\dot{\vs}_\mathrm{ar}(\mathrm{bloque}) = (-\mu\gs)-(-\as)=\as-\mu\gs \quad \Rightarrow \quad \dot{\Ts}^\mathrm{bloque}_\mathrm{camión}=\ms\vs'(\as-\mu\gs)</math>

:<math>\left.\begin{array}{l}
\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{bloque}}=\dot{\Ws}_\mathrm{camión}^{\mathrm{ext}} + \dot{\Ws}_\mathrm{camión}^{\mathrm{ar}}= \ms(\as-\mu\gs)\vs'\\
\dot{\Ts}^\mathrm{bloque}_\mathrm{camión}=\ms\vs'(\as-\mu\gs)
\end{array}\right\} \Rightarrow\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{bloque}}= \dot{\Ts}^\mathrm{bloque}_\mathrm{camión}</math>


====✏️ EJEMPLO E1-3.3: camión frenando con carga====
------
:
{|
|
:[[Archivo:ExE1-3-3-1-esp.png|thumb|left|220px|link=]]
|Un camión frena con aceleración '''a''' respecto al suelo en el instante en el que su velocidad (también respecto al suelo) es v. Los pares de frenada en el eje delantero y en el trasero son iguales y valen <math>\Gamma</math>, y las ruedas, de radio r y masa despreciable, no deslizan sobre el suelo. En el mismo instante, el bloque homogéneo, de masa m, desliz con velocidad v' relativa al camión. Se trata de hacer un balance de potencias para el sistema (bloque+camión) en la referencia del suelo y en la del camión.

|}

:Ya que la potencia de las interacciones internas no es nula en principio (no se trata de un único sólido rígido), es conveniente hacer un DGI del sistema para no olvidar ninguna:
{|
|
[[Archivo:ExE1-3-3-2-esp.png|thumb|right|250px|link=]]

|
|
c.p.s.d.: contacto puntual sin deslizamiento

c.m.c.d.: contacto multipuntual con deslizamiento

|}
:El sistema sobre el que se hace el balance de potencias es todo menos el suelo. Teniendo en cuenta que la potencia de los enlaces internos es siempre nula, las únicas interacciones que hay que tener en cuenta son el peso, los enlaces de las ruedas con el suelo, los pares de frenada y la fricción entre bloque y camión:

[[Archivo:ExE1-3-3-3-esp.png|thumb|center|300px|link=]]

:Referencia suelo:
:Las fuerzas de enlace del suelo sobre las ruedas están aplicadas a puntos de velocidad cero, y por tanto no hacen potencia. El peso es una fuerza vertical, y el movimiento del camión y del bloque es horizontal, por tanto su potencia también es cero.

:<math>\dot{\Ws}_\Ts^{\rightarrow \mathrm{sist}}=\dot{\Ws}_\Ts^{\mathrm{ext}}+\dot{\Ws}^{\mathrm{int}}=\dot{\Ws}_\Ts^{\mathrm{peso}}+\dot{\Ws}_\Ts^{\mathrm{enlace}}+\dot{\Ws}^{\mathrm{fricción}}+\dot{\Ws}^{\Gamma}=\dot{\Ws}^{\mathrm{fricción}}+\dot{\Ws}^{\Gamma}.</math>

:Las potencias de interacciones internas se pueden calcular en cualquier referencia. Por tanto:

:<math>\dot{\Ws}^{\mathrm{fricción}}= \dot{\Ws}_\mathrm{camión}^{\mathrm{fric}\rightarrow \mathrm{bloque}} +\dot{\Ws}_\mathrm{camió}^{\mathrm{fric}\rightarrow \mathrm{camión}} = \dot{\Ws}_\mathrm{camión}^{\mathrm{fric}\rightarrow \mathrm{bloque}}= (\leftarrow \mu\ms\gs)\cdot (\rightarrow \vs')=-\mu\ms\gs\vs'</math>

:<math>\dot{\Ws}^{\Gamma}= \dot{\Ws}_\mathrm{camión}^{\Gamma\rightarrow \mathrm{ruedas}} +\dot{\Ws}_\mathrm{camión}^{\Gamma\rightarrow \mathrm{camión}} = \dot{\Ws}_\mathrm{camión}^{\Gamma\rightarrow \mathrm{ruedas}}= 2 \left(\odot \Gamma \right)\cdot \left(\otimes \frac{\vs}{\rs}\right)=-2\Gamma\frac{\vs}{\rs}.</math>

:Por otro lado: <math>\Ts_\Ts^\mathrm{sist}=\frac{1}{2}\ms\vs_\Ts^2(\mathrm{bloque})+\frac{1}{2}\ms\vs_\Ts^2(\mathrm{camión})\quad \Rightarrow \quad \dot{\Ts}_\Ts^\mathrm{sist}=-\ms(\vs+\vs')\mu\gs-\Ms\vs\as. </math>

:Finalmente: <math>\dot{\Ws}_\Ts^{\rightarrow \mathrm{sist}}=\dot{\Ts}_\Ts^\mathrm{sist} \quad \Rightarrow \quad -\mu\ms\gs\vs'-2\Gamma\frac{\vs}{\rs}=-\ms(\vs+\vs')\mu\gs-\Ms\vs\as.</math>

:Esta ecuación permite calcular el par de frenada: <math>\Gamma=\frac{1}{2}(\mu\ms\gs+\Ms\as)\rs.</math>

:Referencia camión:

:En esta referencia hay que tener en cuenta las fuerzas de arrastre. Por otro lado, las fuerzas de enlace del suelo sobre las ruedas están aplicadas a puntos que ya no tienen velocidad cero, y por tanto hacen potencia. Como antes, la potencia del peso es cero.

:<math>\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{sist}}=\dot{\Ws}_\mathrm{camión}^{\mathrm{ext}}+\dot{\Ws}^{\mathrm{int}}+\dot{\Ws}_\mathrm{camión}^{\mathrm{ar}}=\dot{\Ws}_\mathrm{camión}^{\mathrm{enlace}}+\dot{\Ws}^{\mathrm{fricción}}+\dot{\Ws}^{\Gamma}+\dot{\Ws}_\mathrm{camión}^{\mathrm{ar}}.</math>

:Las fuerzas de arrastre sobre el camión no hacen potencia porque el camión está parado en esta referencia. La de las fuerzas de arrastre sobre el bloque se ha calculado en el [[E1. Teorema de la energía: versión diferencial#✏️ ✏️ EJEMPLO E1-3.2: bloque deslizando sobre un camión|'''ejemplo E1-3.2''']]. La potencia de la fricción y de los pares de frenada es la misma de antes ya que no depende de la referencia.

:<math>\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{sist}}=\dot{\Ws}_\mathrm{camión}^{\mathrm{enllaç}}+\dot{\Ws}^{\mathrm{fricción}}+\dot{\Ws}^{\Gamma}+\dot{\Ws}_\mathrm{camión}^{\mathrm{ar} \rightarrow \mathrm{bloque}}= \dot{\Ws}_\mathrm{camión}^{\mathrm{enlace}} -\mu\ms\gs\vs'-2\Gamma\frac{\vs}{\rs} +\ms\as\vs'.</math>

:Las descripciones cinemática y dinámica de la rueda son:

[[Archivo:ExE1-3-3-4-esp.png|thumb|center|430px|link=]]

:<math>\dot{\Ws}_\mathrm{camión}^{\mathrm{enlace}}=2\Bigr[(\leftarrow \Ts)+(\uparrow \Ns)\Bigr]\cdot(\leftarrow\vs)=2\Ts\vs.</math>

La fuerza tangencial de enlace se puede calcular a partir del TMC aplicado al centro '''C''' de la rueda. Como no tiene masa: <math>\sum\overline{\Ms}_\mathrm{ext}(\Cs)=\overline{0} \quad \Rightarrow \quad (\odot \Gamma)+(\otimes \Ts\rs)=0\quad \Rightarrow \quad \Ts=\Gamma/\rs.</math>

:Por tanto: <math>\dot{\Ws}_\mathrm{camió}^{\rightarrow \mathrm{sist}} =2\frac{\Gamma}{\rs}\vs-\mu\ms\gs\vs'-2\Gamma\frac{\vs}{\rs}+\ms\as\vs'=\ms(\as-\mu\gs)\vs'.</math>

:Por otro lado, como se ha visto en el [[E1. Teorema de la energía: versión diferencial#✏️ ✏️ EJEMPLO E1-3.2: bloque deslizando sobre un camión|'''ejemplo E1-3.2''']]:

:<math>\Ts_\mathrm{camión}^\mathrm{sist}= \Ts_\mathrm{camión}^\mathrm{bloque}=\frac{1}{2}\ms\vs_\mathrm{camión}^2(\mathrm{bloque}) \quad \Rightarrow \quad \dot{\Ts}_\mathrm{camión}^\mathrm{sist}= \dot{\Ts}_\mathrm{camión}^\mathrm{bloque}=\ms\vs'(\as-\mu\gs).</math>

:El balance de potencias es consistente:

:<math>\left.\begin{array}{l}
\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{sist}}= \ms(\as-\mu\gs)\vs'\\
\dot{\Ts}^\mathrm{sist}_\mathrm{camión}=\ms(\as-\mu\gs)\vs'
\end{array}\right\} \Rightarrow\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{sist}}= \dot{\Ts}^\mathrm{sist}_\mathrm{camión}</math>

-----
-----

==E1.4 Balance de potencias para un sistema multisólido: cálculo directo y cálculo indirecto==

El '''cálculo directo''' (a partir de su definición) de la potencia asociada a una interacción (una fuerza sobre un punto '''P''' o un momento sobre un sólido rígido S) exige el conocimiento de la interacción (fuerza o momento):

<math>\dot{\Ws}_\Rs^{\Fs\rightarrow \Ps}=\overline{\Fs}_{\rightarrow\Ps} \cdot \vel{P}{R} \quad , \quad \dot{\Ws}_\Rs^{\Ms\rightarrow \Ss}=\overline{\Ms}_{\rightarrow\Ss} \cdot \velang{S}{R} </math>

Cuando no se trata de una interacción formulable, hay que recurrir a los teoremas vectoriales para determinarla.

El '''cálculo indirecto''' es una alternativa que no necesita recurrir a estos teoremas. Se basa en la evaluación de un balance de potencias para un sistema y una referencia adecuados de manera que la potencia a determinar sea el único término desconocido de la ecuación (el único que no se puede calcular de manera directa). Cuando el balance incluye más de un término desconocido, se pueden combinar varios balances (para distintos sistemas y referencias) hasta tener un número suficiente de ecuaciones que permitan la determinación de todas las potencias desconocidas.

====✏️ EJEMPLO E1-4.1: polea frenada====
------
:
{|
|
:[[Archivo:ExE1-4-1-1-esp.png|thumb|left|150px|link=]]
|Un bloque de masa m cuelga de un hilo inextensible atado a un punto de la periferia de una polea de radio r. Entre soporte (fijo al suelo) y polea actúa un freno que garantiza que el bloque baja a velocidad constante respecto al suelo. Se trata de calcular la potencia del freno.

|}

:Càlculo directo: <math>\dot{\Ws}^{\mathrm{freno}}= \overline{\Gamma}^{\rightarrow \mathrm{polea}}\cdot\velang{polea}{T}=\left(\odot\Gamma\right)\cdot\left( \otimes \frac{\vs_0}{\rs}\right)=-\Gamma \frac{\vs_0}{\rs}.</math>

:El par de frenada no es un dato, y hay que recurrir al TMC para calcularlo.

:<math>\left.\begin{array}{l}
\text{SISTEMA: polea + bloque}\\
\text{TMC en } \Os
\end{array}\right\} \sum\overline{\Ms}_\mathrm{ext}(\Os)=\dot{\overline{\Hs}}_\mathrm{RTO}(\Os)</math>

:El momento cinético proviene de las dos piezas. Ya que '''O''' no es fijo al bloque, hay que hacer descomposición baricéntrica para calcular el momento cinético del bloque:

:<math>\overline{\Hs}_\mathrm{RTO=T} (\Os) =\overline{\Hs}_\Ts^\mathrm{politja}(\Os) + \overline{\Hs}_\Ts^\mathrm{bloc}(\Os)= \overline{\Hs}_\Ts^\mathrm{politja}(\Os) + \overline{\Hs}_\Ts^\mathrm{bloc}(\Gs_\mathrm{bloc})+ \overline{\Hs}_\Ts^{\oplus\mathrm{bloc}}(\Os) = </math>

:<math>\hspace{2.3cm}=\Is_\mathrm{O}^\mathrm{polea} \velang{polea}{T}+\Is_{\Gs_\mathrm{bloque}}^\mathrm{bloque}+\OGvec_\mathrm{bloque}\times \ms\overline{\vs}_\Ts(\Gs_\mathrm{bloque})=\left(\otimes \Is_\mathrm{O}\frac{\vs_0}{\rs}\right)+(\otimes \ms\vs_0\rs).</math>

:[[Archivo:ExE1-4-1-2-esp.png|thumb|right|180px|link=]]
:Se trata de un momento cinético constante en dirección y valor. Por tanto:

:<math>\dot{\overline{\Hs}}_\mathrm{RTO}(\Os)=0 \quad \Rightarrow \quad \sum{\overline{\Ms}}_\mathrm{ext}(\Os)=(\odot\Gamma)+(\otimes \ms\gs\rs) =0\quad \Rightarrow \quad \Gamma= \ms\gs\rs</math>.

:Finalmente: <math>\dot{\Ws}^{\mathrm{freno}}= -\Gamma \frac{\vs_0}{\rs} = -\ms\gs\vs_0.</math>

:Cálculo indirecto:

:<math>\left.\begin{array}{l}
\text{SISTEMA: polea + soporte + bloque}\\
\text{REF: suelo(Gal)}
\end{array}\right\} \quad \dot{\Ts}^\mathrm{sist}_\Ts=\dot{\Ws}^\mathrm{ext}_\Ts+ \dot{\Ws}^\mathrm{int}</math>

:<math>\left.\begin{array}{l}
\dot{\Ts}^\mathrm{sist}_\Ts=0\\
\dot{\Ws}^\mathrm{ext}_\Ts=\left(\downarrow \ms\gs\right)\cdot\left(\downarrow \vs_0\right)= \ms\gs\vs_0\\
\dot{\Ws}^\mathrm{int}=\dot{\Ws}^{\mathrm{freno}}
\end{array}\right\} \quad \Rightarrow \quad \dot{\Ws}^{\mathrm{freno}} =-\ms\gs\vs_0</math>



====✏️ EJEMPLO E1-4.2: mono y plátanos====
------
:
{|
|
:[[Archivo:ExE1-4-2-1-esp.png|thumb|left|180px|link=]]
|Unos plátanos de masa <math>(2/3)\ms</math> cuelgan de una cuerda inextensible que se apoya sin deslizar sobre una polea de radio r y masa despreciable, articulada respecto al suelo. En el otro extremo de la cuerda se agarra un mono de masa m. La polea gira con velocidad angular constante <math>\Omega_0</math> respecto al suelo bajo la acción de un motor. El mono se mueve respecto a la cuerda sin que sus manos se froten contra la cuerda, y de manera que la distancia de su centro de masas al suelo se mantiene constante. Se trata de calcular la potencia del motor.
|}

:Cálculo directo: <math>\dot{\Ws}^{\mathrm{freno}}= \overline{\Gamma}^{\rightarrow \mathrm{polea}}\cdot\velang{polea}{T}=\left(\odot\Gamma\right)\cdot\left( \odot \Omega_0\right)=\Gamma \Omega_0.</math>

:El par motor no es un dato, y hay que recurrir al TMC para calcularlo.

:<math>\left.\begin{array}{l}
\text{SISTEMA: polea + mono + plátanos}\\
\text{TMC en } \Os
\end{array}\right\} \sum\overline{\Ms}_\mathrm{ext}(\Os)=\dot{\overline{\Hs}}_\mathrm{RTO}(\Os)</math>

:El momento cinético proviene del mono y de los plátanos. En primera aproximación, se puede considerar la masa del mono concentrada en su tronco y su cabeza, de manera que no gira y su velocidad respecto al suelo es nula. Teniendo en cuenta que el punto '''O''' no se mueve respecto al mono pero sí respecto a los plátanos:

:<math>\overline{\Hs}_\mathrm{RTO=T} (\Os) =\overline{\Hs}_\Ts^\mathrm{mono}(\Os) + \overline{\Hs}_\Ts^\mathrm{plátanos}(\Os)= \overline{\Hs}_\Ts^\mathrm{mono}(\Os) + \overline{\Hs}_\Ts^\mathrm{plátanos}(\Gs_\mathrm{plát})+ \OGvec_\mathrm{plát} \times \frac{2}{3} \ms \overline{\vs}_\Ts(\Gs_\mathrm{plát})= </math>

:<math>\hspace{2.3cm}=\OGvec_\mathrm{plát}\times \frac{2}{3} \ms \overline{\vs}_\Ts(\Gs_\mathrm{plát}) = \Bigr[(\rightarrow \rs)+(\downarrow \xs)\Bigr] \times\frac{2}{3} \ms \left(\downarrow \rs\Omega_0\right)=\left(\otimes \frac{2}{3}\ms\rs^2\Omega_0\right).</math>

:Se trata de un momento cinético constante en dirección y valor. Por tanto:

:<math>\dot{\overline{\Hs}}_\mathrm{RTO}(\Os)=0 \quad \Rightarrow \quad \sum{\overline{\Ms}}_\mathrm{ext}(\Os)=(\odot\Gamma)+(\otimes \ms\gs\rs) + \left(\otimes \frac{2}{3}\ms\gs\rs\right)=0\quad \Rightarrow \quad \Gamma= \frac{5}{3}\ms\gs\rs</math>.

:Finalmente: <math>\dot{\Ws}^{\mathrm{motor}}= \Gamma \Omega_0 = \frac{5}{3}\ms\gs\rs\Omega_0.</math>

:Cálculo indirecto

:Hay que tener presente que es un problema con dos tipos de actuadores: el motor y la musculatura del mono. Hay diversos modelos biomecánicos para formular las fuerzas de los músculos, pero quedan fuera del ámbito de este curso. Ya que el mono se mueve, sus músculos desarrollan una potencia no nula.

:<math>\left.\begin{array}{l}
\text{SISTEMA: polea + mono + plátano}\\
\text{REF: suelo(Gal)}
\end{array}\right\} \quad \dot{\Ts}^\mathrm{sist}_\Ts=\dot{\Ws}^\mathrm{ext}_\Ts+ \dot{\Ws}^\mathrm{int}</math>

:<math>\left.\begin{array}{l}
\dot{\Ts}^\mathrm{sist}_\Ts=0\\
\dot{\Ws}^\mathrm{ext}_\Ts=\left(\downarrow \frac{2}{3}\ms\gs\right)\cdot\left(\downarrow \rs\Omega_0\right)= \frac{2}{3}\ms\gs\rs\Omega_0\\
\dot{\Ws}^\mathrm{int}=\dot{\Ws}^{\mathrm{musc}}+\dot{\Ws}^{\mathrm{motor}}
\end{array}\right\} \quad \Rightarrow \quad \dot{\Ws}^{\mathrm{musc}} + \dot{\Ws}^{\mathrm{motor}}=-\frac{2}{3}\ms\gs\rs\Omega_0</math>

:Este balance proporciona una ecuación con dos incógnitas. Se puede plantear un segundo balance para un sistema donde una de las dos no aparezca: un sistema sin el mono o un sistema sin el motor. Si se hace un balance de potencias para el sistema (mono + trozo de cuerda), las fuerzas externas a considerar son el peso y la tensión de la cuerda. Si se sigue trabajando en la referencia del suelo, esta tensión (desconocida) hará potencia no nula ya que está aplicada a un punto de velocidad ascendente.
:[[Archivo:ExE1-4-2-2-cat-esp.png|thumb|right|330px|link=]]

:Para evitar que aparezca un término asociado a la tensión de la cuerda, se puede trabajar en la referencia del trozo de cuerda que se ha considerado, que al tener velocidad ascendente constante es galileana. En esta referencia, el mono baja con velocidad constante:

:<math>\left.\begin{array}{l}
\text{REL: trozo de cuerda}\\
\text{AB: suelo}
\end{array}\right\}</math>

:<math>\overline{\vs}_\mathrm{REL}(\mathrm{mono})=\overline{\vs}_\mathrm{ar}(\mathrm{mono})=-\overline{\vs}_\mathrm{ar}(\mathrm{mono})=(\downarrow \rs\Omega_0)</math>

:<math>\left.\begin{array}{l}
\text{SISTEMA: mono+ trozo de cuerda}\\
\text{REF: trozo de cuerda(Gal)}
\end{array}\right\} \quad \dot{\Ts}^\mathrm{sist}_\mathrm{cuerda} =\dot{\Ws}^\mathrm{ext}_\mathrm{cuerda}+ \dot{\Ws}^\mathrm{int}</math>

:<math>\left.\begin{array}{l}
\dot{\Ts}^\mathrm{sist}_\mathrm{cuerda}=0\\
\dot{\Ws}^\mathrm{ext}_\mathrm{cuerda}=\left(\downarrow \ms\gs\right)\cdot\left(\downarrow \rs\Omega_0\right)= \ms\gs\rs\Omega_0\\
\dot{\Ws}^\mathrm{int}=\dot{\Ws}^{\mathrm{musc}}
\end{array}\right\} \quad \Rightarrow \quad \dot{\Ws}^{\mathrm{musc}} =-\ms\gs\rs\Omega_0</math>

:Combinando este resultado con el del balance anterior:

:<math>\dot{\Ws}^\mathrm{motor}=-\frac{2}{3}\ms\gs\rs\Omega_0-\dot{\Ws}^\mathrm{musc}=-\frac{5}{3}\ms\gs\rs\Omega_0</math>



© Universitat Politècnica de Catalunya. [[Mecánica:Derechos de autor |Todos los derechos reservados]]
-------------

<center>
[[D8. Conservaciones|<<< D8. Conservaciones]]

[[E2. Teorema de la energía: versión integrada|E2. Teorema de la energía: versión integrada>>>]]
</center>

E1. Work-Energy Theorem: differential form

2025-05-20T12:18:31Z

Bros: /* ✏️ EXAMPLE E1-1.2: pendulum */

<div class="noautonum">__TOC__</div>
<math>\newcommand{\uvec}{\overline{\textbf{u}}}
\newcommand{\vvec}{\overline{\textbf{v}}}
\newcommand{\evec}{\overline{\textbf{e}}}
\newcommand{\Omegavec}{\overline{\mathbf{\Omega}}}
\newcommand{\velang}[2]{\Omegavec^{\textrm{#1}}_{\textrm{#2}}}
\newcommand{\Alfavec}{\overline{\mathbf{\alpha}}}
\newcommand{\accang}[2]{\Alfavec^{\textrm{#1}}_{\textrm{#2}}}
\newcommand{\ds}{\textrm{d}}
\newcommand{\ps}{\textrm{p}}
\newcommand{\ns}{\textrm{n}}
\newcommand{\hs}{\textrm{h}}
\newcommand{\cs}{\textrm{c}}
\newcommand{\gs}{\textrm{g}}
\newcommand{\Ns}{\textrm{N}}
\newcommand{\Hs}{\textrm{H}}
\newcommand{\Fs}{\textrm{F}}
\newcommand{\ms}{\textrm{m}}
\newcommand{\ts}{\textrm{t}}
\newcommand{\us}{\textrm{u}}
\newcommand{\vs}{\textrm{v}}
\newcommand{\Rs}{\textrm{R}}
\newcommand{\Es}{\textrm{E}}
\newcommand{\Ws}{\textrm{W}}
\newcommand{\Ts}{\textrm{T}}
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\newcommand{\es}{\textrm{e}}
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\newcommand{\is}{\textrm{i}}
\newcommand{\Is}{\textrm{I}}
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\newcommand{\js}{\textrm{j}}
\newcommand{\rs}{\textrm{r}}
\newcommand{\as}{\textrm{a}}
\newcommand{\ss}{\textrm{s}}
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\newcommand{\Gs}{\textbf{G}}
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\newcommand{\deg}{^\textsf{o}}
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\newcommand{\vec}[1]{\overline{#1}}
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\newcommand{\GPvec}{\vec{\Gs\Ps}}
\newcommand{\PSvec}{\vec{\Ps\Ss}}
\newcommand{\QQvec}{\vec{\Qs\Qs}}
\newcommand{\QGvec}{\vec{\Qs\Gs}}
\newcommand{\QPvec}{\vec{\Qs\Ps}}
\newcommand{\JQvec}{\vec{\Js\Qs}}
\newcommand{\PQvec}{\vec{\Ps\Qs}}
\newcommand{\GJvec}{\vec{\Gs\Js}}
\newcommand{\OPvec}{\vec{\Os\textbf{P}}}
\newcommand{\OCvec}{\vec{\Os\Cs}}
\newcommand{\OGvec}{\vec{\Os\Gs}}
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\newcommand{\GCvec}{\vec{\Gs\Cs}}
\newcommand{\PGvec}{\vec{\Ps\Gs}}
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\begin{bmatrix}
{#1} & {#2} & {#3}\\
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{#7} & {#8} & {#9}
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</math>

The '''Work-Energy Theorem''' plays a prominent role in physics, as it bridges all branches of science (electromagnetism, thermodynamics, etc.): all of them deal with energy.

There is a certain parallel between the Vector Theorems and the '''Work-Energy Theorem''': they all allow us to understand the evolution of a magnitude that depends on the mechanical state of the system. In the case of the former, we are dealing with the instantaneous evolution of vector magnitudes (linear momentum, angular momentum), while the latter deals with the finite change (in a finite time interval) of a scalar magnitude ('''mechanical energy''').

This unit presents the differential form of this theorem (the '''power balance'''). The integrated form (which is the Work-Energy Theorem itself) is presented in [[E2. Teorema de la energía: versión integrada|'''unit E2''']].
 
------
------
==E1.1 Power balance in a system of particles==

:[[File:E1-1-eng.png|thumb|right|300px|link=]]
<center>'''Figure E1.1''' Forces on a particle of a system with constant matter</center>

Let us consider a system of constant matter, and let’s apply Newton's second law to each particle (or each mass differential) of this system in a general reference frame R (which may be non-Galilean) ('''Figure E1.1'''):

<math>\F{\rightarrow \Ps} + \Fcal{}{\text{ar}\rightarrow\Ps} + \Fcal{}{\text{Cor}\rightarrow\Ps} =\ms_\Ps\acc{P}{R}</math>

<math>\F{\rightarrow \Ps} </math> is the resultant interaction force on '''P''', and it comes from internal interactions (with particles of the same system) and external ones:

<math>\F{\rightarrow \Ps}=\F{\rightarrow \Ps}^\text{int}+\F{\rightarrow \Ps}^\text{ext}</math>

If both sides of the equation are multiplied by <math>\vel{P}{R}</math>, all forces (interaction and inertia forces) orthogonal to <math>\vel{P}{R}</math> are filtered out. Specifically, since <math>\acc{P}{R}=\accs{P}{R} + \accn{P}{R} </math>, where <math>\accn{P}{R} \perp \vel{P}{R}</math>, and since the Coriolis force is always orthogonal to <math>\vel{P}{R} (\Fcal{}{\text{Cor}\rightarrow\Ps} = -2\ms_\Ps \velang{R}{Gal} \times \vel{P}{R})</math>:

<math>\vel{P}{R} \cdot \left( \F{\rightarrow \Ps}^\text{int}+\F{\rightarrow \Ps}^\text{ext} + \Fcal{}{\text{ar}\rightarrow\Ps} \right) = \ms_\Ps\accs{P}{R} \cdot \vel{P}{R} = \frac{\ds}{\ds \ts} \left(\frac{1}{2}\ms_\Ps \vvec_\Rs^2(\Ps) \right) \equiv \frac{\ds\Ts_\Rs(\Ps)}{\ds\ts}=\dot{\Ts}_\Rs(\Ps)</math>

The scalar function <math>\Ts_\Rs(\Ps)</math> is the '''kinetic energy''' of the particle, and is a function of the [[D1. Foundational laws of Newtonian dynamics#D1.3 Newton’s Principle of Determinacy|'''mechanical state''']] of '''P'''. The scalar product of a force and the velocity of its application point is the '''force power''': <math>\vel{P}{R} \cdot \F{\rightarrow \Ps} \equiv \dot{\Ws}_\Rs^{\F{\rightarrow \Ps}} </math>. Both scalar quantities <math>(\Ts_\Rs(\Ps), \dot{\Ws}_\Rs^{\F{\rightarrow \Ps}})</math> depend on the reference frame. For the same set of interaction forces acting on '''P''', the reference frame determines which are filtered out and which are not.

The equation can be rewritten as <math>\dot{\Ws}_\Rs^{\text{ext}\rightarrow \Ps}+\dot{\Ws}_\Rs^{\text{int}\rightarrow \Ps}+\dot{\Ws}_\Rs^{\text{ar}\rightarrow \Ps} =\dot{\Ts}_\Rs(\Ps)</math>, and constitutes the '''power balance''' for particle '''P''' in the R reference frame. Summing for all particles in the system:

<math>\dot{\Ws}_\Rs^{\text{ext}}+\dot{\Ws}^{\text{int}}+\dot{\Ws}_\Rs^{\text{ar}} =\dot{\Ts}_\Rs^\text{syst}</math>

The calculation of kinetic energy is covered in [[E2. Teorema de la energía: versión integrada|'''unit E2''']]. This unit focuses on the calculation of powers. [[E1. Work-Energy Theorem: differential form#E1.2 Potencia de una pareja de acción-reacción|'''section E1-2''']] shows that the power of internal interactions does not depend on the reference frame (which is why the subscript R does not appear in the <math>\dot{\Ws}^{\text{int}}</math> term of the previous equation).

====✏️ EXAMPLE E1-1.1: free particle====
------
:
{|
:[[File:ExE1-1-1-1-eng.jpg|thumb|left|180px|link=]]
|A particle '''P''', with mass m, is released from the ring of a Ferris wheel when it is at its highest point. We want to calculate the power, at this instant, of all forces acting on the particle from three different reference frames: the ground, the ring, and the Ferris wheel cabin.
|}

:Ground reference frame: Since this is a Galilean reference frame, all forces on '''P''' are interaction forces.
{|
:[[File:ExE1-1-2- eng.png|thumb|right|230px|link=]]

|<math>\hspace{0.5cm}\sum\dot{\Ws}_\Ts=\dot{\Ws}_\Ts^\text{weight}= \left(\leftarrow \Rs\Omega_0\right) \cdot \left(\downarrow \ms\gs\right)=0</math>
|}

:Ring reference frame: It is a non-Galilean reference, and the transportation force on '''P''' must be taken into account.
{|
:[[File:ExE1-1-3-eng.png|thumb|right|250px|link=]]
|<math>\hspace{0.5cm}</math> AB:ground, REL:ring 

<math>\hspace{0.5cm}\vel{P}{REL}=\vel{P}{AB}-\vel{P}{ar}= \left(\leftarrow \Rs\Omega_0\right)-\left(\leftarrow \Rs\Omega_0\right)=\overline{0}</math> 

<math>\hspace{0.5cm}</math>

Since the velocity is zero, the total power is zero.
|}

:Cabin reference frame: It is also a non-Galilean reference, and the transportation force on '''P''' must be taken into account.

{|
:[[File:ExE1-1-4-eng.png|thumb|right|250px|link=]]
|<math>\hspace{0.5cm}</math>AB:ground, REL:cabin

<math>\hspace{0.5cm}\vel{P}{REL}=\vel{P}{AB}-\vel{P}{ar}= \left(\leftarrow \Rs\Omega_0\right)-\left(\uparrow \Rs\Omega_0\right)</math>

<math>\hspace{0.5cm}\dot{\Ws}_\text{cabin}^\text{weight}=\left[ \left(\leftarrow \Rs\Omega_0\right)+\left(\downarrow \Rs\Omega_0\right) \right] \cdot \left(\downarrow\ms\gs\right) =\ms\gs\Rs\Omega_0</math>

<math>\hspace{0.5cm}\dot{\Ws}_\text{cabin}^\text{ar}=\left[ \left(\leftarrow \Rs\Omega_0\right)+\left(\downarrow \Rs\Omega_0\right) \right] \cdot \left(\rightarrow\ms\Rs^2\Omega_0^2\right) =-\ms\gs\Rs^2 \Omega_0^3</math>

<math>\hspace{0.5cm}\sum\dot{\Ws}_\text{cabin}=\ms\Rs\Omega_0(\gs-\Rs\Omega_0^2)</math>

|}


====✏️ EXAMPLE E1-1.2: pendulum====
------
:
{|
:[[File:ExE1-2-1-eng.png|thumb|left|180px|link=]]
|A particle '''P''' of mass m hangs from an inextensible thread of length L attached to the ceiling. The assembly oscillates in a vertical plane fixed to the ground. We want to calculate the power of all the forces acting on the particle in two different reference frames: the ground and a platform rotating relative to the ground with a constant angular velocity <math>\Omega_0</math>. 
|}
 

 

Ground reference frame: Since it is a Galilean reference frame, '''P''' is only subjected to interaction forces.
{|
:[[File:ExE1-2-2-neut.png|thumb|left|300px|link=]]

|<math>\dot{\Ws}_\Es^\Ts=(\nearrow\Ls\dot{\theta})\cdot(\nwarrow\Ts)=0</math>

<math>\dot{\Ws}_\Es^\text{weight}=(\nearrow\Ls\dot{\theta}\cdot(\downarrow \ms\gs)=-\ms\gs\Ls\dot{\theta}\sin\theta</math>
|}

Platform reference frame: It is a non Galilean reference frame, and the transportation force on '''P''' has to be taken into account.

{|
:[[File:ExE1-2-3-eng.png|thumb|left|400px|link=]]

|AB:ground, REL:platform

<math>\vel{P}{REL}=\vel{P}{AB}-\vel{P}{ar}=(\nearrow\Ls\dot{\theta})-(\otimes \Ls \Omega_0 \sin \theta)</math>

<math>\dot{\Ws}_\Es^\Ts\left[(\nearrow\Ls\dot{\theta})+(\odot\Ls\Omega_0\sin\theta)\right]\cdot(\nwarrow\Ts)=0</math>

<math>\dot{\Ws}_\Es^\text{weight}\left[(\nearrow\Ls\dot{\theta})+(\odot\Ls\Omega_0\sin\theta)\right]\cdot(\downarrow \ms\gs)=-\ms\gs\Ls\dot{\theta}\sin\theta</math>

<math>\dot{\Ws}_\Es^\text{ar}\left[(\nearrow\Ls\dot{\theta})+(\odot\Ls\Omega_0\sin\theta)\right]\cdot(\rightarrow \ms\Ls\Omega_0^2\sin\theta)=\ms\Ls^2\Omega_0^2\dot{\theta}\cos\theta</math>
|}



------
------

==E1.2 Power of an action-reaction pair==

The total power associated with internal interactions in a constant-mass system is not zero in principle (since the two forces of each action-reaction pair are applied to particles that may have different velocities relative to the reference frame in which the calculation is made), and it is independent of the reference frame in which it is evaluated. It only depends on the separation velocity of the two particles. The power balance for the system is therefore written as: <math>\dot{\Ts}_\Rs^\text{syst}=\dot{\Ws}_\Rs^\text{ext} + \dot{\Ws}^\text{int} + \dot{\Ws}_\Rs^\text{ar} </math> .

<div>
=====💭 PROOF ➕=====

{|
:[[File:E1-2-neut.png|thumb|left|200px|link=]]
|<math>\dot{\Ws}^\text{ac-reac}=\overline{\Fs}_{\Ps \rightarrow \Qs}\cdot \vel{Q}{R} + \overline{\Fs}_{\Qs \rightarrow \Ps}\cdot \vel{P}{R}= \overline{\Fs}_{\Ps \rightarrow \Qs}\cdot \left[ \vel{Q}{R}-\vel{P}{R}\right]</math>

By the [[D1. Foundational laws of Newtonian dynamics#D1.6 Newton’s third law (action-reaction principle)|'''action-reaction principle''']], <math>\overline{\Fs}_{\Ps \rightarrow \Qs}=-\overline{\Fs}_{\Qs \rightarrow \Ps}</math>. Therefore: <math>\dot{\Ws}^\text{ac-reac}=\overline{\Fs}_{\Ps \rightarrow \Qs}\cdot \left[ \vel{Q}{R}-\vel{P}{R}\right]</math>

As <math>\overline{\Fs}_{\Ps \rightarrow \Qs}</math> has the <math>\PQvec</math> direction, only the components of the velocities in that direction have to be considered:

<math>\dot{\Ws}^\text{ac-reac}=\overline{\Fs}_{\Ps \rightarrow \Qs}\cdot \biggl( \vel{Q}{R} \Bigr]_{||\Ps\Qs}-\ \vel{P}{R} \Bigr]_{||\Ps\Qs}\biggl)</math>
|}

The difference <math>\biggl( \vel{Q}{R} \Bigr]_{\Ps\Qs}-\ \vel{P}{R} \Bigr]_{\Ps\Qs}\biggl)</math> is the separation velocity <math>\dot{\rho}</math> between '''P''' and '''Q''', and it does not dpend on the reference frame since it is the derivative of a scalar (the distance <math>\dot{\rho}</math> between the two particles). For the previous drawing, <math>\dot{\rho}> 0</math>. Since this is an attractive interaction, <math>\dot{\Ws}^\text{ac-reac} < 0</math>.

With a drawing, it is easy to see that:
* attractive interaction and particle approach <math>(\dot{\rho}<0)</math> : <math>\dot{\Ws}^\text{ac-reac} > 0</math> ;
* repulsive interaction and particle separation <math>(\dot{\rho}>0)</math> : <math>\dot{\Ws}^\text{ac-reac} > 0</math>
* repulsive interaction and particle approach <math>(\dot{\rho}<0)</math> : <math>\dot{\Ws}^\text{ac-reac} < 0</math>


</div>

There are two particular cases of interaction forces whose power always has the same sign:

* Friction action-reaction pair between '''P''' and '''Q''': since the friction force always opposes the relative velocity between the two points, its power is always negative:

<math>\hspace{1cm} \dot{\Ws}^\text{friction}=\dot{\Ws}^{\text{fric}\rightarrow\Ps}_\text{RTQ}+\dot{\Ws}^{\text{fric}\rightarrow\Qs}_\text{RTQ}=\overline{\Fs}^{\text{fric}\rightarrow\Ps} \cdot \vel{P}{RTQ} = - \abs{\overline{\Fs}^{\text{fric}\rightarrow\Ps}} \frac{\vel{P}{RTQ}}{\abs{\vel{P}{RTQ}}} \cdot \vel{P}{RTQ}<0;</math>

* Constraint action-reaction pair between '''P''' and '''Q''': from the [[D3. Interactions between rigid bodies#D3.4 Direct constraint interactions|'''analytical characterization equation''']] of the constraint forces between two points '''P''' and '''Q''' belonging to two rigid bodies <math>\mathrm{S}_\Ps</math> and <math>\mathrm{S}_\Qs</math>, it is easy to see that the associated power is zero:

<math>\hspace{1cm} \dot{\Ws}^\text{constraint}=\dot{\Ws}^{\text{constraint}\rightarrow\Ps}_{\mathrm{S}_\Ps}+\dot{\Ws}^{\text{constraint}\rightarrow\Qs}_{\mathrm{S}_\Ps}=\overline{\Fs}^{\text{constraint}\rightarrow\Ps} \cdot \overline{\mathbf{v}}_{\mathrm{S}_\Qs}(\Ps) = - 0;</math>

====✏️ EXAMPLE E1-2.1: power of an action-reaction friction pair====
------
:
{|
:[[File:ExE1-2-1-2-eng.png|thumb|left|180px|link=]]
|A homogeneous wheel, with radius r and mass m, moves down an inclined plane without sliding until it joins a conveyor belt moving at a constant velocity <math>\vs_0</math> relative to the ground. At the moment it makes contact with the belt, the center of the wheel has a velocity <math>\vs_0</math> relative to the ground in the same direction as the belt. We want to calculate the power of the friction force between the belt and the wheel at this instant.
|}

At that momento, the <math>\mathbf{ICR}^\text{wheel}_\Es</math> is point <math>\Js_\text{wheel}</math> in contact with the belt. This point slides backwards with respect to the belt, therefore the friction force on the wheel is forwards:

<math>\left.\begin{array}{l}
\text { AB: ground } \\
\text { REL: belt }
\end{array}\right\} \quad \bar{\vs}_{\text {REL }}\left(\Js_{\text {wheel }}\right)=\bar{\vs}_{\text{AB}}\left(\Js_{\text {wheel }}\right)-\bar{\vs}_{\mathrm{ar}}\left(\Js_{\text {wheel }}\right)=-\bar{\vs}_{\mathrm{ar}}\left(\Js_{\text {wheel }}\right)=-\left(\rightarrow \mathrm{v}_0\right)=\left(\leftarrow \mathrm{v}_0\right) .</math>

[[File:ExE1-2-1-3-eng.png|thumb|center|500px|link=]]

Calculation of the friction power: <math>\dot{\Ws}^\text{friction}=\dot{\Ws}_\Rs^{\text{friction}\rightarrow \text{wheel}}+ \dot{\Ws}_\Rs^{\text{friction}\rightarrow \text{belt}}</math>, where R may be any reference frame. To quickly calculate this power, it is advisable to choose a reference frame R where one of the two points subjected to the action-reaction pair has zero speed, so that one of the two terms is directly zero:

<math>\dot{\Ws}^\text{friction}=\dot{\Ws}_\text{belt}^{\text{fric}\rightarrow \text{wheel}}+ \dot{\Ws}_\text{belt}^{\text{fric }\rightarrow \text{belt}}=\overline{\Fs}^{\text{fric}\rightarrow \text{wheel}} \cdot \bar{\vs}_{\text {belt}}\left(\Js_{\text {wheel }}\right)=(\rightarrow \mu\ms\gs)\cdot(\leftarrow \vs_0)=-\mu\ms\gs\vs_0.</math>

If the calculation is done in any other reference frame, the result is the same:

<math>\dot{\Ws}^\text{friction}=\dot{\Ws}_\text{E}^{\text{fric}\rightarrow \text{wheel}}+ \dot{\Ws}_\text{E}^{\text{fric} \rightarrow \text{belt}}=(\rightarrow \mu\ms\gs)\cdot 0 + (\leftarrow \mu\ms\gs)\cdot(\rightarrow \vs_0)=-\mu\ms\gs\vs_0.</math>

<math>\dot{\Ws}^\text{friction}=\dot{\Ws}_\text{vehicle}^{\text{fric}\rightarrow \text{wheel}}+ \dot{\Ws}_\text{vehicle}^{\text{fric} \rightarrow \text{belt}}=(\rightarrow \mu\ms\gs)\cdot (\rightarrow\vs_0) + (\leftarrow \mu\ms\gs) \cdot(\rightarrow 3\vs_0)=-\mu\ms\gs\vs_0.</math>

However, if only the power of one of the two forces is calculated (for example, the friction on the wheel), the result depends on the reference frame and can have any sign:

<math>\dot{\Ws}_\text{belt}^{\text{fric }\rightarrow \text{wheel}}=(\rightarrow \mu\ms\gs)\cdot(\leftarrow \vs_0)=-\mu\ms\gs\vs_0 ,</math>

<math>\dot{\Ws}_\text{E}^{\text{fric} \rightarrow \text{wheel}}=(\rightarrow \mu\ms\gs)\cdot 0 = 0 ,</math>

<math>\dot{\Ws}_\text{vehicle}^{\text{fric} \rightarrow \text{wheel}}=(\rightarrow \mu\ms\gs)\cdot (\rightarrow\vs_0)=\mu\ms\gs\vs_0.</math>


-----
-----

==E1.3 Power of a system of forces on a rigid body==

{|
|
:[[File:E1-3-eng.png|thumb|left|180px|link=]]
<center>'''Figure E1.3''': Kinematic torsor of a rigid body in a reference frame R, and forces on different points (dm) of that rigid body</center>

|When calculating the power of a system of forces on a rigid body S, the term associated with internal interactions is zero because those interactions are all constraint (cohesion) forces and, as seen in [[E1. Work-Energy Theorem: differential form#E1.2 Power of an action-reaction pair|'''section E1.2''']], their total power is zero.

The total power of a system of forces on S can be calculated from the torsor of the system at any point '''O''' of the rigid body, <math>(\overline{\Fs},\overline{\Ms}(\Os))</math>. Taking into account rigid body kinematics ('''Figure E1.3'''):

<math>\dot{\Ws}_\Rs^{\rightarrow \Ss} =\sum_\Ps\overline{\Fs}^{\rightarrow \Ps} \cdot \vel{P}{R}= \sum_\Ps \left[ \overline{\Fs}^{\rightarrow \Ps} \cdot \vel{O}{R} + \velang{S}{R} \times \OPvec \right] </math>

|}
<math>\dot{\Ws}_\Rs^{\rightarrow \Ss} =\left(\sum_\Ps\overline{\Fs}^{\rightarrow \Ps} \right)\cdot \vel{O}{R} + \sum_\Ps\overline{\Fs}^{\rightarrow \Ps} \cdot \left(\velang{S}{R} \times \OPvec \right) = \left( \sum_\Ps\overline{\Fs}^{\rightarrow \Ps} \right) \cdot \vel{O}{R} + \left[ \sum_\Ps \left( \OPvec \times \overline{\Fs}^{\rightarrow \Ps} \right)\right] \cdot \velang{S}{R} = \overline{\Fs}\cdot \vel{O}{R} + \overline{\Ms}(\Os)\cdot \velang{S}{R} </math>

This calculation also shows that the power of a moment (or a torque <math>\Gamma</math>) on a rigid body in a reference frame R is the product of the moment by the angular velocity of the rigid body: <math>\dot{\Ws}_\Rs^{\Gamma\rightarrow \mathrm{S}} = \overline{\Gamma} \cdot \velang{S}{R}</math>.

The power balance for a rigid body is: <math>\dot{\Ws}_\Rs^{ \rightarrow \mathrm{S}} = \dot{\Ws}_\Rs^{ \mathrm{ext}} + \dot{\Ws}_\Rs^{ \mathrm{ar}}.</math>

====✏️ EXAMPLE E1-3.1: roller sliding on a slope====
------
:
{|
|
:[[File:ExE1-3-1-1-eng.png|thumb|left|180px|link=]]
|A homogeneous roller, with mass m and radius r, slides down a slope. The friction coefficient between the ground and the roller is <math>\mu</math>. We want to calculate the power of every force acting on the roller in the ground reference frame.
|}

:It is a 2D problem. The kinematic and dynamic descriptions are:
{|
|
[[File:ExE1-3-1-2-eng.png|thumb|right|250px|link=]]

|<math> \hspace{0.5cm}\vel{J}{E}=\vel{C}{T} + \velang{}{E} \times \CJvec= </math>

<math>\hspace{1.7cm}=\left( \rightarrow \vs_\Es \right) + \left( \otimes \Omega_\Es\right) \times \left( \downarrow \rs \right)</math>

<math> \hspace{0.5cm}\vel{J}{E} = \left[\rightarrow (\vs_\Es -\rs\Omega_\Es)\right]</math>

<math>\hspace{0.5cm} \dot{\Ws}_\Es^{\mathrm{ext}} = \dot{\Ws}_\Es^{\mathrm{weight}} + \dot{\Ws}_\Es^{\mathrm{friction}} + \dot{\Ws}_\Es^{\mathrm{constraint}}= \Bigr[\left(\downarrow \ms\gs\cos\beta\right)+\left(\rightarrow \ms\gs\sin\beta\right)\Bigr] \cdot (\rightarrow \vs_\Es) + \Bigr[(\leftarrow \mu \Ns) + (\uparrow \Ns ) \Bigr] \cdot \Bigr[\rightarrow (\vs_\Es -\rs\Omega_\Es)\Bigr]</math>

|}
:Since the roller center has no motion perpendicular to the plane, the AMT leads to <math>\Ns =\ms\gs\cos\beta</math>. Therefore:

{|
|
[[File:ExE1-3-1-3-eng.png|thumb|right|140px|link=]]
|<math>\hspace{0.5cm} \dot{\Ws}_\Ts^{\mathrm{ext}} = \ms\gs\vs_\Ts \sin \beta - \ms\gs(\vs_\Ts -\rs\Omega_\Ts)\cos\beta=\ms\gs \Bigr[ \vs_\Ts (\sin\beta-\cos\beta)+\rs\Omega_\Ts \cos\beta \Bigr].</math>

<math>\hspace{0.5cm}</math>

If instead of considering the friction at '''J''' we consider the equivalent torsor at point '''C''', we must consider the force applied in '''C''' and the moment about '''C''', which is not zero:

<math>\hspace{0.5cm} \dot{\Ws}_\Es^{ \mathrm{friction}} = (\leftarrow \mu\Ns)\cdot(\rightarrow\vs_\Es)+(\otimes \mu\Ns\rs)\cdot(\otimes \Omega_\Es) =\mu\ms\gs\cos\beta(\rs\Omega_\Es-\vs_\Es)</math>,

<math>\hspace{0.5cm} \dot{\Ws}_\Es^{\mathrm{ext}} = \dot{\Ws}_\Es^{ \mathrm{weight}} + \dot{\Ws}_\Es^{ \mathrm{friction}}+ \dot{\Ws}_\Es^{ \mathrm{constraint}}= \ms\gs\vs_\Es\sin\beta+\mu\ms\gs\cos\beta (\rs\Omega_\Es-\vs_\Es)</math>.

<math>\hspace{0.5cm}</math>

The result is the same.
|}


====✏️ EXAMPLE E1-3.2: block sliding on a truck====
------
:
{|
|
:[[File:ExE1-3-2-1-eng.png|thumb|left|220px|link=]]
|A truck brakes with acceleration '''a''' relative to the ground at the instant in which its velocity (also relative to the ground) is v. At the same instant, the homogeneous block, of mass m, slides with velocity v' relative to the truck. We want to perform a power balance for the block in two different reference frames: the ground and the truck.
|}

:Ground reference frame: It is a 2D problem. The kinematic and dynamic descriptions are:
{|
|
[[File:ExE1-3-2-2-eng.png|thumb|right|200px|link=]]

| The constraint between the truck and the block allows only one translational DoF between the two. Therefore, in the 2D approximation, the associated torsor has two independent components. If characterized at the midpoint of the contact surface, these components are a normal force and a moment perpendicular to the plane. The AMT applied to the block leads to <math>\Ns=\ms\gs</math>.

|}
:The powers associated with all interactions on the block in the ground reference are:

:<math>\dot{\Ws}_\Es^{\rightarrow \mathrm{bloque}}=\dot{\Ws}_\Ts^{\mathrm{ext}}=\dot{\Ws}_\Ts^{\mathrm{peso}}+\dot{\Ws}_\Ts^{\mathrm{fricción}}+\dot{\Ws}_\Ts^{\mathrm{enlace}}= \Bigr[(\downarrow \ms\gs)+(\leftarrow \mu\ms\gs)+(\uparrow \Ns) \Bigr] \cdot \Bigr[\rightarrow(\vs+\vs')\Bigr] + (\odot \Ms_\es)\cdot 0 = -\mu\ms\gs(\vs+\vs')</math>

:Ya que el bloque se traslada respecto al suelo, su energía cinética total es inmediata, y se puede plantear fácilmente el balance de potencias:

:<math>\Ts_\Ts^\mathrm{bloque}=\frac{1}{2}\int_\mathrm{bloque}\ds\ms \vs_\Ts^2(\mathrm{bloque})=\frac{1}{2}\ms \vs_\Ts^2(\mathrm{bloc})\quad \Rightarrow \quad \dot{\Ts}_\Ts^\mathrm{bloque}=\ms\vs_\Ts( \mathrm{bloque})\dot{\vs}_\Ts( \mathrm{bloque}) </math>

:Como la única fuerza horizontal sobre el bloque es la fricción:

<math>\dot{\vs}_\Ts(\mathrm{bloque})=-\mu\gs\quad \Rightarrow \quad \dot{\Ts}_\Ts^\mathrm{bloque}=-\mu\ms\gs(\vs+\vs').</math>

:El balance de potencias es consistente:

:<math>\left.\begin{array}{l}
\dot{\mathrm{W}}_\Ts^{\rightarrow \text { bloque }}=\dot{\mathrm{W}}_{\mathrm{T}}^{\text {ext }}=-\mu \mathrm{mg}\left(\mathrm{v}+\mathrm{v}^{\prime}\right) \\
\dot{\mathrm{T}}_{\mathrm{T}}^{\text {bloque }}=-\mu \mathrm{mg}\left(\mathrm{v}+\mathrm{v}^{\prime}\right)
\end{array}\right\} \Rightarrow \dot{\mathrm{W}}_{\mathrm{T}}^{\rightarrow \text { bloque }}=\dot{\mathrm{T}}_{\mathrm{T}}^{\text {bloque }}</math>

:Referencia camión: Al estar el camión acelerado respecto al suelo, hay que tener en cuenta las fuerzas de arrastre sobre todos los diferenciales de masa (dm) del bloque:

{|
|

[[Archivo:ExE1-3-2-3-cat-esp.jpg|thumb|right|250px|link=]]
|<math>\cal{\Fs}^{\rightarrow \ds\ms}_{\mathrm{ar}}=-\ds\ms\overline{\as}_\mathrm{ar}(\ds\ms)=-\ds\ms (\leftarrow \as)=(\rightarrow \ds\ms \cdot \as)</math>.

El torsor resultante de este sistema de fuerzas en el centro de inercia '''G''' se reduce a una fuerza horizontal.

<math>\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{bloque}}=\dot{\Ws}_\mathrm{camión}^{\mathrm{ext}} + \dot{\Ws}_\mathrm{camión}^{\mathrm{ar}} =\dot{\Ws}_\mathrm{camión}^{\mathrm{peso}} + \dot{\Ws}_\mathrm{camión}^{\mathrm{fricción}} + \dot{\Ws}_\mathrm{camión}^{\mathrm{enlace}} + \dot{\Ws}_\mathrm{camión}^{\mathrm{ar}} , </math>

<math>\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{bloque}}= \Bigr[(\downarrow \ms\gs)+(\leftarrow \mu\ms\gs)+(\uparrow \Ns) + (\rightarrow \ms\as)\Bigr] \cdot(\rightarrow \vs')+(\odot\Ms_\es) \cdot0=\ms(\as-\mu\gs)\vs'.</math>
|}

:Se puede comprobar que el balance de potencias es consistente:

:<math>\Ts^\mathrm{bloque}_\mathrm{camión}=\frac{1}{2}\ms\vs^2_\mathrm{camión}(\mathrm{bloque}) \quad \Rightarrow \quad \dot{\Ts}^\mathrm{bloque}_\mathrm{camión} = \ms\vs_\mathrm{camión}(\mathrm{bloque})\dot{\vs}_\mathrm{camión}(\mathrm{bloque})</math>

:<math>\left.\begin{array}{l}
\mathrm{AB}:\mathrm{suelo} \\
\mathrm{REL}:\mathrm{camión}
\end{array}\right\} \quad \dot{\vs}_\mathrm{camión}(\mathrm{bloque})= \dot{\vs}_\mathrm{T}(\mathrm{bloque})-\dot{\vs}_\mathrm{ar}(\mathrm{bloque}) = (-\mu\gs)-(-\as)=\as-\mu\gs \quad \Rightarrow \quad \dot{\Ts}^\mathrm{bloque}_\mathrm{camión}=\ms\vs'(\as-\mu\gs)</math>

:<math>\left.\begin{array}{l}
\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{bloque}}=\dot{\Ws}_\mathrm{camión}^{\mathrm{ext}} + \dot{\Ws}_\mathrm{camión}^{\mathrm{ar}}= \ms(\as-\mu\gs)\vs'\\
\dot{\Ts}^\mathrm{bloque}_\mathrm{camión}=\ms\vs'(\as-\mu\gs)
\end{array}\right\} \Rightarrow\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{bloque}}= \dot{\Ts}^\mathrm{bloque}_\mathrm{camión}</math>


====✏️ EJEMPLO E1-3.3: camión frenando con carga====
------
:
{|
|
:[[Archivo:ExE1-3-3-1-esp.png|thumb|left|220px|link=]]
|Un camión frena con aceleración '''a''' respecto al suelo en el instante en el que su velocidad (también respecto al suelo) es v. Los pares de frenada en el eje delantero y en el trasero son iguales y valen <math>\Gamma</math>, y las ruedas, de radio r y masa despreciable, no deslizan sobre el suelo. En el mismo instante, el bloque homogéneo, de masa m, desliz con velocidad v' relativa al camión. Se trata de hacer un balance de potencias para el sistema (bloque+camión) en la referencia del suelo y en la del camión.

|}

:Ya que la potencia de las interacciones internas no es nula en principio (no se trata de un único sólido rígido), es conveniente hacer un DGI del sistema para no olvidar ninguna:
{|
|
[[Archivo:ExE1-3-3-2-esp.png|thumb|right|250px|link=]]

|
|
c.p.s.d.: contacto puntual sin deslizamiento

c.m.c.d.: contacto multipuntual con deslizamiento

|}
:El sistema sobre el que se hace el balance de potencias es todo menos el suelo. Teniendo en cuenta que la potencia de los enlaces internos es siempre nula, las únicas interacciones que hay que tener en cuenta son el peso, los enlaces de las ruedas con el suelo, los pares de frenada y la fricción entre bloque y camión:

[[Archivo:ExE1-3-3-3-esp.png|thumb|center|300px|link=]]

:Referencia suelo:
:Las fuerzas de enlace del suelo sobre las ruedas están aplicadas a puntos de velocidad cero, y por tanto no hacen potencia. El peso es una fuerza vertical, y el movimiento del camión y del bloque es horizontal, por tanto su potencia también es cero.

:<math>\dot{\Ws}_\Ts^{\rightarrow \mathrm{sist}}=\dot{\Ws}_\Ts^{\mathrm{ext}}+\dot{\Ws}^{\mathrm{int}}=\dot{\Ws}_\Ts^{\mathrm{peso}}+\dot{\Ws}_\Ts^{\mathrm{enlace}}+\dot{\Ws}^{\mathrm{fricción}}+\dot{\Ws}^{\Gamma}=\dot{\Ws}^{\mathrm{fricción}}+\dot{\Ws}^{\Gamma}.</math>

:Las potencias de interacciones internas se pueden calcular en cualquier referencia. Por tanto:

:<math>\dot{\Ws}^{\mathrm{fricción}}= \dot{\Ws}_\mathrm{camión}^{\mathrm{fric}\rightarrow \mathrm{bloque}} +\dot{\Ws}_\mathrm{camió}^{\mathrm{fric}\rightarrow \mathrm{camión}} = \dot{\Ws}_\mathrm{camión}^{\mathrm{fric}\rightarrow \mathrm{bloque}}= (\leftarrow \mu\ms\gs)\cdot (\rightarrow \vs')=-\mu\ms\gs\vs'</math>

:<math>\dot{\Ws}^{\Gamma}= \dot{\Ws}_\mathrm{camión}^{\Gamma\rightarrow \mathrm{ruedas}} +\dot{\Ws}_\mathrm{camión}^{\Gamma\rightarrow \mathrm{camión}} = \dot{\Ws}_\mathrm{camión}^{\Gamma\rightarrow \mathrm{ruedas}}= 2 \left(\odot \Gamma \right)\cdot \left(\otimes \frac{\vs}{\rs}\right)=-2\Gamma\frac{\vs}{\rs}.</math>

:Por otro lado: <math>\Ts_\Ts^\mathrm{sist}=\frac{1}{2}\ms\vs_\Ts^2(\mathrm{bloque})+\frac{1}{2}\ms\vs_\Ts^2(\mathrm{camión})\quad \Rightarrow \quad \dot{\Ts}_\Ts^\mathrm{sist}=-\ms(\vs+\vs')\mu\gs-\Ms\vs\as. </math>

:Finalmente: <math>\dot{\Ws}_\Ts^{\rightarrow \mathrm{sist}}=\dot{\Ts}_\Ts^\mathrm{sist} \quad \Rightarrow \quad -\mu\ms\gs\vs'-2\Gamma\frac{\vs}{\rs}=-\ms(\vs+\vs')\mu\gs-\Ms\vs\as.</math>

:Esta ecuación permite calcular el par de frenada: <math>\Gamma=\frac{1}{2}(\mu\ms\gs+\Ms\as)\rs.</math>

:Referencia camión:

:En esta referencia hay que tener en cuenta las fuerzas de arrastre. Por otro lado, las fuerzas de enlace del suelo sobre las ruedas están aplicadas a puntos que ya no tienen velocidad cero, y por tanto hacen potencia. Como antes, la potencia del peso es cero.

:<math>\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{sist}}=\dot{\Ws}_\mathrm{camión}^{\mathrm{ext}}+\dot{\Ws}^{\mathrm{int}}+\dot{\Ws}_\mathrm{camión}^{\mathrm{ar}}=\dot{\Ws}_\mathrm{camión}^{\mathrm{enlace}}+\dot{\Ws}^{\mathrm{fricción}}+\dot{\Ws}^{\Gamma}+\dot{\Ws}_\mathrm{camión}^{\mathrm{ar}}.</math>

:Las fuerzas de arrastre sobre el camión no hacen potencia porque el camión está parado en esta referencia. La de las fuerzas de arrastre sobre el bloque se ha calculado en el [[E1. Teorema de la energía: versión diferencial#✏️ ✏️ EJEMPLO E1-3.2: bloque deslizando sobre un camión|'''ejemplo E1-3.2''']]. La potencia de la fricción y de los pares de frenada es la misma de antes ya que no depende de la referencia.

:<math>\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{sist}}=\dot{\Ws}_\mathrm{camión}^{\mathrm{enllaç}}+\dot{\Ws}^{\mathrm{fricción}}+\dot{\Ws}^{\Gamma}+\dot{\Ws}_\mathrm{camión}^{\mathrm{ar} \rightarrow \mathrm{bloque}}= \dot{\Ws}_\mathrm{camión}^{\mathrm{enlace}} -\mu\ms\gs\vs'-2\Gamma\frac{\vs}{\rs} +\ms\as\vs'.</math>

:Las descripciones cinemática y dinámica de la rueda son:

[[Archivo:ExE1-3-3-4-esp.png|thumb|center|430px|link=]]

:<math>\dot{\Ws}_\mathrm{camión}^{\mathrm{enlace}}=2\Bigr[(\leftarrow \Ts)+(\uparrow \Ns)\Bigr]\cdot(\leftarrow\vs)=2\Ts\vs.</math>

La fuerza tangencial de enlace se puede calcular a partir del TMC aplicado al centro '''C''' de la rueda. Como no tiene masa: <math>\sum\overline{\Ms}_\mathrm{ext}(\Cs)=\overline{0} \quad \Rightarrow \quad (\odot \Gamma)+(\otimes \Ts\rs)=0\quad \Rightarrow \quad \Ts=\Gamma/\rs.</math>

:Por tanto: <math>\dot{\Ws}_\mathrm{camió}^{\rightarrow \mathrm{sist}} =2\frac{\Gamma}{\rs}\vs-\mu\ms\gs\vs'-2\Gamma\frac{\vs}{\rs}+\ms\as\vs'=\ms(\as-\mu\gs)\vs'.</math>

:Por otro lado, como se ha visto en el [[E1. Teorema de la energía: versión diferencial#✏️ ✏️ EJEMPLO E1-3.2: bloque deslizando sobre un camión|'''ejemplo E1-3.2''']]:

:<math>\Ts_\mathrm{camión}^\mathrm{sist}= \Ts_\mathrm{camión}^\mathrm{bloque}=\frac{1}{2}\ms\vs_\mathrm{camión}^2(\mathrm{bloque}) \quad \Rightarrow \quad \dot{\Ts}_\mathrm{camión}^\mathrm{sist}= \dot{\Ts}_\mathrm{camión}^\mathrm{bloque}=\ms\vs'(\as-\mu\gs).</math>

:El balance de potencias es consistente:

:<math>\left.\begin{array}{l}
\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{sist}}= \ms(\as-\mu\gs)\vs'\\
\dot{\Ts}^\mathrm{sist}_\mathrm{camión}=\ms(\as-\mu\gs)\vs'
\end{array}\right\} \Rightarrow\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{sist}}= \dot{\Ts}^\mathrm{sist}_\mathrm{camión}</math>

-----
-----

==E1.4 Balance de potencias para un sistema multisólido: cálculo directo y cálculo indirecto==

El '''cálculo directo''' (a partir de su definición) de la potencia asociada a una interacción (una fuerza sobre un punto '''P''' o un momento sobre un sólido rígido S) exige el conocimiento de la interacción (fuerza o momento):

<math>\dot{\Ws}_\Rs^{\Fs\rightarrow \Ps}=\overline{\Fs}_{\rightarrow\Ps} \cdot \vel{P}{R} \quad , \quad \dot{\Ws}_\Rs^{\Ms\rightarrow \Ss}=\overline{\Ms}_{\rightarrow\Ss} \cdot \velang{S}{R} </math>

Cuando no se trata de una interacción formulable, hay que recurrir a los teoremas vectoriales para determinarla.

El '''cálculo indirecto''' es una alternativa que no necesita recurrir a estos teoremas. Se basa en la evaluación de un balance de potencias para un sistema y una referencia adecuados de manera que la potencia a determinar sea el único término desconocido de la ecuación (el único que no se puede calcular de manera directa). Cuando el balance incluye más de un término desconocido, se pueden combinar varios balances (para distintos sistemas y referencias) hasta tener un número suficiente de ecuaciones que permitan la determinación de todas las potencias desconocidas.

====✏️ EJEMPLO E1-4.1: polea frenada====
------
:
{|
|
:[[Archivo:ExE1-4-1-1-esp.png|thumb|left|150px|link=]]
|Un bloque de masa m cuelga de un hilo inextensible atado a un punto de la periferia de una polea de radio r. Entre soporte (fijo al suelo) y polea actúa un freno que garantiza que el bloque baja a velocidad constante respecto al suelo. Se trata de calcular la potencia del freno.

|}

:Càlculo directo: <math>\dot{\Ws}^{\mathrm{freno}}= \overline{\Gamma}^{\rightarrow \mathrm{polea}}\cdot\velang{polea}{T}=\left(\odot\Gamma\right)\cdot\left( \otimes \frac{\vs_0}{\rs}\right)=-\Gamma \frac{\vs_0}{\rs}.</math>

:El par de frenada no es un dato, y hay que recurrir al TMC para calcularlo.

:<math>\left.\begin{array}{l}
\text{SISTEMA: polea + bloque}\\
\text{TMC en } \Os
\end{array}\right\} \sum\overline{\Ms}_\mathrm{ext}(\Os)=\dot{\overline{\Hs}}_\mathrm{RTO}(\Os)</math>

:El momento cinético proviene de las dos piezas. Ya que '''O''' no es fijo al bloque, hay que hacer descomposición baricéntrica para calcular el momento cinético del bloque:

:<math>\overline{\Hs}_\mathrm{RTO=T} (\Os) =\overline{\Hs}_\Ts^\mathrm{politja}(\Os) + \overline{\Hs}_\Ts^\mathrm{bloc}(\Os)= \overline{\Hs}_\Ts^\mathrm{politja}(\Os) + \overline{\Hs}_\Ts^\mathrm{bloc}(\Gs_\mathrm{bloc})+ \overline{\Hs}_\Ts^{\oplus\mathrm{bloc}}(\Os) = </math>

:<math>\hspace{2.3cm}=\Is_\mathrm{O}^\mathrm{polea} \velang{polea}{T}+\Is_{\Gs_\mathrm{bloque}}^\mathrm{bloque}+\OGvec_\mathrm{bloque}\times \ms\overline{\vs}_\Ts(\Gs_\mathrm{bloque})=\left(\otimes \Is_\mathrm{O}\frac{\vs_0}{\rs}\right)+(\otimes \ms\vs_0\rs).</math>

:[[Archivo:ExE1-4-1-2-esp.png|thumb|right|180px|link=]]
:Se trata de un momento cinético constante en dirección y valor. Por tanto:

:<math>\dot{\overline{\Hs}}_\mathrm{RTO}(\Os)=0 \quad \Rightarrow \quad \sum{\overline{\Ms}}_\mathrm{ext}(\Os)=(\odot\Gamma)+(\otimes \ms\gs\rs) =0\quad \Rightarrow \quad \Gamma= \ms\gs\rs</math>.

:Finalmente: <math>\dot{\Ws}^{\mathrm{freno}}= -\Gamma \frac{\vs_0}{\rs} = -\ms\gs\vs_0.</math>

:Cálculo indirecto:

:<math>\left.\begin{array}{l}
\text{SISTEMA: polea + soporte + bloque}\\
\text{REF: suelo(Gal)}
\end{array}\right\} \quad \dot{\Ts}^\mathrm{sist}_\Ts=\dot{\Ws}^\mathrm{ext}_\Ts+ \dot{\Ws}^\mathrm{int}</math>

:<math>\left.\begin{array}{l}
\dot{\Ts}^\mathrm{sist}_\Ts=0\\
\dot{\Ws}^\mathrm{ext}_\Ts=\left(\downarrow \ms\gs\right)\cdot\left(\downarrow \vs_0\right)= \ms\gs\vs_0\\
\dot{\Ws}^\mathrm{int}=\dot{\Ws}^{\mathrm{freno}}
\end{array}\right\} \quad \Rightarrow \quad \dot{\Ws}^{\mathrm{freno}} =-\ms\gs\vs_0</math>



====✏️ EJEMPLO E1-4.2: mono y plátanos====
------
:
{|
|
:[[Archivo:ExE1-4-2-1-esp.png|thumb|left|180px|link=]]
|Unos plátanos de masa <math>(2/3)\ms</math> cuelgan de una cuerda inextensible que se apoya sin deslizar sobre una polea de radio r y masa despreciable, articulada respecto al suelo. En el otro extremo de la cuerda se agarra un mono de masa m. La polea gira con velocidad angular constante <math>\Omega_0</math> respecto al suelo bajo la acción de un motor. El mono se mueve respecto a la cuerda sin que sus manos se froten contra la cuerda, y de manera que la distancia de su centro de masas al suelo se mantiene constante. Se trata de calcular la potencia del motor.
|}

:Cálculo directo: <math>\dot{\Ws}^{\mathrm{freno}}= \overline{\Gamma}^{\rightarrow \mathrm{polea}}\cdot\velang{polea}{T}=\left(\odot\Gamma\right)\cdot\left( \odot \Omega_0\right)=\Gamma \Omega_0.</math>

:El par motor no es un dato, y hay que recurrir al TMC para calcularlo.

:<math>\left.\begin{array}{l}
\text{SISTEMA: polea + mono + plátanos}\\
\text{TMC en } \Os
\end{array}\right\} \sum\overline{\Ms}_\mathrm{ext}(\Os)=\dot{\overline{\Hs}}_\mathrm{RTO}(\Os)</math>

:El momento cinético proviene del mono y de los plátanos. En primera aproximación, se puede considerar la masa del mono concentrada en su tronco y su cabeza, de manera que no gira y su velocidad respecto al suelo es nula. Teniendo en cuenta que el punto '''O''' no se mueve respecto al mono pero sí respecto a los plátanos:

:<math>\overline{\Hs}_\mathrm{RTO=T} (\Os) =\overline{\Hs}_\Ts^\mathrm{mono}(\Os) + \overline{\Hs}_\Ts^\mathrm{plátanos}(\Os)= \overline{\Hs}_\Ts^\mathrm{mono}(\Os) + \overline{\Hs}_\Ts^\mathrm{plátanos}(\Gs_\mathrm{plát})+ \OGvec_\mathrm{plát} \times \frac{2}{3} \ms \overline{\vs}_\Ts(\Gs_\mathrm{plát})= </math>

:<math>\hspace{2.3cm}=\OGvec_\mathrm{plát}\times \frac{2}{3} \ms \overline{\vs}_\Ts(\Gs_\mathrm{plát}) = \Bigr[(\rightarrow \rs)+(\downarrow \xs)\Bigr] \times\frac{2}{3} \ms \left(\downarrow \rs\Omega_0\right)=\left(\otimes \frac{2}{3}\ms\rs^2\Omega_0\right).</math>

:Se trata de un momento cinético constante en dirección y valor. Por tanto:

:<math>\dot{\overline{\Hs}}_\mathrm{RTO}(\Os)=0 \quad \Rightarrow \quad \sum{\overline{\Ms}}_\mathrm{ext}(\Os)=(\odot\Gamma)+(\otimes \ms\gs\rs) + \left(\otimes \frac{2}{3}\ms\gs\rs\right)=0\quad \Rightarrow \quad \Gamma= \frac{5}{3}\ms\gs\rs</math>.

:Finalmente: <math>\dot{\Ws}^{\mathrm{motor}}= \Gamma \Omega_0 = \frac{5}{3}\ms\gs\rs\Omega_0.</math>

:Cálculo indirecto

:Hay que tener presente que es un problema con dos tipos de actuadores: el motor y la musculatura del mono. Hay diversos modelos biomecánicos para formular las fuerzas de los músculos, pero quedan fuera del ámbito de este curso. Ya que el mono se mueve, sus músculos desarrollan una potencia no nula.

:<math>\left.\begin{array}{l}
\text{SISTEMA: polea + mono + plátano}\\
\text{REF: suelo(Gal)}
\end{array}\right\} \quad \dot{\Ts}^\mathrm{sist}_\Ts=\dot{\Ws}^\mathrm{ext}_\Ts+ \dot{\Ws}^\mathrm{int}</math>

:<math>\left.\begin{array}{l}
\dot{\Ts}^\mathrm{sist}_\Ts=0\\
\dot{\Ws}^\mathrm{ext}_\Ts=\left(\downarrow \frac{2}{3}\ms\gs\right)\cdot\left(\downarrow \rs\Omega_0\right)= \frac{2}{3}\ms\gs\rs\Omega_0\\
\dot{\Ws}^\mathrm{int}=\dot{\Ws}^{\mathrm{musc}}+\dot{\Ws}^{\mathrm{motor}}
\end{array}\right\} \quad \Rightarrow \quad \dot{\Ws}^{\mathrm{musc}} + \dot{\Ws}^{\mathrm{motor}}=-\frac{2}{3}\ms\gs\rs\Omega_0</math>

:Este balance proporciona una ecuación con dos incógnitas. Se puede plantear un segundo balance para un sistema donde una de las dos no aparezca: un sistema sin el mono o un sistema sin el motor. Si se hace un balance de potencias para el sistema (mono + trozo de cuerda), las fuerzas externas a considerar son el peso y la tensión de la cuerda. Si se sigue trabajando en la referencia del suelo, esta tensión (desconocida) hará potencia no nula ya que está aplicada a un punto de velocidad ascendente.
:[[Archivo:ExE1-4-2-2-cat-esp.png|thumb|right|330px|link=]]

:Para evitar que aparezca un término asociado a la tensión de la cuerda, se puede trabajar en la referencia del trozo de cuerda que se ha considerado, que al tener velocidad ascendente constante es galileana. En esta referencia, el mono baja con velocidad constante:

:<math>\left.\begin{array}{l}
\text{REL: trozo de cuerda}\\
\text{AB: suelo}
\end{array}\right\}</math>

:<math>\overline{\vs}_\mathrm{REL}(\mathrm{mono})=\overline{\vs}_\mathrm{ar}(\mathrm{mono})=-\overline{\vs}_\mathrm{ar}(\mathrm{mono})=(\downarrow \rs\Omega_0)</math>

:<math>\left.\begin{array}{l}
\text{SISTEMA: mono+ trozo de cuerda}\\
\text{REF: trozo de cuerda(Gal)}
\end{array}\right\} \quad \dot{\Ts}^\mathrm{sist}_\mathrm{cuerda} =\dot{\Ws}^\mathrm{ext}_\mathrm{cuerda}+ \dot{\Ws}^\mathrm{int}</math>

:<math>\left.\begin{array}{l}
\dot{\Ts}^\mathrm{sist}_\mathrm{cuerda}=0\\
\dot{\Ws}^\mathrm{ext}_\mathrm{cuerda}=\left(\downarrow \ms\gs\right)\cdot\left(\downarrow \rs\Omega_0\right)= \ms\gs\rs\Omega_0\\
\dot{\Ws}^\mathrm{int}=\dot{\Ws}^{\mathrm{musc}}
\end{array}\right\} \quad \Rightarrow \quad \dot{\Ws}^{\mathrm{musc}} =-\ms\gs\rs\Omega_0</math>

:Combinando este resultado con el del balance anterior:

:<math>\dot{\Ws}^\mathrm{motor}=-\frac{2}{3}\ms\gs\rs\Omega_0-\dot{\Ws}^\mathrm{musc}=-\frac{5}{3}\ms\gs\rs\Omega_0</math>



© Universitat Politècnica de Catalunya. [[Mecánica:Derechos de autor |Todos los derechos reservados]]
-------------

<center>
[[D8. Conservaciones|<<< D8. Conservaciones]]

[[E2. Teorema de la energía: versión integrada|E2. Teorema de la energía: versión integrada>>>]]
</center>

E1. Work-Energy Theorem: differential form

2025-05-20T12:17:45Z

Bros: /* ✏️ EXAMPLE E1-1.2: pendulum */

<div class="noautonum">__TOC__</div>
<math>\newcommand{\uvec}{\overline{\textbf{u}}}
\newcommand{\vvec}{\overline{\textbf{v}}}
\newcommand{\evec}{\overline{\textbf{e}}}
\newcommand{\Omegavec}{\overline{\mathbf{\Omega}}}
\newcommand{\velang}[2]{\Omegavec^{\textrm{#1}}_{\textrm{#2}}}
\newcommand{\Alfavec}{\overline{\mathbf{\alpha}}}
\newcommand{\accang}[2]{\Alfavec^{\textrm{#1}}_{\textrm{#2}}}
\newcommand{\ds}{\textrm{d}}
\newcommand{\ps}{\textrm{p}}
\newcommand{\ns}{\textrm{n}}
\newcommand{\hs}{\textrm{h}}
\newcommand{\cs}{\textrm{c}}
\newcommand{\gs}{\textrm{g}}
\newcommand{\Ns}{\textrm{N}}
\newcommand{\Hs}{\textrm{H}}
\newcommand{\Fs}{\textrm{F}}
\newcommand{\ms}{\textrm{m}}
\newcommand{\ts}{\textrm{t}}
\newcommand{\us}{\textrm{u}}
\newcommand{\vs}{\textrm{v}}
\newcommand{\Rs}{\textrm{R}}
\newcommand{\Es}{\textrm{E}}
\newcommand{\Ws}{\textrm{W}}
\newcommand{\Ts}{\textrm{T}}
\newcommand{\Ls}{\textrm{L}}
\newcommand{\As}{\textrm{A}}
\newcommand{\Ds}{\textrm{D}}
\newcommand{\Bs}{\textrm{B}}
\newcommand{\es}{\textrm{e}}
\newcommand{\fs}{\textrm{f}}
\newcommand{\is}{\textrm{i}}
\newcommand{\Is}{\textrm{I}}
\newcommand{\ks}{\textrm{k}}
\newcommand{\js}{\textrm{j}}
\newcommand{\rs}{\textrm{r}}
\newcommand{\as}{\textrm{a}}
\newcommand{\ss}{\textrm{s}}
\newcommand{\Os}{\textbf{O}}
\newcommand{\Gs}{\textbf{G}}
\newcommand{\Cbf}{\textbf{C}}
\newcommand{\Or}{\Os_\Rs}
\newcommand{\Qs}{\textbf{Q}}
\newcommand{\Cs}{\textbf{C}}
\newcommand{\Ps}{\textbf{P}}
\newcommand{\Ss}{\textbf{S}}
\newcommand{\Js}{\textbf{J}}
\newcommand{\P}{\textrm{P}}
\newcommand{\Q}{\textrm{Q}}
\newcommand{\Ms}{\textrm{M}}
\newcommand{\deg}{^\textsf{o}}
\newcommand{\xs}{\textsf{x}}
\newcommand{\ys}{\textsf{y}}
\newcommand{\zs}{\textsf{z}}
\newcommand{\dert}[2]{\left.\frac{\ds{#1}}{\ds\ts}\right]_{\textrm{#2}}}
\newcommand{\ddert}[2]{\left.\frac{\ds^2{#1}}{\ds\ts^2}\right]_{\textrm{#2}}}
\newcommand{\vec}[1]{\overline{#1}}
\newcommand{\vecbf}[1]{\overline{\textbf{#1}}}
\newcommand{\vecdot}[1]{\overline{\dot{#1}}}
\newcommand{\OQvec}{\vec{\Os\Qs}}
\newcommand{\GQvec}{\vec{\Gs\Qs}}
\newcommand{\GPvec}{\vec{\Gs\Ps}}
\newcommand{\PSvec}{\vec{\Ps\Ss}}
\newcommand{\QQvec}{\vec{\Qs\Qs}}
\newcommand{\QGvec}{\vec{\Qs\Gs}}
\newcommand{\QPvec}{\vec{\Qs\Ps}}
\newcommand{\JQvec}{\vec{\Js\Qs}}
\newcommand{\PQvec}{\vec{\Ps\Qs}}
\newcommand{\GJvec}{\vec{\Gs\Js}}
\newcommand{\OPvec}{\vec{\Os\textbf{P}}}
\newcommand{\OCvec}{\vec{\Os\Cs}}
\newcommand{\OGvec}{\vec{\Os\Gs}}
\newcommand{\CJvec}{\vec{\Cs\Js}}
\newcommand{\GCvec}{\vec{\Gs\Cs}}
\newcommand{\PGvec}{\vec{\Ps\Gs}}
\newcommand{\abs}[1]{\left|{#1}\right|}
\newcommand{\braq}[2]{\left\{{#1}\right\}_{\textrm{#2}}}
\newcommand{\matriz}[9]{
\begin{bmatrix}
{#1} & {#2} & {#3}\\
{#4} & {#5} & {#6}\\
{#7} & {#8} & {#9}
\end{bmatrix}}
\newcommand{\diag}[3]{
\begin{bmatrix}
{#1} & {0} & {0}\\
{0} & {#2} & {0}\\
{0} & {0} & {#3}
\end{bmatrix}}
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\begin{Bmatrix}
{#1}\\
{#2}\\
{#3}
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{#1}\\
{#2}
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\newcommand{\agal}[1]{\vecbf{a}_{\textrm{Gal}} (#1)}
\newcommand{\angal}[1]{\vecbf{a}_{\textrm{NGal}} (#1)}
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\newcommand{\fvec}[2]{\overline{\mathbf{F}}_{\textrm{#1}\rightarrow\textrm{#2}}}
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\newcommand{\F}[1]{\vec{\Fs}_{#1}}
\newcommand{\Fcal}[2]{\vec{\cal{F}}^{\text{#1}}_{#2}}
</math>

The '''Work-Energy Theorem''' plays a prominent role in physics, as it bridges all branches of science (electromagnetism, thermodynamics, etc.): all of them deal with energy.

There is a certain parallel between the Vector Theorems and the '''Work-Energy Theorem''': they all allow us to understand the evolution of a magnitude that depends on the mechanical state of the system. In the case of the former, we are dealing with the instantaneous evolution of vector magnitudes (linear momentum, angular momentum), while the latter deals with the finite change (in a finite time interval) of a scalar magnitude ('''mechanical energy''').

This unit presents the differential form of this theorem (the '''power balance'''). The integrated form (which is the Work-Energy Theorem itself) is presented in [[E2. Teorema de la energía: versión integrada|'''unit E2''']].
 
------
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==E1.1 Power balance in a system of particles==

:[[File:E1-1-eng.png|thumb|right|300px|link=]]
<center>'''Figure E1.1''' Forces on a particle of a system with constant matter</center>

Let us consider a system of constant matter, and let’s apply Newton's second law to each particle (or each mass differential) of this system in a general reference frame R (which may be non-Galilean) ('''Figure E1.1'''):

<math>\F{\rightarrow \Ps} + \Fcal{}{\text{ar}\rightarrow\Ps} + \Fcal{}{\text{Cor}\rightarrow\Ps} =\ms_\Ps\acc{P}{R}</math>

<math>\F{\rightarrow \Ps} </math> is the resultant interaction force on '''P''', and it comes from internal interactions (with particles of the same system) and external ones:

<math>\F{\rightarrow \Ps}=\F{\rightarrow \Ps}^\text{int}+\F{\rightarrow \Ps}^\text{ext}</math>

If both sides of the equation are multiplied by <math>\vel{P}{R}</math>, all forces (interaction and inertia forces) orthogonal to <math>\vel{P}{R}</math> are filtered out. Specifically, since <math>\acc{P}{R}=\accs{P}{R} + \accn{P}{R} </math>, where <math>\accn{P}{R} \perp \vel{P}{R}</math>, and since the Coriolis force is always orthogonal to <math>\vel{P}{R} (\Fcal{}{\text{Cor}\rightarrow\Ps} = -2\ms_\Ps \velang{R}{Gal} \times \vel{P}{R})</math>:

<math>\vel{P}{R} \cdot \left( \F{\rightarrow \Ps}^\text{int}+\F{\rightarrow \Ps}^\text{ext} + \Fcal{}{\text{ar}\rightarrow\Ps} \right) = \ms_\Ps\accs{P}{R} \cdot \vel{P}{R} = \frac{\ds}{\ds \ts} \left(\frac{1}{2}\ms_\Ps \vvec_\Rs^2(\Ps) \right) \equiv \frac{\ds\Ts_\Rs(\Ps)}{\ds\ts}=\dot{\Ts}_\Rs(\Ps)</math>

The scalar function <math>\Ts_\Rs(\Ps)</math> is the '''kinetic energy''' of the particle, and is a function of the [[D1. Foundational laws of Newtonian dynamics#D1.3 Newton’s Principle of Determinacy|'''mechanical state''']] of '''P'''. The scalar product of a force and the velocity of its application point is the '''force power''': <math>\vel{P}{R} \cdot \F{\rightarrow \Ps} \equiv \dot{\Ws}_\Rs^{\F{\rightarrow \Ps}} </math>. Both scalar quantities <math>(\Ts_\Rs(\Ps), \dot{\Ws}_\Rs^{\F{\rightarrow \Ps}})</math> depend on the reference frame. For the same set of interaction forces acting on '''P''', the reference frame determines which are filtered out and which are not.

The equation can be rewritten as <math>\dot{\Ws}_\Rs^{\text{ext}\rightarrow \Ps}+\dot{\Ws}_\Rs^{\text{int}\rightarrow \Ps}+\dot{\Ws}_\Rs^{\text{ar}\rightarrow \Ps} =\dot{\Ts}_\Rs(\Ps)</math>, and constitutes the '''power balance''' for particle '''P''' in the R reference frame. Summing for all particles in the system:

<math>\dot{\Ws}_\Rs^{\text{ext}}+\dot{\Ws}^{\text{int}}+\dot{\Ws}_\Rs^{\text{ar}} =\dot{\Ts}_\Rs^\text{syst}</math>

The calculation of kinetic energy is covered in [[E2. Teorema de la energía: versión integrada|'''unit E2''']]. This unit focuses on the calculation of powers. [[E1. Work-Energy Theorem: differential form#E1.2 Potencia de una pareja de acción-reacción|'''section E1-2''']] shows that the power of internal interactions does not depend on the reference frame (which is why the subscript R does not appear in the <math>\dot{\Ws}^{\text{int}}</math> term of the previous equation).

====✏️ EXAMPLE E1-1.1: free particle====
------
:
{|
:[[File:ExE1-1-1-1-eng.jpg|thumb|left|180px|link=]]
|A particle '''P''', with mass m, is released from the ring of a Ferris wheel when it is at its highest point. We want to calculate the power, at this instant, of all forces acting on the particle from three different reference frames: the ground, the ring, and the Ferris wheel cabin.
|}

:Ground reference frame: Since this is a Galilean reference frame, all forces on '''P''' are interaction forces.
{|
:[[File:ExE1-1-2- eng.png|thumb|right|230px|link=]]

|<math>\hspace{0.5cm}\sum\dot{\Ws}_\Ts=\dot{\Ws}_\Ts^\text{weight}= \left(\leftarrow \Rs\Omega_0\right) \cdot \left(\downarrow \ms\gs\right)=0</math>
|}

:Ring reference frame: It is a non-Galilean reference, and the transportation force on '''P''' must be taken into account.
{|
:[[File:ExE1-1-3-eng.png|thumb|right|250px|link=]]
|<math>\hspace{0.5cm}</math> AB:ground, REL:ring 

<math>\hspace{0.5cm}\vel{P}{REL}=\vel{P}{AB}-\vel{P}{ar}= \left(\leftarrow \Rs\Omega_0\right)-\left(\leftarrow \Rs\Omega_0\right)=\overline{0}</math> 

<math>\hspace{0.5cm}</math>

Since the velocity is zero, the total power is zero.
|}

:Cabin reference frame: It is also a non-Galilean reference, and the transportation force on '''P''' must be taken into account.

{|
:[[File:ExE1-1-4-eng.png|thumb|right|250px|link=]]
|<math>\hspace{0.5cm}</math>AB:ground, REL:cabin

<math>\hspace{0.5cm}\vel{P}{REL}=\vel{P}{AB}-\vel{P}{ar}= \left(\leftarrow \Rs\Omega_0\right)-\left(\uparrow \Rs\Omega_0\right)</math>

<math>\hspace{0.5cm}\dot{\Ws}_\text{cabin}^\text{weight}=\left[ \left(\leftarrow \Rs\Omega_0\right)+\left(\downarrow \Rs\Omega_0\right) \right] \cdot \left(\downarrow\ms\gs\right) =\ms\gs\Rs\Omega_0</math>

<math>\hspace{0.5cm}\dot{\Ws}_\text{cabin}^\text{ar}=\left[ \left(\leftarrow \Rs\Omega_0\right)+\left(\downarrow \Rs\Omega_0\right) \right] \cdot \left(\rightarrow\ms\Rs^2\Omega_0^2\right) =-\ms\gs\Rs^2 \Omega_0^3</math>

<math>\hspace{0.5cm}\sum\dot{\Ws}_\text{cabin}=\ms\Rs\Omega_0(\gs-\Rs\Omega_0^2)</math>

|}


====✏️ EXAMPLE E1-1.2: pendulum====
------
:
{|
:[[File:ExE1-2-1-eng.png|thumb|left|180px|link=]]
|A particle '''P''' of mass m hangs from an inextensible thread of length L attached to the ceiling. The assembly oscillates in a vertical plane fixed to the ground. We want to calculate the power of all the forces acting on the particle in two different reference frames: the ground and a platform rotating relative to the ground with a constant angular velocity <math>\Omega_0</math>. 
|}

Ground reference frame: Since it is a Galilean reference frame, '''P''' is only subjected to interaction forces.
{|
:[[File:ExE1-2-2-neut.png|thumb|left|300px|link=]]

|<math>\dot{\Ws}_\Es^\Ts=(\nearrow\Ls\dot{\theta})\cdot(\nwarrow\Ts)=0</math>

<math>\dot{\Ws}_\Es^\text{weight}=(\nearrow\Ls\dot{\theta}\cdot(\downarrow \ms\gs)=-\ms\gs\Ls\dot{\theta}\sin\theta</math>
|}

Platform reference frame: It is a non Galilean reference frame, and the transportation force on '''P''' has to be taken into account.

{|
:[[File:ExE1-2-3-eng.png|thumb|left|400px|link=]]

|AB:ground, REL:platform

<math>\vel{P}{REL}=\vel{P}{AB}-\vel{P}{ar}=(\nearrow\Ls\dot{\theta})-(\otimes \Ls \Omega_0 \sin \theta)</math>

<math>\dot{\Ws}_\Es^\Ts\left[(\nearrow\Ls\dot{\theta})+(\odot\Ls\Omega_0\sin\theta)\right]\cdot(\nwarrow\Ts)=0</math>

<math>\dot{\Ws}_\Es^\text{weight}\left[(\nearrow\Ls\dot{\theta})+(\odot\Ls\Omega_0\sin\theta)\right]\cdot(\downarrow \ms\gs)=-\ms\gs\Ls\dot{\theta}\sin\theta</math>

<math>\dot{\Ws}_\Es^\text{ar}\left[(\nearrow\Ls\dot{\theta})+(\odot\Ls\Omega_0\sin\theta)\right]\cdot(\rightarrow \ms\Ls\Omega_0^2\sin\theta)=\ms\Ls^2\Omega_0^2\dot{\theta}\cos\theta</math>
|}



------
------

==E1.2 Power of an action-reaction pair==

The total power associated with internal interactions in a constant-mass system is not zero in principle (since the two forces of each action-reaction pair are applied to particles that may have different velocities relative to the reference frame in which the calculation is made), and it is independent of the reference frame in which it is evaluated. It only depends on the separation velocity of the two particles. The power balance for the system is therefore written as: <math>\dot{\Ts}_\Rs^\text{syst}=\dot{\Ws}_\Rs^\text{ext} + \dot{\Ws}^\text{int} + \dot{\Ws}_\Rs^\text{ar} </math> .

<div>
=====💭 PROOF ➕=====

{|
:[[File:E1-2-neut.png|thumb|left|200px|link=]]
|<math>\dot{\Ws}^\text{ac-reac}=\overline{\Fs}_{\Ps \rightarrow \Qs}\cdot \vel{Q}{R} + \overline{\Fs}_{\Qs \rightarrow \Ps}\cdot \vel{P}{R}= \overline{\Fs}_{\Ps \rightarrow \Qs}\cdot \left[ \vel{Q}{R}-\vel{P}{R}\right]</math>

By the [[D1. Foundational laws of Newtonian dynamics#D1.6 Newton’s third law (action-reaction principle)|'''action-reaction principle''']], <math>\overline{\Fs}_{\Ps \rightarrow \Qs}=-\overline{\Fs}_{\Qs \rightarrow \Ps}</math>. Therefore: <math>\dot{\Ws}^\text{ac-reac}=\overline{\Fs}_{\Ps \rightarrow \Qs}\cdot \left[ \vel{Q}{R}-\vel{P}{R}\right]</math>

As <math>\overline{\Fs}_{\Ps \rightarrow \Qs}</math> has the <math>\PQvec</math> direction, only the components of the velocities in that direction have to be considered:

<math>\dot{\Ws}^\text{ac-reac}=\overline{\Fs}_{\Ps \rightarrow \Qs}\cdot \biggl( \vel{Q}{R} \Bigr]_{||\Ps\Qs}-\ \vel{P}{R} \Bigr]_{||\Ps\Qs}\biggl)</math>
|}

The difference <math>\biggl( \vel{Q}{R} \Bigr]_{\Ps\Qs}-\ \vel{P}{R} \Bigr]_{\Ps\Qs}\biggl)</math> is the separation velocity <math>\dot{\rho}</math> between '''P''' and '''Q''', and it does not dpend on the reference frame since it is the derivative of a scalar (the distance <math>\dot{\rho}</math> between the two particles). For the previous drawing, <math>\dot{\rho}> 0</math>. Since this is an attractive interaction, <math>\dot{\Ws}^\text{ac-reac} < 0</math>.

With a drawing, it is easy to see that:
* attractive interaction and particle approach <math>(\dot{\rho}<0)</math> : <math>\dot{\Ws}^\text{ac-reac} > 0</math> ;
* repulsive interaction and particle separation <math>(\dot{\rho}>0)</math> : <math>\dot{\Ws}^\text{ac-reac} > 0</math>
* repulsive interaction and particle approach <math>(\dot{\rho}<0)</math> : <math>\dot{\Ws}^\text{ac-reac} < 0</math>


</div>

There are two particular cases of interaction forces whose power always has the same sign:

* Friction action-reaction pair between '''P''' and '''Q''': since the friction force always opposes the relative velocity between the two points, its power is always negative:

<math>\hspace{1cm} \dot{\Ws}^\text{friction}=\dot{\Ws}^{\text{fric}\rightarrow\Ps}_\text{RTQ}+\dot{\Ws}^{\text{fric}\rightarrow\Qs}_\text{RTQ}=\overline{\Fs}^{\text{fric}\rightarrow\Ps} \cdot \vel{P}{RTQ} = - \abs{\overline{\Fs}^{\text{fric}\rightarrow\Ps}} \frac{\vel{P}{RTQ}}{\abs{\vel{P}{RTQ}}} \cdot \vel{P}{RTQ}<0;</math>

* Constraint action-reaction pair between '''P''' and '''Q''': from the [[D3. Interactions between rigid bodies#D3.4 Direct constraint interactions|'''analytical characterization equation''']] of the constraint forces between two points '''P''' and '''Q''' belonging to two rigid bodies <math>\mathrm{S}_\Ps</math> and <math>\mathrm{S}_\Qs</math>, it is easy to see that the associated power is zero:

<math>\hspace{1cm} \dot{\Ws}^\text{constraint}=\dot{\Ws}^{\text{constraint}\rightarrow\Ps}_{\mathrm{S}_\Ps}+\dot{\Ws}^{\text{constraint}\rightarrow\Qs}_{\mathrm{S}_\Ps}=\overline{\Fs}^{\text{constraint}\rightarrow\Ps} \cdot \overline{\mathbf{v}}_{\mathrm{S}_\Qs}(\Ps) = - 0;</math>

====✏️ EXAMPLE E1-2.1: power of an action-reaction friction pair====
------
:
{|
:[[File:ExE1-2-1-2-eng.png|thumb|left|180px|link=]]
|A homogeneous wheel, with radius r and mass m, moves down an inclined plane without sliding until it joins a conveyor belt moving at a constant velocity <math>\vs_0</math> relative to the ground. At the moment it makes contact with the belt, the center of the wheel has a velocity <math>\vs_0</math> relative to the ground in the same direction as the belt. We want to calculate the power of the friction force between the belt and the wheel at this instant.
|}

At that momento, the <math>\mathbf{ICR}^\text{wheel}_\Es</math> is point <math>\Js_\text{wheel}</math> in contact with the belt. This point slides backwards with respect to the belt, therefore the friction force on the wheel is forwards:

<math>\left.\begin{array}{l}
\text { AB: ground } \\
\text { REL: belt }
\end{array}\right\} \quad \bar{\vs}_{\text {REL }}\left(\Js_{\text {wheel }}\right)=\bar{\vs}_{\text{AB}}\left(\Js_{\text {wheel }}\right)-\bar{\vs}_{\mathrm{ar}}\left(\Js_{\text {wheel }}\right)=-\bar{\vs}_{\mathrm{ar}}\left(\Js_{\text {wheel }}\right)=-\left(\rightarrow \mathrm{v}_0\right)=\left(\leftarrow \mathrm{v}_0\right) .</math>

[[File:ExE1-2-1-3-eng.png|thumb|center|500px|link=]]

Calculation of the friction power: <math>\dot{\Ws}^\text{friction}=\dot{\Ws}_\Rs^{\text{friction}\rightarrow \text{wheel}}+ \dot{\Ws}_\Rs^{\text{friction}\rightarrow \text{belt}}</math>, where R may be any reference frame. To quickly calculate this power, it is advisable to choose a reference frame R where one of the two points subjected to the action-reaction pair has zero speed, so that one of the two terms is directly zero:

<math>\dot{\Ws}^\text{friction}=\dot{\Ws}_\text{belt}^{\text{fric}\rightarrow \text{wheel}}+ \dot{\Ws}_\text{belt}^{\text{fric }\rightarrow \text{belt}}=\overline{\Fs}^{\text{fric}\rightarrow \text{wheel}} \cdot \bar{\vs}_{\text {belt}}\left(\Js_{\text {wheel }}\right)=(\rightarrow \mu\ms\gs)\cdot(\leftarrow \vs_0)=-\mu\ms\gs\vs_0.</math>

If the calculation is done in any other reference frame, the result is the same:

<math>\dot{\Ws}^\text{friction}=\dot{\Ws}_\text{E}^{\text{fric}\rightarrow \text{wheel}}+ \dot{\Ws}_\text{E}^{\text{fric} \rightarrow \text{belt}}=(\rightarrow \mu\ms\gs)\cdot 0 + (\leftarrow \mu\ms\gs)\cdot(\rightarrow \vs_0)=-\mu\ms\gs\vs_0.</math>

<math>\dot{\Ws}^\text{friction}=\dot{\Ws}_\text{vehicle}^{\text{fric}\rightarrow \text{wheel}}+ \dot{\Ws}_\text{vehicle}^{\text{fric} \rightarrow \text{belt}}=(\rightarrow \mu\ms\gs)\cdot (\rightarrow\vs_0) + (\leftarrow \mu\ms\gs) \cdot(\rightarrow 3\vs_0)=-\mu\ms\gs\vs_0.</math>

However, if only the power of one of the two forces is calculated (for example, the friction on the wheel), the result depends on the reference frame and can have any sign:

<math>\dot{\Ws}_\text{belt}^{\text{fric }\rightarrow \text{wheel}}=(\rightarrow \mu\ms\gs)\cdot(\leftarrow \vs_0)=-\mu\ms\gs\vs_0 ,</math>

<math>\dot{\Ws}_\text{E}^{\text{fric} \rightarrow \text{wheel}}=(\rightarrow \mu\ms\gs)\cdot 0 = 0 ,</math>

<math>\dot{\Ws}_\text{vehicle}^{\text{fric} \rightarrow \text{wheel}}=(\rightarrow \mu\ms\gs)\cdot (\rightarrow\vs_0)=\mu\ms\gs\vs_0.</math>


-----
-----

==E1.3 Power of a system of forces on a rigid body==

{|
|
:[[File:E1-3-eng.png|thumb|left|180px|link=]]
<center>'''Figure E1.3''': Kinematic torsor of a rigid body in a reference frame R, and forces on different points (dm) of that rigid body</center>

|When calculating the power of a system of forces on a rigid body S, the term associated with internal interactions is zero because those interactions are all constraint (cohesion) forces and, as seen in [[E1. Work-Energy Theorem: differential form#E1.2 Power of an action-reaction pair|'''section E1.2''']], their total power is zero.

The total power of a system of forces on S can be calculated from the torsor of the system at any point '''O''' of the rigid body, <math>(\overline{\Fs},\overline{\Ms}(\Os))</math>. Taking into account rigid body kinematics ('''Figure E1.3'''):

<math>\dot{\Ws}_\Rs^{\rightarrow \Ss} =\sum_\Ps\overline{\Fs}^{\rightarrow \Ps} \cdot \vel{P}{R}= \sum_\Ps \left[ \overline{\Fs}^{\rightarrow \Ps} \cdot \vel{O}{R} + \velang{S}{R} \times \OPvec \right] </math>

|}
<math>\dot{\Ws}_\Rs^{\rightarrow \Ss} =\left(\sum_\Ps\overline{\Fs}^{\rightarrow \Ps} \right)\cdot \vel{O}{R} + \sum_\Ps\overline{\Fs}^{\rightarrow \Ps} \cdot \left(\velang{S}{R} \times \OPvec \right) = \left( \sum_\Ps\overline{\Fs}^{\rightarrow \Ps} \right) \cdot \vel{O}{R} + \left[ \sum_\Ps \left( \OPvec \times \overline{\Fs}^{\rightarrow \Ps} \right)\right] \cdot \velang{S}{R} = \overline{\Fs}\cdot \vel{O}{R} + \overline{\Ms}(\Os)\cdot \velang{S}{R} </math>

This calculation also shows that the power of a moment (or a torque <math>\Gamma</math>) on a rigid body in a reference frame R is the product of the moment by the angular velocity of the rigid body: <math>\dot{\Ws}_\Rs^{\Gamma\rightarrow \mathrm{S}} = \overline{\Gamma} \cdot \velang{S}{R}</math>.

The power balance for a rigid body is: <math>\dot{\Ws}_\Rs^{ \rightarrow \mathrm{S}} = \dot{\Ws}_\Rs^{ \mathrm{ext}} + \dot{\Ws}_\Rs^{ \mathrm{ar}}.</math>

====✏️ EXAMPLE E1-3.1: roller sliding on a slope====
------
:
{|
|
:[[File:ExE1-3-1-1-eng.png|thumb|left|180px|link=]]
|A homogeneous roller, with mass m and radius r, slides down a slope. The friction coefficient between the ground and the roller is <math>\mu</math>. We want to calculate the power of every force acting on the roller in the ground reference frame.
|}

:It is a 2D problem. The kinematic and dynamic descriptions are:
{|
|
[[File:ExE1-3-1-2-eng.png|thumb|right|250px|link=]]

|<math> \hspace{0.5cm}\vel{J}{E}=\vel{C}{T} + \velang{}{E} \times \CJvec= </math>

<math>\hspace{1.7cm}=\left( \rightarrow \vs_\Es \right) + \left( \otimes \Omega_\Es\right) \times \left( \downarrow \rs \right)</math>

<math> \hspace{0.5cm}\vel{J}{E} = \left[\rightarrow (\vs_\Es -\rs\Omega_\Es)\right]</math>

<math>\hspace{0.5cm} \dot{\Ws}_\Es^{\mathrm{ext}} = \dot{\Ws}_\Es^{\mathrm{weight}} + \dot{\Ws}_\Es^{\mathrm{friction}} + \dot{\Ws}_\Es^{\mathrm{constraint}}= \Bigr[\left(\downarrow \ms\gs\cos\beta\right)+\left(\rightarrow \ms\gs\sin\beta\right)\Bigr] \cdot (\rightarrow \vs_\Es) + \Bigr[(\leftarrow \mu \Ns) + (\uparrow \Ns ) \Bigr] \cdot \Bigr[\rightarrow (\vs_\Es -\rs\Omega_\Es)\Bigr]</math>

|}
:Since the roller center has no motion perpendicular to the plane, the AMT leads to <math>\Ns =\ms\gs\cos\beta</math>. Therefore:

{|
|
[[File:ExE1-3-1-3-eng.png|thumb|right|140px|link=]]
|<math>\hspace{0.5cm} \dot{\Ws}_\Ts^{\mathrm{ext}} = \ms\gs\vs_\Ts \sin \beta - \ms\gs(\vs_\Ts -\rs\Omega_\Ts)\cos\beta=\ms\gs \Bigr[ \vs_\Ts (\sin\beta-\cos\beta)+\rs\Omega_\Ts \cos\beta \Bigr].</math>

<math>\hspace{0.5cm}</math>

If instead of considering the friction at '''J''' we consider the equivalent torsor at point '''C''', we must consider the force applied in '''C''' and the moment about '''C''', which is not zero:

<math>\hspace{0.5cm} \dot{\Ws}_\Es^{ \mathrm{friction}} = (\leftarrow \mu\Ns)\cdot(\rightarrow\vs_\Es)+(\otimes \mu\Ns\rs)\cdot(\otimes \Omega_\Es) =\mu\ms\gs\cos\beta(\rs\Omega_\Es-\vs_\Es)</math>,

<math>\hspace{0.5cm} \dot{\Ws}_\Es^{\mathrm{ext}} = \dot{\Ws}_\Es^{ \mathrm{weight}} + \dot{\Ws}_\Es^{ \mathrm{friction}}+ \dot{\Ws}_\Es^{ \mathrm{constraint}}= \ms\gs\vs_\Es\sin\beta+\mu\ms\gs\cos\beta (\rs\Omega_\Es-\vs_\Es)</math>.

<math>\hspace{0.5cm}</math>

The result is the same.
|}


====✏️ EXAMPLE E1-3.2: block sliding on a truck====
------
:
{|
|
:[[File:ExE1-3-2-1-eng.png|thumb|left|220px|link=]]
|A truck brakes with acceleration '''a''' relative to the ground at the instant in which its velocity (also relative to the ground) is v. At the same instant, the homogeneous block, of mass m, slides with velocity v' relative to the truck. We want to perform a power balance for the block in two different reference frames: the ground and the truck.
|}

:Ground reference frame: It is a 2D problem. The kinematic and dynamic descriptions are:
{|
|
[[File:ExE1-3-2-2-eng.png|thumb|right|200px|link=]]

| The constraint between the truck and the block allows only one translational DoF between the two. Therefore, in the 2D approximation, the associated torsor has two independent components. If characterized at the midpoint of the contact surface, these components are a normal force and a moment perpendicular to the plane. The AMT applied to the block leads to <math>\Ns=\ms\gs</math>.

|}
:The powers associated with all interactions on the block in the ground reference are:

:<math>\dot{\Ws}_\Es^{\rightarrow \mathrm{bloque}}=\dot{\Ws}_\Ts^{\mathrm{ext}}=\dot{\Ws}_\Ts^{\mathrm{peso}}+\dot{\Ws}_\Ts^{\mathrm{fricción}}+\dot{\Ws}_\Ts^{\mathrm{enlace}}= \Bigr[(\downarrow \ms\gs)+(\leftarrow \mu\ms\gs)+(\uparrow \Ns) \Bigr] \cdot \Bigr[\rightarrow(\vs+\vs')\Bigr] + (\odot \Ms_\es)\cdot 0 = -\mu\ms\gs(\vs+\vs')</math>

:Ya que el bloque se traslada respecto al suelo, su energía cinética total es inmediata, y se puede plantear fácilmente el balance de potencias:

:<math>\Ts_\Ts^\mathrm{bloque}=\frac{1}{2}\int_\mathrm{bloque}\ds\ms \vs_\Ts^2(\mathrm{bloque})=\frac{1}{2}\ms \vs_\Ts^2(\mathrm{bloc})\quad \Rightarrow \quad \dot{\Ts}_\Ts^\mathrm{bloque}=\ms\vs_\Ts( \mathrm{bloque})\dot{\vs}_\Ts( \mathrm{bloque}) </math>

:Como la única fuerza horizontal sobre el bloque es la fricción:

<math>\dot{\vs}_\Ts(\mathrm{bloque})=-\mu\gs\quad \Rightarrow \quad \dot{\Ts}_\Ts^\mathrm{bloque}=-\mu\ms\gs(\vs+\vs').</math>

:El balance de potencias es consistente:

:<math>\left.\begin{array}{l}
\dot{\mathrm{W}}_\Ts^{\rightarrow \text { bloque }}=\dot{\mathrm{W}}_{\mathrm{T}}^{\text {ext }}=-\mu \mathrm{mg}\left(\mathrm{v}+\mathrm{v}^{\prime}\right) \\
\dot{\mathrm{T}}_{\mathrm{T}}^{\text {bloque }}=-\mu \mathrm{mg}\left(\mathrm{v}+\mathrm{v}^{\prime}\right)
\end{array}\right\} \Rightarrow \dot{\mathrm{W}}_{\mathrm{T}}^{\rightarrow \text { bloque }}=\dot{\mathrm{T}}_{\mathrm{T}}^{\text {bloque }}</math>

:Referencia camión: Al estar el camión acelerado respecto al suelo, hay que tener en cuenta las fuerzas de arrastre sobre todos los diferenciales de masa (dm) del bloque:

{|
|

[[Archivo:ExE1-3-2-3-cat-esp.jpg|thumb|right|250px|link=]]
|<math>\cal{\Fs}^{\rightarrow \ds\ms}_{\mathrm{ar}}=-\ds\ms\overline{\as}_\mathrm{ar}(\ds\ms)=-\ds\ms (\leftarrow \as)=(\rightarrow \ds\ms \cdot \as)</math>.

El torsor resultante de este sistema de fuerzas en el centro de inercia '''G''' se reduce a una fuerza horizontal.

<math>\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{bloque}}=\dot{\Ws}_\mathrm{camión}^{\mathrm{ext}} + \dot{\Ws}_\mathrm{camión}^{\mathrm{ar}} =\dot{\Ws}_\mathrm{camión}^{\mathrm{peso}} + \dot{\Ws}_\mathrm{camión}^{\mathrm{fricción}} + \dot{\Ws}_\mathrm{camión}^{\mathrm{enlace}} + \dot{\Ws}_\mathrm{camión}^{\mathrm{ar}} , </math>

<math>\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{bloque}}= \Bigr[(\downarrow \ms\gs)+(\leftarrow \mu\ms\gs)+(\uparrow \Ns) + (\rightarrow \ms\as)\Bigr] \cdot(\rightarrow \vs')+(\odot\Ms_\es) \cdot0=\ms(\as-\mu\gs)\vs'.</math>
|}

:Se puede comprobar que el balance de potencias es consistente:

:<math>\Ts^\mathrm{bloque}_\mathrm{camión}=\frac{1}{2}\ms\vs^2_\mathrm{camión}(\mathrm{bloque}) \quad \Rightarrow \quad \dot{\Ts}^\mathrm{bloque}_\mathrm{camión} = \ms\vs_\mathrm{camión}(\mathrm{bloque})\dot{\vs}_\mathrm{camión}(\mathrm{bloque})</math>

:<math>\left.\begin{array}{l}
\mathrm{AB}:\mathrm{suelo} \\
\mathrm{REL}:\mathrm{camión}
\end{array}\right\} \quad \dot{\vs}_\mathrm{camión}(\mathrm{bloque})= \dot{\vs}_\mathrm{T}(\mathrm{bloque})-\dot{\vs}_\mathrm{ar}(\mathrm{bloque}) = (-\mu\gs)-(-\as)=\as-\mu\gs \quad \Rightarrow \quad \dot{\Ts}^\mathrm{bloque}_\mathrm{camión}=\ms\vs'(\as-\mu\gs)</math>

:<math>\left.\begin{array}{l}
\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{bloque}}=\dot{\Ws}_\mathrm{camión}^{\mathrm{ext}} + \dot{\Ws}_\mathrm{camión}^{\mathrm{ar}}= \ms(\as-\mu\gs)\vs'\\
\dot{\Ts}^\mathrm{bloque}_\mathrm{camión}=\ms\vs'(\as-\mu\gs)
\end{array}\right\} \Rightarrow\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{bloque}}= \dot{\Ts}^\mathrm{bloque}_\mathrm{camión}</math>


====✏️ EJEMPLO E1-3.3: camión frenando con carga====
------
:
{|
|
:[[Archivo:ExE1-3-3-1-esp.png|thumb|left|220px|link=]]
|Un camión frena con aceleración '''a''' respecto al suelo en el instante en el que su velocidad (también respecto al suelo) es v. Los pares de frenada en el eje delantero y en el trasero son iguales y valen <math>\Gamma</math>, y las ruedas, de radio r y masa despreciable, no deslizan sobre el suelo. En el mismo instante, el bloque homogéneo, de masa m, desliz con velocidad v' relativa al camión. Se trata de hacer un balance de potencias para el sistema (bloque+camión) en la referencia del suelo y en la del camión.

|}

:Ya que la potencia de las interacciones internas no es nula en principio (no se trata de un único sólido rígido), es conveniente hacer un DGI del sistema para no olvidar ninguna:
{|
|
[[Archivo:ExE1-3-3-2-esp.png|thumb|right|250px|link=]]

|
|
c.p.s.d.: contacto puntual sin deslizamiento

c.m.c.d.: contacto multipuntual con deslizamiento

|}
:El sistema sobre el que se hace el balance de potencias es todo menos el suelo. Teniendo en cuenta que la potencia de los enlaces internos es siempre nula, las únicas interacciones que hay que tener en cuenta son el peso, los enlaces de las ruedas con el suelo, los pares de frenada y la fricción entre bloque y camión:

[[Archivo:ExE1-3-3-3-esp.png|thumb|center|300px|link=]]

:Referencia suelo:
:Las fuerzas de enlace del suelo sobre las ruedas están aplicadas a puntos de velocidad cero, y por tanto no hacen potencia. El peso es una fuerza vertical, y el movimiento del camión y del bloque es horizontal, por tanto su potencia también es cero.

:<math>\dot{\Ws}_\Ts^{\rightarrow \mathrm{sist}}=\dot{\Ws}_\Ts^{\mathrm{ext}}+\dot{\Ws}^{\mathrm{int}}=\dot{\Ws}_\Ts^{\mathrm{peso}}+\dot{\Ws}_\Ts^{\mathrm{enlace}}+\dot{\Ws}^{\mathrm{fricción}}+\dot{\Ws}^{\Gamma}=\dot{\Ws}^{\mathrm{fricción}}+\dot{\Ws}^{\Gamma}.</math>

:Las potencias de interacciones internas se pueden calcular en cualquier referencia. Por tanto:

:<math>\dot{\Ws}^{\mathrm{fricción}}= \dot{\Ws}_\mathrm{camión}^{\mathrm{fric}\rightarrow \mathrm{bloque}} +\dot{\Ws}_\mathrm{camió}^{\mathrm{fric}\rightarrow \mathrm{camión}} = \dot{\Ws}_\mathrm{camión}^{\mathrm{fric}\rightarrow \mathrm{bloque}}= (\leftarrow \mu\ms\gs)\cdot (\rightarrow \vs')=-\mu\ms\gs\vs'</math>

:<math>\dot{\Ws}^{\Gamma}= \dot{\Ws}_\mathrm{camión}^{\Gamma\rightarrow \mathrm{ruedas}} +\dot{\Ws}_\mathrm{camión}^{\Gamma\rightarrow \mathrm{camión}} = \dot{\Ws}_\mathrm{camión}^{\Gamma\rightarrow \mathrm{ruedas}}= 2 \left(\odot \Gamma \right)\cdot \left(\otimes \frac{\vs}{\rs}\right)=-2\Gamma\frac{\vs}{\rs}.</math>

:Por otro lado: <math>\Ts_\Ts^\mathrm{sist}=\frac{1}{2}\ms\vs_\Ts^2(\mathrm{bloque})+\frac{1}{2}\ms\vs_\Ts^2(\mathrm{camión})\quad \Rightarrow \quad \dot{\Ts}_\Ts^\mathrm{sist}=-\ms(\vs+\vs')\mu\gs-\Ms\vs\as. </math>

:Finalmente: <math>\dot{\Ws}_\Ts^{\rightarrow \mathrm{sist}}=\dot{\Ts}_\Ts^\mathrm{sist} \quad \Rightarrow \quad -\mu\ms\gs\vs'-2\Gamma\frac{\vs}{\rs}=-\ms(\vs+\vs')\mu\gs-\Ms\vs\as.</math>

:Esta ecuación permite calcular el par de frenada: <math>\Gamma=\frac{1}{2}(\mu\ms\gs+\Ms\as)\rs.</math>

:Referencia camión:

:En esta referencia hay que tener en cuenta las fuerzas de arrastre. Por otro lado, las fuerzas de enlace del suelo sobre las ruedas están aplicadas a puntos que ya no tienen velocidad cero, y por tanto hacen potencia. Como antes, la potencia del peso es cero.

:<math>\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{sist}}=\dot{\Ws}_\mathrm{camión}^{\mathrm{ext}}+\dot{\Ws}^{\mathrm{int}}+\dot{\Ws}_\mathrm{camión}^{\mathrm{ar}}=\dot{\Ws}_\mathrm{camión}^{\mathrm{enlace}}+\dot{\Ws}^{\mathrm{fricción}}+\dot{\Ws}^{\Gamma}+\dot{\Ws}_\mathrm{camión}^{\mathrm{ar}}.</math>

:Las fuerzas de arrastre sobre el camión no hacen potencia porque el camión está parado en esta referencia. La de las fuerzas de arrastre sobre el bloque se ha calculado en el [[E1. Teorema de la energía: versión diferencial#✏️ ✏️ EJEMPLO E1-3.2: bloque deslizando sobre un camión|'''ejemplo E1-3.2''']]. La potencia de la fricción y de los pares de frenada es la misma de antes ya que no depende de la referencia.

:<math>\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{sist}}=\dot{\Ws}_\mathrm{camión}^{\mathrm{enllaç}}+\dot{\Ws}^{\mathrm{fricción}}+\dot{\Ws}^{\Gamma}+\dot{\Ws}_\mathrm{camión}^{\mathrm{ar} \rightarrow \mathrm{bloque}}= \dot{\Ws}_\mathrm{camión}^{\mathrm{enlace}} -\mu\ms\gs\vs'-2\Gamma\frac{\vs}{\rs} +\ms\as\vs'.</math>

:Las descripciones cinemática y dinámica de la rueda son:

[[Archivo:ExE1-3-3-4-esp.png|thumb|center|430px|link=]]

:<math>\dot{\Ws}_\mathrm{camión}^{\mathrm{enlace}}=2\Bigr[(\leftarrow \Ts)+(\uparrow \Ns)\Bigr]\cdot(\leftarrow\vs)=2\Ts\vs.</math>

La fuerza tangencial de enlace se puede calcular a partir del TMC aplicado al centro '''C''' de la rueda. Como no tiene masa: <math>\sum\overline{\Ms}_\mathrm{ext}(\Cs)=\overline{0} \quad \Rightarrow \quad (\odot \Gamma)+(\otimes \Ts\rs)=0\quad \Rightarrow \quad \Ts=\Gamma/\rs.</math>

:Por tanto: <math>\dot{\Ws}_\mathrm{camió}^{\rightarrow \mathrm{sist}} =2\frac{\Gamma}{\rs}\vs-\mu\ms\gs\vs'-2\Gamma\frac{\vs}{\rs}+\ms\as\vs'=\ms(\as-\mu\gs)\vs'.</math>

:Por otro lado, como se ha visto en el [[E1. Teorema de la energía: versión diferencial#✏️ ✏️ EJEMPLO E1-3.2: bloque deslizando sobre un camión|'''ejemplo E1-3.2''']]:

:<math>\Ts_\mathrm{camión}^\mathrm{sist}= \Ts_\mathrm{camión}^\mathrm{bloque}=\frac{1}{2}\ms\vs_\mathrm{camión}^2(\mathrm{bloque}) \quad \Rightarrow \quad \dot{\Ts}_\mathrm{camión}^\mathrm{sist}= \dot{\Ts}_\mathrm{camión}^\mathrm{bloque}=\ms\vs'(\as-\mu\gs).</math>

:El balance de potencias es consistente:

:<math>\left.\begin{array}{l}
\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{sist}}= \ms(\as-\mu\gs)\vs'\\
\dot{\Ts}^\mathrm{sist}_\mathrm{camión}=\ms(\as-\mu\gs)\vs'
\end{array}\right\} \Rightarrow\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{sist}}= \dot{\Ts}^\mathrm{sist}_\mathrm{camión}</math>

-----
-----

==E1.4 Balance de potencias para un sistema multisólido: cálculo directo y cálculo indirecto==

El '''cálculo directo''' (a partir de su definición) de la potencia asociada a una interacción (una fuerza sobre un punto '''P''' o un momento sobre un sólido rígido S) exige el conocimiento de la interacción (fuerza o momento):

<math>\dot{\Ws}_\Rs^{\Fs\rightarrow \Ps}=\overline{\Fs}_{\rightarrow\Ps} \cdot \vel{P}{R} \quad , \quad \dot{\Ws}_\Rs^{\Ms\rightarrow \Ss}=\overline{\Ms}_{\rightarrow\Ss} \cdot \velang{S}{R} </math>

Cuando no se trata de una interacción formulable, hay que recurrir a los teoremas vectoriales para determinarla.

El '''cálculo indirecto''' es una alternativa que no necesita recurrir a estos teoremas. Se basa en la evaluación de un balance de potencias para un sistema y una referencia adecuados de manera que la potencia a determinar sea el único término desconocido de la ecuación (el único que no se puede calcular de manera directa). Cuando el balance incluye más de un término desconocido, se pueden combinar varios balances (para distintos sistemas y referencias) hasta tener un número suficiente de ecuaciones que permitan la determinación de todas las potencias desconocidas.

====✏️ EJEMPLO E1-4.1: polea frenada====
------
:
{|
|
:[[Archivo:ExE1-4-1-1-esp.png|thumb|left|150px|link=]]
|Un bloque de masa m cuelga de un hilo inextensible atado a un punto de la periferia de una polea de radio r. Entre soporte (fijo al suelo) y polea actúa un freno que garantiza que el bloque baja a velocidad constante respecto al suelo. Se trata de calcular la potencia del freno.

|}

:Càlculo directo: <math>\dot{\Ws}^{\mathrm{freno}}= \overline{\Gamma}^{\rightarrow \mathrm{polea}}\cdot\velang{polea}{T}=\left(\odot\Gamma\right)\cdot\left( \otimes \frac{\vs_0}{\rs}\right)=-\Gamma \frac{\vs_0}{\rs}.</math>

:El par de frenada no es un dato, y hay que recurrir al TMC para calcularlo.

:<math>\left.\begin{array}{l}
\text{SISTEMA: polea + bloque}\\
\text{TMC en } \Os
\end{array}\right\} \sum\overline{\Ms}_\mathrm{ext}(\Os)=\dot{\overline{\Hs}}_\mathrm{RTO}(\Os)</math>

:El momento cinético proviene de las dos piezas. Ya que '''O''' no es fijo al bloque, hay que hacer descomposición baricéntrica para calcular el momento cinético del bloque:

:<math>\overline{\Hs}_\mathrm{RTO=T} (\Os) =\overline{\Hs}_\Ts^\mathrm{politja}(\Os) + \overline{\Hs}_\Ts^\mathrm{bloc}(\Os)= \overline{\Hs}_\Ts^\mathrm{politja}(\Os) + \overline{\Hs}_\Ts^\mathrm{bloc}(\Gs_\mathrm{bloc})+ \overline{\Hs}_\Ts^{\oplus\mathrm{bloc}}(\Os) = </math>

:<math>\hspace{2.3cm}=\Is_\mathrm{O}^\mathrm{polea} \velang{polea}{T}+\Is_{\Gs_\mathrm{bloque}}^\mathrm{bloque}+\OGvec_\mathrm{bloque}\times \ms\overline{\vs}_\Ts(\Gs_\mathrm{bloque})=\left(\otimes \Is_\mathrm{O}\frac{\vs_0}{\rs}\right)+(\otimes \ms\vs_0\rs).</math>

:[[Archivo:ExE1-4-1-2-esp.png|thumb|right|180px|link=]]
:Se trata de un momento cinético constante en dirección y valor. Por tanto:

:<math>\dot{\overline{\Hs}}_\mathrm{RTO}(\Os)=0 \quad \Rightarrow \quad \sum{\overline{\Ms}}_\mathrm{ext}(\Os)=(\odot\Gamma)+(\otimes \ms\gs\rs) =0\quad \Rightarrow \quad \Gamma= \ms\gs\rs</math>.

:Finalmente: <math>\dot{\Ws}^{\mathrm{freno}}= -\Gamma \frac{\vs_0}{\rs} = -\ms\gs\vs_0.</math>

:Cálculo indirecto:

:<math>\left.\begin{array}{l}
\text{SISTEMA: polea + soporte + bloque}\\
\text{REF: suelo(Gal)}
\end{array}\right\} \quad \dot{\Ts}^\mathrm{sist}_\Ts=\dot{\Ws}^\mathrm{ext}_\Ts+ \dot{\Ws}^\mathrm{int}</math>

:<math>\left.\begin{array}{l}
\dot{\Ts}^\mathrm{sist}_\Ts=0\\
\dot{\Ws}^\mathrm{ext}_\Ts=\left(\downarrow \ms\gs\right)\cdot\left(\downarrow \vs_0\right)= \ms\gs\vs_0\\
\dot{\Ws}^\mathrm{int}=\dot{\Ws}^{\mathrm{freno}}
\end{array}\right\} \quad \Rightarrow \quad \dot{\Ws}^{\mathrm{freno}} =-\ms\gs\vs_0</math>



====✏️ EJEMPLO E1-4.2: mono y plátanos====
------
:
{|
|
:[[Archivo:ExE1-4-2-1-esp.png|thumb|left|180px|link=]]
|Unos plátanos de masa <math>(2/3)\ms</math> cuelgan de una cuerda inextensible que se apoya sin deslizar sobre una polea de radio r y masa despreciable, articulada respecto al suelo. En el otro extremo de la cuerda se agarra un mono de masa m. La polea gira con velocidad angular constante <math>\Omega_0</math> respecto al suelo bajo la acción de un motor. El mono se mueve respecto a la cuerda sin que sus manos se froten contra la cuerda, y de manera que la distancia de su centro de masas al suelo se mantiene constante. Se trata de calcular la potencia del motor.
|}

:Cálculo directo: <math>\dot{\Ws}^{\mathrm{freno}}= \overline{\Gamma}^{\rightarrow \mathrm{polea}}\cdot\velang{polea}{T}=\left(\odot\Gamma\right)\cdot\left( \odot \Omega_0\right)=\Gamma \Omega_0.</math>

:El par motor no es un dato, y hay que recurrir al TMC para calcularlo.

:<math>\left.\begin{array}{l}
\text{SISTEMA: polea + mono + plátanos}\\
\text{TMC en } \Os
\end{array}\right\} \sum\overline{\Ms}_\mathrm{ext}(\Os)=\dot{\overline{\Hs}}_\mathrm{RTO}(\Os)</math>

:El momento cinético proviene del mono y de los plátanos. En primera aproximación, se puede considerar la masa del mono concentrada en su tronco y su cabeza, de manera que no gira y su velocidad respecto al suelo es nula. Teniendo en cuenta que el punto '''O''' no se mueve respecto al mono pero sí respecto a los plátanos:

:<math>\overline{\Hs}_\mathrm{RTO=T} (\Os) =\overline{\Hs}_\Ts^\mathrm{mono}(\Os) + \overline{\Hs}_\Ts^\mathrm{plátanos}(\Os)= \overline{\Hs}_\Ts^\mathrm{mono}(\Os) + \overline{\Hs}_\Ts^\mathrm{plátanos}(\Gs_\mathrm{plát})+ \OGvec_\mathrm{plát} \times \frac{2}{3} \ms \overline{\vs}_\Ts(\Gs_\mathrm{plát})= </math>

:<math>\hspace{2.3cm}=\OGvec_\mathrm{plát}\times \frac{2}{3} \ms \overline{\vs}_\Ts(\Gs_\mathrm{plát}) = \Bigr[(\rightarrow \rs)+(\downarrow \xs)\Bigr] \times\frac{2}{3} \ms \left(\downarrow \rs\Omega_0\right)=\left(\otimes \frac{2}{3}\ms\rs^2\Omega_0\right).</math>

:Se trata de un momento cinético constante en dirección y valor. Por tanto:

:<math>\dot{\overline{\Hs}}_\mathrm{RTO}(\Os)=0 \quad \Rightarrow \quad \sum{\overline{\Ms}}_\mathrm{ext}(\Os)=(\odot\Gamma)+(\otimes \ms\gs\rs) + \left(\otimes \frac{2}{3}\ms\gs\rs\right)=0\quad \Rightarrow \quad \Gamma= \frac{5}{3}\ms\gs\rs</math>.

:Finalmente: <math>\dot{\Ws}^{\mathrm{motor}}= \Gamma \Omega_0 = \frac{5}{3}\ms\gs\rs\Omega_0.</math>

:Cálculo indirecto

:Hay que tener presente que es un problema con dos tipos de actuadores: el motor y la musculatura del mono. Hay diversos modelos biomecánicos para formular las fuerzas de los músculos, pero quedan fuera del ámbito de este curso. Ya que el mono se mueve, sus músculos desarrollan una potencia no nula.

:<math>\left.\begin{array}{l}
\text{SISTEMA: polea + mono + plátano}\\
\text{REF: suelo(Gal)}
\end{array}\right\} \quad \dot{\Ts}^\mathrm{sist}_\Ts=\dot{\Ws}^\mathrm{ext}_\Ts+ \dot{\Ws}^\mathrm{int}</math>

:<math>\left.\begin{array}{l}
\dot{\Ts}^\mathrm{sist}_\Ts=0\\
\dot{\Ws}^\mathrm{ext}_\Ts=\left(\downarrow \frac{2}{3}\ms\gs\right)\cdot\left(\downarrow \rs\Omega_0\right)= \frac{2}{3}\ms\gs\rs\Omega_0\\
\dot{\Ws}^\mathrm{int}=\dot{\Ws}^{\mathrm{musc}}+\dot{\Ws}^{\mathrm{motor}}
\end{array}\right\} \quad \Rightarrow \quad \dot{\Ws}^{\mathrm{musc}} + \dot{\Ws}^{\mathrm{motor}}=-\frac{2}{3}\ms\gs\rs\Omega_0</math>

:Este balance proporciona una ecuación con dos incógnitas. Se puede plantear un segundo balance para un sistema donde una de las dos no aparezca: un sistema sin el mono o un sistema sin el motor. Si se hace un balance de potencias para el sistema (mono + trozo de cuerda), las fuerzas externas a considerar son el peso y la tensión de la cuerda. Si se sigue trabajando en la referencia del suelo, esta tensión (desconocida) hará potencia no nula ya que está aplicada a un punto de velocidad ascendente.
:[[Archivo:ExE1-4-2-2-cat-esp.png|thumb|right|330px|link=]]

:Para evitar que aparezca un término asociado a la tensión de la cuerda, se puede trabajar en la referencia del trozo de cuerda que se ha considerado, que al tener velocidad ascendente constante es galileana. En esta referencia, el mono baja con velocidad constante:

:<math>\left.\begin{array}{l}
\text{REL: trozo de cuerda}\\
\text{AB: suelo}
\end{array}\right\}</math>

:<math>\overline{\vs}_\mathrm{REL}(\mathrm{mono})=\overline{\vs}_\mathrm{ar}(\mathrm{mono})=-\overline{\vs}_\mathrm{ar}(\mathrm{mono})=(\downarrow \rs\Omega_0)</math>

:<math>\left.\begin{array}{l}
\text{SISTEMA: mono+ trozo de cuerda}\\
\text{REF: trozo de cuerda(Gal)}
\end{array}\right\} \quad \dot{\Ts}^\mathrm{sist}_\mathrm{cuerda} =\dot{\Ws}^\mathrm{ext}_\mathrm{cuerda}+ \dot{\Ws}^\mathrm{int}</math>

:<math>\left.\begin{array}{l}
\dot{\Ts}^\mathrm{sist}_\mathrm{cuerda}=0\\
\dot{\Ws}^\mathrm{ext}_\mathrm{cuerda}=\left(\downarrow \ms\gs\right)\cdot\left(\downarrow \rs\Omega_0\right)= \ms\gs\rs\Omega_0\\
\dot{\Ws}^\mathrm{int}=\dot{\Ws}^{\mathrm{musc}}
\end{array}\right\} \quad \Rightarrow \quad \dot{\Ws}^{\mathrm{musc}} =-\ms\gs\rs\Omega_0</math>

:Combinando este resultado con el del balance anterior:

:<math>\dot{\Ws}^\mathrm{motor}=-\frac{2}{3}\ms\gs\rs\Omega_0-\dot{\Ws}^\mathrm{musc}=-\frac{5}{3}\ms\gs\rs\Omega_0</math>



© Universitat Politècnica de Catalunya. [[Mecánica:Derechos de autor |Todos los derechos reservados]]
-------------

<center>
[[D8. Conservaciones|<<< D8. Conservaciones]]

[[E2. Teorema de la energía: versión integrada|E2. Teorema de la energía: versión integrada>>>]]
</center>

E1. Work-Energy Theorem: differential form

2025-05-20T12:17:33Z

Bros: /* ✏️ EXAMPLE E1-1.2: pendulum */

<div class="noautonum">__TOC__</div>
<math>\newcommand{\uvec}{\overline{\textbf{u}}}
\newcommand{\vvec}{\overline{\textbf{v}}}
\newcommand{\evec}{\overline{\textbf{e}}}
\newcommand{\Omegavec}{\overline{\mathbf{\Omega}}}
\newcommand{\velang}[2]{\Omegavec^{\textrm{#1}}_{\textrm{#2}}}
\newcommand{\Alfavec}{\overline{\mathbf{\alpha}}}
\newcommand{\accang}[2]{\Alfavec^{\textrm{#1}}_{\textrm{#2}}}
\newcommand{\ds}{\textrm{d}}
\newcommand{\ps}{\textrm{p}}
\newcommand{\ns}{\textrm{n}}
\newcommand{\hs}{\textrm{h}}
\newcommand{\cs}{\textrm{c}}
\newcommand{\gs}{\textrm{g}}
\newcommand{\Ns}{\textrm{N}}
\newcommand{\Hs}{\textrm{H}}
\newcommand{\Fs}{\textrm{F}}
\newcommand{\ms}{\textrm{m}}
\newcommand{\ts}{\textrm{t}}
\newcommand{\us}{\textrm{u}}
\newcommand{\vs}{\textrm{v}}
\newcommand{\Rs}{\textrm{R}}
\newcommand{\Es}{\textrm{E}}
\newcommand{\Ws}{\textrm{W}}
\newcommand{\Ts}{\textrm{T}}
\newcommand{\Ls}{\textrm{L}}
\newcommand{\As}{\textrm{A}}
\newcommand{\Ds}{\textrm{D}}
\newcommand{\Bs}{\textrm{B}}
\newcommand{\es}{\textrm{e}}
\newcommand{\fs}{\textrm{f}}
\newcommand{\is}{\textrm{i}}
\newcommand{\Is}{\textrm{I}}
\newcommand{\ks}{\textrm{k}}
\newcommand{\js}{\textrm{j}}
\newcommand{\rs}{\textrm{r}}
\newcommand{\as}{\textrm{a}}
\newcommand{\ss}{\textrm{s}}
\newcommand{\Os}{\textbf{O}}
\newcommand{\Gs}{\textbf{G}}
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\newcommand{\Cs}{\textbf{C}}
\newcommand{\Ps}{\textbf{P}}
\newcommand{\Ss}{\textbf{S}}
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\newcommand{\P}{\textrm{P}}
\newcommand{\Q}{\textrm{Q}}
\newcommand{\Ms}{\textrm{M}}
\newcommand{\deg}{^\textsf{o}}
\newcommand{\xs}{\textsf{x}}
\newcommand{\ys}{\textsf{y}}
\newcommand{\zs}{\textsf{z}}
\newcommand{\dert}[2]{\left.\frac{\ds{#1}}{\ds\ts}\right]_{\textrm{#2}}}
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\newcommand{\vec}[1]{\overline{#1}}
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\newcommand{\vecdot}[1]{\overline{\dot{#1}}}
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\newcommand{\GPvec}{\vec{\Gs\Ps}}
\newcommand{\PSvec}{\vec{\Ps\Ss}}
\newcommand{\QQvec}{\vec{\Qs\Qs}}
\newcommand{\QGvec}{\vec{\Qs\Gs}}
\newcommand{\QPvec}{\vec{\Qs\Ps}}
\newcommand{\JQvec}{\vec{\Js\Qs}}
\newcommand{\PQvec}{\vec{\Ps\Qs}}
\newcommand{\GJvec}{\vec{\Gs\Js}}
\newcommand{\OPvec}{\vec{\Os\textbf{P}}}
\newcommand{\OCvec}{\vec{\Os\Cs}}
\newcommand{\OGvec}{\vec{\Os\Gs}}
\newcommand{\CJvec}{\vec{\Cs\Js}}
\newcommand{\GCvec}{\vec{\Gs\Cs}}
\newcommand{\PGvec}{\vec{\Ps\Gs}}
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\begin{bmatrix}
{#1} & {#2} & {#3}\\
{#4} & {#5} & {#6}\\
{#7} & {#8} & {#9}
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</math>

The '''Work-Energy Theorem''' plays a prominent role in physics, as it bridges all branches of science (electromagnetism, thermodynamics, etc.): all of them deal with energy.

There is a certain parallel between the Vector Theorems and the '''Work-Energy Theorem''': they all allow us to understand the evolution of a magnitude that depends on the mechanical state of the system. In the case of the former, we are dealing with the instantaneous evolution of vector magnitudes (linear momentum, angular momentum), while the latter deals with the finite change (in a finite time interval) of a scalar magnitude ('''mechanical energy''').

This unit presents the differential form of this theorem (the '''power balance'''). The integrated form (which is the Work-Energy Theorem itself) is presented in [[E2. Teorema de la energía: versión integrada|'''unit E2''']].
 
------
------
==E1.1 Power balance in a system of particles==

:[[File:E1-1-eng.png|thumb|right|300px|link=]]
<center>'''Figure E1.1''' Forces on a particle of a system with constant matter</center>

Let us consider a system of constant matter, and let’s apply Newton's second law to each particle (or each mass differential) of this system in a general reference frame R (which may be non-Galilean) ('''Figure E1.1'''):

<math>\F{\rightarrow \Ps} + \Fcal{}{\text{ar}\rightarrow\Ps} + \Fcal{}{\text{Cor}\rightarrow\Ps} =\ms_\Ps\acc{P}{R}</math>

<math>\F{\rightarrow \Ps} </math> is the resultant interaction force on '''P''', and it comes from internal interactions (with particles of the same system) and external ones:

<math>\F{\rightarrow \Ps}=\F{\rightarrow \Ps}^\text{int}+\F{\rightarrow \Ps}^\text{ext}</math>

If both sides of the equation are multiplied by <math>\vel{P}{R}</math>, all forces (interaction and inertia forces) orthogonal to <math>\vel{P}{R}</math> are filtered out. Specifically, since <math>\acc{P}{R}=\accs{P}{R} + \accn{P}{R} </math>, where <math>\accn{P}{R} \perp \vel{P}{R}</math>, and since the Coriolis force is always orthogonal to <math>\vel{P}{R} (\Fcal{}{\text{Cor}\rightarrow\Ps} = -2\ms_\Ps \velang{R}{Gal} \times \vel{P}{R})</math>:

<math>\vel{P}{R} \cdot \left( \F{\rightarrow \Ps}^\text{int}+\F{\rightarrow \Ps}^\text{ext} + \Fcal{}{\text{ar}\rightarrow\Ps} \right) = \ms_\Ps\accs{P}{R} \cdot \vel{P}{R} = \frac{\ds}{\ds \ts} \left(\frac{1}{2}\ms_\Ps \vvec_\Rs^2(\Ps) \right) \equiv \frac{\ds\Ts_\Rs(\Ps)}{\ds\ts}=\dot{\Ts}_\Rs(\Ps)</math>

The scalar function <math>\Ts_\Rs(\Ps)</math> is the '''kinetic energy''' of the particle, and is a function of the [[D1. Foundational laws of Newtonian dynamics#D1.3 Newton’s Principle of Determinacy|'''mechanical state''']] of '''P'''. The scalar product of a force and the velocity of its application point is the '''force power''': <math>\vel{P}{R} \cdot \F{\rightarrow \Ps} \equiv \dot{\Ws}_\Rs^{\F{\rightarrow \Ps}} </math>. Both scalar quantities <math>(\Ts_\Rs(\Ps), \dot{\Ws}_\Rs^{\F{\rightarrow \Ps}})</math> depend on the reference frame. For the same set of interaction forces acting on '''P''', the reference frame determines which are filtered out and which are not.

The equation can be rewritten as <math>\dot{\Ws}_\Rs^{\text{ext}\rightarrow \Ps}+\dot{\Ws}_\Rs^{\text{int}\rightarrow \Ps}+\dot{\Ws}_\Rs^{\text{ar}\rightarrow \Ps} =\dot{\Ts}_\Rs(\Ps)</math>, and constitutes the '''power balance''' for particle '''P''' in the R reference frame. Summing for all particles in the system:

<math>\dot{\Ws}_\Rs^{\text{ext}}+\dot{\Ws}^{\text{int}}+\dot{\Ws}_\Rs^{\text{ar}} =\dot{\Ts}_\Rs^\text{syst}</math>

The calculation of kinetic energy is covered in [[E2. Teorema de la energía: versión integrada|'''unit E2''']]. This unit focuses on the calculation of powers. [[E1. Work-Energy Theorem: differential form#E1.2 Potencia de una pareja de acción-reacción|'''section E1-2''']] shows that the power of internal interactions does not depend on the reference frame (which is why the subscript R does not appear in the <math>\dot{\Ws}^{\text{int}}</math> term of the previous equation).

====✏️ EXAMPLE E1-1.1: free particle====
------
:
{|
:[[File:ExE1-1-1-1-eng.jpg|thumb|left|180px|link=]]
|A particle '''P''', with mass m, is released from the ring of a Ferris wheel when it is at its highest point. We want to calculate the power, at this instant, of all forces acting on the particle from three different reference frames: the ground, the ring, and the Ferris wheel cabin.
|}

:Ground reference frame: Since this is a Galilean reference frame, all forces on '''P''' are interaction forces.
{|
:[[File:ExE1-1-2- eng.png|thumb|right|230px|link=]]

|<math>\hspace{0.5cm}\sum\dot{\Ws}_\Ts=\dot{\Ws}_\Ts^\text{weight}= \left(\leftarrow \Rs\Omega_0\right) \cdot \left(\downarrow \ms\gs\right)=0</math>
|}

:Ring reference frame: It is a non-Galilean reference, and the transportation force on '''P''' must be taken into account.
{|
:[[File:ExE1-1-3-eng.png|thumb|right|250px|link=]]
|<math>\hspace{0.5cm}</math> AB:ground, REL:ring 

<math>\hspace{0.5cm}\vel{P}{REL}=\vel{P}{AB}-\vel{P}{ar}= \left(\leftarrow \Rs\Omega_0\right)-\left(\leftarrow \Rs\Omega_0\right)=\overline{0}</math> 

<math>\hspace{0.5cm}</math>

Since the velocity is zero, the total power is zero.
|}

:Cabin reference frame: It is also a non-Galilean reference, and the transportation force on '''P''' must be taken into account.

{|
:[[File:ExE1-1-4-eng.png|thumb|right|250px|link=]]
|<math>\hspace{0.5cm}</math>AB:ground, REL:cabin

<math>\hspace{0.5cm}\vel{P}{REL}=\vel{P}{AB}-\vel{P}{ar}= \left(\leftarrow \Rs\Omega_0\right)-\left(\uparrow \Rs\Omega_0\right)</math>

<math>\hspace{0.5cm}\dot{\Ws}_\text{cabin}^\text{weight}=\left[ \left(\leftarrow \Rs\Omega_0\right)+\left(\downarrow \Rs\Omega_0\right) \right] \cdot \left(\downarrow\ms\gs\right) =\ms\gs\Rs\Omega_0</math>

<math>\hspace{0.5cm}\dot{\Ws}_\text{cabin}^\text{ar}=\left[ \left(\leftarrow \Rs\Omega_0\right)+\left(\downarrow \Rs\Omega_0\right) \right] \cdot \left(\rightarrow\ms\Rs^2\Omega_0^2\right) =-\ms\gs\Rs^2 \Omega_0^3</math>

<math>\hspace{0.5cm}\sum\dot{\Ws}_\text{cabin}=\ms\Rs\Omega_0(\gs-\Rs\Omega_0^2)</math>

|}


====✏️ EXAMPLE E1-1.2: pendulum====
------
:
{|
:[[File:ExE1-2-1-eng.png|thumb|left|180px|link=]]
|A particle '''P''' of mass m hangs from an inextensible thread of length L attached to the ceiling. The assembly oscillates in a vertical plane fixed to the ground. We want to calculate the power of all the forces acting on the particle in two different reference frames: the ground and a platform rotating relative to the ground with a constant angular velocity <math>\Omega_0</math>. 
|}

Ground reference frame: Since it is a Galilean reference frame, '''P''' is only subjected to interaction forces.
{|
:[[File:ExE1-2-2-neut.png|thumb|left|300px|link=]]

|<math>\dot{\Ws}_\Es^\Ts=(\nearrow\Ls\dot{\theta})\cdot(\nwarrow\Ts)=0</math>

<math>\dot{\Ws}_\Es^\text{weight}=(\nearrow\Ls\dot{\theta}\cdot(\downarrow \ms\gs)=-\ms\gs\Ls\dot{\theta}\sin\theta</math>
|}

<math>\vspace{8cm}</math>

Platform reference frame: It is a non Galilean reference frame, and the transportation force on '''P''' has to be taken into account.

{|
:[[File:ExE1-2-3-eng.png|thumb|left|400px|link=]]

|AB:ground, REL:platform

<math>\vel{P}{REL}=\vel{P}{AB}-\vel{P}{ar}=(\nearrow\Ls\dot{\theta})-(\otimes \Ls \Omega_0 \sin \theta)</math>

<math>\dot{\Ws}_\Es^\Ts\left[(\nearrow\Ls\dot{\theta})+(\odot\Ls\Omega_0\sin\theta)\right]\cdot(\nwarrow\Ts)=0</math>

<math>\dot{\Ws}_\Es^\text{weight}\left[(\nearrow\Ls\dot{\theta})+(\odot\Ls\Omega_0\sin\theta)\right]\cdot(\downarrow \ms\gs)=-\ms\gs\Ls\dot{\theta}\sin\theta</math>

<math>\dot{\Ws}_\Es^\text{ar}\left[(\nearrow\Ls\dot{\theta})+(\odot\Ls\Omega_0\sin\theta)\right]\cdot(\rightarrow \ms\Ls\Omega_0^2\sin\theta)=\ms\Ls^2\Omega_0^2\dot{\theta}\cos\theta</math>
|}



------
------

==E1.2 Power of an action-reaction pair==

The total power associated with internal interactions in a constant-mass system is not zero in principle (since the two forces of each action-reaction pair are applied to particles that may have different velocities relative to the reference frame in which the calculation is made), and it is independent of the reference frame in which it is evaluated. It only depends on the separation velocity of the two particles. The power balance for the system is therefore written as: <math>\dot{\Ts}_\Rs^\text{syst}=\dot{\Ws}_\Rs^\text{ext} + \dot{\Ws}^\text{int} + \dot{\Ws}_\Rs^\text{ar} </math> .

<div>
=====💭 PROOF ➕=====

{|
:[[File:E1-2-neut.png|thumb|left|200px|link=]]
|<math>\dot{\Ws}^\text{ac-reac}=\overline{\Fs}_{\Ps \rightarrow \Qs}\cdot \vel{Q}{R} + \overline{\Fs}_{\Qs \rightarrow \Ps}\cdot \vel{P}{R}= \overline{\Fs}_{\Ps \rightarrow \Qs}\cdot \left[ \vel{Q}{R}-\vel{P}{R}\right]</math>

By the [[D1. Foundational laws of Newtonian dynamics#D1.6 Newton’s third law (action-reaction principle)|'''action-reaction principle''']], <math>\overline{\Fs}_{\Ps \rightarrow \Qs}=-\overline{\Fs}_{\Qs \rightarrow \Ps}</math>. Therefore: <math>\dot{\Ws}^\text{ac-reac}=\overline{\Fs}_{\Ps \rightarrow \Qs}\cdot \left[ \vel{Q}{R}-\vel{P}{R}\right]</math>

As <math>\overline{\Fs}_{\Ps \rightarrow \Qs}</math> has the <math>\PQvec</math> direction, only the components of the velocities in that direction have to be considered:

<math>\dot{\Ws}^\text{ac-reac}=\overline{\Fs}_{\Ps \rightarrow \Qs}\cdot \biggl( \vel{Q}{R} \Bigr]_{||\Ps\Qs}-\ \vel{P}{R} \Bigr]_{||\Ps\Qs}\biggl)</math>
|}

The difference <math>\biggl( \vel{Q}{R} \Bigr]_{\Ps\Qs}-\ \vel{P}{R} \Bigr]_{\Ps\Qs}\biggl)</math> is the separation velocity <math>\dot{\rho}</math> between '''P''' and '''Q''', and it does not dpend on the reference frame since it is the derivative of a scalar (the distance <math>\dot{\rho}</math> between the two particles). For the previous drawing, <math>\dot{\rho}> 0</math>. Since this is an attractive interaction, <math>\dot{\Ws}^\text{ac-reac} < 0</math>.

With a drawing, it is easy to see that:
* attractive interaction and particle approach <math>(\dot{\rho}<0)</math> : <math>\dot{\Ws}^\text{ac-reac} > 0</math> ;
* repulsive interaction and particle separation <math>(\dot{\rho}>0)</math> : <math>\dot{\Ws}^\text{ac-reac} > 0</math>
* repulsive interaction and particle approach <math>(\dot{\rho}<0)</math> : <math>\dot{\Ws}^\text{ac-reac} < 0</math>


</div>

There are two particular cases of interaction forces whose power always has the same sign:

* Friction action-reaction pair between '''P''' and '''Q''': since the friction force always opposes the relative velocity between the two points, its power is always negative:

<math>\hspace{1cm} \dot{\Ws}^\text{friction}=\dot{\Ws}^{\text{fric}\rightarrow\Ps}_\text{RTQ}+\dot{\Ws}^{\text{fric}\rightarrow\Qs}_\text{RTQ}=\overline{\Fs}^{\text{fric}\rightarrow\Ps} \cdot \vel{P}{RTQ} = - \abs{\overline{\Fs}^{\text{fric}\rightarrow\Ps}} \frac{\vel{P}{RTQ}}{\abs{\vel{P}{RTQ}}} \cdot \vel{P}{RTQ}<0;</math>

* Constraint action-reaction pair between '''P''' and '''Q''': from the [[D3. Interactions between rigid bodies#D3.4 Direct constraint interactions|'''analytical characterization equation''']] of the constraint forces between two points '''P''' and '''Q''' belonging to two rigid bodies <math>\mathrm{S}_\Ps</math> and <math>\mathrm{S}_\Qs</math>, it is easy to see that the associated power is zero:

<math>\hspace{1cm} \dot{\Ws}^\text{constraint}=\dot{\Ws}^{\text{constraint}\rightarrow\Ps}_{\mathrm{S}_\Ps}+\dot{\Ws}^{\text{constraint}\rightarrow\Qs}_{\mathrm{S}_\Ps}=\overline{\Fs}^{\text{constraint}\rightarrow\Ps} \cdot \overline{\mathbf{v}}_{\mathrm{S}_\Qs}(\Ps) = - 0;</math>

====✏️ EXAMPLE E1-2.1: power of an action-reaction friction pair====
------
:
{|
:[[File:ExE1-2-1-2-eng.png|thumb|left|180px|link=]]
|A homogeneous wheel, with radius r and mass m, moves down an inclined plane without sliding until it joins a conveyor belt moving at a constant velocity <math>\vs_0</math> relative to the ground. At the moment it makes contact with the belt, the center of the wheel has a velocity <math>\vs_0</math> relative to the ground in the same direction as the belt. We want to calculate the power of the friction force between the belt and the wheel at this instant.
|}

At that momento, the <math>\mathbf{ICR}^\text{wheel}_\Es</math> is point <math>\Js_\text{wheel}</math> in contact with the belt. This point slides backwards with respect to the belt, therefore the friction force on the wheel is forwards:

<math>\left.\begin{array}{l}
\text { AB: ground } \\
\text { REL: belt }
\end{array}\right\} \quad \bar{\vs}_{\text {REL }}\left(\Js_{\text {wheel }}\right)=\bar{\vs}_{\text{AB}}\left(\Js_{\text {wheel }}\right)-\bar{\vs}_{\mathrm{ar}}\left(\Js_{\text {wheel }}\right)=-\bar{\vs}_{\mathrm{ar}}\left(\Js_{\text {wheel }}\right)=-\left(\rightarrow \mathrm{v}_0\right)=\left(\leftarrow \mathrm{v}_0\right) .</math>

[[File:ExE1-2-1-3-eng.png|thumb|center|500px|link=]]

Calculation of the friction power: <math>\dot{\Ws}^\text{friction}=\dot{\Ws}_\Rs^{\text{friction}\rightarrow \text{wheel}}+ \dot{\Ws}_\Rs^{\text{friction}\rightarrow \text{belt}}</math>, where R may be any reference frame. To quickly calculate this power, it is advisable to choose a reference frame R where one of the two points subjected to the action-reaction pair has zero speed, so that one of the two terms is directly zero:

<math>\dot{\Ws}^\text{friction}=\dot{\Ws}_\text{belt}^{\text{fric}\rightarrow \text{wheel}}+ \dot{\Ws}_\text{belt}^{\text{fric }\rightarrow \text{belt}}=\overline{\Fs}^{\text{fric}\rightarrow \text{wheel}} \cdot \bar{\vs}_{\text {belt}}\left(\Js_{\text {wheel }}\right)=(\rightarrow \mu\ms\gs)\cdot(\leftarrow \vs_0)=-\mu\ms\gs\vs_0.</math>

If the calculation is done in any other reference frame, the result is the same:

<math>\dot{\Ws}^\text{friction}=\dot{\Ws}_\text{E}^{\text{fric}\rightarrow \text{wheel}}+ \dot{\Ws}_\text{E}^{\text{fric} \rightarrow \text{belt}}=(\rightarrow \mu\ms\gs)\cdot 0 + (\leftarrow \mu\ms\gs)\cdot(\rightarrow \vs_0)=-\mu\ms\gs\vs_0.</math>

<math>\dot{\Ws}^\text{friction}=\dot{\Ws}_\text{vehicle}^{\text{fric}\rightarrow \text{wheel}}+ \dot{\Ws}_\text{vehicle}^{\text{fric} \rightarrow \text{belt}}=(\rightarrow \mu\ms\gs)\cdot (\rightarrow\vs_0) + (\leftarrow \mu\ms\gs) \cdot(\rightarrow 3\vs_0)=-\mu\ms\gs\vs_0.</math>

However, if only the power of one of the two forces is calculated (for example, the friction on the wheel), the result depends on the reference frame and can have any sign:

<math>\dot{\Ws}_\text{belt}^{\text{fric }\rightarrow \text{wheel}}=(\rightarrow \mu\ms\gs)\cdot(\leftarrow \vs_0)=-\mu\ms\gs\vs_0 ,</math>

<math>\dot{\Ws}_\text{E}^{\text{fric} \rightarrow \text{wheel}}=(\rightarrow \mu\ms\gs)\cdot 0 = 0 ,</math>

<math>\dot{\Ws}_\text{vehicle}^{\text{fric} \rightarrow \text{wheel}}=(\rightarrow \mu\ms\gs)\cdot (\rightarrow\vs_0)=\mu\ms\gs\vs_0.</math>


-----
-----

==E1.3 Power of a system of forces on a rigid body==

{|
|
:[[File:E1-3-eng.png|thumb|left|180px|link=]]
<center>'''Figure E1.3''': Kinematic torsor of a rigid body in a reference frame R, and forces on different points (dm) of that rigid body</center>

|When calculating the power of a system of forces on a rigid body S, the term associated with internal interactions is zero because those interactions are all constraint (cohesion) forces and, as seen in [[E1. Work-Energy Theorem: differential form#E1.2 Power of an action-reaction pair|'''section E1.2''']], their total power is zero.

The total power of a system of forces on S can be calculated from the torsor of the system at any point '''O''' of the rigid body, <math>(\overline{\Fs},\overline{\Ms}(\Os))</math>. Taking into account rigid body kinematics ('''Figure E1.3'''):

<math>\dot{\Ws}_\Rs^{\rightarrow \Ss} =\sum_\Ps\overline{\Fs}^{\rightarrow \Ps} \cdot \vel{P}{R}= \sum_\Ps \left[ \overline{\Fs}^{\rightarrow \Ps} \cdot \vel{O}{R} + \velang{S}{R} \times \OPvec \right] </math>

|}
<math>\dot{\Ws}_\Rs^{\rightarrow \Ss} =\left(\sum_\Ps\overline{\Fs}^{\rightarrow \Ps} \right)\cdot \vel{O}{R} + \sum_\Ps\overline{\Fs}^{\rightarrow \Ps} \cdot \left(\velang{S}{R} \times \OPvec \right) = \left( \sum_\Ps\overline{\Fs}^{\rightarrow \Ps} \right) \cdot \vel{O}{R} + \left[ \sum_\Ps \left( \OPvec \times \overline{\Fs}^{\rightarrow \Ps} \right)\right] \cdot \velang{S}{R} = \overline{\Fs}\cdot \vel{O}{R} + \overline{\Ms}(\Os)\cdot \velang{S}{R} </math>

This calculation also shows that the power of a moment (or a torque <math>\Gamma</math>) on a rigid body in a reference frame R is the product of the moment by the angular velocity of the rigid body: <math>\dot{\Ws}_\Rs^{\Gamma\rightarrow \mathrm{S}} = \overline{\Gamma} \cdot \velang{S}{R}</math>.

The power balance for a rigid body is: <math>\dot{\Ws}_\Rs^{ \rightarrow \mathrm{S}} = \dot{\Ws}_\Rs^{ \mathrm{ext}} + \dot{\Ws}_\Rs^{ \mathrm{ar}}.</math>

====✏️ EXAMPLE E1-3.1: roller sliding on a slope====
------
:
{|
|
:[[File:ExE1-3-1-1-eng.png|thumb|left|180px|link=]]
|A homogeneous roller, with mass m and radius r, slides down a slope. The friction coefficient between the ground and the roller is <math>\mu</math>. We want to calculate the power of every force acting on the roller in the ground reference frame.
|}

:It is a 2D problem. The kinematic and dynamic descriptions are:
{|
|
[[File:ExE1-3-1-2-eng.png|thumb|right|250px|link=]]

|<math> \hspace{0.5cm}\vel{J}{E}=\vel{C}{T} + \velang{}{E} \times \CJvec= </math>

<math>\hspace{1.7cm}=\left( \rightarrow \vs_\Es \right) + \left( \otimes \Omega_\Es\right) \times \left( \downarrow \rs \right)</math>

<math> \hspace{0.5cm}\vel{J}{E} = \left[\rightarrow (\vs_\Es -\rs\Omega_\Es)\right]</math>

<math>\hspace{0.5cm} \dot{\Ws}_\Es^{\mathrm{ext}} = \dot{\Ws}_\Es^{\mathrm{weight}} + \dot{\Ws}_\Es^{\mathrm{friction}} + \dot{\Ws}_\Es^{\mathrm{constraint}}= \Bigr[\left(\downarrow \ms\gs\cos\beta\right)+\left(\rightarrow \ms\gs\sin\beta\right)\Bigr] \cdot (\rightarrow \vs_\Es) + \Bigr[(\leftarrow \mu \Ns) + (\uparrow \Ns ) \Bigr] \cdot \Bigr[\rightarrow (\vs_\Es -\rs\Omega_\Es)\Bigr]</math>

|}
:Since the roller center has no motion perpendicular to the plane, the AMT leads to <math>\Ns =\ms\gs\cos\beta</math>. Therefore:

{|
|
[[File:ExE1-3-1-3-eng.png|thumb|right|140px|link=]]
|<math>\hspace{0.5cm} \dot{\Ws}_\Ts^{\mathrm{ext}} = \ms\gs\vs_\Ts \sin \beta - \ms\gs(\vs_\Ts -\rs\Omega_\Ts)\cos\beta=\ms\gs \Bigr[ \vs_\Ts (\sin\beta-\cos\beta)+\rs\Omega_\Ts \cos\beta \Bigr].</math>

<math>\hspace{0.5cm}</math>

If instead of considering the friction at '''J''' we consider the equivalent torsor at point '''C''', we must consider the force applied in '''C''' and the moment about '''C''', which is not zero:

<math>\hspace{0.5cm} \dot{\Ws}_\Es^{ \mathrm{friction}} = (\leftarrow \mu\Ns)\cdot(\rightarrow\vs_\Es)+(\otimes \mu\Ns\rs)\cdot(\otimes \Omega_\Es) =\mu\ms\gs\cos\beta(\rs\Omega_\Es-\vs_\Es)</math>,

<math>\hspace{0.5cm} \dot{\Ws}_\Es^{\mathrm{ext}} = \dot{\Ws}_\Es^{ \mathrm{weight}} + \dot{\Ws}_\Es^{ \mathrm{friction}}+ \dot{\Ws}_\Es^{ \mathrm{constraint}}= \ms\gs\vs_\Es\sin\beta+\mu\ms\gs\cos\beta (\rs\Omega_\Es-\vs_\Es)</math>.

<math>\hspace{0.5cm}</math>

The result is the same.
|}


====✏️ EXAMPLE E1-3.2: block sliding on a truck====
------
:
{|
|
:[[File:ExE1-3-2-1-eng.png|thumb|left|220px|link=]]
|A truck brakes with acceleration '''a''' relative to the ground at the instant in which its velocity (also relative to the ground) is v. At the same instant, the homogeneous block, of mass m, slides with velocity v' relative to the truck. We want to perform a power balance for the block in two different reference frames: the ground and the truck.
|}

:Ground reference frame: It is a 2D problem. The kinematic and dynamic descriptions are:
{|
|
[[File:ExE1-3-2-2-eng.png|thumb|right|200px|link=]]

| The constraint between the truck and the block allows only one translational DoF between the two. Therefore, in the 2D approximation, the associated torsor has two independent components. If characterized at the midpoint of the contact surface, these components are a normal force and a moment perpendicular to the plane. The AMT applied to the block leads to <math>\Ns=\ms\gs</math>.

|}
:The powers associated with all interactions on the block in the ground reference are:

:<math>\dot{\Ws}_\Es^{\rightarrow \mathrm{bloque}}=\dot{\Ws}_\Ts^{\mathrm{ext}}=\dot{\Ws}_\Ts^{\mathrm{peso}}+\dot{\Ws}_\Ts^{\mathrm{fricción}}+\dot{\Ws}_\Ts^{\mathrm{enlace}}= \Bigr[(\downarrow \ms\gs)+(\leftarrow \mu\ms\gs)+(\uparrow \Ns) \Bigr] \cdot \Bigr[\rightarrow(\vs+\vs')\Bigr] + (\odot \Ms_\es)\cdot 0 = -\mu\ms\gs(\vs+\vs')</math>

:Ya que el bloque se traslada respecto al suelo, su energía cinética total es inmediata, y se puede plantear fácilmente el balance de potencias:

:<math>\Ts_\Ts^\mathrm{bloque}=\frac{1}{2}\int_\mathrm{bloque}\ds\ms \vs_\Ts^2(\mathrm{bloque})=\frac{1}{2}\ms \vs_\Ts^2(\mathrm{bloc})\quad \Rightarrow \quad \dot{\Ts}_\Ts^\mathrm{bloque}=\ms\vs_\Ts( \mathrm{bloque})\dot{\vs}_\Ts( \mathrm{bloque}) </math>

:Como la única fuerza horizontal sobre el bloque es la fricción:

<math>\dot{\vs}_\Ts(\mathrm{bloque})=-\mu\gs\quad \Rightarrow \quad \dot{\Ts}_\Ts^\mathrm{bloque}=-\mu\ms\gs(\vs+\vs').</math>

:El balance de potencias es consistente:

:<math>\left.\begin{array}{l}
\dot{\mathrm{W}}_\Ts^{\rightarrow \text { bloque }}=\dot{\mathrm{W}}_{\mathrm{T}}^{\text {ext }}=-\mu \mathrm{mg}\left(\mathrm{v}+\mathrm{v}^{\prime}\right) \\
\dot{\mathrm{T}}_{\mathrm{T}}^{\text {bloque }}=-\mu \mathrm{mg}\left(\mathrm{v}+\mathrm{v}^{\prime}\right)
\end{array}\right\} \Rightarrow \dot{\mathrm{W}}_{\mathrm{T}}^{\rightarrow \text { bloque }}=\dot{\mathrm{T}}_{\mathrm{T}}^{\text {bloque }}</math>

:Referencia camión: Al estar el camión acelerado respecto al suelo, hay que tener en cuenta las fuerzas de arrastre sobre todos los diferenciales de masa (dm) del bloque:

{|
|

[[Archivo:ExE1-3-2-3-cat-esp.jpg|thumb|right|250px|link=]]
|<math>\cal{\Fs}^{\rightarrow \ds\ms}_{\mathrm{ar}}=-\ds\ms\overline{\as}_\mathrm{ar}(\ds\ms)=-\ds\ms (\leftarrow \as)=(\rightarrow \ds\ms \cdot \as)</math>.

El torsor resultante de este sistema de fuerzas en el centro de inercia '''G''' se reduce a una fuerza horizontal.

<math>\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{bloque}}=\dot{\Ws}_\mathrm{camión}^{\mathrm{ext}} + \dot{\Ws}_\mathrm{camión}^{\mathrm{ar}} =\dot{\Ws}_\mathrm{camión}^{\mathrm{peso}} + \dot{\Ws}_\mathrm{camión}^{\mathrm{fricción}} + \dot{\Ws}_\mathrm{camión}^{\mathrm{enlace}} + \dot{\Ws}_\mathrm{camión}^{\mathrm{ar}} , </math>

<math>\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{bloque}}= \Bigr[(\downarrow \ms\gs)+(\leftarrow \mu\ms\gs)+(\uparrow \Ns) + (\rightarrow \ms\as)\Bigr] \cdot(\rightarrow \vs')+(\odot\Ms_\es) \cdot0=\ms(\as-\mu\gs)\vs'.</math>
|}

:Se puede comprobar que el balance de potencias es consistente:

:<math>\Ts^\mathrm{bloque}_\mathrm{camión}=\frac{1}{2}\ms\vs^2_\mathrm{camión}(\mathrm{bloque}) \quad \Rightarrow \quad \dot{\Ts}^\mathrm{bloque}_\mathrm{camión} = \ms\vs_\mathrm{camión}(\mathrm{bloque})\dot{\vs}_\mathrm{camión}(\mathrm{bloque})</math>

:<math>\left.\begin{array}{l}
\mathrm{AB}:\mathrm{suelo} \\
\mathrm{REL}:\mathrm{camión}
\end{array}\right\} \quad \dot{\vs}_\mathrm{camión}(\mathrm{bloque})= \dot{\vs}_\mathrm{T}(\mathrm{bloque})-\dot{\vs}_\mathrm{ar}(\mathrm{bloque}) = (-\mu\gs)-(-\as)=\as-\mu\gs \quad \Rightarrow \quad \dot{\Ts}^\mathrm{bloque}_\mathrm{camión}=\ms\vs'(\as-\mu\gs)</math>

:<math>\left.\begin{array}{l}
\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{bloque}}=\dot{\Ws}_\mathrm{camión}^{\mathrm{ext}} + \dot{\Ws}_\mathrm{camión}^{\mathrm{ar}}= \ms(\as-\mu\gs)\vs'\\
\dot{\Ts}^\mathrm{bloque}_\mathrm{camión}=\ms\vs'(\as-\mu\gs)
\end{array}\right\} \Rightarrow\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{bloque}}= \dot{\Ts}^\mathrm{bloque}_\mathrm{camión}</math>


====✏️ EJEMPLO E1-3.3: camión frenando con carga====
------
:
{|
|
:[[Archivo:ExE1-3-3-1-esp.png|thumb|left|220px|link=]]
|Un camión frena con aceleración '''a''' respecto al suelo en el instante en el que su velocidad (también respecto al suelo) es v. Los pares de frenada en el eje delantero y en el trasero son iguales y valen <math>\Gamma</math>, y las ruedas, de radio r y masa despreciable, no deslizan sobre el suelo. En el mismo instante, el bloque homogéneo, de masa m, desliz con velocidad v' relativa al camión. Se trata de hacer un balance de potencias para el sistema (bloque+camión) en la referencia del suelo y en la del camión.

|}

:Ya que la potencia de las interacciones internas no es nula en principio (no se trata de un único sólido rígido), es conveniente hacer un DGI del sistema para no olvidar ninguna:
{|
|
[[Archivo:ExE1-3-3-2-esp.png|thumb|right|250px|link=]]

|
|
c.p.s.d.: contacto puntual sin deslizamiento

c.m.c.d.: contacto multipuntual con deslizamiento

|}
:El sistema sobre el que se hace el balance de potencias es todo menos el suelo. Teniendo en cuenta que la potencia de los enlaces internos es siempre nula, las únicas interacciones que hay que tener en cuenta son el peso, los enlaces de las ruedas con el suelo, los pares de frenada y la fricción entre bloque y camión:

[[Archivo:ExE1-3-3-3-esp.png|thumb|center|300px|link=]]

:Referencia suelo:
:Las fuerzas de enlace del suelo sobre las ruedas están aplicadas a puntos de velocidad cero, y por tanto no hacen potencia. El peso es una fuerza vertical, y el movimiento del camión y del bloque es horizontal, por tanto su potencia también es cero.

:<math>\dot{\Ws}_\Ts^{\rightarrow \mathrm{sist}}=\dot{\Ws}_\Ts^{\mathrm{ext}}+\dot{\Ws}^{\mathrm{int}}=\dot{\Ws}_\Ts^{\mathrm{peso}}+\dot{\Ws}_\Ts^{\mathrm{enlace}}+\dot{\Ws}^{\mathrm{fricción}}+\dot{\Ws}^{\Gamma}=\dot{\Ws}^{\mathrm{fricción}}+\dot{\Ws}^{\Gamma}.</math>

:Las potencias de interacciones internas se pueden calcular en cualquier referencia. Por tanto:

:<math>\dot{\Ws}^{\mathrm{fricción}}= \dot{\Ws}_\mathrm{camión}^{\mathrm{fric}\rightarrow \mathrm{bloque}} +\dot{\Ws}_\mathrm{camió}^{\mathrm{fric}\rightarrow \mathrm{camión}} = \dot{\Ws}_\mathrm{camión}^{\mathrm{fric}\rightarrow \mathrm{bloque}}= (\leftarrow \mu\ms\gs)\cdot (\rightarrow \vs')=-\mu\ms\gs\vs'</math>

:<math>\dot{\Ws}^{\Gamma}= \dot{\Ws}_\mathrm{camión}^{\Gamma\rightarrow \mathrm{ruedas}} +\dot{\Ws}_\mathrm{camión}^{\Gamma\rightarrow \mathrm{camión}} = \dot{\Ws}_\mathrm{camión}^{\Gamma\rightarrow \mathrm{ruedas}}= 2 \left(\odot \Gamma \right)\cdot \left(\otimes \frac{\vs}{\rs}\right)=-2\Gamma\frac{\vs}{\rs}.</math>

:Por otro lado: <math>\Ts_\Ts^\mathrm{sist}=\frac{1}{2}\ms\vs_\Ts^2(\mathrm{bloque})+\frac{1}{2}\ms\vs_\Ts^2(\mathrm{camión})\quad \Rightarrow \quad \dot{\Ts}_\Ts^\mathrm{sist}=-\ms(\vs+\vs')\mu\gs-\Ms\vs\as. </math>

:Finalmente: <math>\dot{\Ws}_\Ts^{\rightarrow \mathrm{sist}}=\dot{\Ts}_\Ts^\mathrm{sist} \quad \Rightarrow \quad -\mu\ms\gs\vs'-2\Gamma\frac{\vs}{\rs}=-\ms(\vs+\vs')\mu\gs-\Ms\vs\as.</math>

:Esta ecuación permite calcular el par de frenada: <math>\Gamma=\frac{1}{2}(\mu\ms\gs+\Ms\as)\rs.</math>

:Referencia camión:

:En esta referencia hay que tener en cuenta las fuerzas de arrastre. Por otro lado, las fuerzas de enlace del suelo sobre las ruedas están aplicadas a puntos que ya no tienen velocidad cero, y por tanto hacen potencia. Como antes, la potencia del peso es cero.

:<math>\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{sist}}=\dot{\Ws}_\mathrm{camión}^{\mathrm{ext}}+\dot{\Ws}^{\mathrm{int}}+\dot{\Ws}_\mathrm{camión}^{\mathrm{ar}}=\dot{\Ws}_\mathrm{camión}^{\mathrm{enlace}}+\dot{\Ws}^{\mathrm{fricción}}+\dot{\Ws}^{\Gamma}+\dot{\Ws}_\mathrm{camión}^{\mathrm{ar}}.</math>

:Las fuerzas de arrastre sobre el camión no hacen potencia porque el camión está parado en esta referencia. La de las fuerzas de arrastre sobre el bloque se ha calculado en el [[E1. Teorema de la energía: versión diferencial#✏️ ✏️ EJEMPLO E1-3.2: bloque deslizando sobre un camión|'''ejemplo E1-3.2''']]. La potencia de la fricción y de los pares de frenada es la misma de antes ya que no depende de la referencia.

:<math>\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{sist}}=\dot{\Ws}_\mathrm{camión}^{\mathrm{enllaç}}+\dot{\Ws}^{\mathrm{fricción}}+\dot{\Ws}^{\Gamma}+\dot{\Ws}_\mathrm{camión}^{\mathrm{ar} \rightarrow \mathrm{bloque}}= \dot{\Ws}_\mathrm{camión}^{\mathrm{enlace}} -\mu\ms\gs\vs'-2\Gamma\frac{\vs}{\rs} +\ms\as\vs'.</math>

:Las descripciones cinemática y dinámica de la rueda son:

[[Archivo:ExE1-3-3-4-esp.png|thumb|center|430px|link=]]

:<math>\dot{\Ws}_\mathrm{camión}^{\mathrm{enlace}}=2\Bigr[(\leftarrow \Ts)+(\uparrow \Ns)\Bigr]\cdot(\leftarrow\vs)=2\Ts\vs.</math>

La fuerza tangencial de enlace se puede calcular a partir del TMC aplicado al centro '''C''' de la rueda. Como no tiene masa: <math>\sum\overline{\Ms}_\mathrm{ext}(\Cs)=\overline{0} \quad \Rightarrow \quad (\odot \Gamma)+(\otimes \Ts\rs)=0\quad \Rightarrow \quad \Ts=\Gamma/\rs.</math>

:Por tanto: <math>\dot{\Ws}_\mathrm{camió}^{\rightarrow \mathrm{sist}} =2\frac{\Gamma}{\rs}\vs-\mu\ms\gs\vs'-2\Gamma\frac{\vs}{\rs}+\ms\as\vs'=\ms(\as-\mu\gs)\vs'.</math>

:Por otro lado, como se ha visto en el [[E1. Teorema de la energía: versión diferencial#✏️ ✏️ EJEMPLO E1-3.2: bloque deslizando sobre un camión|'''ejemplo E1-3.2''']]:

:<math>\Ts_\mathrm{camión}^\mathrm{sist}= \Ts_\mathrm{camión}^\mathrm{bloque}=\frac{1}{2}\ms\vs_\mathrm{camión}^2(\mathrm{bloque}) \quad \Rightarrow \quad \dot{\Ts}_\mathrm{camión}^\mathrm{sist}= \dot{\Ts}_\mathrm{camión}^\mathrm{bloque}=\ms\vs'(\as-\mu\gs).</math>

:El balance de potencias es consistente:

:<math>\left.\begin{array}{l}
\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{sist}}= \ms(\as-\mu\gs)\vs'\\
\dot{\Ts}^\mathrm{sist}_\mathrm{camión}=\ms(\as-\mu\gs)\vs'
\end{array}\right\} \Rightarrow\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{sist}}= \dot{\Ts}^\mathrm{sist}_\mathrm{camión}</math>

-----
-----

==E1.4 Balance de potencias para un sistema multisólido: cálculo directo y cálculo indirecto==

El '''cálculo directo''' (a partir de su definición) de la potencia asociada a una interacción (una fuerza sobre un punto '''P''' o un momento sobre un sólido rígido S) exige el conocimiento de la interacción (fuerza o momento):

<math>\dot{\Ws}_\Rs^{\Fs\rightarrow \Ps}=\overline{\Fs}_{\rightarrow\Ps} \cdot \vel{P}{R} \quad , \quad \dot{\Ws}_\Rs^{\Ms\rightarrow \Ss}=\overline{\Ms}_{\rightarrow\Ss} \cdot \velang{S}{R} </math>

Cuando no se trata de una interacción formulable, hay que recurrir a los teoremas vectoriales para determinarla.

El '''cálculo indirecto''' es una alternativa que no necesita recurrir a estos teoremas. Se basa en la evaluación de un balance de potencias para un sistema y una referencia adecuados de manera que la potencia a determinar sea el único término desconocido de la ecuación (el único que no se puede calcular de manera directa). Cuando el balance incluye más de un término desconocido, se pueden combinar varios balances (para distintos sistemas y referencias) hasta tener un número suficiente de ecuaciones que permitan la determinación de todas las potencias desconocidas.

====✏️ EJEMPLO E1-4.1: polea frenada====
------
:
{|
|
:[[Archivo:ExE1-4-1-1-esp.png|thumb|left|150px|link=]]
|Un bloque de masa m cuelga de un hilo inextensible atado a un punto de la periferia de una polea de radio r. Entre soporte (fijo al suelo) y polea actúa un freno que garantiza que el bloque baja a velocidad constante respecto al suelo. Se trata de calcular la potencia del freno.

|}

:Càlculo directo: <math>\dot{\Ws}^{\mathrm{freno}}= \overline{\Gamma}^{\rightarrow \mathrm{polea}}\cdot\velang{polea}{T}=\left(\odot\Gamma\right)\cdot\left( \otimes \frac{\vs_0}{\rs}\right)=-\Gamma \frac{\vs_0}{\rs}.</math>

:El par de frenada no es un dato, y hay que recurrir al TMC para calcularlo.

:<math>\left.\begin{array}{l}
\text{SISTEMA: polea + bloque}\\
\text{TMC en } \Os
\end{array}\right\} \sum\overline{\Ms}_\mathrm{ext}(\Os)=\dot{\overline{\Hs}}_\mathrm{RTO}(\Os)</math>

:El momento cinético proviene de las dos piezas. Ya que '''O''' no es fijo al bloque, hay que hacer descomposición baricéntrica para calcular el momento cinético del bloque:

:<math>\overline{\Hs}_\mathrm{RTO=T} (\Os) =\overline{\Hs}_\Ts^\mathrm{politja}(\Os) + \overline{\Hs}_\Ts^\mathrm{bloc}(\Os)= \overline{\Hs}_\Ts^\mathrm{politja}(\Os) + \overline{\Hs}_\Ts^\mathrm{bloc}(\Gs_\mathrm{bloc})+ \overline{\Hs}_\Ts^{\oplus\mathrm{bloc}}(\Os) = </math>

:<math>\hspace{2.3cm}=\Is_\mathrm{O}^\mathrm{polea} \velang{polea}{T}+\Is_{\Gs_\mathrm{bloque}}^\mathrm{bloque}+\OGvec_\mathrm{bloque}\times \ms\overline{\vs}_\Ts(\Gs_\mathrm{bloque})=\left(\otimes \Is_\mathrm{O}\frac{\vs_0}{\rs}\right)+(\otimes \ms\vs_0\rs).</math>

:[[Archivo:ExE1-4-1-2-esp.png|thumb|right|180px|link=]]
:Se trata de un momento cinético constante en dirección y valor. Por tanto:

:<math>\dot{\overline{\Hs}}_\mathrm{RTO}(\Os)=0 \quad \Rightarrow \quad \sum{\overline{\Ms}}_\mathrm{ext}(\Os)=(\odot\Gamma)+(\otimes \ms\gs\rs) =0\quad \Rightarrow \quad \Gamma= \ms\gs\rs</math>.

:Finalmente: <math>\dot{\Ws}^{\mathrm{freno}}= -\Gamma \frac{\vs_0}{\rs} = -\ms\gs\vs_0.</math>

:Cálculo indirecto:

:<math>\left.\begin{array}{l}
\text{SISTEMA: polea + soporte + bloque}\\
\text{REF: suelo(Gal)}
\end{array}\right\} \quad \dot{\Ts}^\mathrm{sist}_\Ts=\dot{\Ws}^\mathrm{ext}_\Ts+ \dot{\Ws}^\mathrm{int}</math>

:<math>\left.\begin{array}{l}
\dot{\Ts}^\mathrm{sist}_\Ts=0\\
\dot{\Ws}^\mathrm{ext}_\Ts=\left(\downarrow \ms\gs\right)\cdot\left(\downarrow \vs_0\right)= \ms\gs\vs_0\\
\dot{\Ws}^\mathrm{int}=\dot{\Ws}^{\mathrm{freno}}
\end{array}\right\} \quad \Rightarrow \quad \dot{\Ws}^{\mathrm{freno}} =-\ms\gs\vs_0</math>



====✏️ EJEMPLO E1-4.2: mono y plátanos====
------
:
{|
|
:[[Archivo:ExE1-4-2-1-esp.png|thumb|left|180px|link=]]
|Unos plátanos de masa <math>(2/3)\ms</math> cuelgan de una cuerda inextensible que se apoya sin deslizar sobre una polea de radio r y masa despreciable, articulada respecto al suelo. En el otro extremo de la cuerda se agarra un mono de masa m. La polea gira con velocidad angular constante <math>\Omega_0</math> respecto al suelo bajo la acción de un motor. El mono se mueve respecto a la cuerda sin que sus manos se froten contra la cuerda, y de manera que la distancia de su centro de masas al suelo se mantiene constante. Se trata de calcular la potencia del motor.
|}

:Cálculo directo: <math>\dot{\Ws}^{\mathrm{freno}}= \overline{\Gamma}^{\rightarrow \mathrm{polea}}\cdot\velang{polea}{T}=\left(\odot\Gamma\right)\cdot\left( \odot \Omega_0\right)=\Gamma \Omega_0.</math>

:El par motor no es un dato, y hay que recurrir al TMC para calcularlo.

:<math>\left.\begin{array}{l}
\text{SISTEMA: polea + mono + plátanos}\\
\text{TMC en } \Os
\end{array}\right\} \sum\overline{\Ms}_\mathrm{ext}(\Os)=\dot{\overline{\Hs}}_\mathrm{RTO}(\Os)</math>

:El momento cinético proviene del mono y de los plátanos. En primera aproximación, se puede considerar la masa del mono concentrada en su tronco y su cabeza, de manera que no gira y su velocidad respecto al suelo es nula. Teniendo en cuenta que el punto '''O''' no se mueve respecto al mono pero sí respecto a los plátanos:

:<math>\overline{\Hs}_\mathrm{RTO=T} (\Os) =\overline{\Hs}_\Ts^\mathrm{mono}(\Os) + \overline{\Hs}_\Ts^\mathrm{plátanos}(\Os)= \overline{\Hs}_\Ts^\mathrm{mono}(\Os) + \overline{\Hs}_\Ts^\mathrm{plátanos}(\Gs_\mathrm{plát})+ \OGvec_\mathrm{plát} \times \frac{2}{3} \ms \overline{\vs}_\Ts(\Gs_\mathrm{plát})= </math>

:<math>\hspace{2.3cm}=\OGvec_\mathrm{plát}\times \frac{2}{3} \ms \overline{\vs}_\Ts(\Gs_\mathrm{plát}) = \Bigr[(\rightarrow \rs)+(\downarrow \xs)\Bigr] \times\frac{2}{3} \ms \left(\downarrow \rs\Omega_0\right)=\left(\otimes \frac{2}{3}\ms\rs^2\Omega_0\right).</math>

:Se trata de un momento cinético constante en dirección y valor. Por tanto:

:<math>\dot{\overline{\Hs}}_\mathrm{RTO}(\Os)=0 \quad \Rightarrow \quad \sum{\overline{\Ms}}_\mathrm{ext}(\Os)=(\odot\Gamma)+(\otimes \ms\gs\rs) + \left(\otimes \frac{2}{3}\ms\gs\rs\right)=0\quad \Rightarrow \quad \Gamma= \frac{5}{3}\ms\gs\rs</math>.

:Finalmente: <math>\dot{\Ws}^{\mathrm{motor}}= \Gamma \Omega_0 = \frac{5}{3}\ms\gs\rs\Omega_0.</math>

:Cálculo indirecto

:Hay que tener presente que es un problema con dos tipos de actuadores: el motor y la musculatura del mono. Hay diversos modelos biomecánicos para formular las fuerzas de los músculos, pero quedan fuera del ámbito de este curso. Ya que el mono se mueve, sus músculos desarrollan una potencia no nula.

:<math>\left.\begin{array}{l}
\text{SISTEMA: polea + mono + plátano}\\
\text{REF: suelo(Gal)}
\end{array}\right\} \quad \dot{\Ts}^\mathrm{sist}_\Ts=\dot{\Ws}^\mathrm{ext}_\Ts+ \dot{\Ws}^\mathrm{int}</math>

:<math>\left.\begin{array}{l}
\dot{\Ts}^\mathrm{sist}_\Ts=0\\
\dot{\Ws}^\mathrm{ext}_\Ts=\left(\downarrow \frac{2}{3}\ms\gs\right)\cdot\left(\downarrow \rs\Omega_0\right)= \frac{2}{3}\ms\gs\rs\Omega_0\\
\dot{\Ws}^\mathrm{int}=\dot{\Ws}^{\mathrm{musc}}+\dot{\Ws}^{\mathrm{motor}}
\end{array}\right\} \quad \Rightarrow \quad \dot{\Ws}^{\mathrm{musc}} + \dot{\Ws}^{\mathrm{motor}}=-\frac{2}{3}\ms\gs\rs\Omega_0</math>

:Este balance proporciona una ecuación con dos incógnitas. Se puede plantear un segundo balance para un sistema donde una de las dos no aparezca: un sistema sin el mono o un sistema sin el motor. Si se hace un balance de potencias para el sistema (mono + trozo de cuerda), las fuerzas externas a considerar son el peso y la tensión de la cuerda. Si se sigue trabajando en la referencia del suelo, esta tensión (desconocida) hará potencia no nula ya que está aplicada a un punto de velocidad ascendente.
:[[Archivo:ExE1-4-2-2-cat-esp.png|thumb|right|330px|link=]]

:Para evitar que aparezca un término asociado a la tensión de la cuerda, se puede trabajar en la referencia del trozo de cuerda que se ha considerado, que al tener velocidad ascendente constante es galileana. En esta referencia, el mono baja con velocidad constante:

:<math>\left.\begin{array}{l}
\text{REL: trozo de cuerda}\\
\text{AB: suelo}
\end{array}\right\}</math>

:<math>\overline{\vs}_\mathrm{REL}(\mathrm{mono})=\overline{\vs}_\mathrm{ar}(\mathrm{mono})=-\overline{\vs}_\mathrm{ar}(\mathrm{mono})=(\downarrow \rs\Omega_0)</math>

:<math>\left.\begin{array}{l}
\text{SISTEMA: mono+ trozo de cuerda}\\
\text{REF: trozo de cuerda(Gal)}
\end{array}\right\} \quad \dot{\Ts}^\mathrm{sist}_\mathrm{cuerda} =\dot{\Ws}^\mathrm{ext}_\mathrm{cuerda}+ \dot{\Ws}^\mathrm{int}</math>

:<math>\left.\begin{array}{l}
\dot{\Ts}^\mathrm{sist}_\mathrm{cuerda}=0\\
\dot{\Ws}^\mathrm{ext}_\mathrm{cuerda}=\left(\downarrow \ms\gs\right)\cdot\left(\downarrow \rs\Omega_0\right)= \ms\gs\rs\Omega_0\\
\dot{\Ws}^\mathrm{int}=\dot{\Ws}^{\mathrm{musc}}
\end{array}\right\} \quad \Rightarrow \quad \dot{\Ws}^{\mathrm{musc}} =-\ms\gs\rs\Omega_0</math>

:Combinando este resultado con el del balance anterior:

:<math>\dot{\Ws}^\mathrm{motor}=-\frac{2}{3}\ms\gs\rs\Omega_0-\dot{\Ws}^\mathrm{musc}=-\frac{5}{3}\ms\gs\rs\Omega_0</math>



© Universitat Politècnica de Catalunya. [[Mecánica:Derechos de autor |Todos los derechos reservados]]
-------------

<center>
[[D8. Conservaciones|<<< D8. Conservaciones]]

[[E2. Teorema de la energía: versión integrada|E2. Teorema de la energía: versión integrada>>>]]
</center>

E1. Work-Energy Theorem: differential form

2025-05-20T12:17:07Z

Bros: /* ✏️ EXAMPLE E1-1.2: pendulum */

<div class="noautonum">__TOC__</div>
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\newcommand{\vvec}{\overline{\textbf{v}}}
\newcommand{\evec}{\overline{\textbf{e}}}
\newcommand{\Omegavec}{\overline{\mathbf{\Omega}}}
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\newcommand{\Alfavec}{\overline{\mathbf{\alpha}}}
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\newcommand{\ps}{\textrm{p}}
\newcommand{\ns}{\textrm{n}}
\newcommand{\hs}{\textrm{h}}
\newcommand{\cs}{\textrm{c}}
\newcommand{\gs}{\textrm{g}}
\newcommand{\Ns}{\textrm{N}}
\newcommand{\Hs}{\textrm{H}}
\newcommand{\Fs}{\textrm{F}}
\newcommand{\ms}{\textrm{m}}
\newcommand{\ts}{\textrm{t}}
\newcommand{\us}{\textrm{u}}
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\newcommand{\Rs}{\textrm{R}}
\newcommand{\Es}{\textrm{E}}
\newcommand{\Ws}{\textrm{W}}
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\newcommand{\Is}{\textrm{I}}
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\newcommand{\js}{\textrm{j}}
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\newcommand{\deg}{^\textsf{o}}
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\newcommand{\ys}{\textsf{y}}
\newcommand{\zs}{\textsf{z}}
\newcommand{\dert}[2]{\left.\frac{\ds{#1}}{\ds\ts}\right]_{\textrm{#2}}}
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\newcommand{\vec}[1]{\overline{#1}}
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\newcommand{\GPvec}{\vec{\Gs\Ps}}
\newcommand{\PSvec}{\vec{\Ps\Ss}}
\newcommand{\QQvec}{\vec{\Qs\Qs}}
\newcommand{\QGvec}{\vec{\Qs\Gs}}
\newcommand{\QPvec}{\vec{\Qs\Ps}}
\newcommand{\JQvec}{\vec{\Js\Qs}}
\newcommand{\PQvec}{\vec{\Ps\Qs}}
\newcommand{\GJvec}{\vec{\Gs\Js}}
\newcommand{\OPvec}{\vec{\Os\textbf{P}}}
\newcommand{\OCvec}{\vec{\Os\Cs}}
\newcommand{\OGvec}{\vec{\Os\Gs}}
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\newcommand{\PGvec}{\vec{\Ps\Gs}}
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\begin{bmatrix}
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</math>

The '''Work-Energy Theorem''' plays a prominent role in physics, as it bridges all branches of science (electromagnetism, thermodynamics, etc.): all of them deal with energy.

There is a certain parallel between the Vector Theorems and the '''Work-Energy Theorem''': they all allow us to understand the evolution of a magnitude that depends on the mechanical state of the system. In the case of the former, we are dealing with the instantaneous evolution of vector magnitudes (linear momentum, angular momentum), while the latter deals with the finite change (in a finite time interval) of a scalar magnitude ('''mechanical energy''').

This unit presents the differential form of this theorem (the '''power balance'''). The integrated form (which is the Work-Energy Theorem itself) is presented in [[E2. Teorema de la energía: versión integrada|'''unit E2''']].
 
------
------
==E1.1 Power balance in a system of particles==

:[[File:E1-1-eng.png|thumb|right|300px|link=]]
<center>'''Figure E1.1''' Forces on a particle of a system with constant matter</center>

Let us consider a system of constant matter, and let’s apply Newton's second law to each particle (or each mass differential) of this system in a general reference frame R (which may be non-Galilean) ('''Figure E1.1'''):

<math>\F{\rightarrow \Ps} + \Fcal{}{\text{ar}\rightarrow\Ps} + \Fcal{}{\text{Cor}\rightarrow\Ps} =\ms_\Ps\acc{P}{R}</math>

<math>\F{\rightarrow \Ps} </math> is the resultant interaction force on '''P''', and it comes from internal interactions (with particles of the same system) and external ones:

<math>\F{\rightarrow \Ps}=\F{\rightarrow \Ps}^\text{int}+\F{\rightarrow \Ps}^\text{ext}</math>

If both sides of the equation are multiplied by <math>\vel{P}{R}</math>, all forces (interaction and inertia forces) orthogonal to <math>\vel{P}{R}</math> are filtered out. Specifically, since <math>\acc{P}{R}=\accs{P}{R} + \accn{P}{R} </math>, where <math>\accn{P}{R} \perp \vel{P}{R}</math>, and since the Coriolis force is always orthogonal to <math>\vel{P}{R} (\Fcal{}{\text{Cor}\rightarrow\Ps} = -2\ms_\Ps \velang{R}{Gal} \times \vel{P}{R})</math>:

<math>\vel{P}{R} \cdot \left( \F{\rightarrow \Ps}^\text{int}+\F{\rightarrow \Ps}^\text{ext} + \Fcal{}{\text{ar}\rightarrow\Ps} \right) = \ms_\Ps\accs{P}{R} \cdot \vel{P}{R} = \frac{\ds}{\ds \ts} \left(\frac{1}{2}\ms_\Ps \vvec_\Rs^2(\Ps) \right) \equiv \frac{\ds\Ts_\Rs(\Ps)}{\ds\ts}=\dot{\Ts}_\Rs(\Ps)</math>

The scalar function <math>\Ts_\Rs(\Ps)</math> is the '''kinetic energy''' of the particle, and is a function of the [[D1. Foundational laws of Newtonian dynamics#D1.3 Newton’s Principle of Determinacy|'''mechanical state''']] of '''P'''. The scalar product of a force and the velocity of its application point is the '''force power''': <math>\vel{P}{R} \cdot \F{\rightarrow \Ps} \equiv \dot{\Ws}_\Rs^{\F{\rightarrow \Ps}} </math>. Both scalar quantities <math>(\Ts_\Rs(\Ps), \dot{\Ws}_\Rs^{\F{\rightarrow \Ps}})</math> depend on the reference frame. For the same set of interaction forces acting on '''P''', the reference frame determines which are filtered out and which are not.

The equation can be rewritten as <math>\dot{\Ws}_\Rs^{\text{ext}\rightarrow \Ps}+\dot{\Ws}_\Rs^{\text{int}\rightarrow \Ps}+\dot{\Ws}_\Rs^{\text{ar}\rightarrow \Ps} =\dot{\Ts}_\Rs(\Ps)</math>, and constitutes the '''power balance''' for particle '''P''' in the R reference frame. Summing for all particles in the system:

<math>\dot{\Ws}_\Rs^{\text{ext}}+\dot{\Ws}^{\text{int}}+\dot{\Ws}_\Rs^{\text{ar}} =\dot{\Ts}_\Rs^\text{syst}</math>

The calculation of kinetic energy is covered in [[E2. Teorema de la energía: versión integrada|'''unit E2''']]. This unit focuses on the calculation of powers. [[E1. Work-Energy Theorem: differential form#E1.2 Potencia de una pareja de acción-reacción|'''section E1-2''']] shows that the power of internal interactions does not depend on the reference frame (which is why the subscript R does not appear in the <math>\dot{\Ws}^{\text{int}}</math> term of the previous equation).

====✏️ EXAMPLE E1-1.1: free particle====
------
:
{|
:[[File:ExE1-1-1-1-eng.jpg|thumb|left|180px|link=]]
|A particle '''P''', with mass m, is released from the ring of a Ferris wheel when it is at its highest point. We want to calculate the power, at this instant, of all forces acting on the particle from three different reference frames: the ground, the ring, and the Ferris wheel cabin.
|}

:Ground reference frame: Since this is a Galilean reference frame, all forces on '''P''' are interaction forces.
{|
:[[File:ExE1-1-2- eng.png|thumb|right|230px|link=]]

|<math>\hspace{0.5cm}\sum\dot{\Ws}_\Ts=\dot{\Ws}_\Ts^\text{weight}= \left(\leftarrow \Rs\Omega_0\right) \cdot \left(\downarrow \ms\gs\right)=0</math>
|}

:Ring reference frame: It is a non-Galilean reference, and the transportation force on '''P''' must be taken into account.
{|
:[[File:ExE1-1-3-eng.png|thumb|right|250px|link=]]
|<math>\hspace{0.5cm}</math> AB:ground, REL:ring 

<math>\hspace{0.5cm}\vel{P}{REL}=\vel{P}{AB}-\vel{P}{ar}= \left(\leftarrow \Rs\Omega_0\right)-\left(\leftarrow \Rs\Omega_0\right)=\overline{0}</math> 

<math>\hspace{0.5cm}</math>

Since the velocity is zero, the total power is zero.
|}

:Cabin reference frame: It is also a non-Galilean reference, and the transportation force on '''P''' must be taken into account.

{|
:[[File:ExE1-1-4-eng.png|thumb|right|250px|link=]]
|<math>\hspace{0.5cm}</math>AB:ground, REL:cabin

<math>\hspace{0.5cm}\vel{P}{REL}=\vel{P}{AB}-\vel{P}{ar}= \left(\leftarrow \Rs\Omega_0\right)-\left(\uparrow \Rs\Omega_0\right)</math>

<math>\hspace{0.5cm}\dot{\Ws}_\text{cabin}^\text{weight}=\left[ \left(\leftarrow \Rs\Omega_0\right)+\left(\downarrow \Rs\Omega_0\right) \right] \cdot \left(\downarrow\ms\gs\right) =\ms\gs\Rs\Omega_0</math>

<math>\hspace{0.5cm}\dot{\Ws}_\text{cabin}^\text{ar}=\left[ \left(\leftarrow \Rs\Omega_0\right)+\left(\downarrow \Rs\Omega_0\right) \right] \cdot \left(\rightarrow\ms\Rs^2\Omega_0^2\right) =-\ms\gs\Rs^2 \Omega_0^3</math>

<math>\hspace{0.5cm}\sum\dot{\Ws}_\text{cabin}=\ms\Rs\Omega_0(\gs-\Rs\Omega_0^2)</math>

|}


====✏️ EXAMPLE E1-1.2: pendulum====
------
:
{|
:[[File:ExE1-2-1-eng.png|thumb|left|180px|link=]]
|A particle '''P''' of mass m hangs from an inextensible thread of length L attached to the ceiling. The assembly oscillates in a vertical plane fixed to the ground. We want to calculate the power of all the forces acting on the particle in two different reference frames: the ground and a platform rotating relative to the ground with a constant angular velocity <math>\Omega_0</math>. 

Ground reference frame: Since it is a Galilean reference frame, '''P''' is only subjected to interaction forces.
{|
:[[File:ExE1-2-2-neut.png|thumb|left|300px|link=]]

|<math>\dot{\Ws}_\Es^\Ts=(\nearrow\Ls\dot{\theta})\cdot(\nwarrow\Ts)=0</math>

<math>\dot{\Ws}_\Es^\text{weight}=(\nearrow\Ls\dot{\theta}\cdot(\downarrow \ms\gs)=-\ms\gs\Ls\dot{\theta}\sin\theta</math>
|}

<math>\vspace{8cm}</math>

Platform reference frame: It is a non Galilean reference frame, and the transportation force on '''P''' has to be taken into account.

{|
:[[File:ExE1-2-3-eng.png|thumb|left|400px|link=]]

|AB:ground, REL:platform

<math>\vel{P}{REL}=\vel{P}{AB}-\vel{P}{ar}=(\nearrow\Ls\dot{\theta})-(\otimes \Ls \Omega_0 \sin \theta)</math>

<math>\dot{\Ws}_\Es^\Ts\left[(\nearrow\Ls\dot{\theta})+(\odot\Ls\Omega_0\sin\theta)\right]\cdot(\nwarrow\Ts)=0</math>

<math>\dot{\Ws}_\Es^\text{weight}\left[(\nearrow\Ls\dot{\theta})+(\odot\Ls\Omega_0\sin\theta)\right]\cdot(\downarrow \ms\gs)=-\ms\gs\Ls\dot{\theta}\sin\theta</math>

<math>\dot{\Ws}_\Es^\text{ar}\left[(\nearrow\Ls\dot{\theta})+(\odot\Ls\Omega_0\sin\theta)\right]\cdot(\rightarrow \ms\Ls\Omega_0^2\sin\theta)=\ms\Ls^2\Omega_0^2\dot{\theta}\cos\theta</math>
|}



------
------

==E1.2 Power of an action-reaction pair==

The total power associated with internal interactions in a constant-mass system is not zero in principle (since the two forces of each action-reaction pair are applied to particles that may have different velocities relative to the reference frame in which the calculation is made), and it is independent of the reference frame in which it is evaluated. It only depends on the separation velocity of the two particles. The power balance for the system is therefore written as: <math>\dot{\Ts}_\Rs^\text{syst}=\dot{\Ws}_\Rs^\text{ext} + \dot{\Ws}^\text{int} + \dot{\Ws}_\Rs^\text{ar} </math> .

<div>
=====💭 PROOF ➕=====

{|
:[[File:E1-2-neut.png|thumb|left|200px|link=]]
|<math>\dot{\Ws}^\text{ac-reac}=\overline{\Fs}_{\Ps \rightarrow \Qs}\cdot \vel{Q}{R} + \overline{\Fs}_{\Qs \rightarrow \Ps}\cdot \vel{P}{R}= \overline{\Fs}_{\Ps \rightarrow \Qs}\cdot \left[ \vel{Q}{R}-\vel{P}{R}\right]</math>

By the [[D1. Foundational laws of Newtonian dynamics#D1.6 Newton’s third law (action-reaction principle)|'''action-reaction principle''']], <math>\overline{\Fs}_{\Ps \rightarrow \Qs}=-\overline{\Fs}_{\Qs \rightarrow \Ps}</math>. Therefore: <math>\dot{\Ws}^\text{ac-reac}=\overline{\Fs}_{\Ps \rightarrow \Qs}\cdot \left[ \vel{Q}{R}-\vel{P}{R}\right]</math>

As <math>\overline{\Fs}_{\Ps \rightarrow \Qs}</math> has the <math>\PQvec</math> direction, only the components of the velocities in that direction have to be considered:

<math>\dot{\Ws}^\text{ac-reac}=\overline{\Fs}_{\Ps \rightarrow \Qs}\cdot \biggl( \vel{Q}{R} \Bigr]_{||\Ps\Qs}-\ \vel{P}{R} \Bigr]_{||\Ps\Qs}\biggl)</math>
|}

The difference <math>\biggl( \vel{Q}{R} \Bigr]_{\Ps\Qs}-\ \vel{P}{R} \Bigr]_{\Ps\Qs}\biggl)</math> is the separation velocity <math>\dot{\rho}</math> between '''P''' and '''Q''', and it does not dpend on the reference frame since it is the derivative of a scalar (the distance <math>\dot{\rho}</math> between the two particles). For the previous drawing, <math>\dot{\rho}> 0</math>. Since this is an attractive interaction, <math>\dot{\Ws}^\text{ac-reac} < 0</math>.

With a drawing, it is easy to see that:
* attractive interaction and particle approach <math>(\dot{\rho}<0)</math> : <math>\dot{\Ws}^\text{ac-reac} > 0</math> ;
* repulsive interaction and particle separation <math>(\dot{\rho}>0)</math> : <math>\dot{\Ws}^\text{ac-reac} > 0</math>
* repulsive interaction and particle approach <math>(\dot{\rho}<0)</math> : <math>\dot{\Ws}^\text{ac-reac} < 0</math>


</div>

There are two particular cases of interaction forces whose power always has the same sign:

* Friction action-reaction pair between '''P''' and '''Q''': since the friction force always opposes the relative velocity between the two points, its power is always negative:

<math>\hspace{1cm} \dot{\Ws}^\text{friction}=\dot{\Ws}^{\text{fric}\rightarrow\Ps}_\text{RTQ}+\dot{\Ws}^{\text{fric}\rightarrow\Qs}_\text{RTQ}=\overline{\Fs}^{\text{fric}\rightarrow\Ps} \cdot \vel{P}{RTQ} = - \abs{\overline{\Fs}^{\text{fric}\rightarrow\Ps}} \frac{\vel{P}{RTQ}}{\abs{\vel{P}{RTQ}}} \cdot \vel{P}{RTQ}<0;</math>

* Constraint action-reaction pair between '''P''' and '''Q''': from the [[D3. Interactions between rigid bodies#D3.4 Direct constraint interactions|'''analytical characterization equation''']] of the constraint forces between two points '''P''' and '''Q''' belonging to two rigid bodies <math>\mathrm{S}_\Ps</math> and <math>\mathrm{S}_\Qs</math>, it is easy to see that the associated power is zero:

<math>\hspace{1cm} \dot{\Ws}^\text{constraint}=\dot{\Ws}^{\text{constraint}\rightarrow\Ps}_{\mathrm{S}_\Ps}+\dot{\Ws}^{\text{constraint}\rightarrow\Qs}_{\mathrm{S}_\Ps}=\overline{\Fs}^{\text{constraint}\rightarrow\Ps} \cdot \overline{\mathbf{v}}_{\mathrm{S}_\Qs}(\Ps) = - 0;</math>

====✏️ EXAMPLE E1-2.1: power of an action-reaction friction pair====
------
:
{|
:[[File:ExE1-2-1-2-eng.png|thumb|left|180px|link=]]
|A homogeneous wheel, with radius r and mass m, moves down an inclined plane without sliding until it joins a conveyor belt moving at a constant velocity <math>\vs_0</math> relative to the ground. At the moment it makes contact with the belt, the center of the wheel has a velocity <math>\vs_0</math> relative to the ground in the same direction as the belt. We want to calculate the power of the friction force between the belt and the wheel at this instant.
|}

At that momento, the <math>\mathbf{ICR}^\text{wheel}_\Es</math> is point <math>\Js_\text{wheel}</math> in contact with the belt. This point slides backwards with respect to the belt, therefore the friction force on the wheel is forwards:

<math>\left.\begin{array}{l}
\text { AB: ground } \\
\text { REL: belt }
\end{array}\right\} \quad \bar{\vs}_{\text {REL }}\left(\Js_{\text {wheel }}\right)=\bar{\vs}_{\text{AB}}\left(\Js_{\text {wheel }}\right)-\bar{\vs}_{\mathrm{ar}}\left(\Js_{\text {wheel }}\right)=-\bar{\vs}_{\mathrm{ar}}\left(\Js_{\text {wheel }}\right)=-\left(\rightarrow \mathrm{v}_0\right)=\left(\leftarrow \mathrm{v}_0\right) .</math>

[[File:ExE1-2-1-3-eng.png|thumb|center|500px|link=]]

Calculation of the friction power: <math>\dot{\Ws}^\text{friction}=\dot{\Ws}_\Rs^{\text{friction}\rightarrow \text{wheel}}+ \dot{\Ws}_\Rs^{\text{friction}\rightarrow \text{belt}}</math>, where R may be any reference frame. To quickly calculate this power, it is advisable to choose a reference frame R where one of the two points subjected to the action-reaction pair has zero speed, so that one of the two terms is directly zero:

<math>\dot{\Ws}^\text{friction}=\dot{\Ws}_\text{belt}^{\text{fric}\rightarrow \text{wheel}}+ \dot{\Ws}_\text{belt}^{\text{fric }\rightarrow \text{belt}}=\overline{\Fs}^{\text{fric}\rightarrow \text{wheel}} \cdot \bar{\vs}_{\text {belt}}\left(\Js_{\text {wheel }}\right)=(\rightarrow \mu\ms\gs)\cdot(\leftarrow \vs_0)=-\mu\ms\gs\vs_0.</math>

If the calculation is done in any other reference frame, the result is the same:

<math>\dot{\Ws}^\text{friction}=\dot{\Ws}_\text{E}^{\text{fric}\rightarrow \text{wheel}}+ \dot{\Ws}_\text{E}^{\text{fric} \rightarrow \text{belt}}=(\rightarrow \mu\ms\gs)\cdot 0 + (\leftarrow \mu\ms\gs)\cdot(\rightarrow \vs_0)=-\mu\ms\gs\vs_0.</math>

<math>\dot{\Ws}^\text{friction}=\dot{\Ws}_\text{vehicle}^{\text{fric}\rightarrow \text{wheel}}+ \dot{\Ws}_\text{vehicle}^{\text{fric} \rightarrow \text{belt}}=(\rightarrow \mu\ms\gs)\cdot (\rightarrow\vs_0) + (\leftarrow \mu\ms\gs) \cdot(\rightarrow 3\vs_0)=-\mu\ms\gs\vs_0.</math>

However, if only the power of one of the two forces is calculated (for example, the friction on the wheel), the result depends on the reference frame and can have any sign:

<math>\dot{\Ws}_\text{belt}^{\text{fric }\rightarrow \text{wheel}}=(\rightarrow \mu\ms\gs)\cdot(\leftarrow \vs_0)=-\mu\ms\gs\vs_0 ,</math>

<math>\dot{\Ws}_\text{E}^{\text{fric} \rightarrow \text{wheel}}=(\rightarrow \mu\ms\gs)\cdot 0 = 0 ,</math>

<math>\dot{\Ws}_\text{vehicle}^{\text{fric} \rightarrow \text{wheel}}=(\rightarrow \mu\ms\gs)\cdot (\rightarrow\vs_0)=\mu\ms\gs\vs_0.</math>


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==E1.3 Power of a system of forces on a rigid body==

{|
|
:[[File:E1-3-eng.png|thumb|left|180px|link=]]
<center>'''Figure E1.3''': Kinematic torsor of a rigid body in a reference frame R, and forces on different points (dm) of that rigid body</center>

|When calculating the power of a system of forces on a rigid body S, the term associated with internal interactions is zero because those interactions are all constraint (cohesion) forces and, as seen in [[E1. Work-Energy Theorem: differential form#E1.2 Power of an action-reaction pair|'''section E1.2''']], their total power is zero.

The total power of a system of forces on S can be calculated from the torsor of the system at any point '''O''' of the rigid body, <math>(\overline{\Fs},\overline{\Ms}(\Os))</math>. Taking into account rigid body kinematics ('''Figure E1.3'''):

<math>\dot{\Ws}_\Rs^{\rightarrow \Ss} =\sum_\Ps\overline{\Fs}^{\rightarrow \Ps} \cdot \vel{P}{R}= \sum_\Ps \left[ \overline{\Fs}^{\rightarrow \Ps} \cdot \vel{O}{R} + \velang{S}{R} \times \OPvec \right] </math>

|}
<math>\dot{\Ws}_\Rs^{\rightarrow \Ss} =\left(\sum_\Ps\overline{\Fs}^{\rightarrow \Ps} \right)\cdot \vel{O}{R} + \sum_\Ps\overline{\Fs}^{\rightarrow \Ps} \cdot \left(\velang{S}{R} \times \OPvec \right) = \left( \sum_\Ps\overline{\Fs}^{\rightarrow \Ps} \right) \cdot \vel{O}{R} + \left[ \sum_\Ps \left( \OPvec \times \overline{\Fs}^{\rightarrow \Ps} \right)\right] \cdot \velang{S}{R} = \overline{\Fs}\cdot \vel{O}{R} + \overline{\Ms}(\Os)\cdot \velang{S}{R} </math>

This calculation also shows that the power of a moment (or a torque <math>\Gamma</math>) on a rigid body in a reference frame R is the product of the moment by the angular velocity of the rigid body: <math>\dot{\Ws}_\Rs^{\Gamma\rightarrow \mathrm{S}} = \overline{\Gamma} \cdot \velang{S}{R}</math>.

The power balance for a rigid body is: <math>\dot{\Ws}_\Rs^{ \rightarrow \mathrm{S}} = \dot{\Ws}_\Rs^{ \mathrm{ext}} + \dot{\Ws}_\Rs^{ \mathrm{ar}}.</math>

====✏️ EXAMPLE E1-3.1: roller sliding on a slope====
------
:
{|
|
:[[File:ExE1-3-1-1-eng.png|thumb|left|180px|link=]]
|A homogeneous roller, with mass m and radius r, slides down a slope. The friction coefficient between the ground and the roller is <math>\mu</math>. We want to calculate the power of every force acting on the roller in the ground reference frame.
|}

:It is a 2D problem. The kinematic and dynamic descriptions are:
{|
|
[[File:ExE1-3-1-2-eng.png|thumb|right|250px|link=]]

|<math> \hspace{0.5cm}\vel{J}{E}=\vel{C}{T} + \velang{}{E} \times \CJvec= </math>

<math>\hspace{1.7cm}=\left( \rightarrow \vs_\Es \right) + \left( \otimes \Omega_\Es\right) \times \left( \downarrow \rs \right)</math>

<math> \hspace{0.5cm}\vel{J}{E} = \left[\rightarrow (\vs_\Es -\rs\Omega_\Es)\right]</math>

<math>\hspace{0.5cm} \dot{\Ws}_\Es^{\mathrm{ext}} = \dot{\Ws}_\Es^{\mathrm{weight}} + \dot{\Ws}_\Es^{\mathrm{friction}} + \dot{\Ws}_\Es^{\mathrm{constraint}}= \Bigr[\left(\downarrow \ms\gs\cos\beta\right)+\left(\rightarrow \ms\gs\sin\beta\right)\Bigr] \cdot (\rightarrow \vs_\Es) + \Bigr[(\leftarrow \mu \Ns) + (\uparrow \Ns ) \Bigr] \cdot \Bigr[\rightarrow (\vs_\Es -\rs\Omega_\Es)\Bigr]</math>

|}
:Since the roller center has no motion perpendicular to the plane, the AMT leads to <math>\Ns =\ms\gs\cos\beta</math>. Therefore:

{|
|
[[File:ExE1-3-1-3-eng.png|thumb|right|140px|link=]]
|<math>\hspace{0.5cm} \dot{\Ws}_\Ts^{\mathrm{ext}} = \ms\gs\vs_\Ts \sin \beta - \ms\gs(\vs_\Ts -\rs\Omega_\Ts)\cos\beta=\ms\gs \Bigr[ \vs_\Ts (\sin\beta-\cos\beta)+\rs\Omega_\Ts \cos\beta \Bigr].</math>

<math>\hspace{0.5cm}</math>

If instead of considering the friction at '''J''' we consider the equivalent torsor at point '''C''', we must consider the force applied in '''C''' and the moment about '''C''', which is not zero:

<math>\hspace{0.5cm} \dot{\Ws}_\Es^{ \mathrm{friction}} = (\leftarrow \mu\Ns)\cdot(\rightarrow\vs_\Es)+(\otimes \mu\Ns\rs)\cdot(\otimes \Omega_\Es) =\mu\ms\gs\cos\beta(\rs\Omega_\Es-\vs_\Es)</math>,

<math>\hspace{0.5cm} \dot{\Ws}_\Es^{\mathrm{ext}} = \dot{\Ws}_\Es^{ \mathrm{weight}} + \dot{\Ws}_\Es^{ \mathrm{friction}}+ \dot{\Ws}_\Es^{ \mathrm{constraint}}= \ms\gs\vs_\Es\sin\beta+\mu\ms\gs\cos\beta (\rs\Omega_\Es-\vs_\Es)</math>.

<math>\hspace{0.5cm}</math>

The result is the same.
|}


====✏️ EXAMPLE E1-3.2: block sliding on a truck====
------
:
{|
|
:[[File:ExE1-3-2-1-eng.png|thumb|left|220px|link=]]
|A truck brakes with acceleration '''a''' relative to the ground at the instant in which its velocity (also relative to the ground) is v. At the same instant, the homogeneous block, of mass m, slides with velocity v' relative to the truck. We want to perform a power balance for the block in two different reference frames: the ground and the truck.
|}

:Ground reference frame: It is a 2D problem. The kinematic and dynamic descriptions are:
{|
|
[[File:ExE1-3-2-2-eng.png|thumb|right|200px|link=]]

| The constraint between the truck and the block allows only one translational DoF between the two. Therefore, in the 2D approximation, the associated torsor has two independent components. If characterized at the midpoint of the contact surface, these components are a normal force and a moment perpendicular to the plane. The AMT applied to the block leads to <math>\Ns=\ms\gs</math>.

|}
:The powers associated with all interactions on the block in the ground reference are:

:<math>\dot{\Ws}_\Es^{\rightarrow \mathrm{bloque}}=\dot{\Ws}_\Ts^{\mathrm{ext}}=\dot{\Ws}_\Ts^{\mathrm{peso}}+\dot{\Ws}_\Ts^{\mathrm{fricción}}+\dot{\Ws}_\Ts^{\mathrm{enlace}}= \Bigr[(\downarrow \ms\gs)+(\leftarrow \mu\ms\gs)+(\uparrow \Ns) \Bigr] \cdot \Bigr[\rightarrow(\vs+\vs')\Bigr] + (\odot \Ms_\es)\cdot 0 = -\mu\ms\gs(\vs+\vs')</math>

:Ya que el bloque se traslada respecto al suelo, su energía cinética total es inmediata, y se puede plantear fácilmente el balance de potencias:

:<math>\Ts_\Ts^\mathrm{bloque}=\frac{1}{2}\int_\mathrm{bloque}\ds\ms \vs_\Ts^2(\mathrm{bloque})=\frac{1}{2}\ms \vs_\Ts^2(\mathrm{bloc})\quad \Rightarrow \quad \dot{\Ts}_\Ts^\mathrm{bloque}=\ms\vs_\Ts( \mathrm{bloque})\dot{\vs}_\Ts( \mathrm{bloque}) </math>

:Como la única fuerza horizontal sobre el bloque es la fricción:

<math>\dot{\vs}_\Ts(\mathrm{bloque})=-\mu\gs\quad \Rightarrow \quad \dot{\Ts}_\Ts^\mathrm{bloque}=-\mu\ms\gs(\vs+\vs').</math>

:El balance de potencias es consistente:

:<math>\left.\begin{array}{l}
\dot{\mathrm{W}}_\Ts^{\rightarrow \text { bloque }}=\dot{\mathrm{W}}_{\mathrm{T}}^{\text {ext }}=-\mu \mathrm{mg}\left(\mathrm{v}+\mathrm{v}^{\prime}\right) \\
\dot{\mathrm{T}}_{\mathrm{T}}^{\text {bloque }}=-\mu \mathrm{mg}\left(\mathrm{v}+\mathrm{v}^{\prime}\right)
\end{array}\right\} \Rightarrow \dot{\mathrm{W}}_{\mathrm{T}}^{\rightarrow \text { bloque }}=\dot{\mathrm{T}}_{\mathrm{T}}^{\text {bloque }}</math>

:Referencia camión: Al estar el camión acelerado respecto al suelo, hay que tener en cuenta las fuerzas de arrastre sobre todos los diferenciales de masa (dm) del bloque:

{|
|

[[Archivo:ExE1-3-2-3-cat-esp.jpg|thumb|right|250px|link=]]
|<math>\cal{\Fs}^{\rightarrow \ds\ms}_{\mathrm{ar}}=-\ds\ms\overline{\as}_\mathrm{ar}(\ds\ms)=-\ds\ms (\leftarrow \as)=(\rightarrow \ds\ms \cdot \as)</math>.

El torsor resultante de este sistema de fuerzas en el centro de inercia '''G''' se reduce a una fuerza horizontal.

<math>\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{bloque}}=\dot{\Ws}_\mathrm{camión}^{\mathrm{ext}} + \dot{\Ws}_\mathrm{camión}^{\mathrm{ar}} =\dot{\Ws}_\mathrm{camión}^{\mathrm{peso}} + \dot{\Ws}_\mathrm{camión}^{\mathrm{fricción}} + \dot{\Ws}_\mathrm{camión}^{\mathrm{enlace}} + \dot{\Ws}_\mathrm{camión}^{\mathrm{ar}} , </math>

<math>\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{bloque}}= \Bigr[(\downarrow \ms\gs)+(\leftarrow \mu\ms\gs)+(\uparrow \Ns) + (\rightarrow \ms\as)\Bigr] \cdot(\rightarrow \vs')+(\odot\Ms_\es) \cdot0=\ms(\as-\mu\gs)\vs'.</math>
|}

:Se puede comprobar que el balance de potencias es consistente:

:<math>\Ts^\mathrm{bloque}_\mathrm{camión}=\frac{1}{2}\ms\vs^2_\mathrm{camión}(\mathrm{bloque}) \quad \Rightarrow \quad \dot{\Ts}^\mathrm{bloque}_\mathrm{camión} = \ms\vs_\mathrm{camión}(\mathrm{bloque})\dot{\vs}_\mathrm{camión}(\mathrm{bloque})</math>

:<math>\left.\begin{array}{l}
\mathrm{AB}:\mathrm{suelo} \\
\mathrm{REL}:\mathrm{camión}
\end{array}\right\} \quad \dot{\vs}_\mathrm{camión}(\mathrm{bloque})= \dot{\vs}_\mathrm{T}(\mathrm{bloque})-\dot{\vs}_\mathrm{ar}(\mathrm{bloque}) = (-\mu\gs)-(-\as)=\as-\mu\gs \quad \Rightarrow \quad \dot{\Ts}^\mathrm{bloque}_\mathrm{camión}=\ms\vs'(\as-\mu\gs)</math>

:<math>\left.\begin{array}{l}
\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{bloque}}=\dot{\Ws}_\mathrm{camión}^{\mathrm{ext}} + \dot{\Ws}_\mathrm{camión}^{\mathrm{ar}}= \ms(\as-\mu\gs)\vs'\\
\dot{\Ts}^\mathrm{bloque}_\mathrm{camión}=\ms\vs'(\as-\mu\gs)
\end{array}\right\} \Rightarrow\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{bloque}}= \dot{\Ts}^\mathrm{bloque}_\mathrm{camión}</math>


====✏️ EJEMPLO E1-3.3: camión frenando con carga====
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:
{|
|
:[[Archivo:ExE1-3-3-1-esp.png|thumb|left|220px|link=]]
|Un camión frena con aceleración '''a''' respecto al suelo en el instante en el que su velocidad (también respecto al suelo) es v. Los pares de frenada en el eje delantero y en el trasero son iguales y valen <math>\Gamma</math>, y las ruedas, de radio r y masa despreciable, no deslizan sobre el suelo. En el mismo instante, el bloque homogéneo, de masa m, desliz con velocidad v' relativa al camión. Se trata de hacer un balance de potencias para el sistema (bloque+camión) en la referencia del suelo y en la del camión.

|}

:Ya que la potencia de las interacciones internas no es nula en principio (no se trata de un único sólido rígido), es conveniente hacer un DGI del sistema para no olvidar ninguna:
{|
|
[[Archivo:ExE1-3-3-2-esp.png|thumb|right|250px|link=]]

|
|
c.p.s.d.: contacto puntual sin deslizamiento

c.m.c.d.: contacto multipuntual con deslizamiento

|}
:El sistema sobre el que se hace el balance de potencias es todo menos el suelo. Teniendo en cuenta que la potencia de los enlaces internos es siempre nula, las únicas interacciones que hay que tener en cuenta son el peso, los enlaces de las ruedas con el suelo, los pares de frenada y la fricción entre bloque y camión:

[[Archivo:ExE1-3-3-3-esp.png|thumb|center|300px|link=]]

:Referencia suelo:
:Las fuerzas de enlace del suelo sobre las ruedas están aplicadas a puntos de velocidad cero, y por tanto no hacen potencia. El peso es una fuerza vertical, y el movimiento del camión y del bloque es horizontal, por tanto su potencia también es cero.

:<math>\dot{\Ws}_\Ts^{\rightarrow \mathrm{sist}}=\dot{\Ws}_\Ts^{\mathrm{ext}}+\dot{\Ws}^{\mathrm{int}}=\dot{\Ws}_\Ts^{\mathrm{peso}}+\dot{\Ws}_\Ts^{\mathrm{enlace}}+\dot{\Ws}^{\mathrm{fricción}}+\dot{\Ws}^{\Gamma}=\dot{\Ws}^{\mathrm{fricción}}+\dot{\Ws}^{\Gamma}.</math>

:Las potencias de interacciones internas se pueden calcular en cualquier referencia. Por tanto:

:<math>\dot{\Ws}^{\mathrm{fricción}}= \dot{\Ws}_\mathrm{camión}^{\mathrm{fric}\rightarrow \mathrm{bloque}} +\dot{\Ws}_\mathrm{camió}^{\mathrm{fric}\rightarrow \mathrm{camión}} = \dot{\Ws}_\mathrm{camión}^{\mathrm{fric}\rightarrow \mathrm{bloque}}= (\leftarrow \mu\ms\gs)\cdot (\rightarrow \vs')=-\mu\ms\gs\vs'</math>

:<math>\dot{\Ws}^{\Gamma}= \dot{\Ws}_\mathrm{camión}^{\Gamma\rightarrow \mathrm{ruedas}} +\dot{\Ws}_\mathrm{camión}^{\Gamma\rightarrow \mathrm{camión}} = \dot{\Ws}_\mathrm{camión}^{\Gamma\rightarrow \mathrm{ruedas}}= 2 \left(\odot \Gamma \right)\cdot \left(\otimes \frac{\vs}{\rs}\right)=-2\Gamma\frac{\vs}{\rs}.</math>

:Por otro lado: <math>\Ts_\Ts^\mathrm{sist}=\frac{1}{2}\ms\vs_\Ts^2(\mathrm{bloque})+\frac{1}{2}\ms\vs_\Ts^2(\mathrm{camión})\quad \Rightarrow \quad \dot{\Ts}_\Ts^\mathrm{sist}=-\ms(\vs+\vs')\mu\gs-\Ms\vs\as. </math>

:Finalmente: <math>\dot{\Ws}_\Ts^{\rightarrow \mathrm{sist}}=\dot{\Ts}_\Ts^\mathrm{sist} \quad \Rightarrow \quad -\mu\ms\gs\vs'-2\Gamma\frac{\vs}{\rs}=-\ms(\vs+\vs')\mu\gs-\Ms\vs\as.</math>

:Esta ecuación permite calcular el par de frenada: <math>\Gamma=\frac{1}{2}(\mu\ms\gs+\Ms\as)\rs.</math>

:Referencia camión:

:En esta referencia hay que tener en cuenta las fuerzas de arrastre. Por otro lado, las fuerzas de enlace del suelo sobre las ruedas están aplicadas a puntos que ya no tienen velocidad cero, y por tanto hacen potencia. Como antes, la potencia del peso es cero.

:<math>\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{sist}}=\dot{\Ws}_\mathrm{camión}^{\mathrm{ext}}+\dot{\Ws}^{\mathrm{int}}+\dot{\Ws}_\mathrm{camión}^{\mathrm{ar}}=\dot{\Ws}_\mathrm{camión}^{\mathrm{enlace}}+\dot{\Ws}^{\mathrm{fricción}}+\dot{\Ws}^{\Gamma}+\dot{\Ws}_\mathrm{camión}^{\mathrm{ar}}.</math>

:Las fuerzas de arrastre sobre el camión no hacen potencia porque el camión está parado en esta referencia. La de las fuerzas de arrastre sobre el bloque se ha calculado en el [[E1. Teorema de la energía: versión diferencial#✏️ ✏️ EJEMPLO E1-3.2: bloque deslizando sobre un camión|'''ejemplo E1-3.2''']]. La potencia de la fricción y de los pares de frenada es la misma de antes ya que no depende de la referencia.

:<math>\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{sist}}=\dot{\Ws}_\mathrm{camión}^{\mathrm{enllaç}}+\dot{\Ws}^{\mathrm{fricción}}+\dot{\Ws}^{\Gamma}+\dot{\Ws}_\mathrm{camión}^{\mathrm{ar} \rightarrow \mathrm{bloque}}= \dot{\Ws}_\mathrm{camión}^{\mathrm{enlace}} -\mu\ms\gs\vs'-2\Gamma\frac{\vs}{\rs} +\ms\as\vs'.</math>

:Las descripciones cinemática y dinámica de la rueda son:

[[Archivo:ExE1-3-3-4-esp.png|thumb|center|430px|link=]]

:<math>\dot{\Ws}_\mathrm{camión}^{\mathrm{enlace}}=2\Bigr[(\leftarrow \Ts)+(\uparrow \Ns)\Bigr]\cdot(\leftarrow\vs)=2\Ts\vs.</math>

La fuerza tangencial de enlace se puede calcular a partir del TMC aplicado al centro '''C''' de la rueda. Como no tiene masa: <math>\sum\overline{\Ms}_\mathrm{ext}(\Cs)=\overline{0} \quad \Rightarrow \quad (\odot \Gamma)+(\otimes \Ts\rs)=0\quad \Rightarrow \quad \Ts=\Gamma/\rs.</math>

:Por tanto: <math>\dot{\Ws}_\mathrm{camió}^{\rightarrow \mathrm{sist}} =2\frac{\Gamma}{\rs}\vs-\mu\ms\gs\vs'-2\Gamma\frac{\vs}{\rs}+\ms\as\vs'=\ms(\as-\mu\gs)\vs'.</math>

:Por otro lado, como se ha visto en el [[E1. Teorema de la energía: versión diferencial#✏️ ✏️ EJEMPLO E1-3.2: bloque deslizando sobre un camión|'''ejemplo E1-3.2''']]:

:<math>\Ts_\mathrm{camión}^\mathrm{sist}= \Ts_\mathrm{camión}^\mathrm{bloque}=\frac{1}{2}\ms\vs_\mathrm{camión}^2(\mathrm{bloque}) \quad \Rightarrow \quad \dot{\Ts}_\mathrm{camión}^\mathrm{sist}= \dot{\Ts}_\mathrm{camión}^\mathrm{bloque}=\ms\vs'(\as-\mu\gs).</math>

:El balance de potencias es consistente:

:<math>\left.\begin{array}{l}
\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{sist}}= \ms(\as-\mu\gs)\vs'\\
\dot{\Ts}^\mathrm{sist}_\mathrm{camión}=\ms(\as-\mu\gs)\vs'
\end{array}\right\} \Rightarrow\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{sist}}= \dot{\Ts}^\mathrm{sist}_\mathrm{camión}</math>

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==E1.4 Balance de potencias para un sistema multisólido: cálculo directo y cálculo indirecto==

El '''cálculo directo''' (a partir de su definición) de la potencia asociada a una interacción (una fuerza sobre un punto '''P''' o un momento sobre un sólido rígido S) exige el conocimiento de la interacción (fuerza o momento):

<math>\dot{\Ws}_\Rs^{\Fs\rightarrow \Ps}=\overline{\Fs}_{\rightarrow\Ps} \cdot \vel{P}{R} \quad , \quad \dot{\Ws}_\Rs^{\Ms\rightarrow \Ss}=\overline{\Ms}_{\rightarrow\Ss} \cdot \velang{S}{R} </math>

Cuando no se trata de una interacción formulable, hay que recurrir a los teoremas vectoriales para determinarla.

El '''cálculo indirecto''' es una alternativa que no necesita recurrir a estos teoremas. Se basa en la evaluación de un balance de potencias para un sistema y una referencia adecuados de manera que la potencia a determinar sea el único término desconocido de la ecuación (el único que no se puede calcular de manera directa). Cuando el balance incluye más de un término desconocido, se pueden combinar varios balances (para distintos sistemas y referencias) hasta tener un número suficiente de ecuaciones que permitan la determinación de todas las potencias desconocidas.

====✏️ EJEMPLO E1-4.1: polea frenada====
------
:
{|
|
:[[Archivo:ExE1-4-1-1-esp.png|thumb|left|150px|link=]]
|Un bloque de masa m cuelga de un hilo inextensible atado a un punto de la periferia de una polea de radio r. Entre soporte (fijo al suelo) y polea actúa un freno que garantiza que el bloque baja a velocidad constante respecto al suelo. Se trata de calcular la potencia del freno.

|}

:Càlculo directo: <math>\dot{\Ws}^{\mathrm{freno}}= \overline{\Gamma}^{\rightarrow \mathrm{polea}}\cdot\velang{polea}{T}=\left(\odot\Gamma\right)\cdot\left( \otimes \frac{\vs_0}{\rs}\right)=-\Gamma \frac{\vs_0}{\rs}.</math>

:El par de frenada no es un dato, y hay que recurrir al TMC para calcularlo.

:<math>\left.\begin{array}{l}
\text{SISTEMA: polea + bloque}\\
\text{TMC en } \Os
\end{array}\right\} \sum\overline{\Ms}_\mathrm{ext}(\Os)=\dot{\overline{\Hs}}_\mathrm{RTO}(\Os)</math>

:El momento cinético proviene de las dos piezas. Ya que '''O''' no es fijo al bloque, hay que hacer descomposición baricéntrica para calcular el momento cinético del bloque:

:<math>\overline{\Hs}_\mathrm{RTO=T} (\Os) =\overline{\Hs}_\Ts^\mathrm{politja}(\Os) + \overline{\Hs}_\Ts^\mathrm{bloc}(\Os)= \overline{\Hs}_\Ts^\mathrm{politja}(\Os) + \overline{\Hs}_\Ts^\mathrm{bloc}(\Gs_\mathrm{bloc})+ \overline{\Hs}_\Ts^{\oplus\mathrm{bloc}}(\Os) = </math>

:<math>\hspace{2.3cm}=\Is_\mathrm{O}^\mathrm{polea} \velang{polea}{T}+\Is_{\Gs_\mathrm{bloque}}^\mathrm{bloque}+\OGvec_\mathrm{bloque}\times \ms\overline{\vs}_\Ts(\Gs_\mathrm{bloque})=\left(\otimes \Is_\mathrm{O}\frac{\vs_0}{\rs}\right)+(\otimes \ms\vs_0\rs).</math>

:[[Archivo:ExE1-4-1-2-esp.png|thumb|right|180px|link=]]
:Se trata de un momento cinético constante en dirección y valor. Por tanto:

:<math>\dot{\overline{\Hs}}_\mathrm{RTO}(\Os)=0 \quad \Rightarrow \quad \sum{\overline{\Ms}}_\mathrm{ext}(\Os)=(\odot\Gamma)+(\otimes \ms\gs\rs) =0\quad \Rightarrow \quad \Gamma= \ms\gs\rs</math>.

:Finalmente: <math>\dot{\Ws}^{\mathrm{freno}}= -\Gamma \frac{\vs_0}{\rs} = -\ms\gs\vs_0.</math>

:Cálculo indirecto:

:<math>\left.\begin{array}{l}
\text{SISTEMA: polea + soporte + bloque}\\
\text{REF: suelo(Gal)}
\end{array}\right\} \quad \dot{\Ts}^\mathrm{sist}_\Ts=\dot{\Ws}^\mathrm{ext}_\Ts+ \dot{\Ws}^\mathrm{int}</math>

:<math>\left.\begin{array}{l}
\dot{\Ts}^\mathrm{sist}_\Ts=0\\
\dot{\Ws}^\mathrm{ext}_\Ts=\left(\downarrow \ms\gs\right)\cdot\left(\downarrow \vs_0\right)= \ms\gs\vs_0\\
\dot{\Ws}^\mathrm{int}=\dot{\Ws}^{\mathrm{freno}}
\end{array}\right\} \quad \Rightarrow \quad \dot{\Ws}^{\mathrm{freno}} =-\ms\gs\vs_0</math>



====✏️ EJEMPLO E1-4.2: mono y plátanos====
------
:
{|
|
:[[Archivo:ExE1-4-2-1-esp.png|thumb|left|180px|link=]]
|Unos plátanos de masa <math>(2/3)\ms</math> cuelgan de una cuerda inextensible que se apoya sin deslizar sobre una polea de radio r y masa despreciable, articulada respecto al suelo. En el otro extremo de la cuerda se agarra un mono de masa m. La polea gira con velocidad angular constante <math>\Omega_0</math> respecto al suelo bajo la acción de un motor. El mono se mueve respecto a la cuerda sin que sus manos se froten contra la cuerda, y de manera que la distancia de su centro de masas al suelo se mantiene constante. Se trata de calcular la potencia del motor.
|}

:Cálculo directo: <math>\dot{\Ws}^{\mathrm{freno}}= \overline{\Gamma}^{\rightarrow \mathrm{polea}}\cdot\velang{polea}{T}=\left(\odot\Gamma\right)\cdot\left( \odot \Omega_0\right)=\Gamma \Omega_0.</math>

:El par motor no es un dato, y hay que recurrir al TMC para calcularlo.

:<math>\left.\begin{array}{l}
\text{SISTEMA: polea + mono + plátanos}\\
\text{TMC en } \Os
\end{array}\right\} \sum\overline{\Ms}_\mathrm{ext}(\Os)=\dot{\overline{\Hs}}_\mathrm{RTO}(\Os)</math>

:El momento cinético proviene del mono y de los plátanos. En primera aproximación, se puede considerar la masa del mono concentrada en su tronco y su cabeza, de manera que no gira y su velocidad respecto al suelo es nula. Teniendo en cuenta que el punto '''O''' no se mueve respecto al mono pero sí respecto a los plátanos:

:<math>\overline{\Hs}_\mathrm{RTO=T} (\Os) =\overline{\Hs}_\Ts^\mathrm{mono}(\Os) + \overline{\Hs}_\Ts^\mathrm{plátanos}(\Os)= \overline{\Hs}_\Ts^\mathrm{mono}(\Os) + \overline{\Hs}_\Ts^\mathrm{plátanos}(\Gs_\mathrm{plát})+ \OGvec_\mathrm{plát} \times \frac{2}{3} \ms \overline{\vs}_\Ts(\Gs_\mathrm{plát})= </math>

:<math>\hspace{2.3cm}=\OGvec_\mathrm{plát}\times \frac{2}{3} \ms \overline{\vs}_\Ts(\Gs_\mathrm{plát}) = \Bigr[(\rightarrow \rs)+(\downarrow \xs)\Bigr] \times\frac{2}{3} \ms \left(\downarrow \rs\Omega_0\right)=\left(\otimes \frac{2}{3}\ms\rs^2\Omega_0\right).</math>

:Se trata de un momento cinético constante en dirección y valor. Por tanto:

:<math>\dot{\overline{\Hs}}_\mathrm{RTO}(\Os)=0 \quad \Rightarrow \quad \sum{\overline{\Ms}}_\mathrm{ext}(\Os)=(\odot\Gamma)+(\otimes \ms\gs\rs) + \left(\otimes \frac{2}{3}\ms\gs\rs\right)=0\quad \Rightarrow \quad \Gamma= \frac{5}{3}\ms\gs\rs</math>.

:Finalmente: <math>\dot{\Ws}^{\mathrm{motor}}= \Gamma \Omega_0 = \frac{5}{3}\ms\gs\rs\Omega_0.</math>

:Cálculo indirecto

:Hay que tener presente que es un problema con dos tipos de actuadores: el motor y la musculatura del mono. Hay diversos modelos biomecánicos para formular las fuerzas de los músculos, pero quedan fuera del ámbito de este curso. Ya que el mono se mueve, sus músculos desarrollan una potencia no nula.

:<math>\left.\begin{array}{l}
\text{SISTEMA: polea + mono + plátano}\\
\text{REF: suelo(Gal)}
\end{array}\right\} \quad \dot{\Ts}^\mathrm{sist}_\Ts=\dot{\Ws}^\mathrm{ext}_\Ts+ \dot{\Ws}^\mathrm{int}</math>

:<math>\left.\begin{array}{l}
\dot{\Ts}^\mathrm{sist}_\Ts=0\\
\dot{\Ws}^\mathrm{ext}_\Ts=\left(\downarrow \frac{2}{3}\ms\gs\right)\cdot\left(\downarrow \rs\Omega_0\right)= \frac{2}{3}\ms\gs\rs\Omega_0\\
\dot{\Ws}^\mathrm{int}=\dot{\Ws}^{\mathrm{musc}}+\dot{\Ws}^{\mathrm{motor}}
\end{array}\right\} \quad \Rightarrow \quad \dot{\Ws}^{\mathrm{musc}} + \dot{\Ws}^{\mathrm{motor}}=-\frac{2}{3}\ms\gs\rs\Omega_0</math>

:Este balance proporciona una ecuación con dos incógnitas. Se puede plantear un segundo balance para un sistema donde una de las dos no aparezca: un sistema sin el mono o un sistema sin el motor. Si se hace un balance de potencias para el sistema (mono + trozo de cuerda), las fuerzas externas a considerar son el peso y la tensión de la cuerda. Si se sigue trabajando en la referencia del suelo, esta tensión (desconocida) hará potencia no nula ya que está aplicada a un punto de velocidad ascendente.
:[[Archivo:ExE1-4-2-2-cat-esp.png|thumb|right|330px|link=]]

:Para evitar que aparezca un término asociado a la tensión de la cuerda, se puede trabajar en la referencia del trozo de cuerda que se ha considerado, que al tener velocidad ascendente constante es galileana. En esta referencia, el mono baja con velocidad constante:

:<math>\left.\begin{array}{l}
\text{REL: trozo de cuerda}\\
\text{AB: suelo}
\end{array}\right\}</math>

:<math>\overline{\vs}_\mathrm{REL}(\mathrm{mono})=\overline{\vs}_\mathrm{ar}(\mathrm{mono})=-\overline{\vs}_\mathrm{ar}(\mathrm{mono})=(\downarrow \rs\Omega_0)</math>

:<math>\left.\begin{array}{l}
\text{SISTEMA: mono+ trozo de cuerda}\\
\text{REF: trozo de cuerda(Gal)}
\end{array}\right\} \quad \dot{\Ts}^\mathrm{sist}_\mathrm{cuerda} =\dot{\Ws}^\mathrm{ext}_\mathrm{cuerda}+ \dot{\Ws}^\mathrm{int}</math>

:<math>\left.\begin{array}{l}
\dot{\Ts}^\mathrm{sist}_\mathrm{cuerda}=0\\
\dot{\Ws}^\mathrm{ext}_\mathrm{cuerda}=\left(\downarrow \ms\gs\right)\cdot\left(\downarrow \rs\Omega_0\right)= \ms\gs\rs\Omega_0\\
\dot{\Ws}^\mathrm{int}=\dot{\Ws}^{\mathrm{musc}}
\end{array}\right\} \quad \Rightarrow \quad \dot{\Ws}^{\mathrm{musc}} =-\ms\gs\rs\Omega_0</math>

:Combinando este resultado con el del balance anterior:

:<math>\dot{\Ws}^\mathrm{motor}=-\frac{2}{3}\ms\gs\rs\Omega_0-\dot{\Ws}^\mathrm{musc}=-\frac{5}{3}\ms\gs\rs\Omega_0</math>



© Universitat Politècnica de Catalunya. [[Mecánica:Derechos de autor |Todos los derechos reservados]]
-------------

<center>
[[D8. Conservaciones|<<< D8. Conservaciones]]

[[E2. Teorema de la energía: versión integrada|E2. Teorema de la energía: versión integrada>>>]]
</center>

E1. Work-Energy Theorem: differential form

2025-05-20T12:16:55Z

Bros: /* ✏️ EXAMPLE E1-1.2: pendulum */

<div class="noautonum">__TOC__</div>
<math>\newcommand{\uvec}{\overline{\textbf{u}}}
\newcommand{\vvec}{\overline{\textbf{v}}}
\newcommand{\evec}{\overline{\textbf{e}}}
\newcommand{\Omegavec}{\overline{\mathbf{\Omega}}}
\newcommand{\velang}[2]{\Omegavec^{\textrm{#1}}_{\textrm{#2}}}
\newcommand{\Alfavec}{\overline{\mathbf{\alpha}}}
\newcommand{\accang}[2]{\Alfavec^{\textrm{#1}}_{\textrm{#2}}}
\newcommand{\ds}{\textrm{d}}
\newcommand{\ps}{\textrm{p}}
\newcommand{\ns}{\textrm{n}}
\newcommand{\hs}{\textrm{h}}
\newcommand{\cs}{\textrm{c}}
\newcommand{\gs}{\textrm{g}}
\newcommand{\Ns}{\textrm{N}}
\newcommand{\Hs}{\textrm{H}}
\newcommand{\Fs}{\textrm{F}}
\newcommand{\ms}{\textrm{m}}
\newcommand{\ts}{\textrm{t}}
\newcommand{\us}{\textrm{u}}
\newcommand{\vs}{\textrm{v}}
\newcommand{\Rs}{\textrm{R}}
\newcommand{\Es}{\textrm{E}}
\newcommand{\Ws}{\textrm{W}}
\newcommand{\Ts}{\textrm{T}}
\newcommand{\Ls}{\textrm{L}}
\newcommand{\As}{\textrm{A}}
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\newcommand{\es}{\textrm{e}}
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\newcommand{\is}{\textrm{i}}
\newcommand{\Is}{\textrm{I}}
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\newcommand{\js}{\textrm{j}}
\newcommand{\rs}{\textrm{r}}
\newcommand{\as}{\textrm{a}}
\newcommand{\ss}{\textrm{s}}
\newcommand{\Os}{\textbf{O}}
\newcommand{\Gs}{\textbf{G}}
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\newcommand{\Ms}{\textrm{M}}
\newcommand{\deg}{^\textsf{o}}
\newcommand{\xs}{\textsf{x}}
\newcommand{\ys}{\textsf{y}}
\newcommand{\zs}{\textsf{z}}
\newcommand{\dert}[2]{\left.\frac{\ds{#1}}{\ds\ts}\right]_{\textrm{#2}}}
\newcommand{\ddert}[2]{\left.\frac{\ds^2{#1}}{\ds\ts^2}\right]_{\textrm{#2}}}
\newcommand{\vec}[1]{\overline{#1}}
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\newcommand{\vecdot}[1]{\overline{\dot{#1}}}
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\newcommand{\GPvec}{\vec{\Gs\Ps}}
\newcommand{\PSvec}{\vec{\Ps\Ss}}
\newcommand{\QQvec}{\vec{\Qs\Qs}}
\newcommand{\QGvec}{\vec{\Qs\Gs}}
\newcommand{\QPvec}{\vec{\Qs\Ps}}
\newcommand{\JQvec}{\vec{\Js\Qs}}
\newcommand{\PQvec}{\vec{\Ps\Qs}}
\newcommand{\GJvec}{\vec{\Gs\Js}}
\newcommand{\OPvec}{\vec{\Os\textbf{P}}}
\newcommand{\OCvec}{\vec{\Os\Cs}}
\newcommand{\OGvec}{\vec{\Os\Gs}}
\newcommand{\CJvec}{\vec{\Cs\Js}}
\newcommand{\GCvec}{\vec{\Gs\Cs}}
\newcommand{\PGvec}{\vec{\Ps\Gs}}
\newcommand{\abs}[1]{\left|{#1}\right|}
\newcommand{\braq}[2]{\left\{{#1}\right\}_{\textrm{#2}}}
\newcommand{\matriz}[9]{
\begin{bmatrix}
{#1} & {#2} & {#3}\\
{#4} & {#5} & {#6}\\
{#7} & {#8} & {#9}
\end{bmatrix}}
\newcommand{\diag}[3]{
\begin{bmatrix}
{#1} & {0} & {0}\\
{0} & {#2} & {0}\\
{0} & {0} & {#3}
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\newcommand{\vector}[3]{
\begin{Bmatrix}
{#1}\\
{#2}\\
{#3}
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\begin{Bmatrix}
{#1}\\
{#2}
\end{Bmatrix}}
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\newcommand{\agal}[1]{\vecbf{a}_{\textrm{Gal}} (#1)}
\newcommand{\angal}[1]{\vecbf{a}_{\textrm{NGal}} (#1)}
\newcommand{\vgal}[1]{\vecbf{v}_{\textrm{Gal}} (#1)}
\newcommand{\fvec}[2]{\overline{\mathbf{F}}_{\textrm{#1}\rightarrow\textrm{#2}}}
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\newcommand{\F}[1]{\vec{\Fs}_{#1}}
\newcommand{\Fcal}[2]{\vec{\cal{F}}^{\text{#1}}_{#2}}
</math>

The '''Work-Energy Theorem''' plays a prominent role in physics, as it bridges all branches of science (electromagnetism, thermodynamics, etc.): all of them deal with energy.

There is a certain parallel between the Vector Theorems and the '''Work-Energy Theorem''': they all allow us to understand the evolution of a magnitude that depends on the mechanical state of the system. In the case of the former, we are dealing with the instantaneous evolution of vector magnitudes (linear momentum, angular momentum), while the latter deals with the finite change (in a finite time interval) of a scalar magnitude ('''mechanical energy''').

This unit presents the differential form of this theorem (the '''power balance'''). The integrated form (which is the Work-Energy Theorem itself) is presented in [[E2. Teorema de la energía: versión integrada|'''unit E2''']].
 
------
------
==E1.1 Power balance in a system of particles==

:[[File:E1-1-eng.png|thumb|right|300px|link=]]
<center>'''Figure E1.1''' Forces on a particle of a system with constant matter</center>

Let us consider a system of constant matter, and let’s apply Newton's second law to each particle (or each mass differential) of this system in a general reference frame R (which may be non-Galilean) ('''Figure E1.1'''):

<math>\F{\rightarrow \Ps} + \Fcal{}{\text{ar}\rightarrow\Ps} + \Fcal{}{\text{Cor}\rightarrow\Ps} =\ms_\Ps\acc{P}{R}</math>

<math>\F{\rightarrow \Ps} </math> is the resultant interaction force on '''P''', and it comes from internal interactions (with particles of the same system) and external ones:

<math>\F{\rightarrow \Ps}=\F{\rightarrow \Ps}^\text{int}+\F{\rightarrow \Ps}^\text{ext}</math>

If both sides of the equation are multiplied by <math>\vel{P}{R}</math>, all forces (interaction and inertia forces) orthogonal to <math>\vel{P}{R}</math> are filtered out. Specifically, since <math>\acc{P}{R}=\accs{P}{R} + \accn{P}{R} </math>, where <math>\accn{P}{R} \perp \vel{P}{R}</math>, and since the Coriolis force is always orthogonal to <math>\vel{P}{R} (\Fcal{}{\text{Cor}\rightarrow\Ps} = -2\ms_\Ps \velang{R}{Gal} \times \vel{P}{R})</math>:

<math>\vel{P}{R} \cdot \left( \F{\rightarrow \Ps}^\text{int}+\F{\rightarrow \Ps}^\text{ext} + \Fcal{}{\text{ar}\rightarrow\Ps} \right) = \ms_\Ps\accs{P}{R} \cdot \vel{P}{R} = \frac{\ds}{\ds \ts} \left(\frac{1}{2}\ms_\Ps \vvec_\Rs^2(\Ps) \right) \equiv \frac{\ds\Ts_\Rs(\Ps)}{\ds\ts}=\dot{\Ts}_\Rs(\Ps)</math>

The scalar function <math>\Ts_\Rs(\Ps)</math> is the '''kinetic energy''' of the particle, and is a function of the [[D1. Foundational laws of Newtonian dynamics#D1.3 Newton’s Principle of Determinacy|'''mechanical state''']] of '''P'''. The scalar product of a force and the velocity of its application point is the '''force power''': <math>\vel{P}{R} \cdot \F{\rightarrow \Ps} \equiv \dot{\Ws}_\Rs^{\F{\rightarrow \Ps}} </math>. Both scalar quantities <math>(\Ts_\Rs(\Ps), \dot{\Ws}_\Rs^{\F{\rightarrow \Ps}})</math> depend on the reference frame. For the same set of interaction forces acting on '''P''', the reference frame determines which are filtered out and which are not.

The equation can be rewritten as <math>\dot{\Ws}_\Rs^{\text{ext}\rightarrow \Ps}+\dot{\Ws}_\Rs^{\text{int}\rightarrow \Ps}+\dot{\Ws}_\Rs^{\text{ar}\rightarrow \Ps} =\dot{\Ts}_\Rs(\Ps)</math>, and constitutes the '''power balance''' for particle '''P''' in the R reference frame. Summing for all particles in the system:

<math>\dot{\Ws}_\Rs^{\text{ext}}+\dot{\Ws}^{\text{int}}+\dot{\Ws}_\Rs^{\text{ar}} =\dot{\Ts}_\Rs^\text{syst}</math>

The calculation of kinetic energy is covered in [[E2. Teorema de la energía: versión integrada|'''unit E2''']]. This unit focuses on the calculation of powers. [[E1. Work-Energy Theorem: differential form#E1.2 Potencia de una pareja de acción-reacción|'''section E1-2''']] shows that the power of internal interactions does not depend on the reference frame (which is why the subscript R does not appear in the <math>\dot{\Ws}^{\text{int}}</math> term of the previous equation).

====✏️ EXAMPLE E1-1.1: free particle====
------
:
{|
:[[File:ExE1-1-1-1-eng.jpg|thumb|left|180px|link=]]
|A particle '''P''', with mass m, is released from the ring of a Ferris wheel when it is at its highest point. We want to calculate the power, at this instant, of all forces acting on the particle from three different reference frames: the ground, the ring, and the Ferris wheel cabin.
|}

:Ground reference frame: Since this is a Galilean reference frame, all forces on '''P''' are interaction forces.
{|
:[[File:ExE1-1-2- eng.png|thumb|right|230px|link=]]

|<math>\hspace{0.5cm}\sum\dot{\Ws}_\Ts=\dot{\Ws}_\Ts^\text{weight}= \left(\leftarrow \Rs\Omega_0\right) \cdot \left(\downarrow \ms\gs\right)=0</math>
|}

:Ring reference frame: It is a non-Galilean reference, and the transportation force on '''P''' must be taken into account.
{|
:[[File:ExE1-1-3-eng.png|thumb|right|250px|link=]]
|<math>\hspace{0.5cm}</math> AB:ground, REL:ring 

<math>\hspace{0.5cm}\vel{P}{REL}=\vel{P}{AB}-\vel{P}{ar}= \left(\leftarrow \Rs\Omega_0\right)-\left(\leftarrow \Rs\Omega_0\right)=\overline{0}</math> 

<math>\hspace{0.5cm}</math>

Since the velocity is zero, the total power is zero.
|}

:Cabin reference frame: It is also a non-Galilean reference, and the transportation force on '''P''' must be taken into account.

{|
:[[File:ExE1-1-4-eng.png|thumb|right|250px|link=]]
|<math>\hspace{0.5cm}</math>AB:ground, REL:cabin

<math>\hspace{0.5cm}\vel{P}{REL}=\vel{P}{AB}-\vel{P}{ar}= \left(\leftarrow \Rs\Omega_0\right)-\left(\uparrow \Rs\Omega_0\right)</math>

<math>\hspace{0.5cm}\dot{\Ws}_\text{cabin}^\text{weight}=\left[ \left(\leftarrow \Rs\Omega_0\right)+\left(\downarrow \Rs\Omega_0\right) \right] \cdot \left(\downarrow\ms\gs\right) =\ms\gs\Rs\Omega_0</math>

<math>\hspace{0.5cm}\dot{\Ws}_\text{cabin}^\text{ar}=\left[ \left(\leftarrow \Rs\Omega_0\right)+\left(\downarrow \Rs\Omega_0\right) \right] \cdot \left(\rightarrow\ms\Rs^2\Omega_0^2\right) =-\ms\gs\Rs^2 \Omega_0^3</math>

<math>\hspace{0.5cm}\sum\dot{\Ws}_\text{cabin}=\ms\Rs\Omega_0(\gs-\Rs\Omega_0^2)</math>

|}


====✏️ EXAMPLE E1-1.2: pendulum====
------
:
{|
:[[File:ExE1-2-1-eng.png|thumb|left|180px|link=]]
|A particle '''P''' of mass m hangs from an inextensible thread of length L attached to the ceiling. The assembly oscillates in a vertical plane fixed to the ground. We want to calculate the power of all the forces acting on the particle in two different reference frames: the ground and a platform rotating relative to the ground with a constant angular velocity <math>\Omega_0</math>. 

Ground reference frame: Since it is a Galilean reference frame, '''P''' is only subjected to interaction forces.
{|
:[[File:ExE1-2-2-neut.png|thumb|left|300px|link=]]

|<math>\dot{\Ws}_\Es^\Ts=(\nearrow\Ls\dot{\theta})\cdot(\nwarrow\Ts)=0</math>

<math>\dot{\Ws}_\Es^\text{weight}=(\nearrow\Ls\dot{\theta}\cdot(\downarrow \ms\gs)=-\ms\gs\Ls\dot{\theta}\sin\theta</math>
|}

<math>\hspace{8cm}</math>

Platform reference frame: It is a non Galilean reference frame, and the transportation force on '''P''' has to be taken into account.

{|
:[[File:ExE1-2-3-eng.png|thumb|left|400px|link=]]

|AB:ground, REL:platform

<math>\vel{P}{REL}=\vel{P}{AB}-\vel{P}{ar}=(\nearrow\Ls\dot{\theta})-(\otimes \Ls \Omega_0 \sin \theta)</math>

<math>\dot{\Ws}_\Es^\Ts\left[(\nearrow\Ls\dot{\theta})+(\odot\Ls\Omega_0\sin\theta)\right]\cdot(\nwarrow\Ts)=0</math>

<math>\dot{\Ws}_\Es^\text{weight}\left[(\nearrow\Ls\dot{\theta})+(\odot\Ls\Omega_0\sin\theta)\right]\cdot(\downarrow \ms\gs)=-\ms\gs\Ls\dot{\theta}\sin\theta</math>

<math>\dot{\Ws}_\Es^\text{ar}\left[(\nearrow\Ls\dot{\theta})+(\odot\Ls\Omega_0\sin\theta)\right]\cdot(\rightarrow \ms\Ls\Omega_0^2\sin\theta)=\ms\Ls^2\Omega_0^2\dot{\theta}\cos\theta</math>
|}



------
------

==E1.2 Power of an action-reaction pair==

The total power associated with internal interactions in a constant-mass system is not zero in principle (since the two forces of each action-reaction pair are applied to particles that may have different velocities relative to the reference frame in which the calculation is made), and it is independent of the reference frame in which it is evaluated. It only depends on the separation velocity of the two particles. The power balance for the system is therefore written as: <math>\dot{\Ts}_\Rs^\text{syst}=\dot{\Ws}_\Rs^\text{ext} + \dot{\Ws}^\text{int} + \dot{\Ws}_\Rs^\text{ar} </math> .

<div>
=====💭 PROOF ➕=====

{|
:[[File:E1-2-neut.png|thumb|left|200px|link=]]
|<math>\dot{\Ws}^\text{ac-reac}=\overline{\Fs}_{\Ps \rightarrow \Qs}\cdot \vel{Q}{R} + \overline{\Fs}_{\Qs \rightarrow \Ps}\cdot \vel{P}{R}= \overline{\Fs}_{\Ps \rightarrow \Qs}\cdot \left[ \vel{Q}{R}-\vel{P}{R}\right]</math>

By the [[D1. Foundational laws of Newtonian dynamics#D1.6 Newton’s third law (action-reaction principle)|'''action-reaction principle''']], <math>\overline{\Fs}_{\Ps \rightarrow \Qs}=-\overline{\Fs}_{\Qs \rightarrow \Ps}</math>. Therefore: <math>\dot{\Ws}^\text{ac-reac}=\overline{\Fs}_{\Ps \rightarrow \Qs}\cdot \left[ \vel{Q}{R}-\vel{P}{R}\right]</math>

As <math>\overline{\Fs}_{\Ps \rightarrow \Qs}</math> has the <math>\PQvec</math> direction, only the components of the velocities in that direction have to be considered:

<math>\dot{\Ws}^\text{ac-reac}=\overline{\Fs}_{\Ps \rightarrow \Qs}\cdot \biggl( \vel{Q}{R} \Bigr]_{||\Ps\Qs}-\ \vel{P}{R} \Bigr]_{||\Ps\Qs}\biggl)</math>
|}

The difference <math>\biggl( \vel{Q}{R} \Bigr]_{\Ps\Qs}-\ \vel{P}{R} \Bigr]_{\Ps\Qs}\biggl)</math> is the separation velocity <math>\dot{\rho}</math> between '''P''' and '''Q''', and it does not dpend on the reference frame since it is the derivative of a scalar (the distance <math>\dot{\rho}</math> between the two particles). For the previous drawing, <math>\dot{\rho}> 0</math>. Since this is an attractive interaction, <math>\dot{\Ws}^\text{ac-reac} < 0</math>.

With a drawing, it is easy to see that:
* attractive interaction and particle approach <math>(\dot{\rho}<0)</math> : <math>\dot{\Ws}^\text{ac-reac} > 0</math> ;
* repulsive interaction and particle separation <math>(\dot{\rho}>0)</math> : <math>\dot{\Ws}^\text{ac-reac} > 0</math>
* repulsive interaction and particle approach <math>(\dot{\rho}<0)</math> : <math>\dot{\Ws}^\text{ac-reac} < 0</math>


</div>

There are two particular cases of interaction forces whose power always has the same sign:

* Friction action-reaction pair between '''P''' and '''Q''': since the friction force always opposes the relative velocity between the two points, its power is always negative:

<math>\hspace{1cm} \dot{\Ws}^\text{friction}=\dot{\Ws}^{\text{fric}\rightarrow\Ps}_\text{RTQ}+\dot{\Ws}^{\text{fric}\rightarrow\Qs}_\text{RTQ}=\overline{\Fs}^{\text{fric}\rightarrow\Ps} \cdot \vel{P}{RTQ} = - \abs{\overline{\Fs}^{\text{fric}\rightarrow\Ps}} \frac{\vel{P}{RTQ}}{\abs{\vel{P}{RTQ}}} \cdot \vel{P}{RTQ}<0;</math>

* Constraint action-reaction pair between '''P''' and '''Q''': from the [[D3. Interactions between rigid bodies#D3.4 Direct constraint interactions|'''analytical characterization equation''']] of the constraint forces between two points '''P''' and '''Q''' belonging to two rigid bodies <math>\mathrm{S}_\Ps</math> and <math>\mathrm{S}_\Qs</math>, it is easy to see that the associated power is zero:

<math>\hspace{1cm} \dot{\Ws}^\text{constraint}=\dot{\Ws}^{\text{constraint}\rightarrow\Ps}_{\mathrm{S}_\Ps}+\dot{\Ws}^{\text{constraint}\rightarrow\Qs}_{\mathrm{S}_\Ps}=\overline{\Fs}^{\text{constraint}\rightarrow\Ps} \cdot \overline{\mathbf{v}}_{\mathrm{S}_\Qs}(\Ps) = - 0;</math>

====✏️ EXAMPLE E1-2.1: power of an action-reaction friction pair====
------
:
{|
:[[File:ExE1-2-1-2-eng.png|thumb|left|180px|link=]]
|A homogeneous wheel, with radius r and mass m, moves down an inclined plane without sliding until it joins a conveyor belt moving at a constant velocity <math>\vs_0</math> relative to the ground. At the moment it makes contact with the belt, the center of the wheel has a velocity <math>\vs_0</math> relative to the ground in the same direction as the belt. We want to calculate the power of the friction force between the belt and the wheel at this instant.
|}

At that momento, the <math>\mathbf{ICR}^\text{wheel}_\Es</math> is point <math>\Js_\text{wheel}</math> in contact with the belt. This point slides backwards with respect to the belt, therefore the friction force on the wheel is forwards:

<math>\left.\begin{array}{l}
\text { AB: ground } \\
\text { REL: belt }
\end{array}\right\} \quad \bar{\vs}_{\text {REL }}\left(\Js_{\text {wheel }}\right)=\bar{\vs}_{\text{AB}}\left(\Js_{\text {wheel }}\right)-\bar{\vs}_{\mathrm{ar}}\left(\Js_{\text {wheel }}\right)=-\bar{\vs}_{\mathrm{ar}}\left(\Js_{\text {wheel }}\right)=-\left(\rightarrow \mathrm{v}_0\right)=\left(\leftarrow \mathrm{v}_0\right) .</math>

[[File:ExE1-2-1-3-eng.png|thumb|center|500px|link=]]

Calculation of the friction power: <math>\dot{\Ws}^\text{friction}=\dot{\Ws}_\Rs^{\text{friction}\rightarrow \text{wheel}}+ \dot{\Ws}_\Rs^{\text{friction}\rightarrow \text{belt}}</math>, where R may be any reference frame. To quickly calculate this power, it is advisable to choose a reference frame R where one of the two points subjected to the action-reaction pair has zero speed, so that one of the two terms is directly zero:

<math>\dot{\Ws}^\text{friction}=\dot{\Ws}_\text{belt}^{\text{fric}\rightarrow \text{wheel}}+ \dot{\Ws}_\text{belt}^{\text{fric }\rightarrow \text{belt}}=\overline{\Fs}^{\text{fric}\rightarrow \text{wheel}} \cdot \bar{\vs}_{\text {belt}}\left(\Js_{\text {wheel }}\right)=(\rightarrow \mu\ms\gs)\cdot(\leftarrow \vs_0)=-\mu\ms\gs\vs_0.</math>

If the calculation is done in any other reference frame, the result is the same:

<math>\dot{\Ws}^\text{friction}=\dot{\Ws}_\text{E}^{\text{fric}\rightarrow \text{wheel}}+ \dot{\Ws}_\text{E}^{\text{fric} \rightarrow \text{belt}}=(\rightarrow \mu\ms\gs)\cdot 0 + (\leftarrow \mu\ms\gs)\cdot(\rightarrow \vs_0)=-\mu\ms\gs\vs_0.</math>

<math>\dot{\Ws}^\text{friction}=\dot{\Ws}_\text{vehicle}^{\text{fric}\rightarrow \text{wheel}}+ \dot{\Ws}_\text{vehicle}^{\text{fric} \rightarrow \text{belt}}=(\rightarrow \mu\ms\gs)\cdot (\rightarrow\vs_0) + (\leftarrow \mu\ms\gs) \cdot(\rightarrow 3\vs_0)=-\mu\ms\gs\vs_0.</math>

However, if only the power of one of the two forces is calculated (for example, the friction on the wheel), the result depends on the reference frame and can have any sign:

<math>\dot{\Ws}_\text{belt}^{\text{fric }\rightarrow \text{wheel}}=(\rightarrow \mu\ms\gs)\cdot(\leftarrow \vs_0)=-\mu\ms\gs\vs_0 ,</math>

<math>\dot{\Ws}_\text{E}^{\text{fric} \rightarrow \text{wheel}}=(\rightarrow \mu\ms\gs)\cdot 0 = 0 ,</math>

<math>\dot{\Ws}_\text{vehicle}^{\text{fric} \rightarrow \text{wheel}}=(\rightarrow \mu\ms\gs)\cdot (\rightarrow\vs_0)=\mu\ms\gs\vs_0.</math>


-----
-----

==E1.3 Power of a system of forces on a rigid body==

{|
|
:[[File:E1-3-eng.png|thumb|left|180px|link=]]
<center>'''Figure E1.3''': Kinematic torsor of a rigid body in a reference frame R, and forces on different points (dm) of that rigid body</center>

|When calculating the power of a system of forces on a rigid body S, the term associated with internal interactions is zero because those interactions are all constraint (cohesion) forces and, as seen in [[E1. Work-Energy Theorem: differential form#E1.2 Power of an action-reaction pair|'''section E1.2''']], their total power is zero.

The total power of a system of forces on S can be calculated from the torsor of the system at any point '''O''' of the rigid body, <math>(\overline{\Fs},\overline{\Ms}(\Os))</math>. Taking into account rigid body kinematics ('''Figure E1.3'''):

<math>\dot{\Ws}_\Rs^{\rightarrow \Ss} =\sum_\Ps\overline{\Fs}^{\rightarrow \Ps} \cdot \vel{P}{R}= \sum_\Ps \left[ \overline{\Fs}^{\rightarrow \Ps} \cdot \vel{O}{R} + \velang{S}{R} \times \OPvec \right] </math>

|}
<math>\dot{\Ws}_\Rs^{\rightarrow \Ss} =\left(\sum_\Ps\overline{\Fs}^{\rightarrow \Ps} \right)\cdot \vel{O}{R} + \sum_\Ps\overline{\Fs}^{\rightarrow \Ps} \cdot \left(\velang{S}{R} \times \OPvec \right) = \left( \sum_\Ps\overline{\Fs}^{\rightarrow \Ps} \right) \cdot \vel{O}{R} + \left[ \sum_\Ps \left( \OPvec \times \overline{\Fs}^{\rightarrow \Ps} \right)\right] \cdot \velang{S}{R} = \overline{\Fs}\cdot \vel{O}{R} + \overline{\Ms}(\Os)\cdot \velang{S}{R} </math>

This calculation also shows that the power of a moment (or a torque <math>\Gamma</math>) on a rigid body in a reference frame R is the product of the moment by the angular velocity of the rigid body: <math>\dot{\Ws}_\Rs^{\Gamma\rightarrow \mathrm{S}} = \overline{\Gamma} \cdot \velang{S}{R}</math>.

The power balance for a rigid body is: <math>\dot{\Ws}_\Rs^{ \rightarrow \mathrm{S}} = \dot{\Ws}_\Rs^{ \mathrm{ext}} + \dot{\Ws}_\Rs^{ \mathrm{ar}}.</math>

====✏️ EXAMPLE E1-3.1: roller sliding on a slope====
------
:
{|
|
:[[File:ExE1-3-1-1-eng.png|thumb|left|180px|link=]]
|A homogeneous roller, with mass m and radius r, slides down a slope. The friction coefficient between the ground and the roller is <math>\mu</math>. We want to calculate the power of every force acting on the roller in the ground reference frame.
|}

:It is a 2D problem. The kinematic and dynamic descriptions are:
{|
|
[[File:ExE1-3-1-2-eng.png|thumb|right|250px|link=]]

|<math> \hspace{0.5cm}\vel{J}{E}=\vel{C}{T} + \velang{}{E} \times \CJvec= </math>

<math>\hspace{1.7cm}=\left( \rightarrow \vs_\Es \right) + \left( \otimes \Omega_\Es\right) \times \left( \downarrow \rs \right)</math>

<math> \hspace{0.5cm}\vel{J}{E} = \left[\rightarrow (\vs_\Es -\rs\Omega_\Es)\right]</math>

<math>\hspace{0.5cm} \dot{\Ws}_\Es^{\mathrm{ext}} = \dot{\Ws}_\Es^{\mathrm{weight}} + \dot{\Ws}_\Es^{\mathrm{friction}} + \dot{\Ws}_\Es^{\mathrm{constraint}}= \Bigr[\left(\downarrow \ms\gs\cos\beta\right)+\left(\rightarrow \ms\gs\sin\beta\right)\Bigr] \cdot (\rightarrow \vs_\Es) + \Bigr[(\leftarrow \mu \Ns) + (\uparrow \Ns ) \Bigr] \cdot \Bigr[\rightarrow (\vs_\Es -\rs\Omega_\Es)\Bigr]</math>

|}
:Since the roller center has no motion perpendicular to the plane, the AMT leads to <math>\Ns =\ms\gs\cos\beta</math>. Therefore:

{|
|
[[File:ExE1-3-1-3-eng.png|thumb|right|140px|link=]]
|<math>\hspace{0.5cm} \dot{\Ws}_\Ts^{\mathrm{ext}} = \ms\gs\vs_\Ts \sin \beta - \ms\gs(\vs_\Ts -\rs\Omega_\Ts)\cos\beta=\ms\gs \Bigr[ \vs_\Ts (\sin\beta-\cos\beta)+\rs\Omega_\Ts \cos\beta \Bigr].</math>

<math>\hspace{0.5cm}</math>

If instead of considering the friction at '''J''' we consider the equivalent torsor at point '''C''', we must consider the force applied in '''C''' and the moment about '''C''', which is not zero:

<math>\hspace{0.5cm} \dot{\Ws}_\Es^{ \mathrm{friction}} = (\leftarrow \mu\Ns)\cdot(\rightarrow\vs_\Es)+(\otimes \mu\Ns\rs)\cdot(\otimes \Omega_\Es) =\mu\ms\gs\cos\beta(\rs\Omega_\Es-\vs_\Es)</math>,

<math>\hspace{0.5cm} \dot{\Ws}_\Es^{\mathrm{ext}} = \dot{\Ws}_\Es^{ \mathrm{weight}} + \dot{\Ws}_\Es^{ \mathrm{friction}}+ \dot{\Ws}_\Es^{ \mathrm{constraint}}= \ms\gs\vs_\Es\sin\beta+\mu\ms\gs\cos\beta (\rs\Omega_\Es-\vs_\Es)</math>.

<math>\hspace{0.5cm}</math>

The result is the same.
|}


====✏️ EXAMPLE E1-3.2: block sliding on a truck====
------
:
{|
|
:[[File:ExE1-3-2-1-eng.png|thumb|left|220px|link=]]
|A truck brakes with acceleration '''a''' relative to the ground at the instant in which its velocity (also relative to the ground) is v. At the same instant, the homogeneous block, of mass m, slides with velocity v' relative to the truck. We want to perform a power balance for the block in two different reference frames: the ground and the truck.
|}

:Ground reference frame: It is a 2D problem. The kinematic and dynamic descriptions are:
{|
|
[[File:ExE1-3-2-2-eng.png|thumb|right|200px|link=]]

| The constraint between the truck and the block allows only one translational DoF between the two. Therefore, in the 2D approximation, the associated torsor has two independent components. If characterized at the midpoint of the contact surface, these components are a normal force and a moment perpendicular to the plane. The AMT applied to the block leads to <math>\Ns=\ms\gs</math>.

|}
:The powers associated with all interactions on the block in the ground reference are:

:<math>\dot{\Ws}_\Es^{\rightarrow \mathrm{bloque}}=\dot{\Ws}_\Ts^{\mathrm{ext}}=\dot{\Ws}_\Ts^{\mathrm{peso}}+\dot{\Ws}_\Ts^{\mathrm{fricción}}+\dot{\Ws}_\Ts^{\mathrm{enlace}}= \Bigr[(\downarrow \ms\gs)+(\leftarrow \mu\ms\gs)+(\uparrow \Ns) \Bigr] \cdot \Bigr[\rightarrow(\vs+\vs')\Bigr] + (\odot \Ms_\es)\cdot 0 = -\mu\ms\gs(\vs+\vs')</math>

:Ya que el bloque se traslada respecto al suelo, su energía cinética total es inmediata, y se puede plantear fácilmente el balance de potencias:

:<math>\Ts_\Ts^\mathrm{bloque}=\frac{1}{2}\int_\mathrm{bloque}\ds\ms \vs_\Ts^2(\mathrm{bloque})=\frac{1}{2}\ms \vs_\Ts^2(\mathrm{bloc})\quad \Rightarrow \quad \dot{\Ts}_\Ts^\mathrm{bloque}=\ms\vs_\Ts( \mathrm{bloque})\dot{\vs}_\Ts( \mathrm{bloque}) </math>

:Como la única fuerza horizontal sobre el bloque es la fricción:

<math>\dot{\vs}_\Ts(\mathrm{bloque})=-\mu\gs\quad \Rightarrow \quad \dot{\Ts}_\Ts^\mathrm{bloque}=-\mu\ms\gs(\vs+\vs').</math>

:El balance de potencias es consistente:

:<math>\left.\begin{array}{l}
\dot{\mathrm{W}}_\Ts^{\rightarrow \text { bloque }}=\dot{\mathrm{W}}_{\mathrm{T}}^{\text {ext }}=-\mu \mathrm{mg}\left(\mathrm{v}+\mathrm{v}^{\prime}\right) \\
\dot{\mathrm{T}}_{\mathrm{T}}^{\text {bloque }}=-\mu \mathrm{mg}\left(\mathrm{v}+\mathrm{v}^{\prime}\right)
\end{array}\right\} \Rightarrow \dot{\mathrm{W}}_{\mathrm{T}}^{\rightarrow \text { bloque }}=\dot{\mathrm{T}}_{\mathrm{T}}^{\text {bloque }}</math>

:Referencia camión: Al estar el camión acelerado respecto al suelo, hay que tener en cuenta las fuerzas de arrastre sobre todos los diferenciales de masa (dm) del bloque:

{|
|

[[Archivo:ExE1-3-2-3-cat-esp.jpg|thumb|right|250px|link=]]
|<math>\cal{\Fs}^{\rightarrow \ds\ms}_{\mathrm{ar}}=-\ds\ms\overline{\as}_\mathrm{ar}(\ds\ms)=-\ds\ms (\leftarrow \as)=(\rightarrow \ds\ms \cdot \as)</math>.

El torsor resultante de este sistema de fuerzas en el centro de inercia '''G''' se reduce a una fuerza horizontal.

<math>\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{bloque}}=\dot{\Ws}_\mathrm{camión}^{\mathrm{ext}} + \dot{\Ws}_\mathrm{camión}^{\mathrm{ar}} =\dot{\Ws}_\mathrm{camión}^{\mathrm{peso}} + \dot{\Ws}_\mathrm{camión}^{\mathrm{fricción}} + \dot{\Ws}_\mathrm{camión}^{\mathrm{enlace}} + \dot{\Ws}_\mathrm{camión}^{\mathrm{ar}} , </math>

<math>\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{bloque}}= \Bigr[(\downarrow \ms\gs)+(\leftarrow \mu\ms\gs)+(\uparrow \Ns) + (\rightarrow \ms\as)\Bigr] \cdot(\rightarrow \vs')+(\odot\Ms_\es) \cdot0=\ms(\as-\mu\gs)\vs'.</math>
|}

:Se puede comprobar que el balance de potencias es consistente:

:<math>\Ts^\mathrm{bloque}_\mathrm{camión}=\frac{1}{2}\ms\vs^2_\mathrm{camión}(\mathrm{bloque}) \quad \Rightarrow \quad \dot{\Ts}^\mathrm{bloque}_\mathrm{camión} = \ms\vs_\mathrm{camión}(\mathrm{bloque})\dot{\vs}_\mathrm{camión}(\mathrm{bloque})</math>

:<math>\left.\begin{array}{l}
\mathrm{AB}:\mathrm{suelo} \\
\mathrm{REL}:\mathrm{camión}
\end{array}\right\} \quad \dot{\vs}_\mathrm{camión}(\mathrm{bloque})= \dot{\vs}_\mathrm{T}(\mathrm{bloque})-\dot{\vs}_\mathrm{ar}(\mathrm{bloque}) = (-\mu\gs)-(-\as)=\as-\mu\gs \quad \Rightarrow \quad \dot{\Ts}^\mathrm{bloque}_\mathrm{camión}=\ms\vs'(\as-\mu\gs)</math>

:<math>\left.\begin{array}{l}
\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{bloque}}=\dot{\Ws}_\mathrm{camión}^{\mathrm{ext}} + \dot{\Ws}_\mathrm{camión}^{\mathrm{ar}}= \ms(\as-\mu\gs)\vs'\\
\dot{\Ts}^\mathrm{bloque}_\mathrm{camión}=\ms\vs'(\as-\mu\gs)
\end{array}\right\} \Rightarrow\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{bloque}}= \dot{\Ts}^\mathrm{bloque}_\mathrm{camión}</math>


====✏️ EJEMPLO E1-3.3: camión frenando con carga====
------
:
{|
|
:[[Archivo:ExE1-3-3-1-esp.png|thumb|left|220px|link=]]
|Un camión frena con aceleración '''a''' respecto al suelo en el instante en el que su velocidad (también respecto al suelo) es v. Los pares de frenada en el eje delantero y en el trasero son iguales y valen <math>\Gamma</math>, y las ruedas, de radio r y masa despreciable, no deslizan sobre el suelo. En el mismo instante, el bloque homogéneo, de masa m, desliz con velocidad v' relativa al camión. Se trata de hacer un balance de potencias para el sistema (bloque+camión) en la referencia del suelo y en la del camión.

|}

:Ya que la potencia de las interacciones internas no es nula en principio (no se trata de un único sólido rígido), es conveniente hacer un DGI del sistema para no olvidar ninguna:
{|
|
[[Archivo:ExE1-3-3-2-esp.png|thumb|right|250px|link=]]

|
|
c.p.s.d.: contacto puntual sin deslizamiento

c.m.c.d.: contacto multipuntual con deslizamiento

|}
:El sistema sobre el que se hace el balance de potencias es todo menos el suelo. Teniendo en cuenta que la potencia de los enlaces internos es siempre nula, las únicas interacciones que hay que tener en cuenta son el peso, los enlaces de las ruedas con el suelo, los pares de frenada y la fricción entre bloque y camión:

[[Archivo:ExE1-3-3-3-esp.png|thumb|center|300px|link=]]

:Referencia suelo:
:Las fuerzas de enlace del suelo sobre las ruedas están aplicadas a puntos de velocidad cero, y por tanto no hacen potencia. El peso es una fuerza vertical, y el movimiento del camión y del bloque es horizontal, por tanto su potencia también es cero.

:<math>\dot{\Ws}_\Ts^{\rightarrow \mathrm{sist}}=\dot{\Ws}_\Ts^{\mathrm{ext}}+\dot{\Ws}^{\mathrm{int}}=\dot{\Ws}_\Ts^{\mathrm{peso}}+\dot{\Ws}_\Ts^{\mathrm{enlace}}+\dot{\Ws}^{\mathrm{fricción}}+\dot{\Ws}^{\Gamma}=\dot{\Ws}^{\mathrm{fricción}}+\dot{\Ws}^{\Gamma}.</math>

:Las potencias de interacciones internas se pueden calcular en cualquier referencia. Por tanto:

:<math>\dot{\Ws}^{\mathrm{fricción}}= \dot{\Ws}_\mathrm{camión}^{\mathrm{fric}\rightarrow \mathrm{bloque}} +\dot{\Ws}_\mathrm{camió}^{\mathrm{fric}\rightarrow \mathrm{camión}} = \dot{\Ws}_\mathrm{camión}^{\mathrm{fric}\rightarrow \mathrm{bloque}}= (\leftarrow \mu\ms\gs)\cdot (\rightarrow \vs')=-\mu\ms\gs\vs'</math>

:<math>\dot{\Ws}^{\Gamma}= \dot{\Ws}_\mathrm{camión}^{\Gamma\rightarrow \mathrm{ruedas}} +\dot{\Ws}_\mathrm{camión}^{\Gamma\rightarrow \mathrm{camión}} = \dot{\Ws}_\mathrm{camión}^{\Gamma\rightarrow \mathrm{ruedas}}= 2 \left(\odot \Gamma \right)\cdot \left(\otimes \frac{\vs}{\rs}\right)=-2\Gamma\frac{\vs}{\rs}.</math>

:Por otro lado: <math>\Ts_\Ts^\mathrm{sist}=\frac{1}{2}\ms\vs_\Ts^2(\mathrm{bloque})+\frac{1}{2}\ms\vs_\Ts^2(\mathrm{camión})\quad \Rightarrow \quad \dot{\Ts}_\Ts^\mathrm{sist}=-\ms(\vs+\vs')\mu\gs-\Ms\vs\as. </math>

:Finalmente: <math>\dot{\Ws}_\Ts^{\rightarrow \mathrm{sist}}=\dot{\Ts}_\Ts^\mathrm{sist} \quad \Rightarrow \quad -\mu\ms\gs\vs'-2\Gamma\frac{\vs}{\rs}=-\ms(\vs+\vs')\mu\gs-\Ms\vs\as.</math>

:Esta ecuación permite calcular el par de frenada: <math>\Gamma=\frac{1}{2}(\mu\ms\gs+\Ms\as)\rs.</math>

:Referencia camión:

:En esta referencia hay que tener en cuenta las fuerzas de arrastre. Por otro lado, las fuerzas de enlace del suelo sobre las ruedas están aplicadas a puntos que ya no tienen velocidad cero, y por tanto hacen potencia. Como antes, la potencia del peso es cero.

:<math>\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{sist}}=\dot{\Ws}_\mathrm{camión}^{\mathrm{ext}}+\dot{\Ws}^{\mathrm{int}}+\dot{\Ws}_\mathrm{camión}^{\mathrm{ar}}=\dot{\Ws}_\mathrm{camión}^{\mathrm{enlace}}+\dot{\Ws}^{\mathrm{fricción}}+\dot{\Ws}^{\Gamma}+\dot{\Ws}_\mathrm{camión}^{\mathrm{ar}}.</math>

:Las fuerzas de arrastre sobre el camión no hacen potencia porque el camión está parado en esta referencia. La de las fuerzas de arrastre sobre el bloque se ha calculado en el [[E1. Teorema de la energía: versión diferencial#✏️ ✏️ EJEMPLO E1-3.2: bloque deslizando sobre un camión|'''ejemplo E1-3.2''']]. La potencia de la fricción y de los pares de frenada es la misma de antes ya que no depende de la referencia.

:<math>\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{sist}}=\dot{\Ws}_\mathrm{camión}^{\mathrm{enllaç}}+\dot{\Ws}^{\mathrm{fricción}}+\dot{\Ws}^{\Gamma}+\dot{\Ws}_\mathrm{camión}^{\mathrm{ar} \rightarrow \mathrm{bloque}}= \dot{\Ws}_\mathrm{camión}^{\mathrm{enlace}} -\mu\ms\gs\vs'-2\Gamma\frac{\vs}{\rs} +\ms\as\vs'.</math>

:Las descripciones cinemática y dinámica de la rueda son:

[[Archivo:ExE1-3-3-4-esp.png|thumb|center|430px|link=]]

:<math>\dot{\Ws}_\mathrm{camión}^{\mathrm{enlace}}=2\Bigr[(\leftarrow \Ts)+(\uparrow \Ns)\Bigr]\cdot(\leftarrow\vs)=2\Ts\vs.</math>

La fuerza tangencial de enlace se puede calcular a partir del TMC aplicado al centro '''C''' de la rueda. Como no tiene masa: <math>\sum\overline{\Ms}_\mathrm{ext}(\Cs)=\overline{0} \quad \Rightarrow \quad (\odot \Gamma)+(\otimes \Ts\rs)=0\quad \Rightarrow \quad \Ts=\Gamma/\rs.</math>

:Por tanto: <math>\dot{\Ws}_\mathrm{camió}^{\rightarrow \mathrm{sist}} =2\frac{\Gamma}{\rs}\vs-\mu\ms\gs\vs'-2\Gamma\frac{\vs}{\rs}+\ms\as\vs'=\ms(\as-\mu\gs)\vs'.</math>

:Por otro lado, como se ha visto en el [[E1. Teorema de la energía: versión diferencial#✏️ ✏️ EJEMPLO E1-3.2: bloque deslizando sobre un camión|'''ejemplo E1-3.2''']]:

:<math>\Ts_\mathrm{camión}^\mathrm{sist}= \Ts_\mathrm{camión}^\mathrm{bloque}=\frac{1}{2}\ms\vs_\mathrm{camión}^2(\mathrm{bloque}) \quad \Rightarrow \quad \dot{\Ts}_\mathrm{camión}^\mathrm{sist}= \dot{\Ts}_\mathrm{camión}^\mathrm{bloque}=\ms\vs'(\as-\mu\gs).</math>

:El balance de potencias es consistente:

:<math>\left.\begin{array}{l}
\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{sist}}= \ms(\as-\mu\gs)\vs'\\
\dot{\Ts}^\mathrm{sist}_\mathrm{camión}=\ms(\as-\mu\gs)\vs'
\end{array}\right\} \Rightarrow\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{sist}}= \dot{\Ts}^\mathrm{sist}_\mathrm{camión}</math>

-----
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==E1.4 Balance de potencias para un sistema multisólido: cálculo directo y cálculo indirecto==

El '''cálculo directo''' (a partir de su definición) de la potencia asociada a una interacción (una fuerza sobre un punto '''P''' o un momento sobre un sólido rígido S) exige el conocimiento de la interacción (fuerza o momento):

<math>\dot{\Ws}_\Rs^{\Fs\rightarrow \Ps}=\overline{\Fs}_{\rightarrow\Ps} \cdot \vel{P}{R} \quad , \quad \dot{\Ws}_\Rs^{\Ms\rightarrow \Ss}=\overline{\Ms}_{\rightarrow\Ss} \cdot \velang{S}{R} </math>

Cuando no se trata de una interacción formulable, hay que recurrir a los teoremas vectoriales para determinarla.

El '''cálculo indirecto''' es una alternativa que no necesita recurrir a estos teoremas. Se basa en la evaluación de un balance de potencias para un sistema y una referencia adecuados de manera que la potencia a determinar sea el único término desconocido de la ecuación (el único que no se puede calcular de manera directa). Cuando el balance incluye más de un término desconocido, se pueden combinar varios balances (para distintos sistemas y referencias) hasta tener un número suficiente de ecuaciones que permitan la determinación de todas las potencias desconocidas.

====✏️ EJEMPLO E1-4.1: polea frenada====
------
:
{|
|
:[[Archivo:ExE1-4-1-1-esp.png|thumb|left|150px|link=]]
|Un bloque de masa m cuelga de un hilo inextensible atado a un punto de la periferia de una polea de radio r. Entre soporte (fijo al suelo) y polea actúa un freno que garantiza que el bloque baja a velocidad constante respecto al suelo. Se trata de calcular la potencia del freno.

|}

:Càlculo directo: <math>\dot{\Ws}^{\mathrm{freno}}= \overline{\Gamma}^{\rightarrow \mathrm{polea}}\cdot\velang{polea}{T}=\left(\odot\Gamma\right)\cdot\left( \otimes \frac{\vs_0}{\rs}\right)=-\Gamma \frac{\vs_0}{\rs}.</math>

:El par de frenada no es un dato, y hay que recurrir al TMC para calcularlo.

:<math>\left.\begin{array}{l}
\text{SISTEMA: polea + bloque}\\
\text{TMC en } \Os
\end{array}\right\} \sum\overline{\Ms}_\mathrm{ext}(\Os)=\dot{\overline{\Hs}}_\mathrm{RTO}(\Os)</math>

:El momento cinético proviene de las dos piezas. Ya que '''O''' no es fijo al bloque, hay que hacer descomposición baricéntrica para calcular el momento cinético del bloque:

:<math>\overline{\Hs}_\mathrm{RTO=T} (\Os) =\overline{\Hs}_\Ts^\mathrm{politja}(\Os) + \overline{\Hs}_\Ts^\mathrm{bloc}(\Os)= \overline{\Hs}_\Ts^\mathrm{politja}(\Os) + \overline{\Hs}_\Ts^\mathrm{bloc}(\Gs_\mathrm{bloc})+ \overline{\Hs}_\Ts^{\oplus\mathrm{bloc}}(\Os) = </math>

:<math>\hspace{2.3cm}=\Is_\mathrm{O}^\mathrm{polea} \velang{polea}{T}+\Is_{\Gs_\mathrm{bloque}}^\mathrm{bloque}+\OGvec_\mathrm{bloque}\times \ms\overline{\vs}_\Ts(\Gs_\mathrm{bloque})=\left(\otimes \Is_\mathrm{O}\frac{\vs_0}{\rs}\right)+(\otimes \ms\vs_0\rs).</math>

:[[Archivo:ExE1-4-1-2-esp.png|thumb|right|180px|link=]]
:Se trata de un momento cinético constante en dirección y valor. Por tanto:

:<math>\dot{\overline{\Hs}}_\mathrm{RTO}(\Os)=0 \quad \Rightarrow \quad \sum{\overline{\Ms}}_\mathrm{ext}(\Os)=(\odot\Gamma)+(\otimes \ms\gs\rs) =0\quad \Rightarrow \quad \Gamma= \ms\gs\rs</math>.

:Finalmente: <math>\dot{\Ws}^{\mathrm{freno}}= -\Gamma \frac{\vs_0}{\rs} = -\ms\gs\vs_0.</math>

:Cálculo indirecto:

:<math>\left.\begin{array}{l}
\text{SISTEMA: polea + soporte + bloque}\\
\text{REF: suelo(Gal)}
\end{array}\right\} \quad \dot{\Ts}^\mathrm{sist}_\Ts=\dot{\Ws}^\mathrm{ext}_\Ts+ \dot{\Ws}^\mathrm{int}</math>

:<math>\left.\begin{array}{l}
\dot{\Ts}^\mathrm{sist}_\Ts=0\\
\dot{\Ws}^\mathrm{ext}_\Ts=\left(\downarrow \ms\gs\right)\cdot\left(\downarrow \vs_0\right)= \ms\gs\vs_0\\
\dot{\Ws}^\mathrm{int}=\dot{\Ws}^{\mathrm{freno}}
\end{array}\right\} \quad \Rightarrow \quad \dot{\Ws}^{\mathrm{freno}} =-\ms\gs\vs_0</math>



====✏️ EJEMPLO E1-4.2: mono y plátanos====
------
:
{|
|
:[[Archivo:ExE1-4-2-1-esp.png|thumb|left|180px|link=]]
|Unos plátanos de masa <math>(2/3)\ms</math> cuelgan de una cuerda inextensible que se apoya sin deslizar sobre una polea de radio r y masa despreciable, articulada respecto al suelo. En el otro extremo de la cuerda se agarra un mono de masa m. La polea gira con velocidad angular constante <math>\Omega_0</math> respecto al suelo bajo la acción de un motor. El mono se mueve respecto a la cuerda sin que sus manos se froten contra la cuerda, y de manera que la distancia de su centro de masas al suelo se mantiene constante. Se trata de calcular la potencia del motor.
|}

:Cálculo directo: <math>\dot{\Ws}^{\mathrm{freno}}= \overline{\Gamma}^{\rightarrow \mathrm{polea}}\cdot\velang{polea}{T}=\left(\odot\Gamma\right)\cdot\left( \odot \Omega_0\right)=\Gamma \Omega_0.</math>

:El par motor no es un dato, y hay que recurrir al TMC para calcularlo.

:<math>\left.\begin{array}{l}
\text{SISTEMA: polea + mono + plátanos}\\
\text{TMC en } \Os
\end{array}\right\} \sum\overline{\Ms}_\mathrm{ext}(\Os)=\dot{\overline{\Hs}}_\mathrm{RTO}(\Os)</math>

:El momento cinético proviene del mono y de los plátanos. En primera aproximación, se puede considerar la masa del mono concentrada en su tronco y su cabeza, de manera que no gira y su velocidad respecto al suelo es nula. Teniendo en cuenta que el punto '''O''' no se mueve respecto al mono pero sí respecto a los plátanos:

:<math>\overline{\Hs}_\mathrm{RTO=T} (\Os) =\overline{\Hs}_\Ts^\mathrm{mono}(\Os) + \overline{\Hs}_\Ts^\mathrm{plátanos}(\Os)= \overline{\Hs}_\Ts^\mathrm{mono}(\Os) + \overline{\Hs}_\Ts^\mathrm{plátanos}(\Gs_\mathrm{plát})+ \OGvec_\mathrm{plát} \times \frac{2}{3} \ms \overline{\vs}_\Ts(\Gs_\mathrm{plát})= </math>

:<math>\hspace{2.3cm}=\OGvec_\mathrm{plát}\times \frac{2}{3} \ms \overline{\vs}_\Ts(\Gs_\mathrm{plát}) = \Bigr[(\rightarrow \rs)+(\downarrow \xs)\Bigr] \times\frac{2}{3} \ms \left(\downarrow \rs\Omega_0\right)=\left(\otimes \frac{2}{3}\ms\rs^2\Omega_0\right).</math>

:Se trata de un momento cinético constante en dirección y valor. Por tanto:

:<math>\dot{\overline{\Hs}}_\mathrm{RTO}(\Os)=0 \quad \Rightarrow \quad \sum{\overline{\Ms}}_\mathrm{ext}(\Os)=(\odot\Gamma)+(\otimes \ms\gs\rs) + \left(\otimes \frac{2}{3}\ms\gs\rs\right)=0\quad \Rightarrow \quad \Gamma= \frac{5}{3}\ms\gs\rs</math>.

:Finalmente: <math>\dot{\Ws}^{\mathrm{motor}}= \Gamma \Omega_0 = \frac{5}{3}\ms\gs\rs\Omega_0.</math>

:Cálculo indirecto

:Hay que tener presente que es un problema con dos tipos de actuadores: el motor y la musculatura del mono. Hay diversos modelos biomecánicos para formular las fuerzas de los músculos, pero quedan fuera del ámbito de este curso. Ya que el mono se mueve, sus músculos desarrollan una potencia no nula.

:<math>\left.\begin{array}{l}
\text{SISTEMA: polea + mono + plátano}\\
\text{REF: suelo(Gal)}
\end{array}\right\} \quad \dot{\Ts}^\mathrm{sist}_\Ts=\dot{\Ws}^\mathrm{ext}_\Ts+ \dot{\Ws}^\mathrm{int}</math>

:<math>\left.\begin{array}{l}
\dot{\Ts}^\mathrm{sist}_\Ts=0\\
\dot{\Ws}^\mathrm{ext}_\Ts=\left(\downarrow \frac{2}{3}\ms\gs\right)\cdot\left(\downarrow \rs\Omega_0\right)= \frac{2}{3}\ms\gs\rs\Omega_0\\
\dot{\Ws}^\mathrm{int}=\dot{\Ws}^{\mathrm{musc}}+\dot{\Ws}^{\mathrm{motor}}
\end{array}\right\} \quad \Rightarrow \quad \dot{\Ws}^{\mathrm{musc}} + \dot{\Ws}^{\mathrm{motor}}=-\frac{2}{3}\ms\gs\rs\Omega_0</math>

:Este balance proporciona una ecuación con dos incógnitas. Se puede plantear un segundo balance para un sistema donde una de las dos no aparezca: un sistema sin el mono o un sistema sin el motor. Si se hace un balance de potencias para el sistema (mono + trozo de cuerda), las fuerzas externas a considerar son el peso y la tensión de la cuerda. Si se sigue trabajando en la referencia del suelo, esta tensión (desconocida) hará potencia no nula ya que está aplicada a un punto de velocidad ascendente.
:[[Archivo:ExE1-4-2-2-cat-esp.png|thumb|right|330px|link=]]

:Para evitar que aparezca un término asociado a la tensión de la cuerda, se puede trabajar en la referencia del trozo de cuerda que se ha considerado, que al tener velocidad ascendente constante es galileana. En esta referencia, el mono baja con velocidad constante:

:<math>\left.\begin{array}{l}
\text{REL: trozo de cuerda}\\
\text{AB: suelo}
\end{array}\right\}</math>

:<math>\overline{\vs}_\mathrm{REL}(\mathrm{mono})=\overline{\vs}_\mathrm{ar}(\mathrm{mono})=-\overline{\vs}_\mathrm{ar}(\mathrm{mono})=(\downarrow \rs\Omega_0)</math>

:<math>\left.\begin{array}{l}
\text{SISTEMA: mono+ trozo de cuerda}\\
\text{REF: trozo de cuerda(Gal)}
\end{array}\right\} \quad \dot{\Ts}^\mathrm{sist}_\mathrm{cuerda} =\dot{\Ws}^\mathrm{ext}_\mathrm{cuerda}+ \dot{\Ws}^\mathrm{int}</math>

:<math>\left.\begin{array}{l}
\dot{\Ts}^\mathrm{sist}_\mathrm{cuerda}=0\\
\dot{\Ws}^\mathrm{ext}_\mathrm{cuerda}=\left(\downarrow \ms\gs\right)\cdot\left(\downarrow \rs\Omega_0\right)= \ms\gs\rs\Omega_0\\
\dot{\Ws}^\mathrm{int}=\dot{\Ws}^{\mathrm{musc}}
\end{array}\right\} \quad \Rightarrow \quad \dot{\Ws}^{\mathrm{musc}} =-\ms\gs\rs\Omega_0</math>

:Combinando este resultado con el del balance anterior:

:<math>\dot{\Ws}^\mathrm{motor}=-\frac{2}{3}\ms\gs\rs\Omega_0-\dot{\Ws}^\mathrm{musc}=-\frac{5}{3}\ms\gs\rs\Omega_0</math>



© Universitat Politècnica de Catalunya. [[Mecánica:Derechos de autor |Todos los derechos reservados]]
-------------

<center>
[[D8. Conservaciones|<<< D8. Conservaciones]]

[[E2. Teorema de la energía: versión integrada|E2. Teorema de la energía: versión integrada>>>]]
</center>

E1. Work-Energy Theorem: differential form

2025-05-20T12:16:29Z

Bros: /* ✏️ EXAMPLE E1-1.2: pendulum */

<div class="noautonum">__TOC__</div>
<math>\newcommand{\uvec}{\overline{\textbf{u}}}
\newcommand{\vvec}{\overline{\textbf{v}}}
\newcommand{\evec}{\overline{\textbf{e}}}
\newcommand{\Omegavec}{\overline{\mathbf{\Omega}}}
\newcommand{\velang}[2]{\Omegavec^{\textrm{#1}}_{\textrm{#2}}}
\newcommand{\Alfavec}{\overline{\mathbf{\alpha}}}
\newcommand{\accang}[2]{\Alfavec^{\textrm{#1}}_{\textrm{#2}}}
\newcommand{\ds}{\textrm{d}}
\newcommand{\ps}{\textrm{p}}
\newcommand{\ns}{\textrm{n}}
\newcommand{\hs}{\textrm{h}}
\newcommand{\cs}{\textrm{c}}
\newcommand{\gs}{\textrm{g}}
\newcommand{\Ns}{\textrm{N}}
\newcommand{\Hs}{\textrm{H}}
\newcommand{\Fs}{\textrm{F}}
\newcommand{\ms}{\textrm{m}}
\newcommand{\ts}{\textrm{t}}
\newcommand{\us}{\textrm{u}}
\newcommand{\vs}{\textrm{v}}
\newcommand{\Rs}{\textrm{R}}
\newcommand{\Es}{\textrm{E}}
\newcommand{\Ws}{\textrm{W}}
\newcommand{\Ts}{\textrm{T}}
\newcommand{\Ls}{\textrm{L}}
\newcommand{\As}{\textrm{A}}
\newcommand{\Ds}{\textrm{D}}
\newcommand{\Bs}{\textrm{B}}
\newcommand{\es}{\textrm{e}}
\newcommand{\fs}{\textrm{f}}
\newcommand{\is}{\textrm{i}}
\newcommand{\Is}{\textrm{I}}
\newcommand{\ks}{\textrm{k}}
\newcommand{\js}{\textrm{j}}
\newcommand{\rs}{\textrm{r}}
\newcommand{\as}{\textrm{a}}
\newcommand{\ss}{\textrm{s}}
\newcommand{\Os}{\textbf{O}}
\newcommand{\Gs}{\textbf{G}}
\newcommand{\Cbf}{\textbf{C}}
\newcommand{\Or}{\Os_\Rs}
\newcommand{\Qs}{\textbf{Q}}
\newcommand{\Cs}{\textbf{C}}
\newcommand{\Ps}{\textbf{P}}
\newcommand{\Ss}{\textbf{S}}
\newcommand{\Js}{\textbf{J}}
\newcommand{\P}{\textrm{P}}
\newcommand{\Q}{\textrm{Q}}
\newcommand{\Ms}{\textrm{M}}
\newcommand{\deg}{^\textsf{o}}
\newcommand{\xs}{\textsf{x}}
\newcommand{\ys}{\textsf{y}}
\newcommand{\zs}{\textsf{z}}
\newcommand{\dert}[2]{\left.\frac{\ds{#1}}{\ds\ts}\right]_{\textrm{#2}}}
\newcommand{\ddert}[2]{\left.\frac{\ds^2{#1}}{\ds\ts^2}\right]_{\textrm{#2}}}
\newcommand{\vec}[1]{\overline{#1}}
\newcommand{\vecbf}[1]{\overline{\textbf{#1}}}
\newcommand{\vecdot}[1]{\overline{\dot{#1}}}
\newcommand{\OQvec}{\vec{\Os\Qs}}
\newcommand{\GQvec}{\vec{\Gs\Qs}}
\newcommand{\GPvec}{\vec{\Gs\Ps}}
\newcommand{\PSvec}{\vec{\Ps\Ss}}
\newcommand{\QQvec}{\vec{\Qs\Qs}}
\newcommand{\QGvec}{\vec{\Qs\Gs}}
\newcommand{\QPvec}{\vec{\Qs\Ps}}
\newcommand{\JQvec}{\vec{\Js\Qs}}
\newcommand{\PQvec}{\vec{\Ps\Qs}}
\newcommand{\GJvec}{\vec{\Gs\Js}}
\newcommand{\OPvec}{\vec{\Os\textbf{P}}}
\newcommand{\OCvec}{\vec{\Os\Cs}}
\newcommand{\OGvec}{\vec{\Os\Gs}}
\newcommand{\CJvec}{\vec{\Cs\Js}}
\newcommand{\GCvec}{\vec{\Gs\Cs}}
\newcommand{\PGvec}{\vec{\Ps\Gs}}
\newcommand{\abs}[1]{\left|{#1}\right|}
\newcommand{\braq}[2]{\left\{{#1}\right\}_{\textrm{#2}}}
\newcommand{\matriz}[9]{
\begin{bmatrix}
{#1} & {#2} & {#3}\\
{#4} & {#5} & {#6}\\
{#7} & {#8} & {#9}
\end{bmatrix}}
\newcommand{\diag}[3]{
\begin{bmatrix}
{#1} & {0} & {0}\\
{0} & {#2} & {0}\\
{0} & {0} & {#3}
\end{bmatrix}}
\newcommand{\vector}[3]{
\begin{Bmatrix}
{#1}\\
{#2}\\
{#3}
\end{Bmatrix}}
\newcommand{\vecdosd}[2]{
\begin{Bmatrix}
{#1}\\
{#2}
\end{Bmatrix}}
\newcommand{\vel}[2]{\vvec_{\textrm{#2}} (\textbf{#1})}
\newcommand{\acc}[2]{\vecbf{a}_{\textrm{#2}} (\textbf{#1})}
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\newcommand{\psio}{\dot{\psi}_0}
\newcommand{\Pll}{\textbf{P}_\textrm{lliure}}
\newcommand{\agal}[1]{\vecbf{a}_{\textrm{Gal}} (#1)}
\newcommand{\angal}[1]{\vecbf{a}_{\textrm{NGal}} (#1)}
\newcommand{\vgal}[1]{\vecbf{v}_{\textrm{Gal}} (#1)}
\newcommand{\fvec}[2]{\overline{\mathbf{F}}_{\textrm{#1}\rightarrow\textrm{#2}}}
\newcommand{\mvec}[2]{\overline{\mathbf{M}}_{\textrm{#1}\rightarrow\textrm{#2}}}
\newcommand{\F}[1]{\vec{\Fs}_{#1}}
\newcommand{\Fcal}[2]{\vec{\cal{F}}^{\text{#1}}_{#2}}
</math>

The '''Work-Energy Theorem''' plays a prominent role in physics, as it bridges all branches of science (electromagnetism, thermodynamics, etc.): all of them deal with energy.

There is a certain parallel between the Vector Theorems and the '''Work-Energy Theorem''': they all allow us to understand the evolution of a magnitude that depends on the mechanical state of the system. In the case of the former, we are dealing with the instantaneous evolution of vector magnitudes (linear momentum, angular momentum), while the latter deals with the finite change (in a finite time interval) of a scalar magnitude ('''mechanical energy''').

This unit presents the differential form of this theorem (the '''power balance'''). The integrated form (which is the Work-Energy Theorem itself) is presented in [[E2. Teorema de la energía: versión integrada|'''unit E2''']].
 
------
------
==E1.1 Power balance in a system of particles==

:[[File:E1-1-eng.png|thumb|right|300px|link=]]
<center>'''Figure E1.1''' Forces on a particle of a system with constant matter</center>

Let us consider a system of constant matter, and let’s apply Newton's second law to each particle (or each mass differential) of this system in a general reference frame R (which may be non-Galilean) ('''Figure E1.1'''):

<math>\F{\rightarrow \Ps} + \Fcal{}{\text{ar}\rightarrow\Ps} + \Fcal{}{\text{Cor}\rightarrow\Ps} =\ms_\Ps\acc{P}{R}</math>

<math>\F{\rightarrow \Ps} </math> is the resultant interaction force on '''P''', and it comes from internal interactions (with particles of the same system) and external ones:

<math>\F{\rightarrow \Ps}=\F{\rightarrow \Ps}^\text{int}+\F{\rightarrow \Ps}^\text{ext}</math>

If both sides of the equation are multiplied by <math>\vel{P}{R}</math>, all forces (interaction and inertia forces) orthogonal to <math>\vel{P}{R}</math> are filtered out. Specifically, since <math>\acc{P}{R}=\accs{P}{R} + \accn{P}{R} </math>, where <math>\accn{P}{R} \perp \vel{P}{R}</math>, and since the Coriolis force is always orthogonal to <math>\vel{P}{R} (\Fcal{}{\text{Cor}\rightarrow\Ps} = -2\ms_\Ps \velang{R}{Gal} \times \vel{P}{R})</math>:

<math>\vel{P}{R} \cdot \left( \F{\rightarrow \Ps}^\text{int}+\F{\rightarrow \Ps}^\text{ext} + \Fcal{}{\text{ar}\rightarrow\Ps} \right) = \ms_\Ps\accs{P}{R} \cdot \vel{P}{R} = \frac{\ds}{\ds \ts} \left(\frac{1}{2}\ms_\Ps \vvec_\Rs^2(\Ps) \right) \equiv \frac{\ds\Ts_\Rs(\Ps)}{\ds\ts}=\dot{\Ts}_\Rs(\Ps)</math>

The scalar function <math>\Ts_\Rs(\Ps)</math> is the '''kinetic energy''' of the particle, and is a function of the [[D1. Foundational laws of Newtonian dynamics#D1.3 Newton’s Principle of Determinacy|'''mechanical state''']] of '''P'''. The scalar product of a force and the velocity of its application point is the '''force power''': <math>\vel{P}{R} \cdot \F{\rightarrow \Ps} \equiv \dot{\Ws}_\Rs^{\F{\rightarrow \Ps}} </math>. Both scalar quantities <math>(\Ts_\Rs(\Ps), \dot{\Ws}_\Rs^{\F{\rightarrow \Ps}})</math> depend on the reference frame. For the same set of interaction forces acting on '''P''', the reference frame determines which are filtered out and which are not.

The equation can be rewritten as <math>\dot{\Ws}_\Rs^{\text{ext}\rightarrow \Ps}+\dot{\Ws}_\Rs^{\text{int}\rightarrow \Ps}+\dot{\Ws}_\Rs^{\text{ar}\rightarrow \Ps} =\dot{\Ts}_\Rs(\Ps)</math>, and constitutes the '''power balance''' for particle '''P''' in the R reference frame. Summing for all particles in the system:

<math>\dot{\Ws}_\Rs^{\text{ext}}+\dot{\Ws}^{\text{int}}+\dot{\Ws}_\Rs^{\text{ar}} =\dot{\Ts}_\Rs^\text{syst}</math>

The calculation of kinetic energy is covered in [[E2. Teorema de la energía: versión integrada|'''unit E2''']]. This unit focuses on the calculation of powers. [[E1. Work-Energy Theorem: differential form#E1.2 Potencia de una pareja de acción-reacción|'''section E1-2''']] shows that the power of internal interactions does not depend on the reference frame (which is why the subscript R does not appear in the <math>\dot{\Ws}^{\text{int}}</math> term of the previous equation).

====✏️ EXAMPLE E1-1.1: free particle====
------
:
{|
:[[File:ExE1-1-1-1-eng.jpg|thumb|left|180px|link=]]
|A particle '''P''', with mass m, is released from the ring of a Ferris wheel when it is at its highest point. We want to calculate the power, at this instant, of all forces acting on the particle from three different reference frames: the ground, the ring, and the Ferris wheel cabin.
|}

:Ground reference frame: Since this is a Galilean reference frame, all forces on '''P''' are interaction forces.
{|
:[[File:ExE1-1-2- eng.png|thumb|right|230px|link=]]

|<math>\hspace{0.5cm}\sum\dot{\Ws}_\Ts=\dot{\Ws}_\Ts^\text{weight}= \left(\leftarrow \Rs\Omega_0\right) \cdot \left(\downarrow \ms\gs\right)=0</math>
|}

:Ring reference frame: It is a non-Galilean reference, and the transportation force on '''P''' must be taken into account.
{|
:[[File:ExE1-1-3-eng.png|thumb|right|250px|link=]]
|<math>\hspace{0.5cm}</math> AB:ground, REL:ring 

<math>\hspace{0.5cm}\vel{P}{REL}=\vel{P}{AB}-\vel{P}{ar}= \left(\leftarrow \Rs\Omega_0\right)-\left(\leftarrow \Rs\Omega_0\right)=\overline{0}</math> 

<math>\hspace{0.5cm}</math>

Since the velocity is zero, the total power is zero.
|}

:Cabin reference frame: It is also a non-Galilean reference, and the transportation force on '''P''' must be taken into account.

{|
:[[File:ExE1-1-4-eng.png|thumb|right|250px|link=]]
|<math>\hspace{0.5cm}</math>AB:ground, REL:cabin

<math>\hspace{0.5cm}\vel{P}{REL}=\vel{P}{AB}-\vel{P}{ar}= \left(\leftarrow \Rs\Omega_0\right)-\left(\uparrow \Rs\Omega_0\right)</math>

<math>\hspace{0.5cm}\dot{\Ws}_\text{cabin}^\text{weight}=\left[ \left(\leftarrow \Rs\Omega_0\right)+\left(\downarrow \Rs\Omega_0\right) \right] \cdot \left(\downarrow\ms\gs\right) =\ms\gs\Rs\Omega_0</math>

<math>\hspace{0.5cm}\dot{\Ws}_\text{cabin}^\text{ar}=\left[ \left(\leftarrow \Rs\Omega_0\right)+\left(\downarrow \Rs\Omega_0\right) \right] \cdot \left(\rightarrow\ms\Rs^2\Omega_0^2\right) =-\ms\gs\Rs^2 \Omega_0^3</math>

<math>\hspace{0.5cm}\sum\dot{\Ws}_\text{cabin}=\ms\Rs\Omega_0(\gs-\Rs\Omega_0^2)</math>

|}


====✏️ EXAMPLE E1-1.2: pendulum====
------
:
{|
:[[File:ExE1-2-1-eng.png|thumb|left|180px|link=]]
|A particle '''P''' of mass m hangs from an inextensible thread of length L attached to the ceiling. The assembly oscillates in a vertical plane fixed to the ground. We want to calculate the power of all the forces acting on the particle in two different reference frames: the ground and a platform rotating relative to the ground with a constant angular velocity <math>\Omega_0</math>. 

Ground reference frame: Since it is a Galilean reference frame, '''P''' is only subjected to interaction forces.
{|
:[[File:ExE1-2-2-neut.png|thumb|left|300px|link=]]

|<math>\dot{\Ws}_\Es^\Ts=(\nearrow\Ls\dot{\theta})\cdot(\nwarrow\Ts)=0</math>

<math>\dot{\Ws}_\Es^\text{weight}=(\nearrow\Ls\dot{\theta}\cdot(\downarrow \ms\gs)=-\ms\gs\Ls\dot{\theta}\sin\theta</math>
|}

<math>\hspace{2cm}</math>

Platform reference frame: It is a non Galilean reference frame, and the transportation force on '''P''' has to be taken into account.

{|
:[[File:ExE1-2-3-eng.png|thumb|left|400px|link=]]

|AB:ground, REL:platform

<math>\vel{P}{REL}=\vel{P}{AB}-\vel{P}{ar}=(\nearrow\Ls\dot{\theta})-(\otimes \Ls \Omega_0 \sin \theta)</math>

<math>\dot{\Ws}_\Es^\Ts\left[(\nearrow\Ls\dot{\theta})+(\odot\Ls\Omega_0\sin\theta)\right]\cdot(\nwarrow\Ts)=0</math>

<math>\dot{\Ws}_\Es^\text{weight}\left[(\nearrow\Ls\dot{\theta})+(\odot\Ls\Omega_0\sin\theta)\right]\cdot(\downarrow \ms\gs)=-\ms\gs\Ls\dot{\theta}\sin\theta</math>

<math>\dot{\Ws}_\Es^\text{ar}\left[(\nearrow\Ls\dot{\theta})+(\odot\Ls\Omega_0\sin\theta)\right]\cdot(\rightarrow \ms\Ls\Omega_0^2\sin\theta)=\ms\Ls^2\Omega_0^2\dot{\theta}\cos\theta</math>
|}



------
------

==E1.2 Power of an action-reaction pair==

The total power associated with internal interactions in a constant-mass system is not zero in principle (since the two forces of each action-reaction pair are applied to particles that may have different velocities relative to the reference frame in which the calculation is made), and it is independent of the reference frame in which it is evaluated. It only depends on the separation velocity of the two particles. The power balance for the system is therefore written as: <math>\dot{\Ts}_\Rs^\text{syst}=\dot{\Ws}_\Rs^\text{ext} + \dot{\Ws}^\text{int} + \dot{\Ws}_\Rs^\text{ar} </math> .

<div>
=====💭 PROOF ➕=====

{|
:[[File:E1-2-neut.png|thumb|left|200px|link=]]
|<math>\dot{\Ws}^\text{ac-reac}=\overline{\Fs}_{\Ps \rightarrow \Qs}\cdot \vel{Q}{R} + \overline{\Fs}_{\Qs \rightarrow \Ps}\cdot \vel{P}{R}= \overline{\Fs}_{\Ps \rightarrow \Qs}\cdot \left[ \vel{Q}{R}-\vel{P}{R}\right]</math>

By the [[D1. Foundational laws of Newtonian dynamics#D1.6 Newton’s third law (action-reaction principle)|'''action-reaction principle''']], <math>\overline{\Fs}_{\Ps \rightarrow \Qs}=-\overline{\Fs}_{\Qs \rightarrow \Ps}</math>. Therefore: <math>\dot{\Ws}^\text{ac-reac}=\overline{\Fs}_{\Ps \rightarrow \Qs}\cdot \left[ \vel{Q}{R}-\vel{P}{R}\right]</math>

As <math>\overline{\Fs}_{\Ps \rightarrow \Qs}</math> has the <math>\PQvec</math> direction, only the components of the velocities in that direction have to be considered:

<math>\dot{\Ws}^\text{ac-reac}=\overline{\Fs}_{\Ps \rightarrow \Qs}\cdot \biggl( \vel{Q}{R} \Bigr]_{||\Ps\Qs}-\ \vel{P}{R} \Bigr]_{||\Ps\Qs}\biggl)</math>
|}

The difference <math>\biggl( \vel{Q}{R} \Bigr]_{\Ps\Qs}-\ \vel{P}{R} \Bigr]_{\Ps\Qs}\biggl)</math> is the separation velocity <math>\dot{\rho}</math> between '''P''' and '''Q''', and it does not dpend on the reference frame since it is the derivative of a scalar (the distance <math>\dot{\rho}</math> between the two particles). For the previous drawing, <math>\dot{\rho}> 0</math>. Since this is an attractive interaction, <math>\dot{\Ws}^\text{ac-reac} < 0</math>.

With a drawing, it is easy to see that:
* attractive interaction and particle approach <math>(\dot{\rho}<0)</math> : <math>\dot{\Ws}^\text{ac-reac} > 0</math> ;
* repulsive interaction and particle separation <math>(\dot{\rho}>0)</math> : <math>\dot{\Ws}^\text{ac-reac} > 0</math>
* repulsive interaction and particle approach <math>(\dot{\rho}<0)</math> : <math>\dot{\Ws}^\text{ac-reac} < 0</math>


</div>

There are two particular cases of interaction forces whose power always has the same sign:

* Friction action-reaction pair between '''P''' and '''Q''': since the friction force always opposes the relative velocity between the two points, its power is always negative:

<math>\hspace{1cm} \dot{\Ws}^\text{friction}=\dot{\Ws}^{\text{fric}\rightarrow\Ps}_\text{RTQ}+\dot{\Ws}^{\text{fric}\rightarrow\Qs}_\text{RTQ}=\overline{\Fs}^{\text{fric}\rightarrow\Ps} \cdot \vel{P}{RTQ} = - \abs{\overline{\Fs}^{\text{fric}\rightarrow\Ps}} \frac{\vel{P}{RTQ}}{\abs{\vel{P}{RTQ}}} \cdot \vel{P}{RTQ}<0;</math>

* Constraint action-reaction pair between '''P''' and '''Q''': from the [[D3. Interactions between rigid bodies#D3.4 Direct constraint interactions|'''analytical characterization equation''']] of the constraint forces between two points '''P''' and '''Q''' belonging to two rigid bodies <math>\mathrm{S}_\Ps</math> and <math>\mathrm{S}_\Qs</math>, it is easy to see that the associated power is zero:

<math>\hspace{1cm} \dot{\Ws}^\text{constraint}=\dot{\Ws}^{\text{constraint}\rightarrow\Ps}_{\mathrm{S}_\Ps}+\dot{\Ws}^{\text{constraint}\rightarrow\Qs}_{\mathrm{S}_\Ps}=\overline{\Fs}^{\text{constraint}\rightarrow\Ps} \cdot \overline{\mathbf{v}}_{\mathrm{S}_\Qs}(\Ps) = - 0;</math>

====✏️ EXAMPLE E1-2.1: power of an action-reaction friction pair====
------
:
{|
:[[File:ExE1-2-1-2-eng.png|thumb|left|180px|link=]]
|A homogeneous wheel, with radius r and mass m, moves down an inclined plane without sliding until it joins a conveyor belt moving at a constant velocity <math>\vs_0</math> relative to the ground. At the moment it makes contact with the belt, the center of the wheel has a velocity <math>\vs_0</math> relative to the ground in the same direction as the belt. We want to calculate the power of the friction force between the belt and the wheel at this instant.
|}

At that momento, the <math>\mathbf{ICR}^\text{wheel}_\Es</math> is point <math>\Js_\text{wheel}</math> in contact with the belt. This point slides backwards with respect to the belt, therefore the friction force on the wheel is forwards:

<math>\left.\begin{array}{l}
\text { AB: ground } \\
\text { REL: belt }
\end{array}\right\} \quad \bar{\vs}_{\text {REL }}\left(\Js_{\text {wheel }}\right)=\bar{\vs}_{\text{AB}}\left(\Js_{\text {wheel }}\right)-\bar{\vs}_{\mathrm{ar}}\left(\Js_{\text {wheel }}\right)=-\bar{\vs}_{\mathrm{ar}}\left(\Js_{\text {wheel }}\right)=-\left(\rightarrow \mathrm{v}_0\right)=\left(\leftarrow \mathrm{v}_0\right) .</math>

[[File:ExE1-2-1-3-eng.png|thumb|center|500px|link=]]

Calculation of the friction power: <math>\dot{\Ws}^\text{friction}=\dot{\Ws}_\Rs^{\text{friction}\rightarrow \text{wheel}}+ \dot{\Ws}_\Rs^{\text{friction}\rightarrow \text{belt}}</math>, where R may be any reference frame. To quickly calculate this power, it is advisable to choose a reference frame R where one of the two points subjected to the action-reaction pair has zero speed, so that one of the two terms is directly zero:

<math>\dot{\Ws}^\text{friction}=\dot{\Ws}_\text{belt}^{\text{fric}\rightarrow \text{wheel}}+ \dot{\Ws}_\text{belt}^{\text{fric }\rightarrow \text{belt}}=\overline{\Fs}^{\text{fric}\rightarrow \text{wheel}} \cdot \bar{\vs}_{\text {belt}}\left(\Js_{\text {wheel }}\right)=(\rightarrow \mu\ms\gs)\cdot(\leftarrow \vs_0)=-\mu\ms\gs\vs_0.</math>

If the calculation is done in any other reference frame, the result is the same:

<math>\dot{\Ws}^\text{friction}=\dot{\Ws}_\text{E}^{\text{fric}\rightarrow \text{wheel}}+ \dot{\Ws}_\text{E}^{\text{fric} \rightarrow \text{belt}}=(\rightarrow \mu\ms\gs)\cdot 0 + (\leftarrow \mu\ms\gs)\cdot(\rightarrow \vs_0)=-\mu\ms\gs\vs_0.</math>

<math>\dot{\Ws}^\text{friction}=\dot{\Ws}_\text{vehicle}^{\text{fric}\rightarrow \text{wheel}}+ \dot{\Ws}_\text{vehicle}^{\text{fric} \rightarrow \text{belt}}=(\rightarrow \mu\ms\gs)\cdot (\rightarrow\vs_0) + (\leftarrow \mu\ms\gs) \cdot(\rightarrow 3\vs_0)=-\mu\ms\gs\vs_0.</math>

However, if only the power of one of the two forces is calculated (for example, the friction on the wheel), the result depends on the reference frame and can have any sign:

<math>\dot{\Ws}_\text{belt}^{\text{fric }\rightarrow \text{wheel}}=(\rightarrow \mu\ms\gs)\cdot(\leftarrow \vs_0)=-\mu\ms\gs\vs_0 ,</math>

<math>\dot{\Ws}_\text{E}^{\text{fric} \rightarrow \text{wheel}}=(\rightarrow \mu\ms\gs)\cdot 0 = 0 ,</math>

<math>\dot{\Ws}_\text{vehicle}^{\text{fric} \rightarrow \text{wheel}}=(\rightarrow \mu\ms\gs)\cdot (\rightarrow\vs_0)=\mu\ms\gs\vs_0.</math>


-----
-----

==E1.3 Power of a system of forces on a rigid body==

{|
|
:[[File:E1-3-eng.png|thumb|left|180px|link=]]
<center>'''Figure E1.3''': Kinematic torsor of a rigid body in a reference frame R, and forces on different points (dm) of that rigid body</center>

|When calculating the power of a system of forces on a rigid body S, the term associated with internal interactions is zero because those interactions are all constraint (cohesion) forces and, as seen in [[E1. Work-Energy Theorem: differential form#E1.2 Power of an action-reaction pair|'''section E1.2''']], their total power is zero.

The total power of a system of forces on S can be calculated from the torsor of the system at any point '''O''' of the rigid body, <math>(\overline{\Fs},\overline{\Ms}(\Os))</math>. Taking into account rigid body kinematics ('''Figure E1.3'''):

<math>\dot{\Ws}_\Rs^{\rightarrow \Ss} =\sum_\Ps\overline{\Fs}^{\rightarrow \Ps} \cdot \vel{P}{R}= \sum_\Ps \left[ \overline{\Fs}^{\rightarrow \Ps} \cdot \vel{O}{R} + \velang{S}{R} \times \OPvec \right] </math>

|}
<math>\dot{\Ws}_\Rs^{\rightarrow \Ss} =\left(\sum_\Ps\overline{\Fs}^{\rightarrow \Ps} \right)\cdot \vel{O}{R} + \sum_\Ps\overline{\Fs}^{\rightarrow \Ps} \cdot \left(\velang{S}{R} \times \OPvec \right) = \left( \sum_\Ps\overline{\Fs}^{\rightarrow \Ps} \right) \cdot \vel{O}{R} + \left[ \sum_\Ps \left( \OPvec \times \overline{\Fs}^{\rightarrow \Ps} \right)\right] \cdot \velang{S}{R} = \overline{\Fs}\cdot \vel{O}{R} + \overline{\Ms}(\Os)\cdot \velang{S}{R} </math>

This calculation also shows that the power of a moment (or a torque <math>\Gamma</math>) on a rigid body in a reference frame R is the product of the moment by the angular velocity of the rigid body: <math>\dot{\Ws}_\Rs^{\Gamma\rightarrow \mathrm{S}} = \overline{\Gamma} \cdot \velang{S}{R}</math>.

The power balance for a rigid body is: <math>\dot{\Ws}_\Rs^{ \rightarrow \mathrm{S}} = \dot{\Ws}_\Rs^{ \mathrm{ext}} + \dot{\Ws}_\Rs^{ \mathrm{ar}}.</math>

====✏️ EXAMPLE E1-3.1: roller sliding on a slope====
------
:
{|
|
:[[File:ExE1-3-1-1-eng.png|thumb|left|180px|link=]]
|A homogeneous roller, with mass m and radius r, slides down a slope. The friction coefficient between the ground and the roller is <math>\mu</math>. We want to calculate the power of every force acting on the roller in the ground reference frame.
|}

:It is a 2D problem. The kinematic and dynamic descriptions are:
{|
|
[[File:ExE1-3-1-2-eng.png|thumb|right|250px|link=]]

|<math> \hspace{0.5cm}\vel{J}{E}=\vel{C}{T} + \velang{}{E} \times \CJvec= </math>

<math>\hspace{1.7cm}=\left( \rightarrow \vs_\Es \right) + \left( \otimes \Omega_\Es\right) \times \left( \downarrow \rs \right)</math>

<math> \hspace{0.5cm}\vel{J}{E} = \left[\rightarrow (\vs_\Es -\rs\Omega_\Es)\right]</math>

<math>\hspace{0.5cm} \dot{\Ws}_\Es^{\mathrm{ext}} = \dot{\Ws}_\Es^{\mathrm{weight}} + \dot{\Ws}_\Es^{\mathrm{friction}} + \dot{\Ws}_\Es^{\mathrm{constraint}}= \Bigr[\left(\downarrow \ms\gs\cos\beta\right)+\left(\rightarrow \ms\gs\sin\beta\right)\Bigr] \cdot (\rightarrow \vs_\Es) + \Bigr[(\leftarrow \mu \Ns) + (\uparrow \Ns ) \Bigr] \cdot \Bigr[\rightarrow (\vs_\Es -\rs\Omega_\Es)\Bigr]</math>

|}
:Since the roller center has no motion perpendicular to the plane, the AMT leads to <math>\Ns =\ms\gs\cos\beta</math>. Therefore:

{|
|
[[File:ExE1-3-1-3-eng.png|thumb|right|140px|link=]]
|<math>\hspace{0.5cm} \dot{\Ws}_\Ts^{\mathrm{ext}} = \ms\gs\vs_\Ts \sin \beta - \ms\gs(\vs_\Ts -\rs\Omega_\Ts)\cos\beta=\ms\gs \Bigr[ \vs_\Ts (\sin\beta-\cos\beta)+\rs\Omega_\Ts \cos\beta \Bigr].</math>

<math>\hspace{0.5cm}</math>

If instead of considering the friction at '''J''' we consider the equivalent torsor at point '''C''', we must consider the force applied in '''C''' and the moment about '''C''', which is not zero:

<math>\hspace{0.5cm} \dot{\Ws}_\Es^{ \mathrm{friction}} = (\leftarrow \mu\Ns)\cdot(\rightarrow\vs_\Es)+(\otimes \mu\Ns\rs)\cdot(\otimes \Omega_\Es) =\mu\ms\gs\cos\beta(\rs\Omega_\Es-\vs_\Es)</math>,

<math>\hspace{0.5cm} \dot{\Ws}_\Es^{\mathrm{ext}} = \dot{\Ws}_\Es^{ \mathrm{weight}} + \dot{\Ws}_\Es^{ \mathrm{friction}}+ \dot{\Ws}_\Es^{ \mathrm{constraint}}= \ms\gs\vs_\Es\sin\beta+\mu\ms\gs\cos\beta (\rs\Omega_\Es-\vs_\Es)</math>.

<math>\hspace{0.5cm}</math>

The result is the same.
|}


====✏️ EXAMPLE E1-3.2: block sliding on a truck====
------
:
{|
|
:[[File:ExE1-3-2-1-eng.png|thumb|left|220px|link=]]
|A truck brakes with acceleration '''a''' relative to the ground at the instant in which its velocity (also relative to the ground) is v. At the same instant, the homogeneous block, of mass m, slides with velocity v' relative to the truck. We want to perform a power balance for the block in two different reference frames: the ground and the truck.
|}

:Ground reference frame: It is a 2D problem. The kinematic and dynamic descriptions are:
{|
|
[[File:ExE1-3-2-2-eng.png|thumb|right|200px|link=]]

| The constraint between the truck and the block allows only one translational DoF between the two. Therefore, in the 2D approximation, the associated torsor has two independent components. If characterized at the midpoint of the contact surface, these components are a normal force and a moment perpendicular to the plane. The AMT applied to the block leads to <math>\Ns=\ms\gs</math>.

|}
:The powers associated with all interactions on the block in the ground reference are:

:<math>\dot{\Ws}_\Es^{\rightarrow \mathrm{bloque}}=\dot{\Ws}_\Ts^{\mathrm{ext}}=\dot{\Ws}_\Ts^{\mathrm{peso}}+\dot{\Ws}_\Ts^{\mathrm{fricción}}+\dot{\Ws}_\Ts^{\mathrm{enlace}}= \Bigr[(\downarrow \ms\gs)+(\leftarrow \mu\ms\gs)+(\uparrow \Ns) \Bigr] \cdot \Bigr[\rightarrow(\vs+\vs')\Bigr] + (\odot \Ms_\es)\cdot 0 = -\mu\ms\gs(\vs+\vs')</math>

:Ya que el bloque se traslada respecto al suelo, su energía cinética total es inmediata, y se puede plantear fácilmente el balance de potencias:

:<math>\Ts_\Ts^\mathrm{bloque}=\frac{1}{2}\int_\mathrm{bloque}\ds\ms \vs_\Ts^2(\mathrm{bloque})=\frac{1}{2}\ms \vs_\Ts^2(\mathrm{bloc})\quad \Rightarrow \quad \dot{\Ts}_\Ts^\mathrm{bloque}=\ms\vs_\Ts( \mathrm{bloque})\dot{\vs}_\Ts( \mathrm{bloque}) </math>

:Como la única fuerza horizontal sobre el bloque es la fricción:

<math>\dot{\vs}_\Ts(\mathrm{bloque})=-\mu\gs\quad \Rightarrow \quad \dot{\Ts}_\Ts^\mathrm{bloque}=-\mu\ms\gs(\vs+\vs').</math>

:El balance de potencias es consistente:

:<math>\left.\begin{array}{l}
\dot{\mathrm{W}}_\Ts^{\rightarrow \text { bloque }}=\dot{\mathrm{W}}_{\mathrm{T}}^{\text {ext }}=-\mu \mathrm{mg}\left(\mathrm{v}+\mathrm{v}^{\prime}\right) \\
\dot{\mathrm{T}}_{\mathrm{T}}^{\text {bloque }}=-\mu \mathrm{mg}\left(\mathrm{v}+\mathrm{v}^{\prime}\right)
\end{array}\right\} \Rightarrow \dot{\mathrm{W}}_{\mathrm{T}}^{\rightarrow \text { bloque }}=\dot{\mathrm{T}}_{\mathrm{T}}^{\text {bloque }}</math>

:Referencia camión: Al estar el camión acelerado respecto al suelo, hay que tener en cuenta las fuerzas de arrastre sobre todos los diferenciales de masa (dm) del bloque:

{|
|

[[Archivo:ExE1-3-2-3-cat-esp.jpg|thumb|right|250px|link=]]
|<math>\cal{\Fs}^{\rightarrow \ds\ms}_{\mathrm{ar}}=-\ds\ms\overline{\as}_\mathrm{ar}(\ds\ms)=-\ds\ms (\leftarrow \as)=(\rightarrow \ds\ms \cdot \as)</math>.

El torsor resultante de este sistema de fuerzas en el centro de inercia '''G''' se reduce a una fuerza horizontal.

<math>\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{bloque}}=\dot{\Ws}_\mathrm{camión}^{\mathrm{ext}} + \dot{\Ws}_\mathrm{camión}^{\mathrm{ar}} =\dot{\Ws}_\mathrm{camión}^{\mathrm{peso}} + \dot{\Ws}_\mathrm{camión}^{\mathrm{fricción}} + \dot{\Ws}_\mathrm{camión}^{\mathrm{enlace}} + \dot{\Ws}_\mathrm{camión}^{\mathrm{ar}} , </math>

<math>\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{bloque}}= \Bigr[(\downarrow \ms\gs)+(\leftarrow \mu\ms\gs)+(\uparrow \Ns) + (\rightarrow \ms\as)\Bigr] \cdot(\rightarrow \vs')+(\odot\Ms_\es) \cdot0=\ms(\as-\mu\gs)\vs'.</math>
|}

:Se puede comprobar que el balance de potencias es consistente:

:<math>\Ts^\mathrm{bloque}_\mathrm{camión}=\frac{1}{2}\ms\vs^2_\mathrm{camión}(\mathrm{bloque}) \quad \Rightarrow \quad \dot{\Ts}^\mathrm{bloque}_\mathrm{camión} = \ms\vs_\mathrm{camión}(\mathrm{bloque})\dot{\vs}_\mathrm{camión}(\mathrm{bloque})</math>

:<math>\left.\begin{array}{l}
\mathrm{AB}:\mathrm{suelo} \\
\mathrm{REL}:\mathrm{camión}
\end{array}\right\} \quad \dot{\vs}_\mathrm{camión}(\mathrm{bloque})= \dot{\vs}_\mathrm{T}(\mathrm{bloque})-\dot{\vs}_\mathrm{ar}(\mathrm{bloque}) = (-\mu\gs)-(-\as)=\as-\mu\gs \quad \Rightarrow \quad \dot{\Ts}^\mathrm{bloque}_\mathrm{camión}=\ms\vs'(\as-\mu\gs)</math>

:<math>\left.\begin{array}{l}
\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{bloque}}=\dot{\Ws}_\mathrm{camión}^{\mathrm{ext}} + \dot{\Ws}_\mathrm{camión}^{\mathrm{ar}}= \ms(\as-\mu\gs)\vs'\\
\dot{\Ts}^\mathrm{bloque}_\mathrm{camión}=\ms\vs'(\as-\mu\gs)
\end{array}\right\} \Rightarrow\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{bloque}}= \dot{\Ts}^\mathrm{bloque}_\mathrm{camión}</math>


====✏️ EJEMPLO E1-3.3: camión frenando con carga====
------
:
{|
|
:[[Archivo:ExE1-3-3-1-esp.png|thumb|left|220px|link=]]
|Un camión frena con aceleración '''a''' respecto al suelo en el instante en el que su velocidad (también respecto al suelo) es v. Los pares de frenada en el eje delantero y en el trasero son iguales y valen <math>\Gamma</math>, y las ruedas, de radio r y masa despreciable, no deslizan sobre el suelo. En el mismo instante, el bloque homogéneo, de masa m, desliz con velocidad v' relativa al camión. Se trata de hacer un balance de potencias para el sistema (bloque+camión) en la referencia del suelo y en la del camión.

|}

:Ya que la potencia de las interacciones internas no es nula en principio (no se trata de un único sólido rígido), es conveniente hacer un DGI del sistema para no olvidar ninguna:
{|
|
[[Archivo:ExE1-3-3-2-esp.png|thumb|right|250px|link=]]

|
|
c.p.s.d.: contacto puntual sin deslizamiento

c.m.c.d.: contacto multipuntual con deslizamiento

|}
:El sistema sobre el que se hace el balance de potencias es todo menos el suelo. Teniendo en cuenta que la potencia de los enlaces internos es siempre nula, las únicas interacciones que hay que tener en cuenta son el peso, los enlaces de las ruedas con el suelo, los pares de frenada y la fricción entre bloque y camión:

[[Archivo:ExE1-3-3-3-esp.png|thumb|center|300px|link=]]

:Referencia suelo:
:Las fuerzas de enlace del suelo sobre las ruedas están aplicadas a puntos de velocidad cero, y por tanto no hacen potencia. El peso es una fuerza vertical, y el movimiento del camión y del bloque es horizontal, por tanto su potencia también es cero.

:<math>\dot{\Ws}_\Ts^{\rightarrow \mathrm{sist}}=\dot{\Ws}_\Ts^{\mathrm{ext}}+\dot{\Ws}^{\mathrm{int}}=\dot{\Ws}_\Ts^{\mathrm{peso}}+\dot{\Ws}_\Ts^{\mathrm{enlace}}+\dot{\Ws}^{\mathrm{fricción}}+\dot{\Ws}^{\Gamma}=\dot{\Ws}^{\mathrm{fricción}}+\dot{\Ws}^{\Gamma}.</math>

:Las potencias de interacciones internas se pueden calcular en cualquier referencia. Por tanto:

:<math>\dot{\Ws}^{\mathrm{fricción}}= \dot{\Ws}_\mathrm{camión}^{\mathrm{fric}\rightarrow \mathrm{bloque}} +\dot{\Ws}_\mathrm{camió}^{\mathrm{fric}\rightarrow \mathrm{camión}} = \dot{\Ws}_\mathrm{camión}^{\mathrm{fric}\rightarrow \mathrm{bloque}}= (\leftarrow \mu\ms\gs)\cdot (\rightarrow \vs')=-\mu\ms\gs\vs'</math>

:<math>\dot{\Ws}^{\Gamma}= \dot{\Ws}_\mathrm{camión}^{\Gamma\rightarrow \mathrm{ruedas}} +\dot{\Ws}_\mathrm{camión}^{\Gamma\rightarrow \mathrm{camión}} = \dot{\Ws}_\mathrm{camión}^{\Gamma\rightarrow \mathrm{ruedas}}= 2 \left(\odot \Gamma \right)\cdot \left(\otimes \frac{\vs}{\rs}\right)=-2\Gamma\frac{\vs}{\rs}.</math>

:Por otro lado: <math>\Ts_\Ts^\mathrm{sist}=\frac{1}{2}\ms\vs_\Ts^2(\mathrm{bloque})+\frac{1}{2}\ms\vs_\Ts^2(\mathrm{camión})\quad \Rightarrow \quad \dot{\Ts}_\Ts^\mathrm{sist}=-\ms(\vs+\vs')\mu\gs-\Ms\vs\as. </math>

:Finalmente: <math>\dot{\Ws}_\Ts^{\rightarrow \mathrm{sist}}=\dot{\Ts}_\Ts^\mathrm{sist} \quad \Rightarrow \quad -\mu\ms\gs\vs'-2\Gamma\frac{\vs}{\rs}=-\ms(\vs+\vs')\mu\gs-\Ms\vs\as.</math>

:Esta ecuación permite calcular el par de frenada: <math>\Gamma=\frac{1}{2}(\mu\ms\gs+\Ms\as)\rs.</math>

:Referencia camión:

:En esta referencia hay que tener en cuenta las fuerzas de arrastre. Por otro lado, las fuerzas de enlace del suelo sobre las ruedas están aplicadas a puntos que ya no tienen velocidad cero, y por tanto hacen potencia. Como antes, la potencia del peso es cero.

:<math>\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{sist}}=\dot{\Ws}_\mathrm{camión}^{\mathrm{ext}}+\dot{\Ws}^{\mathrm{int}}+\dot{\Ws}_\mathrm{camión}^{\mathrm{ar}}=\dot{\Ws}_\mathrm{camión}^{\mathrm{enlace}}+\dot{\Ws}^{\mathrm{fricción}}+\dot{\Ws}^{\Gamma}+\dot{\Ws}_\mathrm{camión}^{\mathrm{ar}}.</math>

:Las fuerzas de arrastre sobre el camión no hacen potencia porque el camión está parado en esta referencia. La de las fuerzas de arrastre sobre el bloque se ha calculado en el [[E1. Teorema de la energía: versión diferencial#✏️ ✏️ EJEMPLO E1-3.2: bloque deslizando sobre un camión|'''ejemplo E1-3.2''']]. La potencia de la fricción y de los pares de frenada es la misma de antes ya que no depende de la referencia.

:<math>\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{sist}}=\dot{\Ws}_\mathrm{camión}^{\mathrm{enllaç}}+\dot{\Ws}^{\mathrm{fricción}}+\dot{\Ws}^{\Gamma}+\dot{\Ws}_\mathrm{camión}^{\mathrm{ar} \rightarrow \mathrm{bloque}}= \dot{\Ws}_\mathrm{camión}^{\mathrm{enlace}} -\mu\ms\gs\vs'-2\Gamma\frac{\vs}{\rs} +\ms\as\vs'.</math>

:Las descripciones cinemática y dinámica de la rueda son:

[[Archivo:ExE1-3-3-4-esp.png|thumb|center|430px|link=]]

:<math>\dot{\Ws}_\mathrm{camión}^{\mathrm{enlace}}=2\Bigr[(\leftarrow \Ts)+(\uparrow \Ns)\Bigr]\cdot(\leftarrow\vs)=2\Ts\vs.</math>

La fuerza tangencial de enlace se puede calcular a partir del TMC aplicado al centro '''C''' de la rueda. Como no tiene masa: <math>\sum\overline{\Ms}_\mathrm{ext}(\Cs)=\overline{0} \quad \Rightarrow \quad (\odot \Gamma)+(\otimes \Ts\rs)=0\quad \Rightarrow \quad \Ts=\Gamma/\rs.</math>

:Por tanto: <math>\dot{\Ws}_\mathrm{camió}^{\rightarrow \mathrm{sist}} =2\frac{\Gamma}{\rs}\vs-\mu\ms\gs\vs'-2\Gamma\frac{\vs}{\rs}+\ms\as\vs'=\ms(\as-\mu\gs)\vs'.</math>

:Por otro lado, como se ha visto en el [[E1. Teorema de la energía: versión diferencial#✏️ ✏️ EJEMPLO E1-3.2: bloque deslizando sobre un camión|'''ejemplo E1-3.2''']]:

:<math>\Ts_\mathrm{camión}^\mathrm{sist}= \Ts_\mathrm{camión}^\mathrm{bloque}=\frac{1}{2}\ms\vs_\mathrm{camión}^2(\mathrm{bloque}) \quad \Rightarrow \quad \dot{\Ts}_\mathrm{camión}^\mathrm{sist}= \dot{\Ts}_\mathrm{camión}^\mathrm{bloque}=\ms\vs'(\as-\mu\gs).</math>

:El balance de potencias es consistente:

:<math>\left.\begin{array}{l}
\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{sist}}= \ms(\as-\mu\gs)\vs'\\
\dot{\Ts}^\mathrm{sist}_\mathrm{camión}=\ms(\as-\mu\gs)\vs'
\end{array}\right\} \Rightarrow\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{sist}}= \dot{\Ts}^\mathrm{sist}_\mathrm{camión}</math>

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==E1.4 Balance de potencias para un sistema multisólido: cálculo directo y cálculo indirecto==

El '''cálculo directo''' (a partir de su definición) de la potencia asociada a una interacción (una fuerza sobre un punto '''P''' o un momento sobre un sólido rígido S) exige el conocimiento de la interacción (fuerza o momento):

<math>\dot{\Ws}_\Rs^{\Fs\rightarrow \Ps}=\overline{\Fs}_{\rightarrow\Ps} \cdot \vel{P}{R} \quad , \quad \dot{\Ws}_\Rs^{\Ms\rightarrow \Ss}=\overline{\Ms}_{\rightarrow\Ss} \cdot \velang{S}{R} </math>

Cuando no se trata de una interacción formulable, hay que recurrir a los teoremas vectoriales para determinarla.

El '''cálculo indirecto''' es una alternativa que no necesita recurrir a estos teoremas. Se basa en la evaluación de un balance de potencias para un sistema y una referencia adecuados de manera que la potencia a determinar sea el único término desconocido de la ecuación (el único que no se puede calcular de manera directa). Cuando el balance incluye más de un término desconocido, se pueden combinar varios balances (para distintos sistemas y referencias) hasta tener un número suficiente de ecuaciones que permitan la determinación de todas las potencias desconocidas.

====✏️ EJEMPLO E1-4.1: polea frenada====
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:
{|
|
:[[Archivo:ExE1-4-1-1-esp.png|thumb|left|150px|link=]]
|Un bloque de masa m cuelga de un hilo inextensible atado a un punto de la periferia de una polea de radio r. Entre soporte (fijo al suelo) y polea actúa un freno que garantiza que el bloque baja a velocidad constante respecto al suelo. Se trata de calcular la potencia del freno.

|}

:Càlculo directo: <math>\dot{\Ws}^{\mathrm{freno}}= \overline{\Gamma}^{\rightarrow \mathrm{polea}}\cdot\velang{polea}{T}=\left(\odot\Gamma\right)\cdot\left( \otimes \frac{\vs_0}{\rs}\right)=-\Gamma \frac{\vs_0}{\rs}.</math>

:El par de frenada no es un dato, y hay que recurrir al TMC para calcularlo.

:<math>\left.\begin{array}{l}
\text{SISTEMA: polea + bloque}\\
\text{TMC en } \Os
\end{array}\right\} \sum\overline{\Ms}_\mathrm{ext}(\Os)=\dot{\overline{\Hs}}_\mathrm{RTO}(\Os)</math>

:El momento cinético proviene de las dos piezas. Ya que '''O''' no es fijo al bloque, hay que hacer descomposición baricéntrica para calcular el momento cinético del bloque:

:<math>\overline{\Hs}_\mathrm{RTO=T} (\Os) =\overline{\Hs}_\Ts^\mathrm{politja}(\Os) + \overline{\Hs}_\Ts^\mathrm{bloc}(\Os)= \overline{\Hs}_\Ts^\mathrm{politja}(\Os) + \overline{\Hs}_\Ts^\mathrm{bloc}(\Gs_\mathrm{bloc})+ \overline{\Hs}_\Ts^{\oplus\mathrm{bloc}}(\Os) = </math>

:<math>\hspace{2.3cm}=\Is_\mathrm{O}^\mathrm{polea} \velang{polea}{T}+\Is_{\Gs_\mathrm{bloque}}^\mathrm{bloque}+\OGvec_\mathrm{bloque}\times \ms\overline{\vs}_\Ts(\Gs_\mathrm{bloque})=\left(\otimes \Is_\mathrm{O}\frac{\vs_0}{\rs}\right)+(\otimes \ms\vs_0\rs).</math>

:[[Archivo:ExE1-4-1-2-esp.png|thumb|right|180px|link=]]
:Se trata de un momento cinético constante en dirección y valor. Por tanto:

:<math>\dot{\overline{\Hs}}_\mathrm{RTO}(\Os)=0 \quad \Rightarrow \quad \sum{\overline{\Ms}}_\mathrm{ext}(\Os)=(\odot\Gamma)+(\otimes \ms\gs\rs) =0\quad \Rightarrow \quad \Gamma= \ms\gs\rs</math>.

:Finalmente: <math>\dot{\Ws}^{\mathrm{freno}}= -\Gamma \frac{\vs_0}{\rs} = -\ms\gs\vs_0.</math>

:Cálculo indirecto:

:<math>\left.\begin{array}{l}
\text{SISTEMA: polea + soporte + bloque}\\
\text{REF: suelo(Gal)}
\end{array}\right\} \quad \dot{\Ts}^\mathrm{sist}_\Ts=\dot{\Ws}^\mathrm{ext}_\Ts+ \dot{\Ws}^\mathrm{int}</math>

:<math>\left.\begin{array}{l}
\dot{\Ts}^\mathrm{sist}_\Ts=0\\
\dot{\Ws}^\mathrm{ext}_\Ts=\left(\downarrow \ms\gs\right)\cdot\left(\downarrow \vs_0\right)= \ms\gs\vs_0\\
\dot{\Ws}^\mathrm{int}=\dot{\Ws}^{\mathrm{freno}}
\end{array}\right\} \quad \Rightarrow \quad \dot{\Ws}^{\mathrm{freno}} =-\ms\gs\vs_0</math>



====✏️ EJEMPLO E1-4.2: mono y plátanos====
------
:
{|
|
:[[Archivo:ExE1-4-2-1-esp.png|thumb|left|180px|link=]]
|Unos plátanos de masa <math>(2/3)\ms</math> cuelgan de una cuerda inextensible que se apoya sin deslizar sobre una polea de radio r y masa despreciable, articulada respecto al suelo. En el otro extremo de la cuerda se agarra un mono de masa m. La polea gira con velocidad angular constante <math>\Omega_0</math> respecto al suelo bajo la acción de un motor. El mono se mueve respecto a la cuerda sin que sus manos se froten contra la cuerda, y de manera que la distancia de su centro de masas al suelo se mantiene constante. Se trata de calcular la potencia del motor.
|}

:Cálculo directo: <math>\dot{\Ws}^{\mathrm{freno}}= \overline{\Gamma}^{\rightarrow \mathrm{polea}}\cdot\velang{polea}{T}=\left(\odot\Gamma\right)\cdot\left( \odot \Omega_0\right)=\Gamma \Omega_0.</math>

:El par motor no es un dato, y hay que recurrir al TMC para calcularlo.

:<math>\left.\begin{array}{l}
\text{SISTEMA: polea + mono + plátanos}\\
\text{TMC en } \Os
\end{array}\right\} \sum\overline{\Ms}_\mathrm{ext}(\Os)=\dot{\overline{\Hs}}_\mathrm{RTO}(\Os)</math>

:El momento cinético proviene del mono y de los plátanos. En primera aproximación, se puede considerar la masa del mono concentrada en su tronco y su cabeza, de manera que no gira y su velocidad respecto al suelo es nula. Teniendo en cuenta que el punto '''O''' no se mueve respecto al mono pero sí respecto a los plátanos:

:<math>\overline{\Hs}_\mathrm{RTO=T} (\Os) =\overline{\Hs}_\Ts^\mathrm{mono}(\Os) + \overline{\Hs}_\Ts^\mathrm{plátanos}(\Os)= \overline{\Hs}_\Ts^\mathrm{mono}(\Os) + \overline{\Hs}_\Ts^\mathrm{plátanos}(\Gs_\mathrm{plát})+ \OGvec_\mathrm{plát} \times \frac{2}{3} \ms \overline{\vs}_\Ts(\Gs_\mathrm{plát})= </math>

:<math>\hspace{2.3cm}=\OGvec_\mathrm{plát}\times \frac{2}{3} \ms \overline{\vs}_\Ts(\Gs_\mathrm{plát}) = \Bigr[(\rightarrow \rs)+(\downarrow \xs)\Bigr] \times\frac{2}{3} \ms \left(\downarrow \rs\Omega_0\right)=\left(\otimes \frac{2}{3}\ms\rs^2\Omega_0\right).</math>

:Se trata de un momento cinético constante en dirección y valor. Por tanto:

:<math>\dot{\overline{\Hs}}_\mathrm{RTO}(\Os)=0 \quad \Rightarrow \quad \sum{\overline{\Ms}}_\mathrm{ext}(\Os)=(\odot\Gamma)+(\otimes \ms\gs\rs) + \left(\otimes \frac{2}{3}\ms\gs\rs\right)=0\quad \Rightarrow \quad \Gamma= \frac{5}{3}\ms\gs\rs</math>.

:Finalmente: <math>\dot{\Ws}^{\mathrm{motor}}= \Gamma \Omega_0 = \frac{5}{3}\ms\gs\rs\Omega_0.</math>

:Cálculo indirecto

:Hay que tener presente que es un problema con dos tipos de actuadores: el motor y la musculatura del mono. Hay diversos modelos biomecánicos para formular las fuerzas de los músculos, pero quedan fuera del ámbito de este curso. Ya que el mono se mueve, sus músculos desarrollan una potencia no nula.

:<math>\left.\begin{array}{l}
\text{SISTEMA: polea + mono + plátano}\\
\text{REF: suelo(Gal)}
\end{array}\right\} \quad \dot{\Ts}^\mathrm{sist}_\Ts=\dot{\Ws}^\mathrm{ext}_\Ts+ \dot{\Ws}^\mathrm{int}</math>

:<math>\left.\begin{array}{l}
\dot{\Ts}^\mathrm{sist}_\Ts=0\\
\dot{\Ws}^\mathrm{ext}_\Ts=\left(\downarrow \frac{2}{3}\ms\gs\right)\cdot\left(\downarrow \rs\Omega_0\right)= \frac{2}{3}\ms\gs\rs\Omega_0\\
\dot{\Ws}^\mathrm{int}=\dot{\Ws}^{\mathrm{musc}}+\dot{\Ws}^{\mathrm{motor}}
\end{array}\right\} \quad \Rightarrow \quad \dot{\Ws}^{\mathrm{musc}} + \dot{\Ws}^{\mathrm{motor}}=-\frac{2}{3}\ms\gs\rs\Omega_0</math>

:Este balance proporciona una ecuación con dos incógnitas. Se puede plantear un segundo balance para un sistema donde una de las dos no aparezca: un sistema sin el mono o un sistema sin el motor. Si se hace un balance de potencias para el sistema (mono + trozo de cuerda), las fuerzas externas a considerar son el peso y la tensión de la cuerda. Si se sigue trabajando en la referencia del suelo, esta tensión (desconocida) hará potencia no nula ya que está aplicada a un punto de velocidad ascendente.
:[[Archivo:ExE1-4-2-2-cat-esp.png|thumb|right|330px|link=]]

:Para evitar que aparezca un término asociado a la tensión de la cuerda, se puede trabajar en la referencia del trozo de cuerda que se ha considerado, que al tener velocidad ascendente constante es galileana. En esta referencia, el mono baja con velocidad constante:

:<math>\left.\begin{array}{l}
\text{REL: trozo de cuerda}\\
\text{AB: suelo}
\end{array}\right\}</math>

:<math>\overline{\vs}_\mathrm{REL}(\mathrm{mono})=\overline{\vs}_\mathrm{ar}(\mathrm{mono})=-\overline{\vs}_\mathrm{ar}(\mathrm{mono})=(\downarrow \rs\Omega_0)</math>

:<math>\left.\begin{array}{l}
\text{SISTEMA: mono+ trozo de cuerda}\\
\text{REF: trozo de cuerda(Gal)}
\end{array}\right\} \quad \dot{\Ts}^\mathrm{sist}_\mathrm{cuerda} =\dot{\Ws}^\mathrm{ext}_\mathrm{cuerda}+ \dot{\Ws}^\mathrm{int}</math>

:<math>\left.\begin{array}{l}
\dot{\Ts}^\mathrm{sist}_\mathrm{cuerda}=0\\
\dot{\Ws}^\mathrm{ext}_\mathrm{cuerda}=\left(\downarrow \ms\gs\right)\cdot\left(\downarrow \rs\Omega_0\right)= \ms\gs\rs\Omega_0\\
\dot{\Ws}^\mathrm{int}=\dot{\Ws}^{\mathrm{musc}}
\end{array}\right\} \quad \Rightarrow \quad \dot{\Ws}^{\mathrm{musc}} =-\ms\gs\rs\Omega_0</math>

:Combinando este resultado con el del balance anterior:

:<math>\dot{\Ws}^\mathrm{motor}=-\frac{2}{3}\ms\gs\rs\Omega_0-\dot{\Ws}^\mathrm{musc}=-\frac{5}{3}\ms\gs\rs\Omega_0</math>



© Universitat Politècnica de Catalunya. [[Mecánica:Derechos de autor |Todos los derechos reservados]]
-------------

<center>
[[D8. Conservaciones|<<< D8. Conservaciones]]

[[E2. Teorema de la energía: versión integrada|E2. Teorema de la energía: versión integrada>>>]]
</center>

E1. Work-Energy Theorem: differential form

2025-05-20T12:15:55Z

Bros: /* ✏️ EXAMPLE E1-1.2: pendulum */

<div class="noautonum">__TOC__</div>
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\newcommand{\evec}{\overline{\textbf{e}}}
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\newcommand{\dert}[2]{\left.\frac{\ds{#1}}{\ds\ts}\right]_{\textrm{#2}}}
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{#7} & {#8} & {#9}
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\newcommand{\agal}[1]{\vecbf{a}_{\textrm{Gal}} (#1)}
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\newcommand{\F}[1]{\vec{\Fs}_{#1}}
\newcommand{\Fcal}[2]{\vec{\cal{F}}^{\text{#1}}_{#2}}
</math>

The '''Work-Energy Theorem''' plays a prominent role in physics, as it bridges all branches of science (electromagnetism, thermodynamics, etc.): all of them deal with energy.

There is a certain parallel between the Vector Theorems and the '''Work-Energy Theorem''': they all allow us to understand the evolution of a magnitude that depends on the mechanical state of the system. In the case of the former, we are dealing with the instantaneous evolution of vector magnitudes (linear momentum, angular momentum), while the latter deals with the finite change (in a finite time interval) of a scalar magnitude ('''mechanical energy''').

This unit presents the differential form of this theorem (the '''power balance'''). The integrated form (which is the Work-Energy Theorem itself) is presented in [[E2. Teorema de la energía: versión integrada|'''unit E2''']].
 
------
------
==E1.1 Power balance in a system of particles==

:[[File:E1-1-eng.png|thumb|right|300px|link=]]
<center>'''Figure E1.1''' Forces on a particle of a system with constant matter</center>

Let us consider a system of constant matter, and let’s apply Newton's second law to each particle (or each mass differential) of this system in a general reference frame R (which may be non-Galilean) ('''Figure E1.1'''):

<math>\F{\rightarrow \Ps} + \Fcal{}{\text{ar}\rightarrow\Ps} + \Fcal{}{\text{Cor}\rightarrow\Ps} =\ms_\Ps\acc{P}{R}</math>

<math>\F{\rightarrow \Ps} </math> is the resultant interaction force on '''P''', and it comes from internal interactions (with particles of the same system) and external ones:

<math>\F{\rightarrow \Ps}=\F{\rightarrow \Ps}^\text{int}+\F{\rightarrow \Ps}^\text{ext}</math>

If both sides of the equation are multiplied by <math>\vel{P}{R}</math>, all forces (interaction and inertia forces) orthogonal to <math>\vel{P}{R}</math> are filtered out. Specifically, since <math>\acc{P}{R}=\accs{P}{R} + \accn{P}{R} </math>, where <math>\accn{P}{R} \perp \vel{P}{R}</math>, and since the Coriolis force is always orthogonal to <math>\vel{P}{R} (\Fcal{}{\text{Cor}\rightarrow\Ps} = -2\ms_\Ps \velang{R}{Gal} \times \vel{P}{R})</math>:

<math>\vel{P}{R} \cdot \left( \F{\rightarrow \Ps}^\text{int}+\F{\rightarrow \Ps}^\text{ext} + \Fcal{}{\text{ar}\rightarrow\Ps} \right) = \ms_\Ps\accs{P}{R} \cdot \vel{P}{R} = \frac{\ds}{\ds \ts} \left(\frac{1}{2}\ms_\Ps \vvec_\Rs^2(\Ps) \right) \equiv \frac{\ds\Ts_\Rs(\Ps)}{\ds\ts}=\dot{\Ts}_\Rs(\Ps)</math>

The scalar function <math>\Ts_\Rs(\Ps)</math> is the '''kinetic energy''' of the particle, and is a function of the [[D1. Foundational laws of Newtonian dynamics#D1.3 Newton’s Principle of Determinacy|'''mechanical state''']] of '''P'''. The scalar product of a force and the velocity of its application point is the '''force power''': <math>\vel{P}{R} \cdot \F{\rightarrow \Ps} \equiv \dot{\Ws}_\Rs^{\F{\rightarrow \Ps}} </math>. Both scalar quantities <math>(\Ts_\Rs(\Ps), \dot{\Ws}_\Rs^{\F{\rightarrow \Ps}})</math> depend on the reference frame. For the same set of interaction forces acting on '''P''', the reference frame determines which are filtered out and which are not.

The equation can be rewritten as <math>\dot{\Ws}_\Rs^{\text{ext}\rightarrow \Ps}+\dot{\Ws}_\Rs^{\text{int}\rightarrow \Ps}+\dot{\Ws}_\Rs^{\text{ar}\rightarrow \Ps} =\dot{\Ts}_\Rs(\Ps)</math>, and constitutes the '''power balance''' for particle '''P''' in the R reference frame. Summing for all particles in the system:

<math>\dot{\Ws}_\Rs^{\text{ext}}+\dot{\Ws}^{\text{int}}+\dot{\Ws}_\Rs^{\text{ar}} =\dot{\Ts}_\Rs^\text{syst}</math>

The calculation of kinetic energy is covered in [[E2. Teorema de la energía: versión integrada|'''unit E2''']]. This unit focuses on the calculation of powers. [[E1. Work-Energy Theorem: differential form#E1.2 Potencia de una pareja de acción-reacción|'''section E1-2''']] shows that the power of internal interactions does not depend on the reference frame (which is why the subscript R does not appear in the <math>\dot{\Ws}^{\text{int}}</math> term of the previous equation).

====✏️ EXAMPLE E1-1.1: free particle====
------
:
{|
:[[File:ExE1-1-1-1-eng.jpg|thumb|left|180px|link=]]
|A particle '''P''', with mass m, is released from the ring of a Ferris wheel when it is at its highest point. We want to calculate the power, at this instant, of all forces acting on the particle from three different reference frames: the ground, the ring, and the Ferris wheel cabin.
|}

:Ground reference frame: Since this is a Galilean reference frame, all forces on '''P''' are interaction forces.
{|
:[[File:ExE1-1-2- eng.png|thumb|right|230px|link=]]

|<math>\hspace{0.5cm}\sum\dot{\Ws}_\Ts=\dot{\Ws}_\Ts^\text{weight}= \left(\leftarrow \Rs\Omega_0\right) \cdot \left(\downarrow \ms\gs\right)=0</math>
|}

:Ring reference frame: It is a non-Galilean reference, and the transportation force on '''P''' must be taken into account.
{|
:[[File:ExE1-1-3-eng.png|thumb|right|250px|link=]]
|<math>\hspace{0.5cm}</math> AB:ground, REL:ring 

<math>\hspace{0.5cm}\vel{P}{REL}=\vel{P}{AB}-\vel{P}{ar}= \left(\leftarrow \Rs\Omega_0\right)-\left(\leftarrow \Rs\Omega_0\right)=\overline{0}</math> 

<math>\hspace{0.5cm}</math>

Since the velocity is zero, the total power is zero.
|}

:Cabin reference frame: It is also a non-Galilean reference, and the transportation force on '''P''' must be taken into account.

{|
:[[File:ExE1-1-4-eng.png|thumb|right|250px|link=]]
|<math>\hspace{0.5cm}</math>AB:ground, REL:cabin

<math>\hspace{0.5cm}\vel{P}{REL}=\vel{P}{AB}-\vel{P}{ar}= \left(\leftarrow \Rs\Omega_0\right)-\left(\uparrow \Rs\Omega_0\right)</math>

<math>\hspace{0.5cm}\dot{\Ws}_\text{cabin}^\text{weight}=\left[ \left(\leftarrow \Rs\Omega_0\right)+\left(\downarrow \Rs\Omega_0\right) \right] \cdot \left(\downarrow\ms\gs\right) =\ms\gs\Rs\Omega_0</math>

<math>\hspace{0.5cm}\dot{\Ws}_\text{cabin}^\text{ar}=\left[ \left(\leftarrow \Rs\Omega_0\right)+\left(\downarrow \Rs\Omega_0\right) \right] \cdot \left(\rightarrow\ms\Rs^2\Omega_0^2\right) =-\ms\gs\Rs^2 \Omega_0^3</math>

<math>\hspace{0.5cm}\sum\dot{\Ws}_\text{cabin}=\ms\Rs\Omega_0(\gs-\Rs\Omega_0^2)</math>

|}


====✏️ EXAMPLE E1-1.2: pendulum====
------
:
{|
:[[File:ExE1-2-1-eng.png|thumb|left|180px|link=]]
|A particle '''P''' of mass m hangs from an inextensible thread of length L attached to the ceiling. The assembly oscillates in a vertical plane fixed to the ground. We want to calculate the power of all the forces acting on the particle in two different reference frames: the ground and a platform rotating relative to the ground with a constant angular velocity <math>\Omega_0</math>. 

Ground reference frame: Since it is a Galilean reference frame, '''P''' is only subjected to interaction forces.
{|
:[[File:ExE1-2-2-neut.png|thumb|left|300px|link=]]

|<math>\dot{\Ws}_\Es^\Ts=(\nearrow\Ls\dot{\theta})\cdot(\nwarrow\Ts)=0</math>

<math>\dot{\Ws}_\Es^\text{weight}=(\nearrow\Ls\dot{\theta}\cdot(\downarrow \ms\gs)=-\ms\gs\Ls\dot{\theta}\sin\theta</math>
|}

Platform reference frame: It is a non Galilean reference frame, and the transportation force on '''P''' has to be taken into account.

{|
:[[File:ExE1-2-3-eng.png|thumb|left|400px|link=]]

|AB:ground, REL:platform

<math>\vel{P}{REL}=\vel{P}{AB}-\vel{P}{ar}=(\nearrow\Ls\dot{\theta})-(\otimes \Ls \Omega_0 \sin \theta)</math>

<math>\dot{\Ws}_\Es^\Ts\left[(\nearrow\Ls\dot{\theta})+(\odot\Ls\Omega_0\sin\theta)\right]\cdot(\nwarrow\Ts)=0</math>

<math>\dot{\Ws}_\Es^\text{weight}\left[(\nearrow\Ls\dot{\theta})+(\odot\Ls\Omega_0\sin\theta)\right]\cdot(\downarrow \ms\gs)=-\ms\gs\Ls\dot{\theta}\sin\theta</math>

<math>\dot{\Ws}_\Es^\text{ar}\left[(\nearrow\Ls\dot{\theta})+(\odot\Ls\Omega_0\sin\theta)\right]\cdot(\rightarrow \ms\Ls\Omega_0^2\sin\theta)=\ms\Ls^2\Omega_0^2\dot{\theta}\cos\theta</math>
|}



------
------

==E1.2 Power of an action-reaction pair==

The total power associated with internal interactions in a constant-mass system is not zero in principle (since the two forces of each action-reaction pair are applied to particles that may have different velocities relative to the reference frame in which the calculation is made), and it is independent of the reference frame in which it is evaluated. It only depends on the separation velocity of the two particles. The power balance for the system is therefore written as: <math>\dot{\Ts}_\Rs^\text{syst}=\dot{\Ws}_\Rs^\text{ext} + \dot{\Ws}^\text{int} + \dot{\Ws}_\Rs^\text{ar} </math> .

<div>
=====💭 PROOF ➕=====

{|
:[[File:E1-2-neut.png|thumb|left|200px|link=]]
|<math>\dot{\Ws}^\text{ac-reac}=\overline{\Fs}_{\Ps \rightarrow \Qs}\cdot \vel{Q}{R} + \overline{\Fs}_{\Qs \rightarrow \Ps}\cdot \vel{P}{R}= \overline{\Fs}_{\Ps \rightarrow \Qs}\cdot \left[ \vel{Q}{R}-\vel{P}{R}\right]</math>

By the [[D1. Foundational laws of Newtonian dynamics#D1.6 Newton’s third law (action-reaction principle)|'''action-reaction principle''']], <math>\overline{\Fs}_{\Ps \rightarrow \Qs}=-\overline{\Fs}_{\Qs \rightarrow \Ps}</math>. Therefore: <math>\dot{\Ws}^\text{ac-reac}=\overline{\Fs}_{\Ps \rightarrow \Qs}\cdot \left[ \vel{Q}{R}-\vel{P}{R}\right]</math>

As <math>\overline{\Fs}_{\Ps \rightarrow \Qs}</math> has the <math>\PQvec</math> direction, only the components of the velocities in that direction have to be considered:

<math>\dot{\Ws}^\text{ac-reac}=\overline{\Fs}_{\Ps \rightarrow \Qs}\cdot \biggl( \vel{Q}{R} \Bigr]_{||\Ps\Qs}-\ \vel{P}{R} \Bigr]_{||\Ps\Qs}\biggl)</math>
|}

The difference <math>\biggl( \vel{Q}{R} \Bigr]_{\Ps\Qs}-\ \vel{P}{R} \Bigr]_{\Ps\Qs}\biggl)</math> is the separation velocity <math>\dot{\rho}</math> between '''P''' and '''Q''', and it does not dpend on the reference frame since it is the derivative of a scalar (the distance <math>\dot{\rho}</math> between the two particles). For the previous drawing, <math>\dot{\rho}> 0</math>. Since this is an attractive interaction, <math>\dot{\Ws}^\text{ac-reac} < 0</math>.

With a drawing, it is easy to see that:
* attractive interaction and particle approach <math>(\dot{\rho}<0)</math> : <math>\dot{\Ws}^\text{ac-reac} > 0</math> ;
* repulsive interaction and particle separation <math>(\dot{\rho}>0)</math> : <math>\dot{\Ws}^\text{ac-reac} > 0</math>
* repulsive interaction and particle approach <math>(\dot{\rho}<0)</math> : <math>\dot{\Ws}^\text{ac-reac} < 0</math>


</div>

There are two particular cases of interaction forces whose power always has the same sign:

* Friction action-reaction pair between '''P''' and '''Q''': since the friction force always opposes the relative velocity between the two points, its power is always negative:

<math>\hspace{1cm} \dot{\Ws}^\text{friction}=\dot{\Ws}^{\text{fric}\rightarrow\Ps}_\text{RTQ}+\dot{\Ws}^{\text{fric}\rightarrow\Qs}_\text{RTQ}=\overline{\Fs}^{\text{fric}\rightarrow\Ps} \cdot \vel{P}{RTQ} = - \abs{\overline{\Fs}^{\text{fric}\rightarrow\Ps}} \frac{\vel{P}{RTQ}}{\abs{\vel{P}{RTQ}}} \cdot \vel{P}{RTQ}<0;</math>

* Constraint action-reaction pair between '''P''' and '''Q''': from the [[D3. Interactions between rigid bodies#D3.4 Direct constraint interactions|'''analytical characterization equation''']] of the constraint forces between two points '''P''' and '''Q''' belonging to two rigid bodies <math>\mathrm{S}_\Ps</math> and <math>\mathrm{S}_\Qs</math>, it is easy to see that the associated power is zero:

<math>\hspace{1cm} \dot{\Ws}^\text{constraint}=\dot{\Ws}^{\text{constraint}\rightarrow\Ps}_{\mathrm{S}_\Ps}+\dot{\Ws}^{\text{constraint}\rightarrow\Qs}_{\mathrm{S}_\Ps}=\overline{\Fs}^{\text{constraint}\rightarrow\Ps} \cdot \overline{\mathbf{v}}_{\mathrm{S}_\Qs}(\Ps) = - 0;</math>

====✏️ EXAMPLE E1-2.1: power of an action-reaction friction pair====
------
:
{|
:[[File:ExE1-2-1-2-eng.png|thumb|left|180px|link=]]
|A homogeneous wheel, with radius r and mass m, moves down an inclined plane without sliding until it joins a conveyor belt moving at a constant velocity <math>\vs_0</math> relative to the ground. At the moment it makes contact with the belt, the center of the wheel has a velocity <math>\vs_0</math> relative to the ground in the same direction as the belt. We want to calculate the power of the friction force between the belt and the wheel at this instant.
|}

At that momento, the <math>\mathbf{ICR}^\text{wheel}_\Es</math> is point <math>\Js_\text{wheel}</math> in contact with the belt. This point slides backwards with respect to the belt, therefore the friction force on the wheel is forwards:

<math>\left.\begin{array}{l}
\text { AB: ground } \\
\text { REL: belt }
\end{array}\right\} \quad \bar{\vs}_{\text {REL }}\left(\Js_{\text {wheel }}\right)=\bar{\vs}_{\text{AB}}\left(\Js_{\text {wheel }}\right)-\bar{\vs}_{\mathrm{ar}}\left(\Js_{\text {wheel }}\right)=-\bar{\vs}_{\mathrm{ar}}\left(\Js_{\text {wheel }}\right)=-\left(\rightarrow \mathrm{v}_0\right)=\left(\leftarrow \mathrm{v}_0\right) .</math>

[[File:ExE1-2-1-3-eng.png|thumb|center|500px|link=]]

Calculation of the friction power: <math>\dot{\Ws}^\text{friction}=\dot{\Ws}_\Rs^{\text{friction}\rightarrow \text{wheel}}+ \dot{\Ws}_\Rs^{\text{friction}\rightarrow \text{belt}}</math>, where R may be any reference frame. To quickly calculate this power, it is advisable to choose a reference frame R where one of the two points subjected to the action-reaction pair has zero speed, so that one of the two terms is directly zero:

<math>\dot{\Ws}^\text{friction}=\dot{\Ws}_\text{belt}^{\text{fric}\rightarrow \text{wheel}}+ \dot{\Ws}_\text{belt}^{\text{fric }\rightarrow \text{belt}}=\overline{\Fs}^{\text{fric}\rightarrow \text{wheel}} \cdot \bar{\vs}_{\text {belt}}\left(\Js_{\text {wheel }}\right)=(\rightarrow \mu\ms\gs)\cdot(\leftarrow \vs_0)=-\mu\ms\gs\vs_0.</math>

If the calculation is done in any other reference frame, the result is the same:

<math>\dot{\Ws}^\text{friction}=\dot{\Ws}_\text{E}^{\text{fric}\rightarrow \text{wheel}}+ \dot{\Ws}_\text{E}^{\text{fric} \rightarrow \text{belt}}=(\rightarrow \mu\ms\gs)\cdot 0 + (\leftarrow \mu\ms\gs)\cdot(\rightarrow \vs_0)=-\mu\ms\gs\vs_0.</math>

<math>\dot{\Ws}^\text{friction}=\dot{\Ws}_\text{vehicle}^{\text{fric}\rightarrow \text{wheel}}+ \dot{\Ws}_\text{vehicle}^{\text{fric} \rightarrow \text{belt}}=(\rightarrow \mu\ms\gs)\cdot (\rightarrow\vs_0) + (\leftarrow \mu\ms\gs) \cdot(\rightarrow 3\vs_0)=-\mu\ms\gs\vs_0.</math>

However, if only the power of one of the two forces is calculated (for example, the friction on the wheel), the result depends on the reference frame and can have any sign:

<math>\dot{\Ws}_\text{belt}^{\text{fric }\rightarrow \text{wheel}}=(\rightarrow \mu\ms\gs)\cdot(\leftarrow \vs_0)=-\mu\ms\gs\vs_0 ,</math>

<math>\dot{\Ws}_\text{E}^{\text{fric} \rightarrow \text{wheel}}=(\rightarrow \mu\ms\gs)\cdot 0 = 0 ,</math>

<math>\dot{\Ws}_\text{vehicle}^{\text{fric} \rightarrow \text{wheel}}=(\rightarrow \mu\ms\gs)\cdot (\rightarrow\vs_0)=\mu\ms\gs\vs_0.</math>


-----
-----

==E1.3 Power of a system of forces on a rigid body==

{|
|
:[[File:E1-3-eng.png|thumb|left|180px|link=]]
<center>'''Figure E1.3''': Kinematic torsor of a rigid body in a reference frame R, and forces on different points (dm) of that rigid body</center>

|When calculating the power of a system of forces on a rigid body S, the term associated with internal interactions is zero because those interactions are all constraint (cohesion) forces and, as seen in [[E1. Work-Energy Theorem: differential form#E1.2 Power of an action-reaction pair|'''section E1.2''']], their total power is zero.

The total power of a system of forces on S can be calculated from the torsor of the system at any point '''O''' of the rigid body, <math>(\overline{\Fs},\overline{\Ms}(\Os))</math>. Taking into account rigid body kinematics ('''Figure E1.3'''):

<math>\dot{\Ws}_\Rs^{\rightarrow \Ss} =\sum_\Ps\overline{\Fs}^{\rightarrow \Ps} \cdot \vel{P}{R}= \sum_\Ps \left[ \overline{\Fs}^{\rightarrow \Ps} \cdot \vel{O}{R} + \velang{S}{R} \times \OPvec \right] </math>

|}
<math>\dot{\Ws}_\Rs^{\rightarrow \Ss} =\left(\sum_\Ps\overline{\Fs}^{\rightarrow \Ps} \right)\cdot \vel{O}{R} + \sum_\Ps\overline{\Fs}^{\rightarrow \Ps} \cdot \left(\velang{S}{R} \times \OPvec \right) = \left( \sum_\Ps\overline{\Fs}^{\rightarrow \Ps} \right) \cdot \vel{O}{R} + \left[ \sum_\Ps \left( \OPvec \times \overline{\Fs}^{\rightarrow \Ps} \right)\right] \cdot \velang{S}{R} = \overline{\Fs}\cdot \vel{O}{R} + \overline{\Ms}(\Os)\cdot \velang{S}{R} </math>

This calculation also shows that the power of a moment (or a torque <math>\Gamma</math>) on a rigid body in a reference frame R is the product of the moment by the angular velocity of the rigid body: <math>\dot{\Ws}_\Rs^{\Gamma\rightarrow \mathrm{S}} = \overline{\Gamma} \cdot \velang{S}{R}</math>.

The power balance for a rigid body is: <math>\dot{\Ws}_\Rs^{ \rightarrow \mathrm{S}} = \dot{\Ws}_\Rs^{ \mathrm{ext}} + \dot{\Ws}_\Rs^{ \mathrm{ar}}.</math>

====✏️ EXAMPLE E1-3.1: roller sliding on a slope====
------
:
{|
|
:[[File:ExE1-3-1-1-eng.png|thumb|left|180px|link=]]
|A homogeneous roller, with mass m and radius r, slides down a slope. The friction coefficient between the ground and the roller is <math>\mu</math>. We want to calculate the power of every force acting on the roller in the ground reference frame.
|}

:It is a 2D problem. The kinematic and dynamic descriptions are:
{|
|
[[File:ExE1-3-1-2-eng.png|thumb|right|250px|link=]]

|<math> \hspace{0.5cm}\vel{J}{E}=\vel{C}{T} + \velang{}{E} \times \CJvec= </math>

<math>\hspace{1.7cm}=\left( \rightarrow \vs_\Es \right) + \left( \otimes \Omega_\Es\right) \times \left( \downarrow \rs \right)</math>

<math> \hspace{0.5cm}\vel{J}{E} = \left[\rightarrow (\vs_\Es -\rs\Omega_\Es)\right]</math>

<math>\hspace{0.5cm} \dot{\Ws}_\Es^{\mathrm{ext}} = \dot{\Ws}_\Es^{\mathrm{weight}} + \dot{\Ws}_\Es^{\mathrm{friction}} + \dot{\Ws}_\Es^{\mathrm{constraint}}= \Bigr[\left(\downarrow \ms\gs\cos\beta\right)+\left(\rightarrow \ms\gs\sin\beta\right)\Bigr] \cdot (\rightarrow \vs_\Es) + \Bigr[(\leftarrow \mu \Ns) + (\uparrow \Ns ) \Bigr] \cdot \Bigr[\rightarrow (\vs_\Es -\rs\Omega_\Es)\Bigr]</math>

|}
:Since the roller center has no motion perpendicular to the plane, the AMT leads to <math>\Ns =\ms\gs\cos\beta</math>. Therefore:

{|
|
[[File:ExE1-3-1-3-eng.png|thumb|right|140px|link=]]
|<math>\hspace{0.5cm} \dot{\Ws}_\Ts^{\mathrm{ext}} = \ms\gs\vs_\Ts \sin \beta - \ms\gs(\vs_\Ts -\rs\Omega_\Ts)\cos\beta=\ms\gs \Bigr[ \vs_\Ts (\sin\beta-\cos\beta)+\rs\Omega_\Ts \cos\beta \Bigr].</math>

<math>\hspace{0.5cm}</math>

If instead of considering the friction at '''J''' we consider the equivalent torsor at point '''C''', we must consider the force applied in '''C''' and the moment about '''C''', which is not zero:

<math>\hspace{0.5cm} \dot{\Ws}_\Es^{ \mathrm{friction}} = (\leftarrow \mu\Ns)\cdot(\rightarrow\vs_\Es)+(\otimes \mu\Ns\rs)\cdot(\otimes \Omega_\Es) =\mu\ms\gs\cos\beta(\rs\Omega_\Es-\vs_\Es)</math>,

<math>\hspace{0.5cm} \dot{\Ws}_\Es^{\mathrm{ext}} = \dot{\Ws}_\Es^{ \mathrm{weight}} + \dot{\Ws}_\Es^{ \mathrm{friction}}+ \dot{\Ws}_\Es^{ \mathrm{constraint}}= \ms\gs\vs_\Es\sin\beta+\mu\ms\gs\cos\beta (\rs\Omega_\Es-\vs_\Es)</math>.

<math>\hspace{0.5cm}</math>

The result is the same.
|}


====✏️ EXAMPLE E1-3.2: block sliding on a truck====
------
:
{|
|
:[[File:ExE1-3-2-1-eng.png|thumb|left|220px|link=]]
|A truck brakes with acceleration '''a''' relative to the ground at the instant in which its velocity (also relative to the ground) is v. At the same instant, the homogeneous block, of mass m, slides with velocity v' relative to the truck. We want to perform a power balance for the block in two different reference frames: the ground and the truck.
|}

:Ground reference frame: It is a 2D problem. The kinematic and dynamic descriptions are:
{|
|
[[File:ExE1-3-2-2-eng.png|thumb|right|200px|link=]]

| The constraint between the truck and the block allows only one translational DoF between the two. Therefore, in the 2D approximation, the associated torsor has two independent components. If characterized at the midpoint of the contact surface, these components are a normal force and a moment perpendicular to the plane. The AMT applied to the block leads to <math>\Ns=\ms\gs</math>.

|}
:The powers associated with all interactions on the block in the ground reference are:

:<math>\dot{\Ws}_\Es^{\rightarrow \mathrm{bloque}}=\dot{\Ws}_\Ts^{\mathrm{ext}}=\dot{\Ws}_\Ts^{\mathrm{peso}}+\dot{\Ws}_\Ts^{\mathrm{fricción}}+\dot{\Ws}_\Ts^{\mathrm{enlace}}= \Bigr[(\downarrow \ms\gs)+(\leftarrow \mu\ms\gs)+(\uparrow \Ns) \Bigr] \cdot \Bigr[\rightarrow(\vs+\vs')\Bigr] + (\odot \Ms_\es)\cdot 0 = -\mu\ms\gs(\vs+\vs')</math>

:Ya que el bloque se traslada respecto al suelo, su energía cinética total es inmediata, y se puede plantear fácilmente el balance de potencias:

:<math>\Ts_\Ts^\mathrm{bloque}=\frac{1}{2}\int_\mathrm{bloque}\ds\ms \vs_\Ts^2(\mathrm{bloque})=\frac{1}{2}\ms \vs_\Ts^2(\mathrm{bloc})\quad \Rightarrow \quad \dot{\Ts}_\Ts^\mathrm{bloque}=\ms\vs_\Ts( \mathrm{bloque})\dot{\vs}_\Ts( \mathrm{bloque}) </math>

:Como la única fuerza horizontal sobre el bloque es la fricción:

<math>\dot{\vs}_\Ts(\mathrm{bloque})=-\mu\gs\quad \Rightarrow \quad \dot{\Ts}_\Ts^\mathrm{bloque}=-\mu\ms\gs(\vs+\vs').</math>

:El balance de potencias es consistente:

:<math>\left.\begin{array}{l}
\dot{\mathrm{W}}_\Ts^{\rightarrow \text { bloque }}=\dot{\mathrm{W}}_{\mathrm{T}}^{\text {ext }}=-\mu \mathrm{mg}\left(\mathrm{v}+\mathrm{v}^{\prime}\right) \\
\dot{\mathrm{T}}_{\mathrm{T}}^{\text {bloque }}=-\mu \mathrm{mg}\left(\mathrm{v}+\mathrm{v}^{\prime}\right)
\end{array}\right\} \Rightarrow \dot{\mathrm{W}}_{\mathrm{T}}^{\rightarrow \text { bloque }}=\dot{\mathrm{T}}_{\mathrm{T}}^{\text {bloque }}</math>

:Referencia camión: Al estar el camión acelerado respecto al suelo, hay que tener en cuenta las fuerzas de arrastre sobre todos los diferenciales de masa (dm) del bloque:

{|
|

[[Archivo:ExE1-3-2-3-cat-esp.jpg|thumb|right|250px|link=]]
|<math>\cal{\Fs}^{\rightarrow \ds\ms}_{\mathrm{ar}}=-\ds\ms\overline{\as}_\mathrm{ar}(\ds\ms)=-\ds\ms (\leftarrow \as)=(\rightarrow \ds\ms \cdot \as)</math>.

El torsor resultante de este sistema de fuerzas en el centro de inercia '''G''' se reduce a una fuerza horizontal.

<math>\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{bloque}}=\dot{\Ws}_\mathrm{camión}^{\mathrm{ext}} + \dot{\Ws}_\mathrm{camión}^{\mathrm{ar}} =\dot{\Ws}_\mathrm{camión}^{\mathrm{peso}} + \dot{\Ws}_\mathrm{camión}^{\mathrm{fricción}} + \dot{\Ws}_\mathrm{camión}^{\mathrm{enlace}} + \dot{\Ws}_\mathrm{camión}^{\mathrm{ar}} , </math>

<math>\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{bloque}}= \Bigr[(\downarrow \ms\gs)+(\leftarrow \mu\ms\gs)+(\uparrow \Ns) + (\rightarrow \ms\as)\Bigr] \cdot(\rightarrow \vs')+(\odot\Ms_\es) \cdot0=\ms(\as-\mu\gs)\vs'.</math>
|}

:Se puede comprobar que el balance de potencias es consistente:

:<math>\Ts^\mathrm{bloque}_\mathrm{camión}=\frac{1}{2}\ms\vs^2_\mathrm{camión}(\mathrm{bloque}) \quad \Rightarrow \quad \dot{\Ts}^\mathrm{bloque}_\mathrm{camión} = \ms\vs_\mathrm{camión}(\mathrm{bloque})\dot{\vs}_\mathrm{camión}(\mathrm{bloque})</math>

:<math>\left.\begin{array}{l}
\mathrm{AB}:\mathrm{suelo} \\
\mathrm{REL}:\mathrm{camión}
\end{array}\right\} \quad \dot{\vs}_\mathrm{camión}(\mathrm{bloque})= \dot{\vs}_\mathrm{T}(\mathrm{bloque})-\dot{\vs}_\mathrm{ar}(\mathrm{bloque}) = (-\mu\gs)-(-\as)=\as-\mu\gs \quad \Rightarrow \quad \dot{\Ts}^\mathrm{bloque}_\mathrm{camión}=\ms\vs'(\as-\mu\gs)</math>

:<math>\left.\begin{array}{l}
\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{bloque}}=\dot{\Ws}_\mathrm{camión}^{\mathrm{ext}} + \dot{\Ws}_\mathrm{camión}^{\mathrm{ar}}= \ms(\as-\mu\gs)\vs'\\
\dot{\Ts}^\mathrm{bloque}_\mathrm{camión}=\ms\vs'(\as-\mu\gs)
\end{array}\right\} \Rightarrow\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{bloque}}= \dot{\Ts}^\mathrm{bloque}_\mathrm{camión}</math>


====✏️ EJEMPLO E1-3.3: camión frenando con carga====
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:
{|
|
:[[Archivo:ExE1-3-3-1-esp.png|thumb|left|220px|link=]]
|Un camión frena con aceleración '''a''' respecto al suelo en el instante en el que su velocidad (también respecto al suelo) es v. Los pares de frenada en el eje delantero y en el trasero son iguales y valen <math>\Gamma</math>, y las ruedas, de radio r y masa despreciable, no deslizan sobre el suelo. En el mismo instante, el bloque homogéneo, de masa m, desliz con velocidad v' relativa al camión. Se trata de hacer un balance de potencias para el sistema (bloque+camión) en la referencia del suelo y en la del camión.

|}

:Ya que la potencia de las interacciones internas no es nula en principio (no se trata de un único sólido rígido), es conveniente hacer un DGI del sistema para no olvidar ninguna:
{|
|
[[Archivo:ExE1-3-3-2-esp.png|thumb|right|250px|link=]]

|
|
c.p.s.d.: contacto puntual sin deslizamiento

c.m.c.d.: contacto multipuntual con deslizamiento

|}
:El sistema sobre el que se hace el balance de potencias es todo menos el suelo. Teniendo en cuenta que la potencia de los enlaces internos es siempre nula, las únicas interacciones que hay que tener en cuenta son el peso, los enlaces de las ruedas con el suelo, los pares de frenada y la fricción entre bloque y camión:

[[Archivo:ExE1-3-3-3-esp.png|thumb|center|300px|link=]]

:Referencia suelo:
:Las fuerzas de enlace del suelo sobre las ruedas están aplicadas a puntos de velocidad cero, y por tanto no hacen potencia. El peso es una fuerza vertical, y el movimiento del camión y del bloque es horizontal, por tanto su potencia también es cero.

:<math>\dot{\Ws}_\Ts^{\rightarrow \mathrm{sist}}=\dot{\Ws}_\Ts^{\mathrm{ext}}+\dot{\Ws}^{\mathrm{int}}=\dot{\Ws}_\Ts^{\mathrm{peso}}+\dot{\Ws}_\Ts^{\mathrm{enlace}}+\dot{\Ws}^{\mathrm{fricción}}+\dot{\Ws}^{\Gamma}=\dot{\Ws}^{\mathrm{fricción}}+\dot{\Ws}^{\Gamma}.</math>

:Las potencias de interacciones internas se pueden calcular en cualquier referencia. Por tanto:

:<math>\dot{\Ws}^{\mathrm{fricción}}= \dot{\Ws}_\mathrm{camión}^{\mathrm{fric}\rightarrow \mathrm{bloque}} +\dot{\Ws}_\mathrm{camió}^{\mathrm{fric}\rightarrow \mathrm{camión}} = \dot{\Ws}_\mathrm{camión}^{\mathrm{fric}\rightarrow \mathrm{bloque}}= (\leftarrow \mu\ms\gs)\cdot (\rightarrow \vs')=-\mu\ms\gs\vs'</math>

:<math>\dot{\Ws}^{\Gamma}= \dot{\Ws}_\mathrm{camión}^{\Gamma\rightarrow \mathrm{ruedas}} +\dot{\Ws}_\mathrm{camión}^{\Gamma\rightarrow \mathrm{camión}} = \dot{\Ws}_\mathrm{camión}^{\Gamma\rightarrow \mathrm{ruedas}}= 2 \left(\odot \Gamma \right)\cdot \left(\otimes \frac{\vs}{\rs}\right)=-2\Gamma\frac{\vs}{\rs}.</math>

:Por otro lado: <math>\Ts_\Ts^\mathrm{sist}=\frac{1}{2}\ms\vs_\Ts^2(\mathrm{bloque})+\frac{1}{2}\ms\vs_\Ts^2(\mathrm{camión})\quad \Rightarrow \quad \dot{\Ts}_\Ts^\mathrm{sist}=-\ms(\vs+\vs')\mu\gs-\Ms\vs\as. </math>

:Finalmente: <math>\dot{\Ws}_\Ts^{\rightarrow \mathrm{sist}}=\dot{\Ts}_\Ts^\mathrm{sist} \quad \Rightarrow \quad -\mu\ms\gs\vs'-2\Gamma\frac{\vs}{\rs}=-\ms(\vs+\vs')\mu\gs-\Ms\vs\as.</math>

:Esta ecuación permite calcular el par de frenada: <math>\Gamma=\frac{1}{2}(\mu\ms\gs+\Ms\as)\rs.</math>

:Referencia camión:

:En esta referencia hay que tener en cuenta las fuerzas de arrastre. Por otro lado, las fuerzas de enlace del suelo sobre las ruedas están aplicadas a puntos que ya no tienen velocidad cero, y por tanto hacen potencia. Como antes, la potencia del peso es cero.

:<math>\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{sist}}=\dot{\Ws}_\mathrm{camión}^{\mathrm{ext}}+\dot{\Ws}^{\mathrm{int}}+\dot{\Ws}_\mathrm{camión}^{\mathrm{ar}}=\dot{\Ws}_\mathrm{camión}^{\mathrm{enlace}}+\dot{\Ws}^{\mathrm{fricción}}+\dot{\Ws}^{\Gamma}+\dot{\Ws}_\mathrm{camión}^{\mathrm{ar}}.</math>

:Las fuerzas de arrastre sobre el camión no hacen potencia porque el camión está parado en esta referencia. La de las fuerzas de arrastre sobre el bloque se ha calculado en el [[E1. Teorema de la energía: versión diferencial#✏️ ✏️ EJEMPLO E1-3.2: bloque deslizando sobre un camión|'''ejemplo E1-3.2''']]. La potencia de la fricción y de los pares de frenada es la misma de antes ya que no depende de la referencia.

:<math>\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{sist}}=\dot{\Ws}_\mathrm{camión}^{\mathrm{enllaç}}+\dot{\Ws}^{\mathrm{fricción}}+\dot{\Ws}^{\Gamma}+\dot{\Ws}_\mathrm{camión}^{\mathrm{ar} \rightarrow \mathrm{bloque}}= \dot{\Ws}_\mathrm{camión}^{\mathrm{enlace}} -\mu\ms\gs\vs'-2\Gamma\frac{\vs}{\rs} +\ms\as\vs'.</math>

:Las descripciones cinemática y dinámica de la rueda son:

[[Archivo:ExE1-3-3-4-esp.png|thumb|center|430px|link=]]

:<math>\dot{\Ws}_\mathrm{camión}^{\mathrm{enlace}}=2\Bigr[(\leftarrow \Ts)+(\uparrow \Ns)\Bigr]\cdot(\leftarrow\vs)=2\Ts\vs.</math>

La fuerza tangencial de enlace se puede calcular a partir del TMC aplicado al centro '''C''' de la rueda. Como no tiene masa: <math>\sum\overline{\Ms}_\mathrm{ext}(\Cs)=\overline{0} \quad \Rightarrow \quad (\odot \Gamma)+(\otimes \Ts\rs)=0\quad \Rightarrow \quad \Ts=\Gamma/\rs.</math>

:Por tanto: <math>\dot{\Ws}_\mathrm{camió}^{\rightarrow \mathrm{sist}} =2\frac{\Gamma}{\rs}\vs-\mu\ms\gs\vs'-2\Gamma\frac{\vs}{\rs}+\ms\as\vs'=\ms(\as-\mu\gs)\vs'.</math>

:Por otro lado, como se ha visto en el [[E1. Teorema de la energía: versión diferencial#✏️ ✏️ EJEMPLO E1-3.2: bloque deslizando sobre un camión|'''ejemplo E1-3.2''']]:

:<math>\Ts_\mathrm{camión}^\mathrm{sist}= \Ts_\mathrm{camión}^\mathrm{bloque}=\frac{1}{2}\ms\vs_\mathrm{camión}^2(\mathrm{bloque}) \quad \Rightarrow \quad \dot{\Ts}_\mathrm{camión}^\mathrm{sist}= \dot{\Ts}_\mathrm{camión}^\mathrm{bloque}=\ms\vs'(\as-\mu\gs).</math>

:El balance de potencias es consistente:

:<math>\left.\begin{array}{l}
\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{sist}}= \ms(\as-\mu\gs)\vs'\\
\dot{\Ts}^\mathrm{sist}_\mathrm{camión}=\ms(\as-\mu\gs)\vs'
\end{array}\right\} \Rightarrow\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{sist}}= \dot{\Ts}^\mathrm{sist}_\mathrm{camión}</math>

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==E1.4 Balance de potencias para un sistema multisólido: cálculo directo y cálculo indirecto==

El '''cálculo directo''' (a partir de su definición) de la potencia asociada a una interacción (una fuerza sobre un punto '''P''' o un momento sobre un sólido rígido S) exige el conocimiento de la interacción (fuerza o momento):

<math>\dot{\Ws}_\Rs^{\Fs\rightarrow \Ps}=\overline{\Fs}_{\rightarrow\Ps} \cdot \vel{P}{R} \quad , \quad \dot{\Ws}_\Rs^{\Ms\rightarrow \Ss}=\overline{\Ms}_{\rightarrow\Ss} \cdot \velang{S}{R} </math>

Cuando no se trata de una interacción formulable, hay que recurrir a los teoremas vectoriales para determinarla.

El '''cálculo indirecto''' es una alternativa que no necesita recurrir a estos teoremas. Se basa en la evaluación de un balance de potencias para un sistema y una referencia adecuados de manera que la potencia a determinar sea el único término desconocido de la ecuación (el único que no se puede calcular de manera directa). Cuando el balance incluye más de un término desconocido, se pueden combinar varios balances (para distintos sistemas y referencias) hasta tener un número suficiente de ecuaciones que permitan la determinación de todas las potencias desconocidas.

====✏️ EJEMPLO E1-4.1: polea frenada====
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:
{|
|
:[[Archivo:ExE1-4-1-1-esp.png|thumb|left|150px|link=]]
|Un bloque de masa m cuelga de un hilo inextensible atado a un punto de la periferia de una polea de radio r. Entre soporte (fijo al suelo) y polea actúa un freno que garantiza que el bloque baja a velocidad constante respecto al suelo. Se trata de calcular la potencia del freno.

|}

:Càlculo directo: <math>\dot{\Ws}^{\mathrm{freno}}= \overline{\Gamma}^{\rightarrow \mathrm{polea}}\cdot\velang{polea}{T}=\left(\odot\Gamma\right)\cdot\left( \otimes \frac{\vs_0}{\rs}\right)=-\Gamma \frac{\vs_0}{\rs}.</math>

:El par de frenada no es un dato, y hay que recurrir al TMC para calcularlo.

:<math>\left.\begin{array}{l}
\text{SISTEMA: polea + bloque}\\
\text{TMC en } \Os
\end{array}\right\} \sum\overline{\Ms}_\mathrm{ext}(\Os)=\dot{\overline{\Hs}}_\mathrm{RTO}(\Os)</math>

:El momento cinético proviene de las dos piezas. Ya que '''O''' no es fijo al bloque, hay que hacer descomposición baricéntrica para calcular el momento cinético del bloque:

:<math>\overline{\Hs}_\mathrm{RTO=T} (\Os) =\overline{\Hs}_\Ts^\mathrm{politja}(\Os) + \overline{\Hs}_\Ts^\mathrm{bloc}(\Os)= \overline{\Hs}_\Ts^\mathrm{politja}(\Os) + \overline{\Hs}_\Ts^\mathrm{bloc}(\Gs_\mathrm{bloc})+ \overline{\Hs}_\Ts^{\oplus\mathrm{bloc}}(\Os) = </math>

:<math>\hspace{2.3cm}=\Is_\mathrm{O}^\mathrm{polea} \velang{polea}{T}+\Is_{\Gs_\mathrm{bloque}}^\mathrm{bloque}+\OGvec_\mathrm{bloque}\times \ms\overline{\vs}_\Ts(\Gs_\mathrm{bloque})=\left(\otimes \Is_\mathrm{O}\frac{\vs_0}{\rs}\right)+(\otimes \ms\vs_0\rs).</math>

:[[Archivo:ExE1-4-1-2-esp.png|thumb|right|180px|link=]]
:Se trata de un momento cinético constante en dirección y valor. Por tanto:

:<math>\dot{\overline{\Hs}}_\mathrm{RTO}(\Os)=0 \quad \Rightarrow \quad \sum{\overline{\Ms}}_\mathrm{ext}(\Os)=(\odot\Gamma)+(\otimes \ms\gs\rs) =0\quad \Rightarrow \quad \Gamma= \ms\gs\rs</math>.

:Finalmente: <math>\dot{\Ws}^{\mathrm{freno}}= -\Gamma \frac{\vs_0}{\rs} = -\ms\gs\vs_0.</math>

:Cálculo indirecto:

:<math>\left.\begin{array}{l}
\text{SISTEMA: polea + soporte + bloque}\\
\text{REF: suelo(Gal)}
\end{array}\right\} \quad \dot{\Ts}^\mathrm{sist}_\Ts=\dot{\Ws}^\mathrm{ext}_\Ts+ \dot{\Ws}^\mathrm{int}</math>

:<math>\left.\begin{array}{l}
\dot{\Ts}^\mathrm{sist}_\Ts=0\\
\dot{\Ws}^\mathrm{ext}_\Ts=\left(\downarrow \ms\gs\right)\cdot\left(\downarrow \vs_0\right)= \ms\gs\vs_0\\
\dot{\Ws}^\mathrm{int}=\dot{\Ws}^{\mathrm{freno}}
\end{array}\right\} \quad \Rightarrow \quad \dot{\Ws}^{\mathrm{freno}} =-\ms\gs\vs_0</math>



====✏️ EJEMPLO E1-4.2: mono y plátanos====
------
:
{|
|
:[[Archivo:ExE1-4-2-1-esp.png|thumb|left|180px|link=]]
|Unos plátanos de masa <math>(2/3)\ms</math> cuelgan de una cuerda inextensible que se apoya sin deslizar sobre una polea de radio r y masa despreciable, articulada respecto al suelo. En el otro extremo de la cuerda se agarra un mono de masa m. La polea gira con velocidad angular constante <math>\Omega_0</math> respecto al suelo bajo la acción de un motor. El mono se mueve respecto a la cuerda sin que sus manos se froten contra la cuerda, y de manera que la distancia de su centro de masas al suelo se mantiene constante. Se trata de calcular la potencia del motor.
|}

:Cálculo directo: <math>\dot{\Ws}^{\mathrm{freno}}= \overline{\Gamma}^{\rightarrow \mathrm{polea}}\cdot\velang{polea}{T}=\left(\odot\Gamma\right)\cdot\left( \odot \Omega_0\right)=\Gamma \Omega_0.</math>

:El par motor no es un dato, y hay que recurrir al TMC para calcularlo.

:<math>\left.\begin{array}{l}
\text{SISTEMA: polea + mono + plátanos}\\
\text{TMC en } \Os
\end{array}\right\} \sum\overline{\Ms}_\mathrm{ext}(\Os)=\dot{\overline{\Hs}}_\mathrm{RTO}(\Os)</math>

:El momento cinético proviene del mono y de los plátanos. En primera aproximación, se puede considerar la masa del mono concentrada en su tronco y su cabeza, de manera que no gira y su velocidad respecto al suelo es nula. Teniendo en cuenta que el punto '''O''' no se mueve respecto al mono pero sí respecto a los plátanos:

:<math>\overline{\Hs}_\mathrm{RTO=T} (\Os) =\overline{\Hs}_\Ts^\mathrm{mono}(\Os) + \overline{\Hs}_\Ts^\mathrm{plátanos}(\Os)= \overline{\Hs}_\Ts^\mathrm{mono}(\Os) + \overline{\Hs}_\Ts^\mathrm{plátanos}(\Gs_\mathrm{plát})+ \OGvec_\mathrm{plát} \times \frac{2}{3} \ms \overline{\vs}_\Ts(\Gs_\mathrm{plát})= </math>

:<math>\hspace{2.3cm}=\OGvec_\mathrm{plát}\times \frac{2}{3} \ms \overline{\vs}_\Ts(\Gs_\mathrm{plát}) = \Bigr[(\rightarrow \rs)+(\downarrow \xs)\Bigr] \times\frac{2}{3} \ms \left(\downarrow \rs\Omega_0\right)=\left(\otimes \frac{2}{3}\ms\rs^2\Omega_0\right).</math>

:Se trata de un momento cinético constante en dirección y valor. Por tanto:

:<math>\dot{\overline{\Hs}}_\mathrm{RTO}(\Os)=0 \quad \Rightarrow \quad \sum{\overline{\Ms}}_\mathrm{ext}(\Os)=(\odot\Gamma)+(\otimes \ms\gs\rs) + \left(\otimes \frac{2}{3}\ms\gs\rs\right)=0\quad \Rightarrow \quad \Gamma= \frac{5}{3}\ms\gs\rs</math>.

:Finalmente: <math>\dot{\Ws}^{\mathrm{motor}}= \Gamma \Omega_0 = \frac{5}{3}\ms\gs\rs\Omega_0.</math>

:Cálculo indirecto

:Hay que tener presente que es un problema con dos tipos de actuadores: el motor y la musculatura del mono. Hay diversos modelos biomecánicos para formular las fuerzas de los músculos, pero quedan fuera del ámbito de este curso. Ya que el mono se mueve, sus músculos desarrollan una potencia no nula.

:<math>\left.\begin{array}{l}
\text{SISTEMA: polea + mono + plátano}\\
\text{REF: suelo(Gal)}
\end{array}\right\} \quad \dot{\Ts}^\mathrm{sist}_\Ts=\dot{\Ws}^\mathrm{ext}_\Ts+ \dot{\Ws}^\mathrm{int}</math>

:<math>\left.\begin{array}{l}
\dot{\Ts}^\mathrm{sist}_\Ts=0\\
\dot{\Ws}^\mathrm{ext}_\Ts=\left(\downarrow \frac{2}{3}\ms\gs\right)\cdot\left(\downarrow \rs\Omega_0\right)= \frac{2}{3}\ms\gs\rs\Omega_0\\
\dot{\Ws}^\mathrm{int}=\dot{\Ws}^{\mathrm{musc}}+\dot{\Ws}^{\mathrm{motor}}
\end{array}\right\} \quad \Rightarrow \quad \dot{\Ws}^{\mathrm{musc}} + \dot{\Ws}^{\mathrm{motor}}=-\frac{2}{3}\ms\gs\rs\Omega_0</math>

:Este balance proporciona una ecuación con dos incógnitas. Se puede plantear un segundo balance para un sistema donde una de las dos no aparezca: un sistema sin el mono o un sistema sin el motor. Si se hace un balance de potencias para el sistema (mono + trozo de cuerda), las fuerzas externas a considerar son el peso y la tensión de la cuerda. Si se sigue trabajando en la referencia del suelo, esta tensión (desconocida) hará potencia no nula ya que está aplicada a un punto de velocidad ascendente.
:[[Archivo:ExE1-4-2-2-cat-esp.png|thumb|right|330px|link=]]

:Para evitar que aparezca un término asociado a la tensión de la cuerda, se puede trabajar en la referencia del trozo de cuerda que se ha considerado, que al tener velocidad ascendente constante es galileana. En esta referencia, el mono baja con velocidad constante:

:<math>\left.\begin{array}{l}
\text{REL: trozo de cuerda}\\
\text{AB: suelo}
\end{array}\right\}</math>

:<math>\overline{\vs}_\mathrm{REL}(\mathrm{mono})=\overline{\vs}_\mathrm{ar}(\mathrm{mono})=-\overline{\vs}_\mathrm{ar}(\mathrm{mono})=(\downarrow \rs\Omega_0)</math>

:<math>\left.\begin{array}{l}
\text{SISTEMA: mono+ trozo de cuerda}\\
\text{REF: trozo de cuerda(Gal)}
\end{array}\right\} \quad \dot{\Ts}^\mathrm{sist}_\mathrm{cuerda} =\dot{\Ws}^\mathrm{ext}_\mathrm{cuerda}+ \dot{\Ws}^\mathrm{int}</math>

:<math>\left.\begin{array}{l}
\dot{\Ts}^\mathrm{sist}_\mathrm{cuerda}=0\\
\dot{\Ws}^\mathrm{ext}_\mathrm{cuerda}=\left(\downarrow \ms\gs\right)\cdot\left(\downarrow \rs\Omega_0\right)= \ms\gs\rs\Omega_0\\
\dot{\Ws}^\mathrm{int}=\dot{\Ws}^{\mathrm{musc}}
\end{array}\right\} \quad \Rightarrow \quad \dot{\Ws}^{\mathrm{musc}} =-\ms\gs\rs\Omega_0</math>

:Combinando este resultado con el del balance anterior:

:<math>\dot{\Ws}^\mathrm{motor}=-\frac{2}{3}\ms\gs\rs\Omega_0-\dot{\Ws}^\mathrm{musc}=-\frac{5}{3}\ms\gs\rs\Omega_0</math>



© Universitat Politècnica de Catalunya. [[Mecánica:Derechos de autor |Todos los derechos reservados]]
-------------

<center>
[[D8. Conservaciones|<<< D8. Conservaciones]]

[[E2. Teorema de la energía: versión integrada|E2. Teorema de la energía: versión integrada>>>]]
</center>

E1. Work-Energy Theorem: differential form

2025-05-20T12:15:26Z

Bros: /* ✏️ EXAMPLE E1-1.2: pendulum */

<div class="noautonum">__TOC__</div>
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</math>

The '''Work-Energy Theorem''' plays a prominent role in physics, as it bridges all branches of science (electromagnetism, thermodynamics, etc.): all of them deal with energy.

There is a certain parallel between the Vector Theorems and the '''Work-Energy Theorem''': they all allow us to understand the evolution of a magnitude that depends on the mechanical state of the system. In the case of the former, we are dealing with the instantaneous evolution of vector magnitudes (linear momentum, angular momentum), while the latter deals with the finite change (in a finite time interval) of a scalar magnitude ('''mechanical energy''').

This unit presents the differential form of this theorem (the '''power balance'''). The integrated form (which is the Work-Energy Theorem itself) is presented in [[E2. Teorema de la energía: versión integrada|'''unit E2''']].
 
------
------
==E1.1 Power balance in a system of particles==

:[[File:E1-1-eng.png|thumb|right|300px|link=]]
<center>'''Figure E1.1''' Forces on a particle of a system with constant matter</center>

Let us consider a system of constant matter, and let’s apply Newton's second law to each particle (or each mass differential) of this system in a general reference frame R (which may be non-Galilean) ('''Figure E1.1'''):

<math>\F{\rightarrow \Ps} + \Fcal{}{\text{ar}\rightarrow\Ps} + \Fcal{}{\text{Cor}\rightarrow\Ps} =\ms_\Ps\acc{P}{R}</math>

<math>\F{\rightarrow \Ps} </math> is the resultant interaction force on '''P''', and it comes from internal interactions (with particles of the same system) and external ones:

<math>\F{\rightarrow \Ps}=\F{\rightarrow \Ps}^\text{int}+\F{\rightarrow \Ps}^\text{ext}</math>

If both sides of the equation are multiplied by <math>\vel{P}{R}</math>, all forces (interaction and inertia forces) orthogonal to <math>\vel{P}{R}</math> are filtered out. Specifically, since <math>\acc{P}{R}=\accs{P}{R} + \accn{P}{R} </math>, where <math>\accn{P}{R} \perp \vel{P}{R}</math>, and since the Coriolis force is always orthogonal to <math>\vel{P}{R} (\Fcal{}{\text{Cor}\rightarrow\Ps} = -2\ms_\Ps \velang{R}{Gal} \times \vel{P}{R})</math>:

<math>\vel{P}{R} \cdot \left( \F{\rightarrow \Ps}^\text{int}+\F{\rightarrow \Ps}^\text{ext} + \Fcal{}{\text{ar}\rightarrow\Ps} \right) = \ms_\Ps\accs{P}{R} \cdot \vel{P}{R} = \frac{\ds}{\ds \ts} \left(\frac{1}{2}\ms_\Ps \vvec_\Rs^2(\Ps) \right) \equiv \frac{\ds\Ts_\Rs(\Ps)}{\ds\ts}=\dot{\Ts}_\Rs(\Ps)</math>

The scalar function <math>\Ts_\Rs(\Ps)</math> is the '''kinetic energy''' of the particle, and is a function of the [[D1. Foundational laws of Newtonian dynamics#D1.3 Newton’s Principle of Determinacy|'''mechanical state''']] of '''P'''. The scalar product of a force and the velocity of its application point is the '''force power''': <math>\vel{P}{R} \cdot \F{\rightarrow \Ps} \equiv \dot{\Ws}_\Rs^{\F{\rightarrow \Ps}} </math>. Both scalar quantities <math>(\Ts_\Rs(\Ps), \dot{\Ws}_\Rs^{\F{\rightarrow \Ps}})</math> depend on the reference frame. For the same set of interaction forces acting on '''P''', the reference frame determines which are filtered out and which are not.

The equation can be rewritten as <math>\dot{\Ws}_\Rs^{\text{ext}\rightarrow \Ps}+\dot{\Ws}_\Rs^{\text{int}\rightarrow \Ps}+\dot{\Ws}_\Rs^{\text{ar}\rightarrow \Ps} =\dot{\Ts}_\Rs(\Ps)</math>, and constitutes the '''power balance''' for particle '''P''' in the R reference frame. Summing for all particles in the system:

<math>\dot{\Ws}_\Rs^{\text{ext}}+\dot{\Ws}^{\text{int}}+\dot{\Ws}_\Rs^{\text{ar}} =\dot{\Ts}_\Rs^\text{syst}</math>

The calculation of kinetic energy is covered in [[E2. Teorema de la energía: versión integrada|'''unit E2''']]. This unit focuses on the calculation of powers. [[E1. Work-Energy Theorem: differential form#E1.2 Potencia de una pareja de acción-reacción|'''section E1-2''']] shows that the power of internal interactions does not depend on the reference frame (which is why the subscript R does not appear in the <math>\dot{\Ws}^{\text{int}}</math> term of the previous equation).

====✏️ EXAMPLE E1-1.1: free particle====
------
:
{|
:[[File:ExE1-1-1-1-eng.jpg|thumb|left|180px|link=]]
|A particle '''P''', with mass m, is released from the ring of a Ferris wheel when it is at its highest point. We want to calculate the power, at this instant, of all forces acting on the particle from three different reference frames: the ground, the ring, and the Ferris wheel cabin.
|}

:Ground reference frame: Since this is a Galilean reference frame, all forces on '''P''' are interaction forces.
{|
:[[File:ExE1-1-2- eng.png|thumb|right|230px|link=]]

|<math>\hspace{0.5cm}\sum\dot{\Ws}_\Ts=\dot{\Ws}_\Ts^\text{weight}= \left(\leftarrow \Rs\Omega_0\right) \cdot \left(\downarrow \ms\gs\right)=0</math>
|}

:Ring reference frame: It is a non-Galilean reference, and the transportation force on '''P''' must be taken into account.
{|
:[[File:ExE1-1-3-eng.png|thumb|right|250px|link=]]
|<math>\hspace{0.5cm}</math> AB:ground, REL:ring 

<math>\hspace{0.5cm}\vel{P}{REL}=\vel{P}{AB}-\vel{P}{ar}= \left(\leftarrow \Rs\Omega_0\right)-\left(\leftarrow \Rs\Omega_0\right)=\overline{0}</math> 

<math>\hspace{0.5cm}</math>

Since the velocity is zero, the total power is zero.
|}

:Cabin reference frame: It is also a non-Galilean reference, and the transportation force on '''P''' must be taken into account.

{|
:[[File:ExE1-1-4-eng.png|thumb|right|250px|link=]]
|<math>\hspace{0.5cm}</math>AB:ground, REL:cabin

<math>\hspace{0.5cm}\vel{P}{REL}=\vel{P}{AB}-\vel{P}{ar}= \left(\leftarrow \Rs\Omega_0\right)-\left(\uparrow \Rs\Omega_0\right)</math>

<math>\hspace{0.5cm}\dot{\Ws}_\text{cabin}^\text{weight}=\left[ \left(\leftarrow \Rs\Omega_0\right)+\left(\downarrow \Rs\Omega_0\right) \right] \cdot \left(\downarrow\ms\gs\right) =\ms\gs\Rs\Omega_0</math>

<math>\hspace{0.5cm}\dot{\Ws}_\text{cabin}^\text{ar}=\left[ \left(\leftarrow \Rs\Omega_0\right)+\left(\downarrow \Rs\Omega_0\right) \right] \cdot \left(\rightarrow\ms\Rs^2\Omega_0^2\right) =-\ms\gs\Rs^2 \Omega_0^3</math>

<math>\hspace{0.5cm}\sum\dot{\Ws}_\text{cabin}=\ms\Rs\Omega_0(\gs-\Rs\Omega_0^2)</math>

|}


====✏️ EXAMPLE E1-1.2: pendulum====
------
:
{|
:[[File:ExE1-2-1-eng.png|thumb|left|180px|link=]]
|A particle '''P''' of mass m hangs from an inextensible thread of length L attached to the ceiling. The assembly oscillates in a vertical plane fixed to the ground. We want to calculate the power of all the forces acting on the particle in two different reference frames: the ground and a platform rotating relative to the ground with a constant angular velocity <math>\Omega_0</math>. 

Ground reference frame: Since it is a Galilean reference frame, '''P''' is only subjected to interaction forces.
{|
:[[File:ExE1-2-2-neut.png|thumb|left|300px|link=]]

|<math>\dot{\Ws}_\Es^\Ts=(\nearrow\Ls\dot{\theta})\cdot(\nwarrow\Ts)=0</math>

<math>\dot{\Ws}_\Es^\text{weight}=(\nearrow\Ls\dot{\theta}\cdot(\downarrow \ms\gs)=-\ms\gs\Ls\dot{\theta}\sin\theta</math>

|}

Platform reference frame: It is a non Galilean reference frame, and the transportation force on '''P''' has to be taken into account.

{|
:[[File:ExE1-2-3-eng.png|thumb|left|400px|link=]]

|AB:ground, REL:platform

<math>\vel{P}{REL}=\vel{P}{AB}-\vel{P}{ar}=(\nearrow\Ls\dot{\theta})-(\otimes \Ls \Omega_0 \sin \theta)</math>

<math>\dot{\Ws}_\Es^\Ts\left[(\nearrow\Ls\dot{\theta})+(\odot\Ls\Omega_0\sin\theta)\right]\cdot(\nwarrow\Ts)=0</math>

<math>\dot{\Ws}_\Es^\text{weight}\left[(\nearrow\Ls\dot{\theta})+(\odot\Ls\Omega_0\sin\theta)\right]\cdot(\downarrow \ms\gs)=-\ms\gs\Ls\dot{\theta}\sin\theta</math>

<math>\dot{\Ws}_\Es^\text{ar}\left[(\nearrow\Ls\dot{\theta})+(\odot\Ls\Omega_0\sin\theta)\right]\cdot(\rightarrow \ms\Ls\Omega_0^2\sin\theta)=\ms\Ls^2\Omega_0^2\dot{\theta}\cos\theta</math>
|}



------
------

==E1.2 Power of an action-reaction pair==

The total power associated with internal interactions in a constant-mass system is not zero in principle (since the two forces of each action-reaction pair are applied to particles that may have different velocities relative to the reference frame in which the calculation is made), and it is independent of the reference frame in which it is evaluated. It only depends on the separation velocity of the two particles. The power balance for the system is therefore written as: <math>\dot{\Ts}_\Rs^\text{syst}=\dot{\Ws}_\Rs^\text{ext} + \dot{\Ws}^\text{int} + \dot{\Ws}_\Rs^\text{ar} </math> .

<div>
=====💭 PROOF ➕=====

{|
:[[File:E1-2-neut.png|thumb|left|200px|link=]]
|<math>\dot{\Ws}^\text{ac-reac}=\overline{\Fs}_{\Ps \rightarrow \Qs}\cdot \vel{Q}{R} + \overline{\Fs}_{\Qs \rightarrow \Ps}\cdot \vel{P}{R}= \overline{\Fs}_{\Ps \rightarrow \Qs}\cdot \left[ \vel{Q}{R}-\vel{P}{R}\right]</math>

By the [[D1. Foundational laws of Newtonian dynamics#D1.6 Newton’s third law (action-reaction principle)|'''action-reaction principle''']], <math>\overline{\Fs}_{\Ps \rightarrow \Qs}=-\overline{\Fs}_{\Qs \rightarrow \Ps}</math>. Therefore: <math>\dot{\Ws}^\text{ac-reac}=\overline{\Fs}_{\Ps \rightarrow \Qs}\cdot \left[ \vel{Q}{R}-\vel{P}{R}\right]</math>

As <math>\overline{\Fs}_{\Ps \rightarrow \Qs}</math> has the <math>\PQvec</math> direction, only the components of the velocities in that direction have to be considered:

<math>\dot{\Ws}^\text{ac-reac}=\overline{\Fs}_{\Ps \rightarrow \Qs}\cdot \biggl( \vel{Q}{R} \Bigr]_{||\Ps\Qs}-\ \vel{P}{R} \Bigr]_{||\Ps\Qs}\biggl)</math>
|}

The difference <math>\biggl( \vel{Q}{R} \Bigr]_{\Ps\Qs}-\ \vel{P}{R} \Bigr]_{\Ps\Qs}\biggl)</math> is the separation velocity <math>\dot{\rho}</math> between '''P''' and '''Q''', and it does not dpend on the reference frame since it is the derivative of a scalar (the distance <math>\dot{\rho}</math> between the two particles). For the previous drawing, <math>\dot{\rho}> 0</math>. Since this is an attractive interaction, <math>\dot{\Ws}^\text{ac-reac} < 0</math>.

With a drawing, it is easy to see that:
* attractive interaction and particle approach <math>(\dot{\rho}<0)</math> : <math>\dot{\Ws}^\text{ac-reac} > 0</math> ;
* repulsive interaction and particle separation <math>(\dot{\rho}>0)</math> : <math>\dot{\Ws}^\text{ac-reac} > 0</math>
* repulsive interaction and particle approach <math>(\dot{\rho}<0)</math> : <math>\dot{\Ws}^\text{ac-reac} < 0</math>


</div>

There are two particular cases of interaction forces whose power always has the same sign:

* Friction action-reaction pair between '''P''' and '''Q''': since the friction force always opposes the relative velocity between the two points, its power is always negative:

<math>\hspace{1cm} \dot{\Ws}^\text{friction}=\dot{\Ws}^{\text{fric}\rightarrow\Ps}_\text{RTQ}+\dot{\Ws}^{\text{fric}\rightarrow\Qs}_\text{RTQ}=\overline{\Fs}^{\text{fric}\rightarrow\Ps} \cdot \vel{P}{RTQ} = - \abs{\overline{\Fs}^{\text{fric}\rightarrow\Ps}} \frac{\vel{P}{RTQ}}{\abs{\vel{P}{RTQ}}} \cdot \vel{P}{RTQ}<0;</math>

* Constraint action-reaction pair between '''P''' and '''Q''': from the [[D3. Interactions between rigid bodies#D3.4 Direct constraint interactions|'''analytical characterization equation''']] of the constraint forces between two points '''P''' and '''Q''' belonging to two rigid bodies <math>\mathrm{S}_\Ps</math> and <math>\mathrm{S}_\Qs</math>, it is easy to see that the associated power is zero:

<math>\hspace{1cm} \dot{\Ws}^\text{constraint}=\dot{\Ws}^{\text{constraint}\rightarrow\Ps}_{\mathrm{S}_\Ps}+\dot{\Ws}^{\text{constraint}\rightarrow\Qs}_{\mathrm{S}_\Ps}=\overline{\Fs}^{\text{constraint}\rightarrow\Ps} \cdot \overline{\mathbf{v}}_{\mathrm{S}_\Qs}(\Ps) = - 0;</math>

====✏️ EXAMPLE E1-2.1: power of an action-reaction friction pair====
------
:
{|
:[[File:ExE1-2-1-2-eng.png|thumb|left|180px|link=]]
|A homogeneous wheel, with radius r and mass m, moves down an inclined plane without sliding until it joins a conveyor belt moving at a constant velocity <math>\vs_0</math> relative to the ground. At the moment it makes contact with the belt, the center of the wheel has a velocity <math>\vs_0</math> relative to the ground in the same direction as the belt. We want to calculate the power of the friction force between the belt and the wheel at this instant.
|}

At that momento, the <math>\mathbf{ICR}^\text{wheel}_\Es</math> is point <math>\Js_\text{wheel}</math> in contact with the belt. This point slides backwards with respect to the belt, therefore the friction force on the wheel is forwards:

<math>\left.\begin{array}{l}
\text { AB: ground } \\
\text { REL: belt }
\end{array}\right\} \quad \bar{\vs}_{\text {REL }}\left(\Js_{\text {wheel }}\right)=\bar{\vs}_{\text{AB}}\left(\Js_{\text {wheel }}\right)-\bar{\vs}_{\mathrm{ar}}\left(\Js_{\text {wheel }}\right)=-\bar{\vs}_{\mathrm{ar}}\left(\Js_{\text {wheel }}\right)=-\left(\rightarrow \mathrm{v}_0\right)=\left(\leftarrow \mathrm{v}_0\right) .</math>

[[File:ExE1-2-1-3-eng.png|thumb|center|500px|link=]]

Calculation of the friction power: <math>\dot{\Ws}^\text{friction}=\dot{\Ws}_\Rs^{\text{friction}\rightarrow \text{wheel}}+ \dot{\Ws}_\Rs^{\text{friction}\rightarrow \text{belt}}</math>, where R may be any reference frame. To quickly calculate this power, it is advisable to choose a reference frame R where one of the two points subjected to the action-reaction pair has zero speed, so that one of the two terms is directly zero:

<math>\dot{\Ws}^\text{friction}=\dot{\Ws}_\text{belt}^{\text{fric}\rightarrow \text{wheel}}+ \dot{\Ws}_\text{belt}^{\text{fric }\rightarrow \text{belt}}=\overline{\Fs}^{\text{fric}\rightarrow \text{wheel}} \cdot \bar{\vs}_{\text {belt}}\left(\Js_{\text {wheel }}\right)=(\rightarrow \mu\ms\gs)\cdot(\leftarrow \vs_0)=-\mu\ms\gs\vs_0.</math>

If the calculation is done in any other reference frame, the result is the same:

<math>\dot{\Ws}^\text{friction}=\dot{\Ws}_\text{E}^{\text{fric}\rightarrow \text{wheel}}+ \dot{\Ws}_\text{E}^{\text{fric} \rightarrow \text{belt}}=(\rightarrow \mu\ms\gs)\cdot 0 + (\leftarrow \mu\ms\gs)\cdot(\rightarrow \vs_0)=-\mu\ms\gs\vs_0.</math>

<math>\dot{\Ws}^\text{friction}=\dot{\Ws}_\text{vehicle}^{\text{fric}\rightarrow \text{wheel}}+ \dot{\Ws}_\text{vehicle}^{\text{fric} \rightarrow \text{belt}}=(\rightarrow \mu\ms\gs)\cdot (\rightarrow\vs_0) + (\leftarrow \mu\ms\gs) \cdot(\rightarrow 3\vs_0)=-\mu\ms\gs\vs_0.</math>

However, if only the power of one of the two forces is calculated (for example, the friction on the wheel), the result depends on the reference frame and can have any sign:

<math>\dot{\Ws}_\text{belt}^{\text{fric }\rightarrow \text{wheel}}=(\rightarrow \mu\ms\gs)\cdot(\leftarrow \vs_0)=-\mu\ms\gs\vs_0 ,</math>

<math>\dot{\Ws}_\text{E}^{\text{fric} \rightarrow \text{wheel}}=(\rightarrow \mu\ms\gs)\cdot 0 = 0 ,</math>

<math>\dot{\Ws}_\text{vehicle}^{\text{fric} \rightarrow \text{wheel}}=(\rightarrow \mu\ms\gs)\cdot (\rightarrow\vs_0)=\mu\ms\gs\vs_0.</math>


-----
-----

==E1.3 Power of a system of forces on a rigid body==

{|
|
:[[File:E1-3-eng.png|thumb|left|180px|link=]]
<center>'''Figure E1.3''': Kinematic torsor of a rigid body in a reference frame R, and forces on different points (dm) of that rigid body</center>

|When calculating the power of a system of forces on a rigid body S, the term associated with internal interactions is zero because those interactions are all constraint (cohesion) forces and, as seen in [[E1. Work-Energy Theorem: differential form#E1.2 Power of an action-reaction pair|'''section E1.2''']], their total power is zero.

The total power of a system of forces on S can be calculated from the torsor of the system at any point '''O''' of the rigid body, <math>(\overline{\Fs},\overline{\Ms}(\Os))</math>. Taking into account rigid body kinematics ('''Figure E1.3'''):

<math>\dot{\Ws}_\Rs^{\rightarrow \Ss} =\sum_\Ps\overline{\Fs}^{\rightarrow \Ps} \cdot \vel{P}{R}= \sum_\Ps \left[ \overline{\Fs}^{\rightarrow \Ps} \cdot \vel{O}{R} + \velang{S}{R} \times \OPvec \right] </math>

|}
<math>\dot{\Ws}_\Rs^{\rightarrow \Ss} =\left(\sum_\Ps\overline{\Fs}^{\rightarrow \Ps} \right)\cdot \vel{O}{R} + \sum_\Ps\overline{\Fs}^{\rightarrow \Ps} \cdot \left(\velang{S}{R} \times \OPvec \right) = \left( \sum_\Ps\overline{\Fs}^{\rightarrow \Ps} \right) \cdot \vel{O}{R} + \left[ \sum_\Ps \left( \OPvec \times \overline{\Fs}^{\rightarrow \Ps} \right)\right] \cdot \velang{S}{R} = \overline{\Fs}\cdot \vel{O}{R} + \overline{\Ms}(\Os)\cdot \velang{S}{R} </math>

This calculation also shows that the power of a moment (or a torque <math>\Gamma</math>) on a rigid body in a reference frame R is the product of the moment by the angular velocity of the rigid body: <math>\dot{\Ws}_\Rs^{\Gamma\rightarrow \mathrm{S}} = \overline{\Gamma} \cdot \velang{S}{R}</math>.

The power balance for a rigid body is: <math>\dot{\Ws}_\Rs^{ \rightarrow \mathrm{S}} = \dot{\Ws}_\Rs^{ \mathrm{ext}} + \dot{\Ws}_\Rs^{ \mathrm{ar}}.</math>

====✏️ EXAMPLE E1-3.1: roller sliding on a slope====
------
:
{|
|
:[[File:ExE1-3-1-1-eng.png|thumb|left|180px|link=]]
|A homogeneous roller, with mass m and radius r, slides down a slope. The friction coefficient between the ground and the roller is <math>\mu</math>. We want to calculate the power of every force acting on the roller in the ground reference frame.
|}

:It is a 2D problem. The kinematic and dynamic descriptions are:
{|
|
[[File:ExE1-3-1-2-eng.png|thumb|right|250px|link=]]

|<math> \hspace{0.5cm}\vel{J}{E}=\vel{C}{T} + \velang{}{E} \times \CJvec= </math>

<math>\hspace{1.7cm}=\left( \rightarrow \vs_\Es \right) + \left( \otimes \Omega_\Es\right) \times \left( \downarrow \rs \right)</math>

<math> \hspace{0.5cm}\vel{J}{E} = \left[\rightarrow (\vs_\Es -\rs\Omega_\Es)\right]</math>

<math>\hspace{0.5cm} \dot{\Ws}_\Es^{\mathrm{ext}} = \dot{\Ws}_\Es^{\mathrm{weight}} + \dot{\Ws}_\Es^{\mathrm{friction}} + \dot{\Ws}_\Es^{\mathrm{constraint}}= \Bigr[\left(\downarrow \ms\gs\cos\beta\right)+\left(\rightarrow \ms\gs\sin\beta\right)\Bigr] \cdot (\rightarrow \vs_\Es) + \Bigr[(\leftarrow \mu \Ns) + (\uparrow \Ns ) \Bigr] \cdot \Bigr[\rightarrow (\vs_\Es -\rs\Omega_\Es)\Bigr]</math>

|}
:Since the roller center has no motion perpendicular to the plane, the AMT leads to <math>\Ns =\ms\gs\cos\beta</math>. Therefore:

{|
|
[[File:ExE1-3-1-3-eng.png|thumb|right|140px|link=]]
|<math>\hspace{0.5cm} \dot{\Ws}_\Ts^{\mathrm{ext}} = \ms\gs\vs_\Ts \sin \beta - \ms\gs(\vs_\Ts -\rs\Omega_\Ts)\cos\beta=\ms\gs \Bigr[ \vs_\Ts (\sin\beta-\cos\beta)+\rs\Omega_\Ts \cos\beta \Bigr].</math>

<math>\hspace{0.5cm}</math>

If instead of considering the friction at '''J''' we consider the equivalent torsor at point '''C''', we must consider the force applied in '''C''' and the moment about '''C''', which is not zero:

<math>\hspace{0.5cm} \dot{\Ws}_\Es^{ \mathrm{friction}} = (\leftarrow \mu\Ns)\cdot(\rightarrow\vs_\Es)+(\otimes \mu\Ns\rs)\cdot(\otimes \Omega_\Es) =\mu\ms\gs\cos\beta(\rs\Omega_\Es-\vs_\Es)</math>,

<math>\hspace{0.5cm} \dot{\Ws}_\Es^{\mathrm{ext}} = \dot{\Ws}_\Es^{ \mathrm{weight}} + \dot{\Ws}_\Es^{ \mathrm{friction}}+ \dot{\Ws}_\Es^{ \mathrm{constraint}}= \ms\gs\vs_\Es\sin\beta+\mu\ms\gs\cos\beta (\rs\Omega_\Es-\vs_\Es)</math>.

<math>\hspace{0.5cm}</math>

The result is the same.
|}


====✏️ EXAMPLE E1-3.2: block sliding on a truck====
------
:
{|
|
:[[File:ExE1-3-2-1-eng.png|thumb|left|220px|link=]]
|A truck brakes with acceleration '''a''' relative to the ground at the instant in which its velocity (also relative to the ground) is v. At the same instant, the homogeneous block, of mass m, slides with velocity v' relative to the truck. We want to perform a power balance for the block in two different reference frames: the ground and the truck.
|}

:Ground reference frame: It is a 2D problem. The kinematic and dynamic descriptions are:
{|
|
[[File:ExE1-3-2-2-eng.png|thumb|right|200px|link=]]

| The constraint between the truck and the block allows only one translational DoF between the two. Therefore, in the 2D approximation, the associated torsor has two independent components. If characterized at the midpoint of the contact surface, these components are a normal force and a moment perpendicular to the plane. The AMT applied to the block leads to <math>\Ns=\ms\gs</math>.

|}
:The powers associated with all interactions on the block in the ground reference are:

:<math>\dot{\Ws}_\Es^{\rightarrow \mathrm{bloque}}=\dot{\Ws}_\Ts^{\mathrm{ext}}=\dot{\Ws}_\Ts^{\mathrm{peso}}+\dot{\Ws}_\Ts^{\mathrm{fricción}}+\dot{\Ws}_\Ts^{\mathrm{enlace}}= \Bigr[(\downarrow \ms\gs)+(\leftarrow \mu\ms\gs)+(\uparrow \Ns) \Bigr] \cdot \Bigr[\rightarrow(\vs+\vs')\Bigr] + (\odot \Ms_\es)\cdot 0 = -\mu\ms\gs(\vs+\vs')</math>

:Ya que el bloque se traslada respecto al suelo, su energía cinética total es inmediata, y se puede plantear fácilmente el balance de potencias:

:<math>\Ts_\Ts^\mathrm{bloque}=\frac{1}{2}\int_\mathrm{bloque}\ds\ms \vs_\Ts^2(\mathrm{bloque})=\frac{1}{2}\ms \vs_\Ts^2(\mathrm{bloc})\quad \Rightarrow \quad \dot{\Ts}_\Ts^\mathrm{bloque}=\ms\vs_\Ts( \mathrm{bloque})\dot{\vs}_\Ts( \mathrm{bloque}) </math>

:Como la única fuerza horizontal sobre el bloque es la fricción:

<math>\dot{\vs}_\Ts(\mathrm{bloque})=-\mu\gs\quad \Rightarrow \quad \dot{\Ts}_\Ts^\mathrm{bloque}=-\mu\ms\gs(\vs+\vs').</math>

:El balance de potencias es consistente:

:<math>\left.\begin{array}{l}
\dot{\mathrm{W}}_\Ts^{\rightarrow \text { bloque }}=\dot{\mathrm{W}}_{\mathrm{T}}^{\text {ext }}=-\mu \mathrm{mg}\left(\mathrm{v}+\mathrm{v}^{\prime}\right) \\
\dot{\mathrm{T}}_{\mathrm{T}}^{\text {bloque }}=-\mu \mathrm{mg}\left(\mathrm{v}+\mathrm{v}^{\prime}\right)
\end{array}\right\} \Rightarrow \dot{\mathrm{W}}_{\mathrm{T}}^{\rightarrow \text { bloque }}=\dot{\mathrm{T}}_{\mathrm{T}}^{\text {bloque }}</math>

:Referencia camión: Al estar el camión acelerado respecto al suelo, hay que tener en cuenta las fuerzas de arrastre sobre todos los diferenciales de masa (dm) del bloque:

{|
|

[[Archivo:ExE1-3-2-3-cat-esp.jpg|thumb|right|250px|link=]]
|<math>\cal{\Fs}^{\rightarrow \ds\ms}_{\mathrm{ar}}=-\ds\ms\overline{\as}_\mathrm{ar}(\ds\ms)=-\ds\ms (\leftarrow \as)=(\rightarrow \ds\ms \cdot \as)</math>.

El torsor resultante de este sistema de fuerzas en el centro de inercia '''G''' se reduce a una fuerza horizontal.

<math>\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{bloque}}=\dot{\Ws}_\mathrm{camión}^{\mathrm{ext}} + \dot{\Ws}_\mathrm{camión}^{\mathrm{ar}} =\dot{\Ws}_\mathrm{camión}^{\mathrm{peso}} + \dot{\Ws}_\mathrm{camión}^{\mathrm{fricción}} + \dot{\Ws}_\mathrm{camión}^{\mathrm{enlace}} + \dot{\Ws}_\mathrm{camión}^{\mathrm{ar}} , </math>

<math>\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{bloque}}= \Bigr[(\downarrow \ms\gs)+(\leftarrow \mu\ms\gs)+(\uparrow \Ns) + (\rightarrow \ms\as)\Bigr] \cdot(\rightarrow \vs')+(\odot\Ms_\es) \cdot0=\ms(\as-\mu\gs)\vs'.</math>
|}

:Se puede comprobar que el balance de potencias es consistente:

:<math>\Ts^\mathrm{bloque}_\mathrm{camión}=\frac{1}{2}\ms\vs^2_\mathrm{camión}(\mathrm{bloque}) \quad \Rightarrow \quad \dot{\Ts}^\mathrm{bloque}_\mathrm{camión} = \ms\vs_\mathrm{camión}(\mathrm{bloque})\dot{\vs}_\mathrm{camión}(\mathrm{bloque})</math>

:<math>\left.\begin{array}{l}
\mathrm{AB}:\mathrm{suelo} \\
\mathrm{REL}:\mathrm{camión}
\end{array}\right\} \quad \dot{\vs}_\mathrm{camión}(\mathrm{bloque})= \dot{\vs}_\mathrm{T}(\mathrm{bloque})-\dot{\vs}_\mathrm{ar}(\mathrm{bloque}) = (-\mu\gs)-(-\as)=\as-\mu\gs \quad \Rightarrow \quad \dot{\Ts}^\mathrm{bloque}_\mathrm{camión}=\ms\vs'(\as-\mu\gs)</math>

:<math>\left.\begin{array}{l}
\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{bloque}}=\dot{\Ws}_\mathrm{camión}^{\mathrm{ext}} + \dot{\Ws}_\mathrm{camión}^{\mathrm{ar}}= \ms(\as-\mu\gs)\vs'\\
\dot{\Ts}^\mathrm{bloque}_\mathrm{camión}=\ms\vs'(\as-\mu\gs)
\end{array}\right\} \Rightarrow\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{bloque}}= \dot{\Ts}^\mathrm{bloque}_\mathrm{camión}</math>


====✏️ EJEMPLO E1-3.3: camión frenando con carga====
------
:
{|
|
:[[Archivo:ExE1-3-3-1-esp.png|thumb|left|220px|link=]]
|Un camión frena con aceleración '''a''' respecto al suelo en el instante en el que su velocidad (también respecto al suelo) es v. Los pares de frenada en el eje delantero y en el trasero son iguales y valen <math>\Gamma</math>, y las ruedas, de radio r y masa despreciable, no deslizan sobre el suelo. En el mismo instante, el bloque homogéneo, de masa m, desliz con velocidad v' relativa al camión. Se trata de hacer un balance de potencias para el sistema (bloque+camión) en la referencia del suelo y en la del camión.

|}

:Ya que la potencia de las interacciones internas no es nula en principio (no se trata de un único sólido rígido), es conveniente hacer un DGI del sistema para no olvidar ninguna:
{|
|
[[Archivo:ExE1-3-3-2-esp.png|thumb|right|250px|link=]]

|
|
c.p.s.d.: contacto puntual sin deslizamiento

c.m.c.d.: contacto multipuntual con deslizamiento

|}
:El sistema sobre el que se hace el balance de potencias es todo menos el suelo. Teniendo en cuenta que la potencia de los enlaces internos es siempre nula, las únicas interacciones que hay que tener en cuenta son el peso, los enlaces de las ruedas con el suelo, los pares de frenada y la fricción entre bloque y camión:

[[Archivo:ExE1-3-3-3-esp.png|thumb|center|300px|link=]]

:Referencia suelo:
:Las fuerzas de enlace del suelo sobre las ruedas están aplicadas a puntos de velocidad cero, y por tanto no hacen potencia. El peso es una fuerza vertical, y el movimiento del camión y del bloque es horizontal, por tanto su potencia también es cero.

:<math>\dot{\Ws}_\Ts^{\rightarrow \mathrm{sist}}=\dot{\Ws}_\Ts^{\mathrm{ext}}+\dot{\Ws}^{\mathrm{int}}=\dot{\Ws}_\Ts^{\mathrm{peso}}+\dot{\Ws}_\Ts^{\mathrm{enlace}}+\dot{\Ws}^{\mathrm{fricción}}+\dot{\Ws}^{\Gamma}=\dot{\Ws}^{\mathrm{fricción}}+\dot{\Ws}^{\Gamma}.</math>

:Las potencias de interacciones internas se pueden calcular en cualquier referencia. Por tanto:

:<math>\dot{\Ws}^{\mathrm{fricción}}= \dot{\Ws}_\mathrm{camión}^{\mathrm{fric}\rightarrow \mathrm{bloque}} +\dot{\Ws}_\mathrm{camió}^{\mathrm{fric}\rightarrow \mathrm{camión}} = \dot{\Ws}_\mathrm{camión}^{\mathrm{fric}\rightarrow \mathrm{bloque}}= (\leftarrow \mu\ms\gs)\cdot (\rightarrow \vs')=-\mu\ms\gs\vs'</math>

:<math>\dot{\Ws}^{\Gamma}= \dot{\Ws}_\mathrm{camión}^{\Gamma\rightarrow \mathrm{ruedas}} +\dot{\Ws}_\mathrm{camión}^{\Gamma\rightarrow \mathrm{camión}} = \dot{\Ws}_\mathrm{camión}^{\Gamma\rightarrow \mathrm{ruedas}}= 2 \left(\odot \Gamma \right)\cdot \left(\otimes \frac{\vs}{\rs}\right)=-2\Gamma\frac{\vs}{\rs}.</math>

:Por otro lado: <math>\Ts_\Ts^\mathrm{sist}=\frac{1}{2}\ms\vs_\Ts^2(\mathrm{bloque})+\frac{1}{2}\ms\vs_\Ts^2(\mathrm{camión})\quad \Rightarrow \quad \dot{\Ts}_\Ts^\mathrm{sist}=-\ms(\vs+\vs')\mu\gs-\Ms\vs\as. </math>

:Finalmente: <math>\dot{\Ws}_\Ts^{\rightarrow \mathrm{sist}}=\dot{\Ts}_\Ts^\mathrm{sist} \quad \Rightarrow \quad -\mu\ms\gs\vs'-2\Gamma\frac{\vs}{\rs}=-\ms(\vs+\vs')\mu\gs-\Ms\vs\as.</math>

:Esta ecuación permite calcular el par de frenada: <math>\Gamma=\frac{1}{2}(\mu\ms\gs+\Ms\as)\rs.</math>

:Referencia camión:

:En esta referencia hay que tener en cuenta las fuerzas de arrastre. Por otro lado, las fuerzas de enlace del suelo sobre las ruedas están aplicadas a puntos que ya no tienen velocidad cero, y por tanto hacen potencia. Como antes, la potencia del peso es cero.

:<math>\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{sist}}=\dot{\Ws}_\mathrm{camión}^{\mathrm{ext}}+\dot{\Ws}^{\mathrm{int}}+\dot{\Ws}_\mathrm{camión}^{\mathrm{ar}}=\dot{\Ws}_\mathrm{camión}^{\mathrm{enlace}}+\dot{\Ws}^{\mathrm{fricción}}+\dot{\Ws}^{\Gamma}+\dot{\Ws}_\mathrm{camión}^{\mathrm{ar}}.</math>

:Las fuerzas de arrastre sobre el camión no hacen potencia porque el camión está parado en esta referencia. La de las fuerzas de arrastre sobre el bloque se ha calculado en el [[E1. Teorema de la energía: versión diferencial#✏️ ✏️ EJEMPLO E1-3.2: bloque deslizando sobre un camión|'''ejemplo E1-3.2''']]. La potencia de la fricción y de los pares de frenada es la misma de antes ya que no depende de la referencia.

:<math>\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{sist}}=\dot{\Ws}_\mathrm{camión}^{\mathrm{enllaç}}+\dot{\Ws}^{\mathrm{fricción}}+\dot{\Ws}^{\Gamma}+\dot{\Ws}_\mathrm{camión}^{\mathrm{ar} \rightarrow \mathrm{bloque}}= \dot{\Ws}_\mathrm{camión}^{\mathrm{enlace}} -\mu\ms\gs\vs'-2\Gamma\frac{\vs}{\rs} +\ms\as\vs'.</math>

:Las descripciones cinemática y dinámica de la rueda son:

[[Archivo:ExE1-3-3-4-esp.png|thumb|center|430px|link=]]

:<math>\dot{\Ws}_\mathrm{camión}^{\mathrm{enlace}}=2\Bigr[(\leftarrow \Ts)+(\uparrow \Ns)\Bigr]\cdot(\leftarrow\vs)=2\Ts\vs.</math>

La fuerza tangencial de enlace se puede calcular a partir del TMC aplicado al centro '''C''' de la rueda. Como no tiene masa: <math>\sum\overline{\Ms}_\mathrm{ext}(\Cs)=\overline{0} \quad \Rightarrow \quad (\odot \Gamma)+(\otimes \Ts\rs)=0\quad \Rightarrow \quad \Ts=\Gamma/\rs.</math>

:Por tanto: <math>\dot{\Ws}_\mathrm{camió}^{\rightarrow \mathrm{sist}} =2\frac{\Gamma}{\rs}\vs-\mu\ms\gs\vs'-2\Gamma\frac{\vs}{\rs}+\ms\as\vs'=\ms(\as-\mu\gs)\vs'.</math>

:Por otro lado, como se ha visto en el [[E1. Teorema de la energía: versión diferencial#✏️ ✏️ EJEMPLO E1-3.2: bloque deslizando sobre un camión|'''ejemplo E1-3.2''']]:

:<math>\Ts_\mathrm{camión}^\mathrm{sist}= \Ts_\mathrm{camión}^\mathrm{bloque}=\frac{1}{2}\ms\vs_\mathrm{camión}^2(\mathrm{bloque}) \quad \Rightarrow \quad \dot{\Ts}_\mathrm{camión}^\mathrm{sist}= \dot{\Ts}_\mathrm{camión}^\mathrm{bloque}=\ms\vs'(\as-\mu\gs).</math>

:El balance de potencias es consistente:

:<math>\left.\begin{array}{l}
\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{sist}}= \ms(\as-\mu\gs)\vs'\\
\dot{\Ts}^\mathrm{sist}_\mathrm{camión}=\ms(\as-\mu\gs)\vs'
\end{array}\right\} \Rightarrow\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{sist}}= \dot{\Ts}^\mathrm{sist}_\mathrm{camión}</math>

-----
-----

==E1.4 Balance de potencias para un sistema multisólido: cálculo directo y cálculo indirecto==

El '''cálculo directo''' (a partir de su definición) de la potencia asociada a una interacción (una fuerza sobre un punto '''P''' o un momento sobre un sólido rígido S) exige el conocimiento de la interacción (fuerza o momento):

<math>\dot{\Ws}_\Rs^{\Fs\rightarrow \Ps}=\overline{\Fs}_{\rightarrow\Ps} \cdot \vel{P}{R} \quad , \quad \dot{\Ws}_\Rs^{\Ms\rightarrow \Ss}=\overline{\Ms}_{\rightarrow\Ss} \cdot \velang{S}{R} </math>

Cuando no se trata de una interacción formulable, hay que recurrir a los teoremas vectoriales para determinarla.

El '''cálculo indirecto''' es una alternativa que no necesita recurrir a estos teoremas. Se basa en la evaluación de un balance de potencias para un sistema y una referencia adecuados de manera que la potencia a determinar sea el único término desconocido de la ecuación (el único que no se puede calcular de manera directa). Cuando el balance incluye más de un término desconocido, se pueden combinar varios balances (para distintos sistemas y referencias) hasta tener un número suficiente de ecuaciones que permitan la determinación de todas las potencias desconocidas.

====✏️ EJEMPLO E1-4.1: polea frenada====
------
:
{|
|
:[[Archivo:ExE1-4-1-1-esp.png|thumb|left|150px|link=]]
|Un bloque de masa m cuelga de un hilo inextensible atado a un punto de la periferia de una polea de radio r. Entre soporte (fijo al suelo) y polea actúa un freno que garantiza que el bloque baja a velocidad constante respecto al suelo. Se trata de calcular la potencia del freno.

|}

:Càlculo directo: <math>\dot{\Ws}^{\mathrm{freno}}= \overline{\Gamma}^{\rightarrow \mathrm{polea}}\cdot\velang{polea}{T}=\left(\odot\Gamma\right)\cdot\left( \otimes \frac{\vs_0}{\rs}\right)=-\Gamma \frac{\vs_0}{\rs}.</math>

:El par de frenada no es un dato, y hay que recurrir al TMC para calcularlo.

:<math>\left.\begin{array}{l}
\text{SISTEMA: polea + bloque}\\
\text{TMC en } \Os
\end{array}\right\} \sum\overline{\Ms}_\mathrm{ext}(\Os)=\dot{\overline{\Hs}}_\mathrm{RTO}(\Os)</math>

:El momento cinético proviene de las dos piezas. Ya que '''O''' no es fijo al bloque, hay que hacer descomposición baricéntrica para calcular el momento cinético del bloque:

:<math>\overline{\Hs}_\mathrm{RTO=T} (\Os) =\overline{\Hs}_\Ts^\mathrm{politja}(\Os) + \overline{\Hs}_\Ts^\mathrm{bloc}(\Os)= \overline{\Hs}_\Ts^\mathrm{politja}(\Os) + \overline{\Hs}_\Ts^\mathrm{bloc}(\Gs_\mathrm{bloc})+ \overline{\Hs}_\Ts^{\oplus\mathrm{bloc}}(\Os) = </math>

:<math>\hspace{2.3cm}=\Is_\mathrm{O}^\mathrm{polea} \velang{polea}{T}+\Is_{\Gs_\mathrm{bloque}}^\mathrm{bloque}+\OGvec_\mathrm{bloque}\times \ms\overline{\vs}_\Ts(\Gs_\mathrm{bloque})=\left(\otimes \Is_\mathrm{O}\frac{\vs_0}{\rs}\right)+(\otimes \ms\vs_0\rs).</math>

:[[Archivo:ExE1-4-1-2-esp.png|thumb|right|180px|link=]]
:Se trata de un momento cinético constante en dirección y valor. Por tanto:

:<math>\dot{\overline{\Hs}}_\mathrm{RTO}(\Os)=0 \quad \Rightarrow \quad \sum{\overline{\Ms}}_\mathrm{ext}(\Os)=(\odot\Gamma)+(\otimes \ms\gs\rs) =0\quad \Rightarrow \quad \Gamma= \ms\gs\rs</math>.

:Finalmente: <math>\dot{\Ws}^{\mathrm{freno}}= -\Gamma \frac{\vs_0}{\rs} = -\ms\gs\vs_0.</math>

:Cálculo indirecto:

:<math>\left.\begin{array}{l}
\text{SISTEMA: polea + soporte + bloque}\\
\text{REF: suelo(Gal)}
\end{array}\right\} \quad \dot{\Ts}^\mathrm{sist}_\Ts=\dot{\Ws}^\mathrm{ext}_\Ts+ \dot{\Ws}^\mathrm{int}</math>

:<math>\left.\begin{array}{l}
\dot{\Ts}^\mathrm{sist}_\Ts=0\\
\dot{\Ws}^\mathrm{ext}_\Ts=\left(\downarrow \ms\gs\right)\cdot\left(\downarrow \vs_0\right)= \ms\gs\vs_0\\
\dot{\Ws}^\mathrm{int}=\dot{\Ws}^{\mathrm{freno}}
\end{array}\right\} \quad \Rightarrow \quad \dot{\Ws}^{\mathrm{freno}} =-\ms\gs\vs_0</math>



====✏️ EJEMPLO E1-4.2: mono y plátanos====
------
:
{|
|
:[[Archivo:ExE1-4-2-1-esp.png|thumb|left|180px|link=]]
|Unos plátanos de masa <math>(2/3)\ms</math> cuelgan de una cuerda inextensible que se apoya sin deslizar sobre una polea de radio r y masa despreciable, articulada respecto al suelo. En el otro extremo de la cuerda se agarra un mono de masa m. La polea gira con velocidad angular constante <math>\Omega_0</math> respecto al suelo bajo la acción de un motor. El mono se mueve respecto a la cuerda sin que sus manos se froten contra la cuerda, y de manera que la distancia de su centro de masas al suelo se mantiene constante. Se trata de calcular la potencia del motor.
|}

:Cálculo directo: <math>\dot{\Ws}^{\mathrm{freno}}= \overline{\Gamma}^{\rightarrow \mathrm{polea}}\cdot\velang{polea}{T}=\left(\odot\Gamma\right)\cdot\left( \odot \Omega_0\right)=\Gamma \Omega_0.</math>

:El par motor no es un dato, y hay que recurrir al TMC para calcularlo.

:<math>\left.\begin{array}{l}
\text{SISTEMA: polea + mono + plátanos}\\
\text{TMC en } \Os
\end{array}\right\} \sum\overline{\Ms}_\mathrm{ext}(\Os)=\dot{\overline{\Hs}}_\mathrm{RTO}(\Os)</math>

:El momento cinético proviene del mono y de los plátanos. En primera aproximación, se puede considerar la masa del mono concentrada en su tronco y su cabeza, de manera que no gira y su velocidad respecto al suelo es nula. Teniendo en cuenta que el punto '''O''' no se mueve respecto al mono pero sí respecto a los plátanos:

:<math>\overline{\Hs}_\mathrm{RTO=T} (\Os) =\overline{\Hs}_\Ts^\mathrm{mono}(\Os) + \overline{\Hs}_\Ts^\mathrm{plátanos}(\Os)= \overline{\Hs}_\Ts^\mathrm{mono}(\Os) + \overline{\Hs}_\Ts^\mathrm{plátanos}(\Gs_\mathrm{plát})+ \OGvec_\mathrm{plát} \times \frac{2}{3} \ms \overline{\vs}_\Ts(\Gs_\mathrm{plát})= </math>

:<math>\hspace{2.3cm}=\OGvec_\mathrm{plát}\times \frac{2}{3} \ms \overline{\vs}_\Ts(\Gs_\mathrm{plát}) = \Bigr[(\rightarrow \rs)+(\downarrow \xs)\Bigr] \times\frac{2}{3} \ms \left(\downarrow \rs\Omega_0\right)=\left(\otimes \frac{2}{3}\ms\rs^2\Omega_0\right).</math>

:Se trata de un momento cinético constante en dirección y valor. Por tanto:

:<math>\dot{\overline{\Hs}}_\mathrm{RTO}(\Os)=0 \quad \Rightarrow \quad \sum{\overline{\Ms}}_\mathrm{ext}(\Os)=(\odot\Gamma)+(\otimes \ms\gs\rs) + \left(\otimes \frac{2}{3}\ms\gs\rs\right)=0\quad \Rightarrow \quad \Gamma= \frac{5}{3}\ms\gs\rs</math>.

:Finalmente: <math>\dot{\Ws}^{\mathrm{motor}}= \Gamma \Omega_0 = \frac{5}{3}\ms\gs\rs\Omega_0.</math>

:Cálculo indirecto

:Hay que tener presente que es un problema con dos tipos de actuadores: el motor y la musculatura del mono. Hay diversos modelos biomecánicos para formular las fuerzas de los músculos, pero quedan fuera del ámbito de este curso. Ya que el mono se mueve, sus músculos desarrollan una potencia no nula.

:<math>\left.\begin{array}{l}
\text{SISTEMA: polea + mono + plátano}\\
\text{REF: suelo(Gal)}
\end{array}\right\} \quad \dot{\Ts}^\mathrm{sist}_\Ts=\dot{\Ws}^\mathrm{ext}_\Ts+ \dot{\Ws}^\mathrm{int}</math>

:<math>\left.\begin{array}{l}
\dot{\Ts}^\mathrm{sist}_\Ts=0\\
\dot{\Ws}^\mathrm{ext}_\Ts=\left(\downarrow \frac{2}{3}\ms\gs\right)\cdot\left(\downarrow \rs\Omega_0\right)= \frac{2}{3}\ms\gs\rs\Omega_0\\
\dot{\Ws}^\mathrm{int}=\dot{\Ws}^{\mathrm{musc}}+\dot{\Ws}^{\mathrm{motor}}
\end{array}\right\} \quad \Rightarrow \quad \dot{\Ws}^{\mathrm{musc}} + \dot{\Ws}^{\mathrm{motor}}=-\frac{2}{3}\ms\gs\rs\Omega_0</math>

:Este balance proporciona una ecuación con dos incógnitas. Se puede plantear un segundo balance para un sistema donde una de las dos no aparezca: un sistema sin el mono o un sistema sin el motor. Si se hace un balance de potencias para el sistema (mono + trozo de cuerda), las fuerzas externas a considerar son el peso y la tensión de la cuerda. Si se sigue trabajando en la referencia del suelo, esta tensión (desconocida) hará potencia no nula ya que está aplicada a un punto de velocidad ascendente.
:[[Archivo:ExE1-4-2-2-cat-esp.png|thumb|right|330px|link=]]

:Para evitar que aparezca un término asociado a la tensión de la cuerda, se puede trabajar en la referencia del trozo de cuerda que se ha considerado, que al tener velocidad ascendente constante es galileana. En esta referencia, el mono baja con velocidad constante:

:<math>\left.\begin{array}{l}
\text{REL: trozo de cuerda}\\
\text{AB: suelo}
\end{array}\right\}</math>

:<math>\overline{\vs}_\mathrm{REL}(\mathrm{mono})=\overline{\vs}_\mathrm{ar}(\mathrm{mono})=-\overline{\vs}_\mathrm{ar}(\mathrm{mono})=(\downarrow \rs\Omega_0)</math>

:<math>\left.\begin{array}{l}
\text{SISTEMA: mono+ trozo de cuerda}\\
\text{REF: trozo de cuerda(Gal)}
\end{array}\right\} \quad \dot{\Ts}^\mathrm{sist}_\mathrm{cuerda} =\dot{\Ws}^\mathrm{ext}_\mathrm{cuerda}+ \dot{\Ws}^\mathrm{int}</math>

:<math>\left.\begin{array}{l}
\dot{\Ts}^\mathrm{sist}_\mathrm{cuerda}=0\\
\dot{\Ws}^\mathrm{ext}_\mathrm{cuerda}=\left(\downarrow \ms\gs\right)\cdot\left(\downarrow \rs\Omega_0\right)= \ms\gs\rs\Omega_0\\
\dot{\Ws}^\mathrm{int}=\dot{\Ws}^{\mathrm{musc}}
\end{array}\right\} \quad \Rightarrow \quad \dot{\Ws}^{\mathrm{musc}} =-\ms\gs\rs\Omega_0</math>

:Combinando este resultado con el del balance anterior:

:<math>\dot{\Ws}^\mathrm{motor}=-\frac{2}{3}\ms\gs\rs\Omega_0-\dot{\Ws}^\mathrm{musc}=-\frac{5}{3}\ms\gs\rs\Omega_0</math>



© Universitat Politècnica de Catalunya. [[Mecánica:Derechos de autor |Todos los derechos reservados]]
-------------

<center>
[[D8. Conservaciones|<<< D8. Conservaciones]]

[[E2. Teorema de la energía: versión integrada|E2. Teorema de la energía: versión integrada>>>]]
</center>

E1. Work-Energy Theorem: differential form

2025-05-20T12:15:02Z

Bros: /* ✏️ EXAMPLE E1-3.1: roller sliding on a slope */

<div class="noautonum">__TOC__</div>
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</math>

The '''Work-Energy Theorem''' plays a prominent role in physics, as it bridges all branches of science (electromagnetism, thermodynamics, etc.): all of them deal with energy.

There is a certain parallel between the Vector Theorems and the '''Work-Energy Theorem''': they all allow us to understand the evolution of a magnitude that depends on the mechanical state of the system. In the case of the former, we are dealing with the instantaneous evolution of vector magnitudes (linear momentum, angular momentum), while the latter deals with the finite change (in a finite time interval) of a scalar magnitude ('''mechanical energy''').

This unit presents the differential form of this theorem (the '''power balance'''). The integrated form (which is the Work-Energy Theorem itself) is presented in [[E2. Teorema de la energía: versión integrada|'''unit E2''']].
 
------
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==E1.1 Power balance in a system of particles==

:[[File:E1-1-eng.png|thumb|right|300px|link=]]
<center>'''Figure E1.1''' Forces on a particle of a system with constant matter</center>

Let us consider a system of constant matter, and let’s apply Newton's second law to each particle (or each mass differential) of this system in a general reference frame R (which may be non-Galilean) ('''Figure E1.1'''):

<math>\F{\rightarrow \Ps} + \Fcal{}{\text{ar}\rightarrow\Ps} + \Fcal{}{\text{Cor}\rightarrow\Ps} =\ms_\Ps\acc{P}{R}</math>

<math>\F{\rightarrow \Ps} </math> is the resultant interaction force on '''P''', and it comes from internal interactions (with particles of the same system) and external ones:

<math>\F{\rightarrow \Ps}=\F{\rightarrow \Ps}^\text{int}+\F{\rightarrow \Ps}^\text{ext}</math>

If both sides of the equation are multiplied by <math>\vel{P}{R}</math>, all forces (interaction and inertia forces) orthogonal to <math>\vel{P}{R}</math> are filtered out. Specifically, since <math>\acc{P}{R}=\accs{P}{R} + \accn{P}{R} </math>, where <math>\accn{P}{R} \perp \vel{P}{R}</math>, and since the Coriolis force is always orthogonal to <math>\vel{P}{R} (\Fcal{}{\text{Cor}\rightarrow\Ps} = -2\ms_\Ps \velang{R}{Gal} \times \vel{P}{R})</math>:

<math>\vel{P}{R} \cdot \left( \F{\rightarrow \Ps}^\text{int}+\F{\rightarrow \Ps}^\text{ext} + \Fcal{}{\text{ar}\rightarrow\Ps} \right) = \ms_\Ps\accs{P}{R} \cdot \vel{P}{R} = \frac{\ds}{\ds \ts} \left(\frac{1}{2}\ms_\Ps \vvec_\Rs^2(\Ps) \right) \equiv \frac{\ds\Ts_\Rs(\Ps)}{\ds\ts}=\dot{\Ts}_\Rs(\Ps)</math>

The scalar function <math>\Ts_\Rs(\Ps)</math> is the '''kinetic energy''' of the particle, and is a function of the [[D1. Foundational laws of Newtonian dynamics#D1.3 Newton’s Principle of Determinacy|'''mechanical state''']] of '''P'''. The scalar product of a force and the velocity of its application point is the '''force power''': <math>\vel{P}{R} \cdot \F{\rightarrow \Ps} \equiv \dot{\Ws}_\Rs^{\F{\rightarrow \Ps}} </math>. Both scalar quantities <math>(\Ts_\Rs(\Ps), \dot{\Ws}_\Rs^{\F{\rightarrow \Ps}})</math> depend on the reference frame. For the same set of interaction forces acting on '''P''', the reference frame determines which are filtered out and which are not.

The equation can be rewritten as <math>\dot{\Ws}_\Rs^{\text{ext}\rightarrow \Ps}+\dot{\Ws}_\Rs^{\text{int}\rightarrow \Ps}+\dot{\Ws}_\Rs^{\text{ar}\rightarrow \Ps} =\dot{\Ts}_\Rs(\Ps)</math>, and constitutes the '''power balance''' for particle '''P''' in the R reference frame. Summing for all particles in the system:

<math>\dot{\Ws}_\Rs^{\text{ext}}+\dot{\Ws}^{\text{int}}+\dot{\Ws}_\Rs^{\text{ar}} =\dot{\Ts}_\Rs^\text{syst}</math>

The calculation of kinetic energy is covered in [[E2. Teorema de la energía: versión integrada|'''unit E2''']]. This unit focuses on the calculation of powers. [[E1. Work-Energy Theorem: differential form#E1.2 Potencia de una pareja de acción-reacción|'''section E1-2''']] shows that the power of internal interactions does not depend on the reference frame (which is why the subscript R does not appear in the <math>\dot{\Ws}^{\text{int}}</math> term of the previous equation).

====✏️ EXAMPLE E1-1.1: free particle====
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:
{|
:[[File:ExE1-1-1-1-eng.jpg|thumb|left|180px|link=]]
|A particle '''P''', with mass m, is released from the ring of a Ferris wheel when it is at its highest point. We want to calculate the power, at this instant, of all forces acting on the particle from three different reference frames: the ground, the ring, and the Ferris wheel cabin.
|}

:Ground reference frame: Since this is a Galilean reference frame, all forces on '''P''' are interaction forces.
{|
:[[File:ExE1-1-2- eng.png|thumb|right|230px|link=]]

|<math>\hspace{0.5cm}\sum\dot{\Ws}_\Ts=\dot{\Ws}_\Ts^\text{weight}= \left(\leftarrow \Rs\Omega_0\right) \cdot \left(\downarrow \ms\gs\right)=0</math>
|}

:Ring reference frame: It is a non-Galilean reference, and the transportation force on '''P''' must be taken into account.
{|
:[[File:ExE1-1-3-eng.png|thumb|right|250px|link=]]
|<math>\hspace{0.5cm}</math> AB:ground, REL:ring 

<math>\hspace{0.5cm}\vel{P}{REL}=\vel{P}{AB}-\vel{P}{ar}= \left(\leftarrow \Rs\Omega_0\right)-\left(\leftarrow \Rs\Omega_0\right)=\overline{0}</math> 

<math>\hspace{0.5cm}</math>

Since the velocity is zero, the total power is zero.
|}

:Cabin reference frame: It is also a non-Galilean reference, and the transportation force on '''P''' must be taken into account.

{|
:[[File:ExE1-1-4-eng.png|thumb|right|250px|link=]]
|<math>\hspace{0.5cm}</math>AB:ground, REL:cabin

<math>\hspace{0.5cm}\vel{P}{REL}=\vel{P}{AB}-\vel{P}{ar}= \left(\leftarrow \Rs\Omega_0\right)-\left(\uparrow \Rs\Omega_0\right)</math>

<math>\hspace{0.5cm}\dot{\Ws}_\text{cabin}^\text{weight}=\left[ \left(\leftarrow \Rs\Omega_0\right)+\left(\downarrow \Rs\Omega_0\right) \right] \cdot \left(\downarrow\ms\gs\right) =\ms\gs\Rs\Omega_0</math>

<math>\hspace{0.5cm}\dot{\Ws}_\text{cabin}^\text{ar}=\left[ \left(\leftarrow \Rs\Omega_0\right)+\left(\downarrow \Rs\Omega_0\right) \right] \cdot \left(\rightarrow\ms\Rs^2\Omega_0^2\right) =-\ms\gs\Rs^2 \Omega_0^3</math>

<math>\hspace{0.5cm}\sum\dot{\Ws}_\text{cabin}=\ms\Rs\Omega_0(\gs-\Rs\Omega_0^2)</math>

|}


====✏️ EXAMPLE E1-1.2: pendulum====
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:
{|
:[[File:ExE1-2-1-eng.png|thumb|left|180px|link=]]
|A particle '''P''' of mass m hangs from an inextensible thread of length L attached to the ceiling. The assembly oscillates in a vertical plane fixed to the ground. We want to calculate the power of all the forces acting on the particle in two different reference frames: the ground and a platform rotating relative to the ground with a constant angular velocity <math>\Omega_0</math>. 

Ground reference frame: Since it is a Galilean reference frame, '''P''' is only subjected to interaction forces.
{|
:[[File:ExE1-2-2-neut.png|thumb|left|300px|link=]]

|<math>\dot{\Ws}_\Es^\Ts=(\nearrow\Ls\dot{\theta})\cdot(\nwarrow\Ts)=0</math>

<math>\dot{\Ws}_\Es^\text{weight}=(\nearrow\Ls\dot{\theta}\cdot(\downarrow \ms\gs)=-\ms\gs\Ls\dot{\theta}\sin\theta</math>

|}

Platform reference frame: It is a non Galilean reference frame, and the transportation force on '''P''' has to be taken into account.

{|
:[[File:ExE1-2-3-eng.png|thumb|left|400px|link=]]

|AB:ground, REL:platform

<math>\vel{P}{REL}=\vel{P}{AB}-\vel{P}{ar}=(\nearrow\Ls\dot{\theta})-(\otimes \Ls \Omega_0 \sin \theta)</math>

<math>\dot{\Ws}_\Es^\Ts\left[(\nearrow\Ls\dot{\theta})+(\odot\Ls\Omega_0\sin\theta)\right]\cdot(\nwarrow\Ts)=0</math>

<math>\dot{\Ws}_\Es^\text{weight}\left[(\nearrow\Ls\dot{\theta})+(\odot\Ls\Omega_0\sin\theta)\right]\cdot(\downarrow \ms\gs)=-\ms\gs\Ls\dot{\theta}\sin\theta</math>

<math>\dot{\Ws}_\Es^\text{ar}\left[(\nearrow\Ls\dot{\theta})+(\odot\Ls\Omega_0\sin\theta)\right]\cdot(\rightarrow \ms\Ls\Omega_0^2\sin\theta)=\ms\Ls^2\Omega_0^2\dot{\theta}\cos\theta</math>
|}


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==E1.2 Power of an action-reaction pair==

The total power associated with internal interactions in a constant-mass system is not zero in principle (since the two forces of each action-reaction pair are applied to particles that may have different velocities relative to the reference frame in which the calculation is made), and it is independent of the reference frame in which it is evaluated. It only depends on the separation velocity of the two particles. The power balance for the system is therefore written as: <math>\dot{\Ts}_\Rs^\text{syst}=\dot{\Ws}_\Rs^\text{ext} + \dot{\Ws}^\text{int} + \dot{\Ws}_\Rs^\text{ar} </math> .

<div>
=====💭 PROOF ➕=====

{|
:[[File:E1-2-neut.png|thumb|left|200px|link=]]
|<math>\dot{\Ws}^\text{ac-reac}=\overline{\Fs}_{\Ps \rightarrow \Qs}\cdot \vel{Q}{R} + \overline{\Fs}_{\Qs \rightarrow \Ps}\cdot \vel{P}{R}= \overline{\Fs}_{\Ps \rightarrow \Qs}\cdot \left[ \vel{Q}{R}-\vel{P}{R}\right]</math>

By the [[D1. Foundational laws of Newtonian dynamics#D1.6 Newton’s third law (action-reaction principle)|'''action-reaction principle''']], <math>\overline{\Fs}_{\Ps \rightarrow \Qs}=-\overline{\Fs}_{\Qs \rightarrow \Ps}</math>. Therefore: <math>\dot{\Ws}^\text{ac-reac}=\overline{\Fs}_{\Ps \rightarrow \Qs}\cdot \left[ \vel{Q}{R}-\vel{P}{R}\right]</math>

As <math>\overline{\Fs}_{\Ps \rightarrow \Qs}</math> has the <math>\PQvec</math> direction, only the components of the velocities in that direction have to be considered:

<math>\dot{\Ws}^\text{ac-reac}=\overline{\Fs}_{\Ps \rightarrow \Qs}\cdot \biggl( \vel{Q}{R} \Bigr]_{||\Ps\Qs}-\ \vel{P}{R} \Bigr]_{||\Ps\Qs}\biggl)</math>
|}

The difference <math>\biggl( \vel{Q}{R} \Bigr]_{\Ps\Qs}-\ \vel{P}{R} \Bigr]_{\Ps\Qs}\biggl)</math> is the separation velocity <math>\dot{\rho}</math> between '''P''' and '''Q''', and it does not dpend on the reference frame since it is the derivative of a scalar (the distance <math>\dot{\rho}</math> between the two particles). For the previous drawing, <math>\dot{\rho}> 0</math>. Since this is an attractive interaction, <math>\dot{\Ws}^\text{ac-reac} < 0</math>.

With a drawing, it is easy to see that:
* attractive interaction and particle approach <math>(\dot{\rho}<0)</math> : <math>\dot{\Ws}^\text{ac-reac} > 0</math> ;
* repulsive interaction and particle separation <math>(\dot{\rho}>0)</math> : <math>\dot{\Ws}^\text{ac-reac} > 0</math>
* repulsive interaction and particle approach <math>(\dot{\rho}<0)</math> : <math>\dot{\Ws}^\text{ac-reac} < 0</math>


</div>

There are two particular cases of interaction forces whose power always has the same sign:

* Friction action-reaction pair between '''P''' and '''Q''': since the friction force always opposes the relative velocity between the two points, its power is always negative:

<math>\hspace{1cm} \dot{\Ws}^\text{friction}=\dot{\Ws}^{\text{fric}\rightarrow\Ps}_\text{RTQ}+\dot{\Ws}^{\text{fric}\rightarrow\Qs}_\text{RTQ}=\overline{\Fs}^{\text{fric}\rightarrow\Ps} \cdot \vel{P}{RTQ} = - \abs{\overline{\Fs}^{\text{fric}\rightarrow\Ps}} \frac{\vel{P}{RTQ}}{\abs{\vel{P}{RTQ}}} \cdot \vel{P}{RTQ}<0;</math>

* Constraint action-reaction pair between '''P''' and '''Q''': from the [[D3. Interactions between rigid bodies#D3.4 Direct constraint interactions|'''analytical characterization equation''']] of the constraint forces between two points '''P''' and '''Q''' belonging to two rigid bodies <math>\mathrm{S}_\Ps</math> and <math>\mathrm{S}_\Qs</math>, it is easy to see that the associated power is zero:

<math>\hspace{1cm} \dot{\Ws}^\text{constraint}=\dot{\Ws}^{\text{constraint}\rightarrow\Ps}_{\mathrm{S}_\Ps}+\dot{\Ws}^{\text{constraint}\rightarrow\Qs}_{\mathrm{S}_\Ps}=\overline{\Fs}^{\text{constraint}\rightarrow\Ps} \cdot \overline{\mathbf{v}}_{\mathrm{S}_\Qs}(\Ps) = - 0;</math>

====✏️ EXAMPLE E1-2.1: power of an action-reaction friction pair====
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:
{|
:[[File:ExE1-2-1-2-eng.png|thumb|left|180px|link=]]
|A homogeneous wheel, with radius r and mass m, moves down an inclined plane without sliding until it joins a conveyor belt moving at a constant velocity <math>\vs_0</math> relative to the ground. At the moment it makes contact with the belt, the center of the wheel has a velocity <math>\vs_0</math> relative to the ground in the same direction as the belt. We want to calculate the power of the friction force between the belt and the wheel at this instant.
|}

At that momento, the <math>\mathbf{ICR}^\text{wheel}_\Es</math> is point <math>\Js_\text{wheel}</math> in contact with the belt. This point slides backwards with respect to the belt, therefore the friction force on the wheel is forwards:

<math>\left.\begin{array}{l}
\text { AB: ground } \\
\text { REL: belt }
\end{array}\right\} \quad \bar{\vs}_{\text {REL }}\left(\Js_{\text {wheel }}\right)=\bar{\vs}_{\text{AB}}\left(\Js_{\text {wheel }}\right)-\bar{\vs}_{\mathrm{ar}}\left(\Js_{\text {wheel }}\right)=-\bar{\vs}_{\mathrm{ar}}\left(\Js_{\text {wheel }}\right)=-\left(\rightarrow \mathrm{v}_0\right)=\left(\leftarrow \mathrm{v}_0\right) .</math>

[[File:ExE1-2-1-3-eng.png|thumb|center|500px|link=]]

Calculation of the friction power: <math>\dot{\Ws}^\text{friction}=\dot{\Ws}_\Rs^{\text{friction}\rightarrow \text{wheel}}+ \dot{\Ws}_\Rs^{\text{friction}\rightarrow \text{belt}}</math>, where R may be any reference frame. To quickly calculate this power, it is advisable to choose a reference frame R where one of the two points subjected to the action-reaction pair has zero speed, so that one of the two terms is directly zero:

<math>\dot{\Ws}^\text{friction}=\dot{\Ws}_\text{belt}^{\text{fric}\rightarrow \text{wheel}}+ \dot{\Ws}_\text{belt}^{\text{fric }\rightarrow \text{belt}}=\overline{\Fs}^{\text{fric}\rightarrow \text{wheel}} \cdot \bar{\vs}_{\text {belt}}\left(\Js_{\text {wheel }}\right)=(\rightarrow \mu\ms\gs)\cdot(\leftarrow \vs_0)=-\mu\ms\gs\vs_0.</math>

If the calculation is done in any other reference frame, the result is the same:

<math>\dot{\Ws}^\text{friction}=\dot{\Ws}_\text{E}^{\text{fric}\rightarrow \text{wheel}}+ \dot{\Ws}_\text{E}^{\text{fric} \rightarrow \text{belt}}=(\rightarrow \mu\ms\gs)\cdot 0 + (\leftarrow \mu\ms\gs)\cdot(\rightarrow \vs_0)=-\mu\ms\gs\vs_0.</math>

<math>\dot{\Ws}^\text{friction}=\dot{\Ws}_\text{vehicle}^{\text{fric}\rightarrow \text{wheel}}+ \dot{\Ws}_\text{vehicle}^{\text{fric} \rightarrow \text{belt}}=(\rightarrow \mu\ms\gs)\cdot (\rightarrow\vs_0) + (\leftarrow \mu\ms\gs) \cdot(\rightarrow 3\vs_0)=-\mu\ms\gs\vs_0.</math>

However, if only the power of one of the two forces is calculated (for example, the friction on the wheel), the result depends on the reference frame and can have any sign:

<math>\dot{\Ws}_\text{belt}^{\text{fric }\rightarrow \text{wheel}}=(\rightarrow \mu\ms\gs)\cdot(\leftarrow \vs_0)=-\mu\ms\gs\vs_0 ,</math>

<math>\dot{\Ws}_\text{E}^{\text{fric} \rightarrow \text{wheel}}=(\rightarrow \mu\ms\gs)\cdot 0 = 0 ,</math>

<math>\dot{\Ws}_\text{vehicle}^{\text{fric} \rightarrow \text{wheel}}=(\rightarrow \mu\ms\gs)\cdot (\rightarrow\vs_0)=\mu\ms\gs\vs_0.</math>


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==E1.3 Power of a system of forces on a rigid body==

{|
|
:[[File:E1-3-eng.png|thumb|left|180px|link=]]
<center>'''Figure E1.3''': Kinematic torsor of a rigid body in a reference frame R, and forces on different points (dm) of that rigid body</center>

|When calculating the power of a system of forces on a rigid body S, the term associated with internal interactions is zero because those interactions are all constraint (cohesion) forces and, as seen in [[E1. Work-Energy Theorem: differential form#E1.2 Power of an action-reaction pair|'''section E1.2''']], their total power is zero.

The total power of a system of forces on S can be calculated from the torsor of the system at any point '''O''' of the rigid body, <math>(\overline{\Fs},\overline{\Ms}(\Os))</math>. Taking into account rigid body kinematics ('''Figure E1.3'''):

<math>\dot{\Ws}_\Rs^{\rightarrow \Ss} =\sum_\Ps\overline{\Fs}^{\rightarrow \Ps} \cdot \vel{P}{R}= \sum_\Ps \left[ \overline{\Fs}^{\rightarrow \Ps} \cdot \vel{O}{R} + \velang{S}{R} \times \OPvec \right] </math>

|}
<math>\dot{\Ws}_\Rs^{\rightarrow \Ss} =\left(\sum_\Ps\overline{\Fs}^{\rightarrow \Ps} \right)\cdot \vel{O}{R} + \sum_\Ps\overline{\Fs}^{\rightarrow \Ps} \cdot \left(\velang{S}{R} \times \OPvec \right) = \left( \sum_\Ps\overline{\Fs}^{\rightarrow \Ps} \right) \cdot \vel{O}{R} + \left[ \sum_\Ps \left( \OPvec \times \overline{\Fs}^{\rightarrow \Ps} \right)\right] \cdot \velang{S}{R} = \overline{\Fs}\cdot \vel{O}{R} + \overline{\Ms}(\Os)\cdot \velang{S}{R} </math>

This calculation also shows that the power of a moment (or a torque <math>\Gamma</math>) on a rigid body in a reference frame R is the product of the moment by the angular velocity of the rigid body: <math>\dot{\Ws}_\Rs^{\Gamma\rightarrow \mathrm{S}} = \overline{\Gamma} \cdot \velang{S}{R}</math>.

The power balance for a rigid body is: <math>\dot{\Ws}_\Rs^{ \rightarrow \mathrm{S}} = \dot{\Ws}_\Rs^{ \mathrm{ext}} + \dot{\Ws}_\Rs^{ \mathrm{ar}}.</math>

====✏️ EXAMPLE E1-3.1: roller sliding on a slope====
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:
{|
|
:[[File:ExE1-3-1-1-eng.png|thumb|left|180px|link=]]
|A homogeneous roller, with mass m and radius r, slides down a slope. The friction coefficient between the ground and the roller is <math>\mu</math>. We want to calculate the power of every force acting on the roller in the ground reference frame.
|}

:It is a 2D problem. The kinematic and dynamic descriptions are:
{|
|
[[File:ExE1-3-1-2-eng.png|thumb|right|250px|link=]]

|<math> \hspace{0.5cm}\vel{J}{E}=\vel{C}{T} + \velang{}{E} \times \CJvec= </math>

<math>\hspace{1.7cm}=\left( \rightarrow \vs_\Es \right) + \left( \otimes \Omega_\Es\right) \times \left( \downarrow \rs \right)</math>

<math> \hspace{0.5cm}\vel{J}{E} = \left[\rightarrow (\vs_\Es -\rs\Omega_\Es)\right]</math>

<math>\hspace{0.5cm} \dot{\Ws}_\Es^{\mathrm{ext}} = \dot{\Ws}_\Es^{\mathrm{weight}} + \dot{\Ws}_\Es^{\mathrm{friction}} + \dot{\Ws}_\Es^{\mathrm{constraint}}= \Bigr[\left(\downarrow \ms\gs\cos\beta\right)+\left(\rightarrow \ms\gs\sin\beta\right)\Bigr] \cdot (\rightarrow \vs_\Es) + \Bigr[(\leftarrow \mu \Ns) + (\uparrow \Ns ) \Bigr] \cdot \Bigr[\rightarrow (\vs_\Es -\rs\Omega_\Es)\Bigr]</math>

|}
:Since the roller center has no motion perpendicular to the plane, the AMT leads to <math>\Ns =\ms\gs\cos\beta</math>. Therefore:

{|
|
[[File:ExE1-3-1-3-eng.png|thumb|right|140px|link=]]
|<math>\hspace{0.5cm} \dot{\Ws}_\Ts^{\mathrm{ext}} = \ms\gs\vs_\Ts \sin \beta - \ms\gs(\vs_\Ts -\rs\Omega_\Ts)\cos\beta=\ms\gs \Bigr[ \vs_\Ts (\sin\beta-\cos\beta)+\rs\Omega_\Ts \cos\beta \Bigr].</math>

<math>\hspace{0.5cm}</math>

If instead of considering the friction at '''J''' we consider the equivalent torsor at point '''C''', we must consider the force applied in '''C''' and the moment about '''C''', which is not zero:

<math>\hspace{0.5cm} \dot{\Ws}_\Es^{ \mathrm{friction}} = (\leftarrow \mu\Ns)\cdot(\rightarrow\vs_\Es)+(\otimes \mu\Ns\rs)\cdot(\otimes \Omega_\Es) =\mu\ms\gs\cos\beta(\rs\Omega_\Es-\vs_\Es)</math>,

<math>\hspace{0.5cm} \dot{\Ws}_\Es^{\mathrm{ext}} = \dot{\Ws}_\Es^{ \mathrm{weight}} + \dot{\Ws}_\Es^{ \mathrm{friction}}+ \dot{\Ws}_\Es^{ \mathrm{constraint}}= \ms\gs\vs_\Es\sin\beta+\mu\ms\gs\cos\beta (\rs\Omega_\Es-\vs_\Es)</math>.

<math>\hspace{0.5cm}</math>

The result is the same.
|}


====✏️ EXAMPLE E1-3.2: block sliding on a truck====
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:
{|
|
:[[File:ExE1-3-2-1-eng.png|thumb|left|220px|link=]]
|A truck brakes with acceleration '''a''' relative to the ground at the instant in which its velocity (also relative to the ground) is v. At the same instant, the homogeneous block, of mass m, slides with velocity v' relative to the truck. We want to perform a power balance for the block in two different reference frames: the ground and the truck.
|}

:Ground reference frame: It is a 2D problem. The kinematic and dynamic descriptions are:
{|
|
[[File:ExE1-3-2-2-eng.png|thumb|right|200px|link=]]

| The constraint between the truck and the block allows only one translational DoF between the two. Therefore, in the 2D approximation, the associated torsor has two independent components. If characterized at the midpoint of the contact surface, these components are a normal force and a moment perpendicular to the plane. The AMT applied to the block leads to <math>\Ns=\ms\gs</math>.

|}
:The powers associated with all interactions on the block in the ground reference are:

:<math>\dot{\Ws}_\Es^{\rightarrow \mathrm{bloque}}=\dot{\Ws}_\Ts^{\mathrm{ext}}=\dot{\Ws}_\Ts^{\mathrm{peso}}+\dot{\Ws}_\Ts^{\mathrm{fricción}}+\dot{\Ws}_\Ts^{\mathrm{enlace}}= \Bigr[(\downarrow \ms\gs)+(\leftarrow \mu\ms\gs)+(\uparrow \Ns) \Bigr] \cdot \Bigr[\rightarrow(\vs+\vs')\Bigr] + (\odot \Ms_\es)\cdot 0 = -\mu\ms\gs(\vs+\vs')</math>

:Ya que el bloque se traslada respecto al suelo, su energía cinética total es inmediata, y se puede plantear fácilmente el balance de potencias:

:<math>\Ts_\Ts^\mathrm{bloque}=\frac{1}{2}\int_\mathrm{bloque}\ds\ms \vs_\Ts^2(\mathrm{bloque})=\frac{1}{2}\ms \vs_\Ts^2(\mathrm{bloc})\quad \Rightarrow \quad \dot{\Ts}_\Ts^\mathrm{bloque}=\ms\vs_\Ts( \mathrm{bloque})\dot{\vs}_\Ts( \mathrm{bloque}) </math>

:Como la única fuerza horizontal sobre el bloque es la fricción:

<math>\dot{\vs}_\Ts(\mathrm{bloque})=-\mu\gs\quad \Rightarrow \quad \dot{\Ts}_\Ts^\mathrm{bloque}=-\mu\ms\gs(\vs+\vs').</math>

:El balance de potencias es consistente:

:<math>\left.\begin{array}{l}
\dot{\mathrm{W}}_\Ts^{\rightarrow \text { bloque }}=\dot{\mathrm{W}}_{\mathrm{T}}^{\text {ext }}=-\mu \mathrm{mg}\left(\mathrm{v}+\mathrm{v}^{\prime}\right) \\
\dot{\mathrm{T}}_{\mathrm{T}}^{\text {bloque }}=-\mu \mathrm{mg}\left(\mathrm{v}+\mathrm{v}^{\prime}\right)
\end{array}\right\} \Rightarrow \dot{\mathrm{W}}_{\mathrm{T}}^{\rightarrow \text { bloque }}=\dot{\mathrm{T}}_{\mathrm{T}}^{\text {bloque }}</math>

:Referencia camión: Al estar el camión acelerado respecto al suelo, hay que tener en cuenta las fuerzas de arrastre sobre todos los diferenciales de masa (dm) del bloque:

{|
|

[[Archivo:ExE1-3-2-3-cat-esp.jpg|thumb|right|250px|link=]]
|<math>\cal{\Fs}^{\rightarrow \ds\ms}_{\mathrm{ar}}=-\ds\ms\overline{\as}_\mathrm{ar}(\ds\ms)=-\ds\ms (\leftarrow \as)=(\rightarrow \ds\ms \cdot \as)</math>.

El torsor resultante de este sistema de fuerzas en el centro de inercia '''G''' se reduce a una fuerza horizontal.

<math>\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{bloque}}=\dot{\Ws}_\mathrm{camión}^{\mathrm{ext}} + \dot{\Ws}_\mathrm{camión}^{\mathrm{ar}} =\dot{\Ws}_\mathrm{camión}^{\mathrm{peso}} + \dot{\Ws}_\mathrm{camión}^{\mathrm{fricción}} + \dot{\Ws}_\mathrm{camión}^{\mathrm{enlace}} + \dot{\Ws}_\mathrm{camión}^{\mathrm{ar}} , </math>

<math>\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{bloque}}= \Bigr[(\downarrow \ms\gs)+(\leftarrow \mu\ms\gs)+(\uparrow \Ns) + (\rightarrow \ms\as)\Bigr] \cdot(\rightarrow \vs')+(\odot\Ms_\es) \cdot0=\ms(\as-\mu\gs)\vs'.</math>
|}

:Se puede comprobar que el balance de potencias es consistente:

:<math>\Ts^\mathrm{bloque}_\mathrm{camión}=\frac{1}{2}\ms\vs^2_\mathrm{camión}(\mathrm{bloque}) \quad \Rightarrow \quad \dot{\Ts}^\mathrm{bloque}_\mathrm{camión} = \ms\vs_\mathrm{camión}(\mathrm{bloque})\dot{\vs}_\mathrm{camión}(\mathrm{bloque})</math>

:<math>\left.\begin{array}{l}
\mathrm{AB}:\mathrm{suelo} \\
\mathrm{REL}:\mathrm{camión}
\end{array}\right\} \quad \dot{\vs}_\mathrm{camión}(\mathrm{bloque})= \dot{\vs}_\mathrm{T}(\mathrm{bloque})-\dot{\vs}_\mathrm{ar}(\mathrm{bloque}) = (-\mu\gs)-(-\as)=\as-\mu\gs \quad \Rightarrow \quad \dot{\Ts}^\mathrm{bloque}_\mathrm{camión}=\ms\vs'(\as-\mu\gs)</math>

:<math>\left.\begin{array}{l}
\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{bloque}}=\dot{\Ws}_\mathrm{camión}^{\mathrm{ext}} + \dot{\Ws}_\mathrm{camión}^{\mathrm{ar}}= \ms(\as-\mu\gs)\vs'\\
\dot{\Ts}^\mathrm{bloque}_\mathrm{camión}=\ms\vs'(\as-\mu\gs)
\end{array}\right\} \Rightarrow\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{bloque}}= \dot{\Ts}^\mathrm{bloque}_\mathrm{camión}</math>


====✏️ EJEMPLO E1-3.3: camión frenando con carga====
------
:
{|
|
:[[Archivo:ExE1-3-3-1-esp.png|thumb|left|220px|link=]]
|Un camión frena con aceleración '''a''' respecto al suelo en el instante en el que su velocidad (también respecto al suelo) es v. Los pares de frenada en el eje delantero y en el trasero son iguales y valen <math>\Gamma</math>, y las ruedas, de radio r y masa despreciable, no deslizan sobre el suelo. En el mismo instante, el bloque homogéneo, de masa m, desliz con velocidad v' relativa al camión. Se trata de hacer un balance de potencias para el sistema (bloque+camión) en la referencia del suelo y en la del camión.

|}

:Ya que la potencia de las interacciones internas no es nula en principio (no se trata de un único sólido rígido), es conveniente hacer un DGI del sistema para no olvidar ninguna:
{|
|
[[Archivo:ExE1-3-3-2-esp.png|thumb|right|250px|link=]]

|
|
c.p.s.d.: contacto puntual sin deslizamiento

c.m.c.d.: contacto multipuntual con deslizamiento

|}
:El sistema sobre el que se hace el balance de potencias es todo menos el suelo. Teniendo en cuenta que la potencia de los enlaces internos es siempre nula, las únicas interacciones que hay que tener en cuenta son el peso, los enlaces de las ruedas con el suelo, los pares de frenada y la fricción entre bloque y camión:

[[Archivo:ExE1-3-3-3-esp.png|thumb|center|300px|link=]]

:Referencia suelo:
:Las fuerzas de enlace del suelo sobre las ruedas están aplicadas a puntos de velocidad cero, y por tanto no hacen potencia. El peso es una fuerza vertical, y el movimiento del camión y del bloque es horizontal, por tanto su potencia también es cero.

:<math>\dot{\Ws}_\Ts^{\rightarrow \mathrm{sist}}=\dot{\Ws}_\Ts^{\mathrm{ext}}+\dot{\Ws}^{\mathrm{int}}=\dot{\Ws}_\Ts^{\mathrm{peso}}+\dot{\Ws}_\Ts^{\mathrm{enlace}}+\dot{\Ws}^{\mathrm{fricción}}+\dot{\Ws}^{\Gamma}=\dot{\Ws}^{\mathrm{fricción}}+\dot{\Ws}^{\Gamma}.</math>

:Las potencias de interacciones internas se pueden calcular en cualquier referencia. Por tanto:

:<math>\dot{\Ws}^{\mathrm{fricción}}= \dot{\Ws}_\mathrm{camión}^{\mathrm{fric}\rightarrow \mathrm{bloque}} +\dot{\Ws}_\mathrm{camió}^{\mathrm{fric}\rightarrow \mathrm{camión}} = \dot{\Ws}_\mathrm{camión}^{\mathrm{fric}\rightarrow \mathrm{bloque}}= (\leftarrow \mu\ms\gs)\cdot (\rightarrow \vs')=-\mu\ms\gs\vs'</math>

:<math>\dot{\Ws}^{\Gamma}= \dot{\Ws}_\mathrm{camión}^{\Gamma\rightarrow \mathrm{ruedas}} +\dot{\Ws}_\mathrm{camión}^{\Gamma\rightarrow \mathrm{camión}} = \dot{\Ws}_\mathrm{camión}^{\Gamma\rightarrow \mathrm{ruedas}}= 2 \left(\odot \Gamma \right)\cdot \left(\otimes \frac{\vs}{\rs}\right)=-2\Gamma\frac{\vs}{\rs}.</math>

:Por otro lado: <math>\Ts_\Ts^\mathrm{sist}=\frac{1}{2}\ms\vs_\Ts^2(\mathrm{bloque})+\frac{1}{2}\ms\vs_\Ts^2(\mathrm{camión})\quad \Rightarrow \quad \dot{\Ts}_\Ts^\mathrm{sist}=-\ms(\vs+\vs')\mu\gs-\Ms\vs\as. </math>

:Finalmente: <math>\dot{\Ws}_\Ts^{\rightarrow \mathrm{sist}}=\dot{\Ts}_\Ts^\mathrm{sist} \quad \Rightarrow \quad -\mu\ms\gs\vs'-2\Gamma\frac{\vs}{\rs}=-\ms(\vs+\vs')\mu\gs-\Ms\vs\as.</math>

:Esta ecuación permite calcular el par de frenada: <math>\Gamma=\frac{1}{2}(\mu\ms\gs+\Ms\as)\rs.</math>

:Referencia camión:

:En esta referencia hay que tener en cuenta las fuerzas de arrastre. Por otro lado, las fuerzas de enlace del suelo sobre las ruedas están aplicadas a puntos que ya no tienen velocidad cero, y por tanto hacen potencia. Como antes, la potencia del peso es cero.

:<math>\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{sist}}=\dot{\Ws}_\mathrm{camión}^{\mathrm{ext}}+\dot{\Ws}^{\mathrm{int}}+\dot{\Ws}_\mathrm{camión}^{\mathrm{ar}}=\dot{\Ws}_\mathrm{camión}^{\mathrm{enlace}}+\dot{\Ws}^{\mathrm{fricción}}+\dot{\Ws}^{\Gamma}+\dot{\Ws}_\mathrm{camión}^{\mathrm{ar}}.</math>

:Las fuerzas de arrastre sobre el camión no hacen potencia porque el camión está parado en esta referencia. La de las fuerzas de arrastre sobre el bloque se ha calculado en el [[E1. Teorema de la energía: versión diferencial#✏️ ✏️ EJEMPLO E1-3.2: bloque deslizando sobre un camión|'''ejemplo E1-3.2''']]. La potencia de la fricción y de los pares de frenada es la misma de antes ya que no depende de la referencia.

:<math>\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{sist}}=\dot{\Ws}_\mathrm{camión}^{\mathrm{enllaç}}+\dot{\Ws}^{\mathrm{fricción}}+\dot{\Ws}^{\Gamma}+\dot{\Ws}_\mathrm{camión}^{\mathrm{ar} \rightarrow \mathrm{bloque}}= \dot{\Ws}_\mathrm{camión}^{\mathrm{enlace}} -\mu\ms\gs\vs'-2\Gamma\frac{\vs}{\rs} +\ms\as\vs'.</math>

:Las descripciones cinemática y dinámica de la rueda son:

[[Archivo:ExE1-3-3-4-esp.png|thumb|center|430px|link=]]

:<math>\dot{\Ws}_\mathrm{camión}^{\mathrm{enlace}}=2\Bigr[(\leftarrow \Ts)+(\uparrow \Ns)\Bigr]\cdot(\leftarrow\vs)=2\Ts\vs.</math>

La fuerza tangencial de enlace se puede calcular a partir del TMC aplicado al centro '''C''' de la rueda. Como no tiene masa: <math>\sum\overline{\Ms}_\mathrm{ext}(\Cs)=\overline{0} \quad \Rightarrow \quad (\odot \Gamma)+(\otimes \Ts\rs)=0\quad \Rightarrow \quad \Ts=\Gamma/\rs.</math>

:Por tanto: <math>\dot{\Ws}_\mathrm{camió}^{\rightarrow \mathrm{sist}} =2\frac{\Gamma}{\rs}\vs-\mu\ms\gs\vs'-2\Gamma\frac{\vs}{\rs}+\ms\as\vs'=\ms(\as-\mu\gs)\vs'.</math>

:Por otro lado, como se ha visto en el [[E1. Teorema de la energía: versión diferencial#✏️ ✏️ EJEMPLO E1-3.2: bloque deslizando sobre un camión|'''ejemplo E1-3.2''']]:

:<math>\Ts_\mathrm{camión}^\mathrm{sist}= \Ts_\mathrm{camión}^\mathrm{bloque}=\frac{1}{2}\ms\vs_\mathrm{camión}^2(\mathrm{bloque}) \quad \Rightarrow \quad \dot{\Ts}_\mathrm{camión}^\mathrm{sist}= \dot{\Ts}_\mathrm{camión}^\mathrm{bloque}=\ms\vs'(\as-\mu\gs).</math>

:El balance de potencias es consistente:

:<math>\left.\begin{array}{l}
\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{sist}}= \ms(\as-\mu\gs)\vs'\\
\dot{\Ts}^\mathrm{sist}_\mathrm{camión}=\ms(\as-\mu\gs)\vs'
\end{array}\right\} \Rightarrow\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{sist}}= \dot{\Ts}^\mathrm{sist}_\mathrm{camión}</math>

-----
-----

==E1.4 Balance de potencias para un sistema multisólido: cálculo directo y cálculo indirecto==

El '''cálculo directo''' (a partir de su definición) de la potencia asociada a una interacción (una fuerza sobre un punto '''P''' o un momento sobre un sólido rígido S) exige el conocimiento de la interacción (fuerza o momento):

<math>\dot{\Ws}_\Rs^{\Fs\rightarrow \Ps}=\overline{\Fs}_{\rightarrow\Ps} \cdot \vel{P}{R} \quad , \quad \dot{\Ws}_\Rs^{\Ms\rightarrow \Ss}=\overline{\Ms}_{\rightarrow\Ss} \cdot \velang{S}{R} </math>

Cuando no se trata de una interacción formulable, hay que recurrir a los teoremas vectoriales para determinarla.

El '''cálculo indirecto''' es una alternativa que no necesita recurrir a estos teoremas. Se basa en la evaluación de un balance de potencias para un sistema y una referencia adecuados de manera que la potencia a determinar sea el único término desconocido de la ecuación (el único que no se puede calcular de manera directa). Cuando el balance incluye más de un término desconocido, se pueden combinar varios balances (para distintos sistemas y referencias) hasta tener un número suficiente de ecuaciones que permitan la determinación de todas las potencias desconocidas.

====✏️ EJEMPLO E1-4.1: polea frenada====
------
:
{|
|
:[[Archivo:ExE1-4-1-1-esp.png|thumb|left|150px|link=]]
|Un bloque de masa m cuelga de un hilo inextensible atado a un punto de la periferia de una polea de radio r. Entre soporte (fijo al suelo) y polea actúa un freno que garantiza que el bloque baja a velocidad constante respecto al suelo. Se trata de calcular la potencia del freno.

|}

:Càlculo directo: <math>\dot{\Ws}^{\mathrm{freno}}= \overline{\Gamma}^{\rightarrow \mathrm{polea}}\cdot\velang{polea}{T}=\left(\odot\Gamma\right)\cdot\left( \otimes \frac{\vs_0}{\rs}\right)=-\Gamma \frac{\vs_0}{\rs}.</math>

:El par de frenada no es un dato, y hay que recurrir al TMC para calcularlo.

:<math>\left.\begin{array}{l}
\text{SISTEMA: polea + bloque}\\
\text{TMC en } \Os
\end{array}\right\} \sum\overline{\Ms}_\mathrm{ext}(\Os)=\dot{\overline{\Hs}}_\mathrm{RTO}(\Os)</math>

:El momento cinético proviene de las dos piezas. Ya que '''O''' no es fijo al bloque, hay que hacer descomposición baricéntrica para calcular el momento cinético del bloque:

:<math>\overline{\Hs}_\mathrm{RTO=T} (\Os) =\overline{\Hs}_\Ts^\mathrm{politja}(\Os) + \overline{\Hs}_\Ts^\mathrm{bloc}(\Os)= \overline{\Hs}_\Ts^\mathrm{politja}(\Os) + \overline{\Hs}_\Ts^\mathrm{bloc}(\Gs_\mathrm{bloc})+ \overline{\Hs}_\Ts^{\oplus\mathrm{bloc}}(\Os) = </math>

:<math>\hspace{2.3cm}=\Is_\mathrm{O}^\mathrm{polea} \velang{polea}{T}+\Is_{\Gs_\mathrm{bloque}}^\mathrm{bloque}+\OGvec_\mathrm{bloque}\times \ms\overline{\vs}_\Ts(\Gs_\mathrm{bloque})=\left(\otimes \Is_\mathrm{O}\frac{\vs_0}{\rs}\right)+(\otimes \ms\vs_0\rs).</math>

:[[Archivo:ExE1-4-1-2-esp.png|thumb|right|180px|link=]]
:Se trata de un momento cinético constante en dirección y valor. Por tanto:

:<math>\dot{\overline{\Hs}}_\mathrm{RTO}(\Os)=0 \quad \Rightarrow \quad \sum{\overline{\Ms}}_\mathrm{ext}(\Os)=(\odot\Gamma)+(\otimes \ms\gs\rs) =0\quad \Rightarrow \quad \Gamma= \ms\gs\rs</math>.

:Finalmente: <math>\dot{\Ws}^{\mathrm{freno}}= -\Gamma \frac{\vs_0}{\rs} = -\ms\gs\vs_0.</math>

:Cálculo indirecto:

:<math>\left.\begin{array}{l}
\text{SISTEMA: polea + soporte + bloque}\\
\text{REF: suelo(Gal)}
\end{array}\right\} \quad \dot{\Ts}^\mathrm{sist}_\Ts=\dot{\Ws}^\mathrm{ext}_\Ts+ \dot{\Ws}^\mathrm{int}</math>

:<math>\left.\begin{array}{l}
\dot{\Ts}^\mathrm{sist}_\Ts=0\\
\dot{\Ws}^\mathrm{ext}_\Ts=\left(\downarrow \ms\gs\right)\cdot\left(\downarrow \vs_0\right)= \ms\gs\vs_0\\
\dot{\Ws}^\mathrm{int}=\dot{\Ws}^{\mathrm{freno}}
\end{array}\right\} \quad \Rightarrow \quad \dot{\Ws}^{\mathrm{freno}} =-\ms\gs\vs_0</math>



====✏️ EJEMPLO E1-4.2: mono y plátanos====
------
:
{|
|
:[[Archivo:ExE1-4-2-1-esp.png|thumb|left|180px|link=]]
|Unos plátanos de masa <math>(2/3)\ms</math> cuelgan de una cuerda inextensible que se apoya sin deslizar sobre una polea de radio r y masa despreciable, articulada respecto al suelo. En el otro extremo de la cuerda se agarra un mono de masa m. La polea gira con velocidad angular constante <math>\Omega_0</math> respecto al suelo bajo la acción de un motor. El mono se mueve respecto a la cuerda sin que sus manos se froten contra la cuerda, y de manera que la distancia de su centro de masas al suelo se mantiene constante. Se trata de calcular la potencia del motor.
|}

:Cálculo directo: <math>\dot{\Ws}^{\mathrm{freno}}= \overline{\Gamma}^{\rightarrow \mathrm{polea}}\cdot\velang{polea}{T}=\left(\odot\Gamma\right)\cdot\left( \odot \Omega_0\right)=\Gamma \Omega_0.</math>

:El par motor no es un dato, y hay que recurrir al TMC para calcularlo.

:<math>\left.\begin{array}{l}
\text{SISTEMA: polea + mono + plátanos}\\
\text{TMC en } \Os
\end{array}\right\} \sum\overline{\Ms}_\mathrm{ext}(\Os)=\dot{\overline{\Hs}}_\mathrm{RTO}(\Os)</math>

:El momento cinético proviene del mono y de los plátanos. En primera aproximación, se puede considerar la masa del mono concentrada en su tronco y su cabeza, de manera que no gira y su velocidad respecto al suelo es nula. Teniendo en cuenta que el punto '''O''' no se mueve respecto al mono pero sí respecto a los plátanos:

:<math>\overline{\Hs}_\mathrm{RTO=T} (\Os) =\overline{\Hs}_\Ts^\mathrm{mono}(\Os) + \overline{\Hs}_\Ts^\mathrm{plátanos}(\Os)= \overline{\Hs}_\Ts^\mathrm{mono}(\Os) + \overline{\Hs}_\Ts^\mathrm{plátanos}(\Gs_\mathrm{plát})+ \OGvec_\mathrm{plát} \times \frac{2}{3} \ms \overline{\vs}_\Ts(\Gs_\mathrm{plát})= </math>

:<math>\hspace{2.3cm}=\OGvec_\mathrm{plát}\times \frac{2}{3} \ms \overline{\vs}_\Ts(\Gs_\mathrm{plát}) = \Bigr[(\rightarrow \rs)+(\downarrow \xs)\Bigr] \times\frac{2}{3} \ms \left(\downarrow \rs\Omega_0\right)=\left(\otimes \frac{2}{3}\ms\rs^2\Omega_0\right).</math>

:Se trata de un momento cinético constante en dirección y valor. Por tanto:

:<math>\dot{\overline{\Hs}}_\mathrm{RTO}(\Os)=0 \quad \Rightarrow \quad \sum{\overline{\Ms}}_\mathrm{ext}(\Os)=(\odot\Gamma)+(\otimes \ms\gs\rs) + \left(\otimes \frac{2}{3}\ms\gs\rs\right)=0\quad \Rightarrow \quad \Gamma= \frac{5}{3}\ms\gs\rs</math>.

:Finalmente: <math>\dot{\Ws}^{\mathrm{motor}}= \Gamma \Omega_0 = \frac{5}{3}\ms\gs\rs\Omega_0.</math>

:Cálculo indirecto

:Hay que tener presente que es un problema con dos tipos de actuadores: el motor y la musculatura del mono. Hay diversos modelos biomecánicos para formular las fuerzas de los músculos, pero quedan fuera del ámbito de este curso. Ya que el mono se mueve, sus músculos desarrollan una potencia no nula.

:<math>\left.\begin{array}{l}
\text{SISTEMA: polea + mono + plátano}\\
\text{REF: suelo(Gal)}
\end{array}\right\} \quad \dot{\Ts}^\mathrm{sist}_\Ts=\dot{\Ws}^\mathrm{ext}_\Ts+ \dot{\Ws}^\mathrm{int}</math>

:<math>\left.\begin{array}{l}
\dot{\Ts}^\mathrm{sist}_\Ts=0\\
\dot{\Ws}^\mathrm{ext}_\Ts=\left(\downarrow \frac{2}{3}\ms\gs\right)\cdot\left(\downarrow \rs\Omega_0\right)= \frac{2}{3}\ms\gs\rs\Omega_0\\
\dot{\Ws}^\mathrm{int}=\dot{\Ws}^{\mathrm{musc}}+\dot{\Ws}^{\mathrm{motor}}
\end{array}\right\} \quad \Rightarrow \quad \dot{\Ws}^{\mathrm{musc}} + \dot{\Ws}^{\mathrm{motor}}=-\frac{2}{3}\ms\gs\rs\Omega_0</math>

:Este balance proporciona una ecuación con dos incógnitas. Se puede plantear un segundo balance para un sistema donde una de las dos no aparezca: un sistema sin el mono o un sistema sin el motor. Si se hace un balance de potencias para el sistema (mono + trozo de cuerda), las fuerzas externas a considerar son el peso y la tensión de la cuerda. Si se sigue trabajando en la referencia del suelo, esta tensión (desconocida) hará potencia no nula ya que está aplicada a un punto de velocidad ascendente.
:[[Archivo:ExE1-4-2-2-cat-esp.png|thumb|right|330px|link=]]

:Para evitar que aparezca un término asociado a la tensión de la cuerda, se puede trabajar en la referencia del trozo de cuerda que se ha considerado, que al tener velocidad ascendente constante es galileana. En esta referencia, el mono baja con velocidad constante:

:<math>\left.\begin{array}{l}
\text{REL: trozo de cuerda}\\
\text{AB: suelo}
\end{array}\right\}</math>

:<math>\overline{\vs}_\mathrm{REL}(\mathrm{mono})=\overline{\vs}_\mathrm{ar}(\mathrm{mono})=-\overline{\vs}_\mathrm{ar}(\mathrm{mono})=(\downarrow \rs\Omega_0)</math>

:<math>\left.\begin{array}{l}
\text{SISTEMA: mono+ trozo de cuerda}\\
\text{REF: trozo de cuerda(Gal)}
\end{array}\right\} \quad \dot{\Ts}^\mathrm{sist}_\mathrm{cuerda} =\dot{\Ws}^\mathrm{ext}_\mathrm{cuerda}+ \dot{\Ws}^\mathrm{int}</math>

:<math>\left.\begin{array}{l}
\dot{\Ts}^\mathrm{sist}_\mathrm{cuerda}=0\\
\dot{\Ws}^\mathrm{ext}_\mathrm{cuerda}=\left(\downarrow \ms\gs\right)\cdot\left(\downarrow \rs\Omega_0\right)= \ms\gs\rs\Omega_0\\
\dot{\Ws}^\mathrm{int}=\dot{\Ws}^{\mathrm{musc}}
\end{array}\right\} \quad \Rightarrow \quad \dot{\Ws}^{\mathrm{musc}} =-\ms\gs\rs\Omega_0</math>

:Combinando este resultado con el del balance anterior:

:<math>\dot{\Ws}^\mathrm{motor}=-\frac{2}{3}\ms\gs\rs\Omega_0-\dot{\Ws}^\mathrm{musc}=-\frac{5}{3}\ms\gs\rs\Omega_0</math>



© Universitat Politècnica de Catalunya. [[Mecánica:Derechos de autor |Todos los derechos reservados]]
-------------

<center>
[[D8. Conservaciones|<<< D8. Conservaciones]]

[[E2. Teorema de la energía: versión integrada|E2. Teorema de la energía: versión integrada>>>]]
</center>

File:ExE1-3-1-3-eng.png

2025-05-20T12:14:47Z

Bros:

ExE1-3-1-3-eng

E1. Work-Energy Theorem: differential form

2025-05-20T12:13:31Z

Bros:

<div class="noautonum">__TOC__</div>
<math>\newcommand{\uvec}{\overline{\textbf{u}}}
\newcommand{\vvec}{\overline{\textbf{v}}}
\newcommand{\evec}{\overline{\textbf{e}}}
\newcommand{\Omegavec}{\overline{\mathbf{\Omega}}}
\newcommand{\velang}[2]{\Omegavec^{\textrm{#1}}_{\textrm{#2}}}
\newcommand{\Alfavec}{\overline{\mathbf{\alpha}}}
\newcommand{\accang}[2]{\Alfavec^{\textrm{#1}}_{\textrm{#2}}}
\newcommand{\ds}{\textrm{d}}
\newcommand{\ps}{\textrm{p}}
\newcommand{\ns}{\textrm{n}}
\newcommand{\hs}{\textrm{h}}
\newcommand{\cs}{\textrm{c}}
\newcommand{\gs}{\textrm{g}}
\newcommand{\Ns}{\textrm{N}}
\newcommand{\Hs}{\textrm{H}}
\newcommand{\Fs}{\textrm{F}}
\newcommand{\ms}{\textrm{m}}
\newcommand{\ts}{\textrm{t}}
\newcommand{\us}{\textrm{u}}
\newcommand{\vs}{\textrm{v}}
\newcommand{\Rs}{\textrm{R}}
\newcommand{\Es}{\textrm{E}}
\newcommand{\Ws}{\textrm{W}}
\newcommand{\Ts}{\textrm{T}}
\newcommand{\Ls}{\textrm{L}}
\newcommand{\As}{\textrm{A}}
\newcommand{\Ds}{\textrm{D}}
\newcommand{\Bs}{\textrm{B}}
\newcommand{\es}{\textrm{e}}
\newcommand{\fs}{\textrm{f}}
\newcommand{\is}{\textrm{i}}
\newcommand{\Is}{\textrm{I}}
\newcommand{\ks}{\textrm{k}}
\newcommand{\js}{\textrm{j}}
\newcommand{\rs}{\textrm{r}}
\newcommand{\as}{\textrm{a}}
\newcommand{\ss}{\textrm{s}}
\newcommand{\Os}{\textbf{O}}
\newcommand{\Gs}{\textbf{G}}
\newcommand{\Cbf}{\textbf{C}}
\newcommand{\Or}{\Os_\Rs}
\newcommand{\Qs}{\textbf{Q}}
\newcommand{\Cs}{\textbf{C}}
\newcommand{\Ps}{\textbf{P}}
\newcommand{\Ss}{\textbf{S}}
\newcommand{\Js}{\textbf{J}}
\newcommand{\P}{\textrm{P}}
\newcommand{\Q}{\textrm{Q}}
\newcommand{\Ms}{\textrm{M}}
\newcommand{\deg}{^\textsf{o}}
\newcommand{\xs}{\textsf{x}}
\newcommand{\ys}{\textsf{y}}
\newcommand{\zs}{\textsf{z}}
\newcommand{\dert}[2]{\left.\frac{\ds{#1}}{\ds\ts}\right]_{\textrm{#2}}}
\newcommand{\ddert}[2]{\left.\frac{\ds^2{#1}}{\ds\ts^2}\right]_{\textrm{#2}}}
\newcommand{\vec}[1]{\overline{#1}}
\newcommand{\vecbf}[1]{\overline{\textbf{#1}}}
\newcommand{\vecdot}[1]{\overline{\dot{#1}}}
\newcommand{\OQvec}{\vec{\Os\Qs}}
\newcommand{\GQvec}{\vec{\Gs\Qs}}
\newcommand{\GPvec}{\vec{\Gs\Ps}}
\newcommand{\PSvec}{\vec{\Ps\Ss}}
\newcommand{\QQvec}{\vec{\Qs\Qs}}
\newcommand{\QGvec}{\vec{\Qs\Gs}}
\newcommand{\QPvec}{\vec{\Qs\Ps}}
\newcommand{\JQvec}{\vec{\Js\Qs}}
\newcommand{\PQvec}{\vec{\Ps\Qs}}
\newcommand{\GJvec}{\vec{\Gs\Js}}
\newcommand{\OPvec}{\vec{\Os\textbf{P}}}
\newcommand{\OCvec}{\vec{\Os\Cs}}
\newcommand{\OGvec}{\vec{\Os\Gs}}
\newcommand{\CJvec}{\vec{\Cs\Js}}
\newcommand{\GCvec}{\vec{\Gs\Cs}}
\newcommand{\PGvec}{\vec{\Ps\Gs}}
\newcommand{\abs}[1]{\left|{#1}\right|}
\newcommand{\braq}[2]{\left\{{#1}\right\}_{\textrm{#2}}}
\newcommand{\matriz}[9]{
\begin{bmatrix}
{#1} & {#2} & {#3}\\
{#4} & {#5} & {#6}\\
{#7} & {#8} & {#9}
\end{bmatrix}}
\newcommand{\diag}[3]{
\begin{bmatrix}
{#1} & {0} & {0}\\
{0} & {#2} & {0}\\
{0} & {0} & {#3}
\end{bmatrix}}
\newcommand{\vector}[3]{
\begin{Bmatrix}
{#1}\\
{#2}\\
{#3}
\end{Bmatrix}}
\newcommand{\vecdosd}[2]{
\begin{Bmatrix}
{#1}\\
{#2}
\end{Bmatrix}}
\newcommand{\vel}[2]{\vvec_{\textrm{#2}} (\textbf{#1})}
\newcommand{\acc}[2]{\vecbf{a}_{\textrm{#2}} (\textbf{#1})}
\newcommand{\accs}[2]{\vecbf{a}_{\textrm{#2}}^{\textrm{s}} (\textbf{#1})}
\newcommand{\accn}[2]{\vecbf{a}_{\textrm{#2}}^{\textrm{n}} (\textbf{#1})}
\newcommand{\velo}[1]{\vvec_{\textrm{#1}}}
\newcommand{\accso}[1]{\vecbf{a}_{\textrm{#1}}^{\textrm{s}}}
\newcommand{\accno}[1]{\vecbf{a}_{\textrm{#1}}^{\textrm{n}}}
\newcommand{\re}[2]{\Re_{\textrm{#2}}(\textbf{#1})}
\newcommand{\psio}{\dot{\psi}_0}
\newcommand{\Pll}{\textbf{P}_\textrm{lliure}}
\newcommand{\agal}[1]{\vecbf{a}_{\textrm{Gal}} (#1)}
\newcommand{\angal}[1]{\vecbf{a}_{\textrm{NGal}} (#1)}
\newcommand{\vgal}[1]{\vecbf{v}_{\textrm{Gal}} (#1)}
\newcommand{\fvec}[2]{\overline{\mathbf{F}}_{\textrm{#1}\rightarrow\textrm{#2}}}
\newcommand{\mvec}[2]{\overline{\mathbf{M}}_{\textrm{#1}\rightarrow\textrm{#2}}}
\newcommand{\F}[1]{\vec{\Fs}_{#1}}
\newcommand{\Fcal}[2]{\vec{\cal{F}}^{\text{#1}}_{#2}}
</math>

The '''Work-Energy Theorem''' plays a prominent role in physics, as it bridges all branches of science (electromagnetism, thermodynamics, etc.): all of them deal with energy.

There is a certain parallel between the Vector Theorems and the '''Work-Energy Theorem''': they all allow us to understand the evolution of a magnitude that depends on the mechanical state of the system. In the case of the former, we are dealing with the instantaneous evolution of vector magnitudes (linear momentum, angular momentum), while the latter deals with the finite change (in a finite time interval) of a scalar magnitude ('''mechanical energy''').

This unit presents the differential form of this theorem (the '''power balance'''). The integrated form (which is the Work-Energy Theorem itself) is presented in [[E2. Teorema de la energía: versión integrada|'''unit E2''']].
 
------
------
==E1.1 Power balance in a system of particles==

:[[File:E1-1-eng.png|thumb|right|300px|link=]]
<center>'''Figure E1.1''' Forces on a particle of a system with constant matter</center>

Let us consider a system of constant matter, and let’s apply Newton's second law to each particle (or each mass differential) of this system in a general reference frame R (which may be non-Galilean) ('''Figure E1.1'''):

<math>\F{\rightarrow \Ps} + \Fcal{}{\text{ar}\rightarrow\Ps} + \Fcal{}{\text{Cor}\rightarrow\Ps} =\ms_\Ps\acc{P}{R}</math>

<math>\F{\rightarrow \Ps} </math> is the resultant interaction force on '''P''', and it comes from internal interactions (with particles of the same system) and external ones:

<math>\F{\rightarrow \Ps}=\F{\rightarrow \Ps}^\text{int}+\F{\rightarrow \Ps}^\text{ext}</math>

If both sides of the equation are multiplied by <math>\vel{P}{R}</math>, all forces (interaction and inertia forces) orthogonal to <math>\vel{P}{R}</math> are filtered out. Specifically, since <math>\acc{P}{R}=\accs{P}{R} + \accn{P}{R} </math>, where <math>\accn{P}{R} \perp \vel{P}{R}</math>, and since the Coriolis force is always orthogonal to <math>\vel{P}{R} (\Fcal{}{\text{Cor}\rightarrow\Ps} = -2\ms_\Ps \velang{R}{Gal} \times \vel{P}{R})</math>:

<math>\vel{P}{R} \cdot \left( \F{\rightarrow \Ps}^\text{int}+\F{\rightarrow \Ps}^\text{ext} + \Fcal{}{\text{ar}\rightarrow\Ps} \right) = \ms_\Ps\accs{P}{R} \cdot \vel{P}{R} = \frac{\ds}{\ds \ts} \left(\frac{1}{2}\ms_\Ps \vvec_\Rs^2(\Ps) \right) \equiv \frac{\ds\Ts_\Rs(\Ps)}{\ds\ts}=\dot{\Ts}_\Rs(\Ps)</math>

The scalar function <math>\Ts_\Rs(\Ps)</math> is the '''kinetic energy''' of the particle, and is a function of the [[D1. Foundational laws of Newtonian dynamics#D1.3 Newton’s Principle of Determinacy|'''mechanical state''']] of '''P'''. The scalar product of a force and the velocity of its application point is the '''force power''': <math>\vel{P}{R} \cdot \F{\rightarrow \Ps} \equiv \dot{\Ws}_\Rs^{\F{\rightarrow \Ps}} </math>. Both scalar quantities <math>(\Ts_\Rs(\Ps), \dot{\Ws}_\Rs^{\F{\rightarrow \Ps}})</math> depend on the reference frame. For the same set of interaction forces acting on '''P''', the reference frame determines which are filtered out and which are not.

The equation can be rewritten as <math>\dot{\Ws}_\Rs^{\text{ext}\rightarrow \Ps}+\dot{\Ws}_\Rs^{\text{int}\rightarrow \Ps}+\dot{\Ws}_\Rs^{\text{ar}\rightarrow \Ps} =\dot{\Ts}_\Rs(\Ps)</math>, and constitutes the '''power balance''' for particle '''P''' in the R reference frame. Summing for all particles in the system:

<math>\dot{\Ws}_\Rs^{\text{ext}}+\dot{\Ws}^{\text{int}}+\dot{\Ws}_\Rs^{\text{ar}} =\dot{\Ts}_\Rs^\text{syst}</math>

The calculation of kinetic energy is covered in [[E2. Teorema de la energía: versión integrada|'''unit E2''']]. This unit focuses on the calculation of powers. [[E1. Work-Energy Theorem: differential form#E1.2 Potencia de una pareja de acción-reacción|'''section E1-2''']] shows that the power of internal interactions does not depend on the reference frame (which is why the subscript R does not appear in the <math>\dot{\Ws}^{\text{int}}</math> term of the previous equation).

====✏️ EXAMPLE E1-1.1: free particle====
------
:
{|
:[[File:ExE1-1-1-1-eng.jpg|thumb|left|180px|link=]]
|A particle '''P''', with mass m, is released from the ring of a Ferris wheel when it is at its highest point. We want to calculate the power, at this instant, of all forces acting on the particle from three different reference frames: the ground, the ring, and the Ferris wheel cabin.
|}

:Ground reference frame: Since this is a Galilean reference frame, all forces on '''P''' are interaction forces.
{|
:[[File:ExE1-1-2- eng.png|thumb|right|230px|link=]]

|<math>\hspace{0.5cm}\sum\dot{\Ws}_\Ts=\dot{\Ws}_\Ts^\text{weight}= \left(\leftarrow \Rs\Omega_0\right) \cdot \left(\downarrow \ms\gs\right)=0</math>
|}

:Ring reference frame: It is a non-Galilean reference, and the transportation force on '''P''' must be taken into account.
{|
:[[File:ExE1-1-3-eng.png|thumb|right|250px|link=]]
|<math>\hspace{0.5cm}</math> AB:ground, REL:ring 

<math>\hspace{0.5cm}\vel{P}{REL}=\vel{P}{AB}-\vel{P}{ar}= \left(\leftarrow \Rs\Omega_0\right)-\left(\leftarrow \Rs\Omega_0\right)=\overline{0}</math> 

<math>\hspace{0.5cm}</math>

Since the velocity is zero, the total power is zero.
|}

:Cabin reference frame: It is also a non-Galilean reference, and the transportation force on '''P''' must be taken into account.

{|
:[[File:ExE1-1-4-eng.png|thumb|right|250px|link=]]
|<math>\hspace{0.5cm}</math>AB:ground, REL:cabin

<math>\hspace{0.5cm}\vel{P}{REL}=\vel{P}{AB}-\vel{P}{ar}= \left(\leftarrow \Rs\Omega_0\right)-\left(\uparrow \Rs\Omega_0\right)</math>

<math>\hspace{0.5cm}\dot{\Ws}_\text{cabin}^\text{weight}=\left[ \left(\leftarrow \Rs\Omega_0\right)+\left(\downarrow \Rs\Omega_0\right) \right] \cdot \left(\downarrow\ms\gs\right) =\ms\gs\Rs\Omega_0</math>

<math>\hspace{0.5cm}\dot{\Ws}_\text{cabin}^\text{ar}=\left[ \left(\leftarrow \Rs\Omega_0\right)+\left(\downarrow \Rs\Omega_0\right) \right] \cdot \left(\rightarrow\ms\Rs^2\Omega_0^2\right) =-\ms\gs\Rs^2 \Omega_0^3</math>

<math>\hspace{0.5cm}\sum\dot{\Ws}_\text{cabin}=\ms\Rs\Omega_0(\gs-\Rs\Omega_0^2)</math>

|}


====✏️ EXAMPLE E1-1.2: pendulum====
------
:
{|
:[[File:ExE1-2-1-eng.png|thumb|left|180px|link=]]
|A particle '''P''' of mass m hangs from an inextensible thread of length L attached to the ceiling. The assembly oscillates in a vertical plane fixed to the ground. We want to calculate the power of all the forces acting on the particle in two different reference frames: the ground and a platform rotating relative to the ground with a constant angular velocity <math>\Omega_0</math>. 

Ground reference frame: Since it is a Galilean reference frame, '''P''' is only subjected to interaction forces.
{|
:[[File:ExE1-2-2-neut.png|thumb|left|300px|link=]]

|<math>\dot{\Ws}_\Es^\Ts=(\nearrow\Ls\dot{\theta})\cdot(\nwarrow\Ts)=0</math>

<math>\dot{\Ws}_\Es^\text{weight}=(\nearrow\Ls\dot{\theta}\cdot(\downarrow \ms\gs)=-\ms\gs\Ls\dot{\theta}\sin\theta</math>

|}

Platform reference frame: It is a non Galilean reference frame, and the transportation force on '''P''' has to be taken into account.

{|
:[[File:ExE1-2-3-eng.png|thumb|left|400px|link=]]

|AB:ground, REL:platform

<math>\vel{P}{REL}=\vel{P}{AB}-\vel{P}{ar}=(\nearrow\Ls\dot{\theta})-(\otimes \Ls \Omega_0 \sin \theta)</math>

<math>\dot{\Ws}_\Es^\Ts\left[(\nearrow\Ls\dot{\theta})+(\odot\Ls\Omega_0\sin\theta)\right]\cdot(\nwarrow\Ts)=0</math>

<math>\dot{\Ws}_\Es^\text{weight}\left[(\nearrow\Ls\dot{\theta})+(\odot\Ls\Omega_0\sin\theta)\right]\cdot(\downarrow \ms\gs)=-\ms\gs\Ls\dot{\theta}\sin\theta</math>

<math>\dot{\Ws}_\Es^\text{ar}\left[(\nearrow\Ls\dot{\theta})+(\odot\Ls\Omega_0\sin\theta)\right]\cdot(\rightarrow \ms\Ls\Omega_0^2\sin\theta)=\ms\Ls^2\Omega_0^2\dot{\theta}\cos\theta</math>
|}


------
------

==E1.2 Power of an action-reaction pair==

The total power associated with internal interactions in a constant-mass system is not zero in principle (since the two forces of each action-reaction pair are applied to particles that may have different velocities relative to the reference frame in which the calculation is made), and it is independent of the reference frame in which it is evaluated. It only depends on the separation velocity of the two particles. The power balance for the system is therefore written as: <math>\dot{\Ts}_\Rs^\text{syst}=\dot{\Ws}_\Rs^\text{ext} + \dot{\Ws}^\text{int} + \dot{\Ws}_\Rs^\text{ar} </math> .

<div>
=====💭 PROOF ➕=====

{|
:[[File:E1-2-neut.png|thumb|left|200px|link=]]
|<math>\dot{\Ws}^\text{ac-reac}=\overline{\Fs}_{\Ps \rightarrow \Qs}\cdot \vel{Q}{R} + \overline{\Fs}_{\Qs \rightarrow \Ps}\cdot \vel{P}{R}= \overline{\Fs}_{\Ps \rightarrow \Qs}\cdot \left[ \vel{Q}{R}-\vel{P}{R}\right]</math>

By the [[D1. Foundational laws of Newtonian dynamics#D1.6 Newton’s third law (action-reaction principle)|'''action-reaction principle''']], <math>\overline{\Fs}_{\Ps \rightarrow \Qs}=-\overline{\Fs}_{\Qs \rightarrow \Ps}</math>. Therefore: <math>\dot{\Ws}^\text{ac-reac}=\overline{\Fs}_{\Ps \rightarrow \Qs}\cdot \left[ \vel{Q}{R}-\vel{P}{R}\right]</math>

As <math>\overline{\Fs}_{\Ps \rightarrow \Qs}</math> has the <math>\PQvec</math> direction, only the components of the velocities in that direction have to be considered:

<math>\dot{\Ws}^\text{ac-reac}=\overline{\Fs}_{\Ps \rightarrow \Qs}\cdot \biggl( \vel{Q}{R} \Bigr]_{||\Ps\Qs}-\ \vel{P}{R} \Bigr]_{||\Ps\Qs}\biggl)</math>
|}

The difference <math>\biggl( \vel{Q}{R} \Bigr]_{\Ps\Qs}-\ \vel{P}{R} \Bigr]_{\Ps\Qs}\biggl)</math> is the separation velocity <math>\dot{\rho}</math> between '''P''' and '''Q''', and it does not dpend on the reference frame since it is the derivative of a scalar (the distance <math>\dot{\rho}</math> between the two particles). For the previous drawing, <math>\dot{\rho}> 0</math>. Since this is an attractive interaction, <math>\dot{\Ws}^\text{ac-reac} < 0</math>.

With a drawing, it is easy to see that:
* attractive interaction and particle approach <math>(\dot{\rho}<0)</math> : <math>\dot{\Ws}^\text{ac-reac} > 0</math> ;
* repulsive interaction and particle separation <math>(\dot{\rho}>0)</math> : <math>\dot{\Ws}^\text{ac-reac} > 0</math>
* repulsive interaction and particle approach <math>(\dot{\rho}<0)</math> : <math>\dot{\Ws}^\text{ac-reac} < 0</math>


</div>

There are two particular cases of interaction forces whose power always has the same sign:

* Friction action-reaction pair between '''P''' and '''Q''': since the friction force always opposes the relative velocity between the two points, its power is always negative:

<math>\hspace{1cm} \dot{\Ws}^\text{friction}=\dot{\Ws}^{\text{fric}\rightarrow\Ps}_\text{RTQ}+\dot{\Ws}^{\text{fric}\rightarrow\Qs}_\text{RTQ}=\overline{\Fs}^{\text{fric}\rightarrow\Ps} \cdot \vel{P}{RTQ} = - \abs{\overline{\Fs}^{\text{fric}\rightarrow\Ps}} \frac{\vel{P}{RTQ}}{\abs{\vel{P}{RTQ}}} \cdot \vel{P}{RTQ}<0;</math>

* Constraint action-reaction pair between '''P''' and '''Q''': from the [[D3. Interactions between rigid bodies#D3.4 Direct constraint interactions|'''analytical characterization equation''']] of the constraint forces between two points '''P''' and '''Q''' belonging to two rigid bodies <math>\mathrm{S}_\Ps</math> and <math>\mathrm{S}_\Qs</math>, it is easy to see that the associated power is zero:

<math>\hspace{1cm} \dot{\Ws}^\text{constraint}=\dot{\Ws}^{\text{constraint}\rightarrow\Ps}_{\mathrm{S}_\Ps}+\dot{\Ws}^{\text{constraint}\rightarrow\Qs}_{\mathrm{S}_\Ps}=\overline{\Fs}^{\text{constraint}\rightarrow\Ps} \cdot \overline{\mathbf{v}}_{\mathrm{S}_\Qs}(\Ps) = - 0;</math>

====✏️ EXAMPLE E1-2.1: power of an action-reaction friction pair====
------
:
{|
:[[File:ExE1-2-1-2-eng.png|thumb|left|180px|link=]]
|A homogeneous wheel, with radius r and mass m, moves down an inclined plane without sliding until it joins a conveyor belt moving at a constant velocity <math>\vs_0</math> relative to the ground. At the moment it makes contact with the belt, the center of the wheel has a velocity <math>\vs_0</math> relative to the ground in the same direction as the belt. We want to calculate the power of the friction force between the belt and the wheel at this instant.
|}

At that momento, the <math>\mathbf{ICR}^\text{wheel}_\Es</math> is point <math>\Js_\text{wheel}</math> in contact with the belt. This point slides backwards with respect to the belt, therefore the friction force on the wheel is forwards:

<math>\left.\begin{array}{l}
\text { AB: ground } \\
\text { REL: belt }
\end{array}\right\} \quad \bar{\vs}_{\text {REL }}\left(\Js_{\text {wheel }}\right)=\bar{\vs}_{\text{AB}}\left(\Js_{\text {wheel }}\right)-\bar{\vs}_{\mathrm{ar}}\left(\Js_{\text {wheel }}\right)=-\bar{\vs}_{\mathrm{ar}}\left(\Js_{\text {wheel }}\right)=-\left(\rightarrow \mathrm{v}_0\right)=\left(\leftarrow \mathrm{v}_0\right) .</math>

[[File:ExE1-2-1-3-eng.png|thumb|center|500px|link=]]

Calculation of the friction power: <math>\dot{\Ws}^\text{friction}=\dot{\Ws}_\Rs^{\text{friction}\rightarrow \text{wheel}}+ \dot{\Ws}_\Rs^{\text{friction}\rightarrow \text{belt}}</math>, where R may be any reference frame. To quickly calculate this power, it is advisable to choose a reference frame R where one of the two points subjected to the action-reaction pair has zero speed, so that one of the two terms is directly zero:

<math>\dot{\Ws}^\text{friction}=\dot{\Ws}_\text{belt}^{\text{fric}\rightarrow \text{wheel}}+ \dot{\Ws}_\text{belt}^{\text{fric }\rightarrow \text{belt}}=\overline{\Fs}^{\text{fric}\rightarrow \text{wheel}} \cdot \bar{\vs}_{\text {belt}}\left(\Js_{\text {wheel }}\right)=(\rightarrow \mu\ms\gs)\cdot(\leftarrow \vs_0)=-\mu\ms\gs\vs_0.</math>

If the calculation is done in any other reference frame, the result is the same:

<math>\dot{\Ws}^\text{friction}=\dot{\Ws}_\text{E}^{\text{fric}\rightarrow \text{wheel}}+ \dot{\Ws}_\text{E}^{\text{fric} \rightarrow \text{belt}}=(\rightarrow \mu\ms\gs)\cdot 0 + (\leftarrow \mu\ms\gs)\cdot(\rightarrow \vs_0)=-\mu\ms\gs\vs_0.</math>

<math>\dot{\Ws}^\text{friction}=\dot{\Ws}_\text{vehicle}^{\text{fric}\rightarrow \text{wheel}}+ \dot{\Ws}_\text{vehicle}^{\text{fric} \rightarrow \text{belt}}=(\rightarrow \mu\ms\gs)\cdot (\rightarrow\vs_0) + (\leftarrow \mu\ms\gs) \cdot(\rightarrow 3\vs_0)=-\mu\ms\gs\vs_0.</math>

However, if only the power of one of the two forces is calculated (for example, the friction on the wheel), the result depends on the reference frame and can have any sign:

<math>\dot{\Ws}_\text{belt}^{\text{fric }\rightarrow \text{wheel}}=(\rightarrow \mu\ms\gs)\cdot(\leftarrow \vs_0)=-\mu\ms\gs\vs_0 ,</math>

<math>\dot{\Ws}_\text{E}^{\text{fric} \rightarrow \text{wheel}}=(\rightarrow \mu\ms\gs)\cdot 0 = 0 ,</math>

<math>\dot{\Ws}_\text{vehicle}^{\text{fric} \rightarrow \text{wheel}}=(\rightarrow \mu\ms\gs)\cdot (\rightarrow\vs_0)=\mu\ms\gs\vs_0.</math>


-----
-----

==E1.3 Power of a system of forces on a rigid body==

{|
|
:[[File:E1-3-eng.png|thumb|left|180px|link=]]
<center>'''Figure E1.3''': Kinematic torsor of a rigid body in a reference frame R, and forces on different points (dm) of that rigid body</center>

|When calculating the power of a system of forces on a rigid body S, the term associated with internal interactions is zero because those interactions are all constraint (cohesion) forces and, as seen in [[E1. Work-Energy Theorem: differential form#E1.2 Power of an action-reaction pair|'''section E1.2''']], their total power is zero.

The total power of a system of forces on S can be calculated from the torsor of the system at any point '''O''' of the rigid body, <math>(\overline{\Fs},\overline{\Ms}(\Os))</math>. Taking into account rigid body kinematics ('''Figure E1.3'''):

<math>\dot{\Ws}_\Rs^{\rightarrow \Ss} =\sum_\Ps\overline{\Fs}^{\rightarrow \Ps} \cdot \vel{P}{R}= \sum_\Ps \left[ \overline{\Fs}^{\rightarrow \Ps} \cdot \vel{O}{R} + \velang{S}{R} \times \OPvec \right] </math>

|}
<math>\dot{\Ws}_\Rs^{\rightarrow \Ss} =\left(\sum_\Ps\overline{\Fs}^{\rightarrow \Ps} \right)\cdot \vel{O}{R} + \sum_\Ps\overline{\Fs}^{\rightarrow \Ps} \cdot \left(\velang{S}{R} \times \OPvec \right) = \left( \sum_\Ps\overline{\Fs}^{\rightarrow \Ps} \right) \cdot \vel{O}{R} + \left[ \sum_\Ps \left( \OPvec \times \overline{\Fs}^{\rightarrow \Ps} \right)\right] \cdot \velang{S}{R} = \overline{\Fs}\cdot \vel{O}{R} + \overline{\Ms}(\Os)\cdot \velang{S}{R} </math>

This calculation also shows that the power of a moment (or a torque <math>\Gamma</math>) on a rigid body in a reference frame R is the product of the moment by the angular velocity of the rigid body: <math>\dot{\Ws}_\Rs^{\Gamma\rightarrow \mathrm{S}} = \overline{\Gamma} \cdot \velang{S}{R}</math>.

The power balance for a rigid body is: <math>\dot{\Ws}_\Rs^{ \rightarrow \mathrm{S}} = \dot{\Ws}_\Rs^{ \mathrm{ext}} + \dot{\Ws}_\Rs^{ \mathrm{ar}}.</math>

====✏️ EXAMPLE E1-3.1: roller sliding on a slope====
------
:
{|
|
:[[File:ExE1-3-1-1-eng.png|thumb|left|180px|link=]]
|A homogeneous roller, with mass m and radius r, slides down a slope. The friction coefficient between the ground and the roller is <math>\mu</math>. We want to calculate the power of every force acting on the roller in the ground reference frame.
|}

:It is a 2D problem. The kinematic and dynamic descriptions are:
{|
|
[[File:ExE1-3-1-2-eng.png|thumb|right|250px|link=]]

|<math> \hspace{0.5cm}\vel{J}{E}=\vel{C}{T} + \velang{}{E} \times \CJvec= </math>

<math>\hspace{1.7cm}=\left( \rightarrow \vs_\Es \right) + \left( \otimes \Omega_\Es\right) \times \left( \downarrow \rs \right)</math>

<math> \hspace{0.5cm}\vel{J}{E} = \left[\rightarrow (\vs_\Es -\rs\Omega_\Es)\right]</math>

<math>\hspace{0.5cm} \dot{\Ws}_\Es^{\mathrm{ext}} = \dot{\Ws}_\Es^{\mathrm{weight}} + \dot{\Ws}_\Es^{\mathrm{friction}} + \dot{\Ws}_\Es^{\mathrm{constraint}}= \Bigr[\left(\downarrow \ms\gs\cos\beta\right)+\left(\rightarrow \ms\gs\sin\beta\right)\Bigr] \cdot (\rightarrow \vs_\Es) + \Bigr[(\leftarrow \mu \Ns) + (\uparrow \Ns ) \Bigr] \cdot \Bigr[\rightarrow (\vs_\Es -\rs\Omega_\Es)\Bigr]</math>

|}
:Since the roller center has no motion perpendicular to the plane, the AMT leads to <math>\Ns =\ms\gs\cos\beta</math>. Therefore:

{|
|
[[Archivo:ExE1-3-1-3-eng.png|thumb|right|140px|link=]]

|<math>\hspace{0.5cm} \dot{\Ws}_\Ts^{\mathrm{ext}} = \ms\gs\vs_\Ts \sin \beta - \ms\gs(\vs_\Ts -\rs\Omega_\Ts)\cos\beta=\ms\gs \Bigr[ \vs_\Ts (\sin\beta-\cos\beta)+\rs\Omega_\Ts \cos\beta \Bigr].</math>

<math>\hspace{0.5cm}</math>

If instead of considering the friction at '''J''' we consider the equivalent torsor at point '''C''', we must consider the force applied in '''C''' and the moment about '''C''', which is not zero:

<math>\hspace{0.5cm} \dot{\Ws}_\Es^{ \mathrm{friction}} = (\leftarrow \mu\Ns)\cdot(\rightarrow\vs_\Es)+(\otimes \mu\Ns\rs)\cdot(\otimes \Omega_\Es) =\mu\ms\gs\cos\beta(\rs\Omega_\Es-\vs_\Es)</math>,

<math>\hspace{0.5cm} \dot{\Ws}_\Es^{\mathrm{ext}} = \dot{\Ws}_\Es^{ \mathrm{weight}} + \dot{\Ws}_\Es^{ \mathrm{friction}}+ \dot{\Ws}_\Es^{ \mathrm{constraint}}= \ms\gs\vs_\Es\sin\beta+\mu\ms\gs\cos\beta (\rs\Omega_\Es-\vs_\Es)</math>.

<math>\hspace{0.5cm}</math>

The result is the same.
|}


====✏️ EXAMPLE E1-3.2: block sliding on a truck====
------
:
{|
|
:[[File:ExE1-3-2-1-eng.png|thumb|left|220px|link=]]
|A truck brakes with acceleration '''a''' relative to the ground at the instant in which its velocity (also relative to the ground) is v. At the same instant, the homogeneous block, of mass m, slides with velocity v' relative to the truck. We want to perform a power balance for the block in two different reference frames: the ground and the truck.
|}

:Ground reference frame: It is a 2D problem. The kinematic and dynamic descriptions are:
{|
|
[[File:ExE1-3-2-2-eng.png|thumb|right|200px|link=]]

| The constraint between the truck and the block allows only one translational DoF between the two. Therefore, in the 2D approximation, the associated torsor has two independent components. If characterized at the midpoint of the contact surface, these components are a normal force and a moment perpendicular to the plane. The AMT applied to the block leads to <math>\Ns=\ms\gs</math>.

|}
:The powers associated with all interactions on the block in the ground reference are:

:<math>\dot{\Ws}_\Es^{\rightarrow \mathrm{bloque}}=\dot{\Ws}_\Ts^{\mathrm{ext}}=\dot{\Ws}_\Ts^{\mathrm{peso}}+\dot{\Ws}_\Ts^{\mathrm{fricción}}+\dot{\Ws}_\Ts^{\mathrm{enlace}}= \Bigr[(\downarrow \ms\gs)+(\leftarrow \mu\ms\gs)+(\uparrow \Ns) \Bigr] \cdot \Bigr[\rightarrow(\vs+\vs')\Bigr] + (\odot \Ms_\es)\cdot 0 = -\mu\ms\gs(\vs+\vs')</math>

:Ya que el bloque se traslada respecto al suelo, su energía cinética total es inmediata, y se puede plantear fácilmente el balance de potencias:

:<math>\Ts_\Ts^\mathrm{bloque}=\frac{1}{2}\int_\mathrm{bloque}\ds\ms \vs_\Ts^2(\mathrm{bloque})=\frac{1}{2}\ms \vs_\Ts^2(\mathrm{bloc})\quad \Rightarrow \quad \dot{\Ts}_\Ts^\mathrm{bloque}=\ms\vs_\Ts( \mathrm{bloque})\dot{\vs}_\Ts( \mathrm{bloque}) </math>

:Como la única fuerza horizontal sobre el bloque es la fricción:

<math>\dot{\vs}_\Ts(\mathrm{bloque})=-\mu\gs\quad \Rightarrow \quad \dot{\Ts}_\Ts^\mathrm{bloque}=-\mu\ms\gs(\vs+\vs').</math>

:El balance de potencias es consistente:

:<math>\left.\begin{array}{l}
\dot{\mathrm{W}}_\Ts^{\rightarrow \text { bloque }}=\dot{\mathrm{W}}_{\mathrm{T}}^{\text {ext }}=-\mu \mathrm{mg}\left(\mathrm{v}+\mathrm{v}^{\prime}\right) \\
\dot{\mathrm{T}}_{\mathrm{T}}^{\text {bloque }}=-\mu \mathrm{mg}\left(\mathrm{v}+\mathrm{v}^{\prime}\right)
\end{array}\right\} \Rightarrow \dot{\mathrm{W}}_{\mathrm{T}}^{\rightarrow \text { bloque }}=\dot{\mathrm{T}}_{\mathrm{T}}^{\text {bloque }}</math>

:Referencia camión: Al estar el camión acelerado respecto al suelo, hay que tener en cuenta las fuerzas de arrastre sobre todos los diferenciales de masa (dm) del bloque:

{|
|

[[Archivo:ExE1-3-2-3-cat-esp.jpg|thumb|right|250px|link=]]
|<math>\cal{\Fs}^{\rightarrow \ds\ms}_{\mathrm{ar}}=-\ds\ms\overline{\as}_\mathrm{ar}(\ds\ms)=-\ds\ms (\leftarrow \as)=(\rightarrow \ds\ms \cdot \as)</math>.

El torsor resultante de este sistema de fuerzas en el centro de inercia '''G''' se reduce a una fuerza horizontal.

<math>\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{bloque}}=\dot{\Ws}_\mathrm{camión}^{\mathrm{ext}} + \dot{\Ws}_\mathrm{camión}^{\mathrm{ar}} =\dot{\Ws}_\mathrm{camión}^{\mathrm{peso}} + \dot{\Ws}_\mathrm{camión}^{\mathrm{fricción}} + \dot{\Ws}_\mathrm{camión}^{\mathrm{enlace}} + \dot{\Ws}_\mathrm{camión}^{\mathrm{ar}} , </math>

<math>\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{bloque}}= \Bigr[(\downarrow \ms\gs)+(\leftarrow \mu\ms\gs)+(\uparrow \Ns) + (\rightarrow \ms\as)\Bigr] \cdot(\rightarrow \vs')+(\odot\Ms_\es) \cdot0=\ms(\as-\mu\gs)\vs'.</math>
|}

:Se puede comprobar que el balance de potencias es consistente:

:<math>\Ts^\mathrm{bloque}_\mathrm{camión}=\frac{1}{2}\ms\vs^2_\mathrm{camión}(\mathrm{bloque}) \quad \Rightarrow \quad \dot{\Ts}^\mathrm{bloque}_\mathrm{camión} = \ms\vs_\mathrm{camión}(\mathrm{bloque})\dot{\vs}_\mathrm{camión}(\mathrm{bloque})</math>

:<math>\left.\begin{array}{l}
\mathrm{AB}:\mathrm{suelo} \\
\mathrm{REL}:\mathrm{camión}
\end{array}\right\} \quad \dot{\vs}_\mathrm{camión}(\mathrm{bloque})= \dot{\vs}_\mathrm{T}(\mathrm{bloque})-\dot{\vs}_\mathrm{ar}(\mathrm{bloque}) = (-\mu\gs)-(-\as)=\as-\mu\gs \quad \Rightarrow \quad \dot{\Ts}^\mathrm{bloque}_\mathrm{camión}=\ms\vs'(\as-\mu\gs)</math>

:<math>\left.\begin{array}{l}
\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{bloque}}=\dot{\Ws}_\mathrm{camión}^{\mathrm{ext}} + \dot{\Ws}_\mathrm{camión}^{\mathrm{ar}}= \ms(\as-\mu\gs)\vs'\\
\dot{\Ts}^\mathrm{bloque}_\mathrm{camión}=\ms\vs'(\as-\mu\gs)
\end{array}\right\} \Rightarrow\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{bloque}}= \dot{\Ts}^\mathrm{bloque}_\mathrm{camión}</math>


====✏️ EJEMPLO E1-3.3: camión frenando con carga====
------
:
{|
|
:[[Archivo:ExE1-3-3-1-esp.png|thumb|left|220px|link=]]
|Un camión frena con aceleración '''a''' respecto al suelo en el instante en el que su velocidad (también respecto al suelo) es v. Los pares de frenada en el eje delantero y en el trasero son iguales y valen <math>\Gamma</math>, y las ruedas, de radio r y masa despreciable, no deslizan sobre el suelo. En el mismo instante, el bloque homogéneo, de masa m, desliz con velocidad v' relativa al camión. Se trata de hacer un balance de potencias para el sistema (bloque+camión) en la referencia del suelo y en la del camión.

|}

:Ya que la potencia de las interacciones internas no es nula en principio (no se trata de un único sólido rígido), es conveniente hacer un DGI del sistema para no olvidar ninguna:
{|
|
[[Archivo:ExE1-3-3-2-esp.png|thumb|right|250px|link=]]

|
|
c.p.s.d.: contacto puntual sin deslizamiento

c.m.c.d.: contacto multipuntual con deslizamiento

|}
:El sistema sobre el que se hace el balance de potencias es todo menos el suelo. Teniendo en cuenta que la potencia de los enlaces internos es siempre nula, las únicas interacciones que hay que tener en cuenta son el peso, los enlaces de las ruedas con el suelo, los pares de frenada y la fricción entre bloque y camión:

[[Archivo:ExE1-3-3-3-esp.png|thumb|center|300px|link=]]

:Referencia suelo:
:Las fuerzas de enlace del suelo sobre las ruedas están aplicadas a puntos de velocidad cero, y por tanto no hacen potencia. El peso es una fuerza vertical, y el movimiento del camión y del bloque es horizontal, por tanto su potencia también es cero.

:<math>\dot{\Ws}_\Ts^{\rightarrow \mathrm{sist}}=\dot{\Ws}_\Ts^{\mathrm{ext}}+\dot{\Ws}^{\mathrm{int}}=\dot{\Ws}_\Ts^{\mathrm{peso}}+\dot{\Ws}_\Ts^{\mathrm{enlace}}+\dot{\Ws}^{\mathrm{fricción}}+\dot{\Ws}^{\Gamma}=\dot{\Ws}^{\mathrm{fricción}}+\dot{\Ws}^{\Gamma}.</math>

:Las potencias de interacciones internas se pueden calcular en cualquier referencia. Por tanto:

:<math>\dot{\Ws}^{\mathrm{fricción}}= \dot{\Ws}_\mathrm{camión}^{\mathrm{fric}\rightarrow \mathrm{bloque}} +\dot{\Ws}_\mathrm{camió}^{\mathrm{fric}\rightarrow \mathrm{camión}} = \dot{\Ws}_\mathrm{camión}^{\mathrm{fric}\rightarrow \mathrm{bloque}}= (\leftarrow \mu\ms\gs)\cdot (\rightarrow \vs')=-\mu\ms\gs\vs'</math>

:<math>\dot{\Ws}^{\Gamma}= \dot{\Ws}_\mathrm{camión}^{\Gamma\rightarrow \mathrm{ruedas}} +\dot{\Ws}_\mathrm{camión}^{\Gamma\rightarrow \mathrm{camión}} = \dot{\Ws}_\mathrm{camión}^{\Gamma\rightarrow \mathrm{ruedas}}= 2 \left(\odot \Gamma \right)\cdot \left(\otimes \frac{\vs}{\rs}\right)=-2\Gamma\frac{\vs}{\rs}.</math>

:Por otro lado: <math>\Ts_\Ts^\mathrm{sist}=\frac{1}{2}\ms\vs_\Ts^2(\mathrm{bloque})+\frac{1}{2}\ms\vs_\Ts^2(\mathrm{camión})\quad \Rightarrow \quad \dot{\Ts}_\Ts^\mathrm{sist}=-\ms(\vs+\vs')\mu\gs-\Ms\vs\as. </math>

:Finalmente: <math>\dot{\Ws}_\Ts^{\rightarrow \mathrm{sist}}=\dot{\Ts}_\Ts^\mathrm{sist} \quad \Rightarrow \quad -\mu\ms\gs\vs'-2\Gamma\frac{\vs}{\rs}=-\ms(\vs+\vs')\mu\gs-\Ms\vs\as.</math>

:Esta ecuación permite calcular el par de frenada: <math>\Gamma=\frac{1}{2}(\mu\ms\gs+\Ms\as)\rs.</math>

:Referencia camión:

:En esta referencia hay que tener en cuenta las fuerzas de arrastre. Por otro lado, las fuerzas de enlace del suelo sobre las ruedas están aplicadas a puntos que ya no tienen velocidad cero, y por tanto hacen potencia. Como antes, la potencia del peso es cero.

:<math>\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{sist}}=\dot{\Ws}_\mathrm{camión}^{\mathrm{ext}}+\dot{\Ws}^{\mathrm{int}}+\dot{\Ws}_\mathrm{camión}^{\mathrm{ar}}=\dot{\Ws}_\mathrm{camión}^{\mathrm{enlace}}+\dot{\Ws}^{\mathrm{fricción}}+\dot{\Ws}^{\Gamma}+\dot{\Ws}_\mathrm{camión}^{\mathrm{ar}}.</math>

:Las fuerzas de arrastre sobre el camión no hacen potencia porque el camión está parado en esta referencia. La de las fuerzas de arrastre sobre el bloque se ha calculado en el [[E1. Teorema de la energía: versión diferencial#✏️ ✏️ EJEMPLO E1-3.2: bloque deslizando sobre un camión|'''ejemplo E1-3.2''']]. La potencia de la fricción y de los pares de frenada es la misma de antes ya que no depende de la referencia.

:<math>\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{sist}}=\dot{\Ws}_\mathrm{camión}^{\mathrm{enllaç}}+\dot{\Ws}^{\mathrm{fricción}}+\dot{\Ws}^{\Gamma}+\dot{\Ws}_\mathrm{camión}^{\mathrm{ar} \rightarrow \mathrm{bloque}}= \dot{\Ws}_\mathrm{camión}^{\mathrm{enlace}} -\mu\ms\gs\vs'-2\Gamma\frac{\vs}{\rs} +\ms\as\vs'.</math>

:Las descripciones cinemática y dinámica de la rueda son:

[[Archivo:ExE1-3-3-4-esp.png|thumb|center|430px|link=]]

:<math>\dot{\Ws}_\mathrm{camión}^{\mathrm{enlace}}=2\Bigr[(\leftarrow \Ts)+(\uparrow \Ns)\Bigr]\cdot(\leftarrow\vs)=2\Ts\vs.</math>

La fuerza tangencial de enlace se puede calcular a partir del TMC aplicado al centro '''C''' de la rueda. Como no tiene masa: <math>\sum\overline{\Ms}_\mathrm{ext}(\Cs)=\overline{0} \quad \Rightarrow \quad (\odot \Gamma)+(\otimes \Ts\rs)=0\quad \Rightarrow \quad \Ts=\Gamma/\rs.</math>

:Por tanto: <math>\dot{\Ws}_\mathrm{camió}^{\rightarrow \mathrm{sist}} =2\frac{\Gamma}{\rs}\vs-\mu\ms\gs\vs'-2\Gamma\frac{\vs}{\rs}+\ms\as\vs'=\ms(\as-\mu\gs)\vs'.</math>

:Por otro lado, como se ha visto en el [[E1. Teorema de la energía: versión diferencial#✏️ ✏️ EJEMPLO E1-3.2: bloque deslizando sobre un camión|'''ejemplo E1-3.2''']]:

:<math>\Ts_\mathrm{camión}^\mathrm{sist}= \Ts_\mathrm{camión}^\mathrm{bloque}=\frac{1}{2}\ms\vs_\mathrm{camión}^2(\mathrm{bloque}) \quad \Rightarrow \quad \dot{\Ts}_\mathrm{camión}^\mathrm{sist}= \dot{\Ts}_\mathrm{camión}^\mathrm{bloque}=\ms\vs'(\as-\mu\gs).</math>

:El balance de potencias es consistente:

:<math>\left.\begin{array}{l}
\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{sist}}= \ms(\as-\mu\gs)\vs'\\
\dot{\Ts}^\mathrm{sist}_\mathrm{camión}=\ms(\as-\mu\gs)\vs'
\end{array}\right\} \Rightarrow\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{sist}}= \dot{\Ts}^\mathrm{sist}_\mathrm{camión}</math>

-----
-----

==E1.4 Balance de potencias para un sistema multisólido: cálculo directo y cálculo indirecto==

El '''cálculo directo''' (a partir de su definición) de la potencia asociada a una interacción (una fuerza sobre un punto '''P''' o un momento sobre un sólido rígido S) exige el conocimiento de la interacción (fuerza o momento):

<math>\dot{\Ws}_\Rs^{\Fs\rightarrow \Ps}=\overline{\Fs}_{\rightarrow\Ps} \cdot \vel{P}{R} \quad , \quad \dot{\Ws}_\Rs^{\Ms\rightarrow \Ss}=\overline{\Ms}_{\rightarrow\Ss} \cdot \velang{S}{R} </math>

Cuando no se trata de una interacción formulable, hay que recurrir a los teoremas vectoriales para determinarla.

El '''cálculo indirecto''' es una alternativa que no necesita recurrir a estos teoremas. Se basa en la evaluación de un balance de potencias para un sistema y una referencia adecuados de manera que la potencia a determinar sea el único término desconocido de la ecuación (el único que no se puede calcular de manera directa). Cuando el balance incluye más de un término desconocido, se pueden combinar varios balances (para distintos sistemas y referencias) hasta tener un número suficiente de ecuaciones que permitan la determinación de todas las potencias desconocidas.

====✏️ EJEMPLO E1-4.1: polea frenada====
------
:
{|
|
:[[Archivo:ExE1-4-1-1-esp.png|thumb|left|150px|link=]]
|Un bloque de masa m cuelga de un hilo inextensible atado a un punto de la periferia de una polea de radio r. Entre soporte (fijo al suelo) y polea actúa un freno que garantiza que el bloque baja a velocidad constante respecto al suelo. Se trata de calcular la potencia del freno.

|}

:Càlculo directo: <math>\dot{\Ws}^{\mathrm{freno}}= \overline{\Gamma}^{\rightarrow \mathrm{polea}}\cdot\velang{polea}{T}=\left(\odot\Gamma\right)\cdot\left( \otimes \frac{\vs_0}{\rs}\right)=-\Gamma \frac{\vs_0}{\rs}.</math>

:El par de frenada no es un dato, y hay que recurrir al TMC para calcularlo.

:<math>\left.\begin{array}{l}
\text{SISTEMA: polea + bloque}\\
\text{TMC en } \Os
\end{array}\right\} \sum\overline{\Ms}_\mathrm{ext}(\Os)=\dot{\overline{\Hs}}_\mathrm{RTO}(\Os)</math>

:El momento cinético proviene de las dos piezas. Ya que '''O''' no es fijo al bloque, hay que hacer descomposición baricéntrica para calcular el momento cinético del bloque:

:<math>\overline{\Hs}_\mathrm{RTO=T} (\Os) =\overline{\Hs}_\Ts^\mathrm{politja}(\Os) + \overline{\Hs}_\Ts^\mathrm{bloc}(\Os)= \overline{\Hs}_\Ts^\mathrm{politja}(\Os) + \overline{\Hs}_\Ts^\mathrm{bloc}(\Gs_\mathrm{bloc})+ \overline{\Hs}_\Ts^{\oplus\mathrm{bloc}}(\Os) = </math>

:<math>\hspace{2.3cm}=\Is_\mathrm{O}^\mathrm{polea} \velang{polea}{T}+\Is_{\Gs_\mathrm{bloque}}^\mathrm{bloque}+\OGvec_\mathrm{bloque}\times \ms\overline{\vs}_\Ts(\Gs_\mathrm{bloque})=\left(\otimes \Is_\mathrm{O}\frac{\vs_0}{\rs}\right)+(\otimes \ms\vs_0\rs).</math>

:[[Archivo:ExE1-4-1-2-esp.png|thumb|right|180px|link=]]
:Se trata de un momento cinético constante en dirección y valor. Por tanto:

:<math>\dot{\overline{\Hs}}_\mathrm{RTO}(\Os)=0 \quad \Rightarrow \quad \sum{\overline{\Ms}}_\mathrm{ext}(\Os)=(\odot\Gamma)+(\otimes \ms\gs\rs) =0\quad \Rightarrow \quad \Gamma= \ms\gs\rs</math>.

:Finalmente: <math>\dot{\Ws}^{\mathrm{freno}}= -\Gamma \frac{\vs_0}{\rs} = -\ms\gs\vs_0.</math>

:Cálculo indirecto:

:<math>\left.\begin{array}{l}
\text{SISTEMA: polea + soporte + bloque}\\
\text{REF: suelo(Gal)}
\end{array}\right\} \quad \dot{\Ts}^\mathrm{sist}_\Ts=\dot{\Ws}^\mathrm{ext}_\Ts+ \dot{\Ws}^\mathrm{int}</math>

:<math>\left.\begin{array}{l}
\dot{\Ts}^\mathrm{sist}_\Ts=0\\
\dot{\Ws}^\mathrm{ext}_\Ts=\left(\downarrow \ms\gs\right)\cdot\left(\downarrow \vs_0\right)= \ms\gs\vs_0\\
\dot{\Ws}^\mathrm{int}=\dot{\Ws}^{\mathrm{freno}}
\end{array}\right\} \quad \Rightarrow \quad \dot{\Ws}^{\mathrm{freno}} =-\ms\gs\vs_0</math>



====✏️ EJEMPLO E1-4.2: mono y plátanos====
------
:
{|
|
:[[Archivo:ExE1-4-2-1-esp.png|thumb|left|180px|link=]]
|Unos plátanos de masa <math>(2/3)\ms</math> cuelgan de una cuerda inextensible que se apoya sin deslizar sobre una polea de radio r y masa despreciable, articulada respecto al suelo. En el otro extremo de la cuerda se agarra un mono de masa m. La polea gira con velocidad angular constante <math>\Omega_0</math> respecto al suelo bajo la acción de un motor. El mono se mueve respecto a la cuerda sin que sus manos se froten contra la cuerda, y de manera que la distancia de su centro de masas al suelo se mantiene constante. Se trata de calcular la potencia del motor.
|}

:Cálculo directo: <math>\dot{\Ws}^{\mathrm{freno}}= \overline{\Gamma}^{\rightarrow \mathrm{polea}}\cdot\velang{polea}{T}=\left(\odot\Gamma\right)\cdot\left( \odot \Omega_0\right)=\Gamma \Omega_0.</math>

:El par motor no es un dato, y hay que recurrir al TMC para calcularlo.

:<math>\left.\begin{array}{l}
\text{SISTEMA: polea + mono + plátanos}\\
\text{TMC en } \Os
\end{array}\right\} \sum\overline{\Ms}_\mathrm{ext}(\Os)=\dot{\overline{\Hs}}_\mathrm{RTO}(\Os)</math>

:El momento cinético proviene del mono y de los plátanos. En primera aproximación, se puede considerar la masa del mono concentrada en su tronco y su cabeza, de manera que no gira y su velocidad respecto al suelo es nula. Teniendo en cuenta que el punto '''O''' no se mueve respecto al mono pero sí respecto a los plátanos:

:<math>\overline{\Hs}_\mathrm{RTO=T} (\Os) =\overline{\Hs}_\Ts^\mathrm{mono}(\Os) + \overline{\Hs}_\Ts^\mathrm{plátanos}(\Os)= \overline{\Hs}_\Ts^\mathrm{mono}(\Os) + \overline{\Hs}_\Ts^\mathrm{plátanos}(\Gs_\mathrm{plát})+ \OGvec_\mathrm{plát} \times \frac{2}{3} \ms \overline{\vs}_\Ts(\Gs_\mathrm{plát})= </math>

:<math>\hspace{2.3cm}=\OGvec_\mathrm{plát}\times \frac{2}{3} \ms \overline{\vs}_\Ts(\Gs_\mathrm{plát}) = \Bigr[(\rightarrow \rs)+(\downarrow \xs)\Bigr] \times\frac{2}{3} \ms \left(\downarrow \rs\Omega_0\right)=\left(\otimes \frac{2}{3}\ms\rs^2\Omega_0\right).</math>

:Se trata de un momento cinético constante en dirección y valor. Por tanto:

:<math>\dot{\overline{\Hs}}_\mathrm{RTO}(\Os)=0 \quad \Rightarrow \quad \sum{\overline{\Ms}}_\mathrm{ext}(\Os)=(\odot\Gamma)+(\otimes \ms\gs\rs) + \left(\otimes \frac{2}{3}\ms\gs\rs\right)=0\quad \Rightarrow \quad \Gamma= \frac{5}{3}\ms\gs\rs</math>.

:Finalmente: <math>\dot{\Ws}^{\mathrm{motor}}= \Gamma \Omega_0 = \frac{5}{3}\ms\gs\rs\Omega_0.</math>

:Cálculo indirecto

:Hay que tener presente que es un problema con dos tipos de actuadores: el motor y la musculatura del mono. Hay diversos modelos biomecánicos para formular las fuerzas de los músculos, pero quedan fuera del ámbito de este curso. Ya que el mono se mueve, sus músculos desarrollan una potencia no nula.

:<math>\left.\begin{array}{l}
\text{SISTEMA: polea + mono + plátano}\\
\text{REF: suelo(Gal)}
\end{array}\right\} \quad \dot{\Ts}^\mathrm{sist}_\Ts=\dot{\Ws}^\mathrm{ext}_\Ts+ \dot{\Ws}^\mathrm{int}</math>

:<math>\left.\begin{array}{l}
\dot{\Ts}^\mathrm{sist}_\Ts=0\\
\dot{\Ws}^\mathrm{ext}_\Ts=\left(\downarrow \frac{2}{3}\ms\gs\right)\cdot\left(\downarrow \rs\Omega_0\right)= \frac{2}{3}\ms\gs\rs\Omega_0\\
\dot{\Ws}^\mathrm{int}=\dot{\Ws}^{\mathrm{musc}}+\dot{\Ws}^{\mathrm{motor}}
\end{array}\right\} \quad \Rightarrow \quad \dot{\Ws}^{\mathrm{musc}} + \dot{\Ws}^{\mathrm{motor}}=-\frac{2}{3}\ms\gs\rs\Omega_0</math>

:Este balance proporciona una ecuación con dos incógnitas. Se puede plantear un segundo balance para un sistema donde una de las dos no aparezca: un sistema sin el mono o un sistema sin el motor. Si se hace un balance de potencias para el sistema (mono + trozo de cuerda), las fuerzas externas a considerar son el peso y la tensión de la cuerda. Si se sigue trabajando en la referencia del suelo, esta tensión (desconocida) hará potencia no nula ya que está aplicada a un punto de velocidad ascendente.
:[[Archivo:ExE1-4-2-2-cat-esp.png|thumb|right|330px|link=]]

:Para evitar que aparezca un término asociado a la tensión de la cuerda, se puede trabajar en la referencia del trozo de cuerda que se ha considerado, que al tener velocidad ascendente constante es galileana. En esta referencia, el mono baja con velocidad constante:

:<math>\left.\begin{array}{l}
\text{REL: trozo de cuerda}\\
\text{AB: suelo}
\end{array}\right\}</math>

:<math>\overline{\vs}_\mathrm{REL}(\mathrm{mono})=\overline{\vs}_\mathrm{ar}(\mathrm{mono})=-\overline{\vs}_\mathrm{ar}(\mathrm{mono})=(\downarrow \rs\Omega_0)</math>

:<math>\left.\begin{array}{l}
\text{SISTEMA: mono+ trozo de cuerda}\\
\text{REF: trozo de cuerda(Gal)}
\end{array}\right\} \quad \dot{\Ts}^\mathrm{sist}_\mathrm{cuerda} =\dot{\Ws}^\mathrm{ext}_\mathrm{cuerda}+ \dot{\Ws}^\mathrm{int}</math>

:<math>\left.\begin{array}{l}
\dot{\Ts}^\mathrm{sist}_\mathrm{cuerda}=0\\
\dot{\Ws}^\mathrm{ext}_\mathrm{cuerda}=\left(\downarrow \ms\gs\right)\cdot\left(\downarrow \rs\Omega_0\right)= \ms\gs\rs\Omega_0\\
\dot{\Ws}^\mathrm{int}=\dot{\Ws}^{\mathrm{musc}}
\end{array}\right\} \quad \Rightarrow \quad \dot{\Ws}^{\mathrm{musc}} =-\ms\gs\rs\Omega_0</math>

:Combinando este resultado con el del balance anterior:

:<math>\dot{\Ws}^\mathrm{motor}=-\frac{2}{3}\ms\gs\rs\Omega_0-\dot{\Ws}^\mathrm{musc}=-\frac{5}{3}\ms\gs\rs\Omega_0</math>



© Universitat Politècnica de Catalunya. [[Mecánica:Derechos de autor |Todos los derechos reservados]]
-------------

<center>
[[D8. Conservaciones|<<< D8. Conservaciones]]

[[E2. Teorema de la energía: versión integrada|E2. Teorema de la energía: versión integrada>>>]]
</center>

E1. Work-Energy Theorem: differential form

2025-05-20T12:12:52Z

Bros:

<div class="noautonum">__TOC__</div>
<math>\newcommand{\uvec}{\overline{\textbf{u}}}
\newcommand{\vvec}{\overline{\textbf{v}}}
\newcommand{\evec}{\overline{\textbf{e}}}
\newcommand{\Omegavec}{\overline{\mathbf{\Omega}}}
\newcommand{\velang}[2]{\Omegavec^{\textrm{#1}}_{\textrm{#2}}}
\newcommand{\Alfavec}{\overline{\mathbf{\alpha}}}
\newcommand{\accang}[2]{\Alfavec^{\textrm{#1}}_{\textrm{#2}}}
\newcommand{\ds}{\textrm{d}}
\newcommand{\ps}{\textrm{p}}
\newcommand{\ns}{\textrm{n}}
\newcommand{\hs}{\textrm{h}}
\newcommand{\cs}{\textrm{c}}
\newcommand{\gs}{\textrm{g}}
\newcommand{\Ns}{\textrm{N}}
\newcommand{\Hs}{\textrm{H}}
\newcommand{\Fs}{\textrm{F}}
\newcommand{\ms}{\textrm{m}}
\newcommand{\ts}{\textrm{t}}
\newcommand{\us}{\textrm{u}}
\newcommand{\vs}{\textrm{v}}
\newcommand{\Rs}{\textrm{R}}
\newcommand{\Ws}{\textrm{W}}
\newcommand{\Ts}{\textrm{T}}
\newcommand{\Ls}{\textrm{L}}
\newcommand{\As}{\textrm{A}}
\newcommand{\Ds}{\textrm{D}}
\newcommand{\Bs}{\textrm{B}}
\newcommand{\es}{\textrm{e}}
\newcommand{\fs}{\textrm{f}}
\newcommand{\is}{\textrm{i}}
\newcommand{\Is}{\textrm{I}}
\newcommand{\ks}{\textrm{k}}
\newcommand{\js}{\textrm{j}}
\newcommand{\rs}{\textrm{r}}
\newcommand{\as}{\textrm{a}}
\newcommand{\ss}{\textrm{s}}
\newcommand{\Os}{\textbf{O}}
\newcommand{\Gs}{\textbf{G}}
\newcommand{\Cbf}{\textbf{C}}
\newcommand{\Or}{\Os_\Rs}
\newcommand{\Qs}{\textbf{Q}}
\newcommand{\Cs}{\textbf{C}}
\newcommand{\Ps}{\textbf{P}}
\newcommand{\Ss}{\textbf{S}}
\newcommand{\Js}{\textbf{J}}
\newcommand{\P}{\textrm{P}}
\newcommand{\Q}{\textrm{Q}}
\newcommand{\Ms}{\textrm{M}}
\newcommand{\deg}{^\textsf{o}}
\newcommand{\xs}{\textsf{x}}
\newcommand{\ys}{\textsf{y}}
\newcommand{\zs}{\textsf{z}}
\newcommand{\dert}[2]{\left.\frac{\ds{#1}}{\ds\ts}\right]_{\textrm{#2}}}
\newcommand{\ddert}[2]{\left.\frac{\ds^2{#1}}{\ds\ts^2}\right]_{\textrm{#2}}}
\newcommand{\vec}[1]{\overline{#1}}
\newcommand{\vecbf}[1]{\overline{\textbf{#1}}}
\newcommand{\vecdot}[1]{\overline{\dot{#1}}}
\newcommand{\OQvec}{\vec{\Os\Qs}}
\newcommand{\GQvec}{\vec{\Gs\Qs}}
\newcommand{\GPvec}{\vec{\Gs\Ps}}
\newcommand{\PSvec}{\vec{\Ps\Ss}}
\newcommand{\QQvec}{\vec{\Qs\Qs}}
\newcommand{\QGvec}{\vec{\Qs\Gs}}
\newcommand{\QPvec}{\vec{\Qs\Ps}}
\newcommand{\JQvec}{\vec{\Js\Qs}}
\newcommand{\PQvec}{\vec{\Ps\Qs}}
\newcommand{\GJvec}{\vec{\Gs\Js}}
\newcommand{\OPvec}{\vec{\Os\textbf{P}}}
\newcommand{\OCvec}{\vec{\Os\Cs}}
\newcommand{\OGvec}{\vec{\Os\Gs}}
\newcommand{\CJvec}{\vec{\Cs\Js}}
\newcommand{\GCvec}{\vec{\Gs\Cs}}
\newcommand{\PGvec}{\vec{\Ps\Gs}}
\newcommand{\abs}[1]{\left|{#1}\right|}
\newcommand{\braq}[2]{\left\{{#1}\right\}_{\textrm{#2}}}
\newcommand{\matriz}[9]{
\begin{bmatrix}
{#1} & {#2} & {#3}\\
{#4} & {#5} & {#6}\\
{#7} & {#8} & {#9}
\end{bmatrix}}
\newcommand{\diag}[3]{
\begin{bmatrix}
{#1} & {0} & {0}\\
{0} & {#2} & {0}\\
{0} & {0} & {#3}
\end{bmatrix}}
\newcommand{\vector}[3]{
\begin{Bmatrix}
{#1}\\
{#2}\\
{#3}
\end{Bmatrix}}
\newcommand{\vecdosd}[2]{
\begin{Bmatrix}
{#1}\\
{#2}
\end{Bmatrix}}
\newcommand{\vel}[2]{\vvec_{\textrm{#2}} (\textbf{#1})}
\newcommand{\acc}[2]{\vecbf{a}_{\textrm{#2}} (\textbf{#1})}
\newcommand{\accs}[2]{\vecbf{a}_{\textrm{#2}}^{\textrm{s}} (\textbf{#1})}
\newcommand{\accn}[2]{\vecbf{a}_{\textrm{#2}}^{\textrm{n}} (\textbf{#1})}
\newcommand{\velo}[1]{\vvec_{\textrm{#1}}}
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\newcommand{\psio}{\dot{\psi}_0}
\newcommand{\Pll}{\textbf{P}_\textrm{lliure}}
\newcommand{\agal}[1]{\vecbf{a}_{\textrm{Gal}} (#1)}
\newcommand{\angal}[1]{\vecbf{a}_{\textrm{NGal}} (#1)}
\newcommand{\vgal}[1]{\vecbf{v}_{\textrm{Gal}} (#1)}
\newcommand{\fvec}[2]{\overline{\mathbf{F}}_{\textrm{#1}\rightarrow\textrm{#2}}}
\newcommand{\mvec}[2]{\overline{\mathbf{M}}_{\textrm{#1}\rightarrow\textrm{#2}}}
\newcommand{\F}[1]{\vec{\Fs}_{#1}}
\newcommand{\Fcal}[2]{\vec{\cal{F}}^{\text{#1}}_{#2}}
</math>

The '''Work-Energy Theorem''' plays a prominent role in physics, as it bridges all branches of science (electromagnetism, thermodynamics, etc.): all of them deal with energy.

There is a certain parallel between the Vector Theorems and the '''Work-Energy Theorem''': they all allow us to understand the evolution of a magnitude that depends on the mechanical state of the system. In the case of the former, we are dealing with the instantaneous evolution of vector magnitudes (linear momentum, angular momentum), while the latter deals with the finite change (in a finite time interval) of a scalar magnitude ('''mechanical energy''').

This unit presents the differential form of this theorem (the '''power balance'''). The integrated form (which is the Work-Energy Theorem itself) is presented in [[E2. Teorema de la energía: versión integrada|'''unit E2''']].
 
------
------
==E1.1 Power balance in a system of particles==

:[[File:E1-1-eng.png|thumb|right|300px|link=]]
<center>'''Figure E1.1''' Forces on a particle of a system with constant matter</center>

Let us consider a system of constant matter, and let’s apply Newton's second law to each particle (or each mass differential) of this system in a general reference frame R (which may be non-Galilean) ('''Figure E1.1'''):

<math>\F{\rightarrow \Ps} + \Fcal{}{\text{ar}\rightarrow\Ps} + \Fcal{}{\text{Cor}\rightarrow\Ps} =\ms_\Ps\acc{P}{R}</math>

<math>\F{\rightarrow \Ps} </math> is the resultant interaction force on '''P''', and it comes from internal interactions (with particles of the same system) and external ones:

<math>\F{\rightarrow \Ps}=\F{\rightarrow \Ps}^\text{int}+\F{\rightarrow \Ps}^\text{ext}</math>

If both sides of the equation are multiplied by <math>\vel{P}{R}</math>, all forces (interaction and inertia forces) orthogonal to <math>\vel{P}{R}</math> are filtered out. Specifically, since <math>\acc{P}{R}=\accs{P}{R} + \accn{P}{R} </math>, where <math>\accn{P}{R} \perp \vel{P}{R}</math>, and since the Coriolis force is always orthogonal to <math>\vel{P}{R} (\Fcal{}{\text{Cor}\rightarrow\Ps} = -2\ms_\Ps \velang{R}{Gal} \times \vel{P}{R})</math>:

<math>\vel{P}{R} \cdot \left( \F{\rightarrow \Ps}^\text{int}+\F{\rightarrow \Ps}^\text{ext} + \Fcal{}{\text{ar}\rightarrow\Ps} \right) = \ms_\Ps\accs{P}{R} \cdot \vel{P}{R} = \frac{\ds}{\ds \ts} \left(\frac{1}{2}\ms_\Ps \vvec_\Rs^2(\Ps) \right) \equiv \frac{\ds\Ts_\Rs(\Ps)}{\ds\ts}=\dot{\Ts}_\Rs(\Ps)</math>

The scalar function <math>\Ts_\Rs(\Ps)</math> is the '''kinetic energy''' of the particle, and is a function of the [[D1. Foundational laws of Newtonian dynamics#D1.3 Newton’s Principle of Determinacy|'''mechanical state''']] of '''P'''. The scalar product of a force and the velocity of its application point is the '''force power''': <math>\vel{P}{R} \cdot \F{\rightarrow \Ps} \equiv \dot{\Ws}_\Rs^{\F{\rightarrow \Ps}} </math>. Both scalar quantities <math>(\Ts_\Rs(\Ps), \dot{\Ws}_\Rs^{\F{\rightarrow \Ps}})</math> depend on the reference frame. For the same set of interaction forces acting on '''P''', the reference frame determines which are filtered out and which are not.

The equation can be rewritten as <math>\dot{\Ws}_\Rs^{\text{ext}\rightarrow \Ps}+\dot{\Ws}_\Rs^{\text{int}\rightarrow \Ps}+\dot{\Ws}_\Rs^{\text{ar}\rightarrow \Ps} =\dot{\Ts}_\Rs(\Ps)</math>, and constitutes the '''power balance''' for particle '''P''' in the R reference frame. Summing for all particles in the system:

<math>\dot{\Ws}_\Rs^{\text{ext}}+\dot{\Ws}^{\text{int}}+\dot{\Ws}_\Rs^{\text{ar}} =\dot{\Ts}_\Rs^\text{syst}</math>

The calculation of kinetic energy is covered in [[E2. Teorema de la energía: versión integrada|'''unit E2''']]. This unit focuses on the calculation of powers. [[E1. Work-Energy Theorem: differential form#E1.2 Potencia de una pareja de acción-reacción|'''section E1-2''']] shows that the power of internal interactions does not depend on the reference frame (which is why the subscript R does not appear in the <math>\dot{\Ws}^{\text{int}}</math> term of the previous equation).

====✏️ EXAMPLE E1-1.1: free particle====
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:
{|
:[[File:ExE1-1-1-1-eng.jpg|thumb|left|180px|link=]]
|A particle '''P''', with mass m, is released from the ring of a Ferris wheel when it is at its highest point. We want to calculate the power, at this instant, of all forces acting on the particle from three different reference frames: the ground, the ring, and the Ferris wheel cabin.
|}

:Ground reference frame: Since this is a Galilean reference frame, all forces on '''P''' are interaction forces.
{|
:[[File:ExE1-1-2- eng.png|thumb|right|230px|link=]]

|<math>\hspace{0.5cm}\sum\dot{\Ws}_\Ts=\dot{\Ws}_\Ts^\text{weight}= \left(\leftarrow \Rs\Omega_0\right) \cdot \left(\downarrow \ms\gs\right)=0</math>
|}

:Ring reference frame: It is a non-Galilean reference, and the transportation force on '''P''' must be taken into account.
{|
:[[File:ExE1-1-3-eng.png|thumb|right|250px|link=]]
|<math>\hspace{0.5cm}</math> AB:ground, REL:ring 

<math>\hspace{0.5cm}\vel{P}{REL}=\vel{P}{AB}-\vel{P}{ar}= \left(\leftarrow \Rs\Omega_0\right)-\left(\leftarrow \Rs\Omega_0\right)=\overline{0}</math> 

<math>\hspace{0.5cm}</math>

Since the velocity is zero, the total power is zero.
|}

:Cabin reference frame: It is also a non-Galilean reference, and the transportation force on '''P''' must be taken into account.

{|
:[[File:ExE1-1-4-eng.png|thumb|right|250px|link=]]
|<math>\hspace{0.5cm}</math>AB:ground, REL:cabin

<math>\hspace{0.5cm}\vel{P}{REL}=\vel{P}{AB}-\vel{P}{ar}= \left(\leftarrow \Rs\Omega_0\right)-\left(\uparrow \Rs\Omega_0\right)</math>

<math>\hspace{0.5cm}\dot{\Ws}_\text{cabin}^\text{weight}=\left[ \left(\leftarrow \Rs\Omega_0\right)+\left(\downarrow \Rs\Omega_0\right) \right] \cdot \left(\downarrow\ms\gs\right) =\ms\gs\Rs\Omega_0</math>

<math>\hspace{0.5cm}\dot{\Ws}_\text{cabin}^\text{ar}=\left[ \left(\leftarrow \Rs\Omega_0\right)+\left(\downarrow \Rs\Omega_0\right) \right] \cdot \left(\rightarrow\ms\Rs^2\Omega_0^2\right) =-\ms\gs\Rs^2 \Omega_0^3</math>

<math>\hspace{0.5cm}\sum\dot{\Ws}_\text{cabin}=\ms\Rs\Omega_0(\gs-\Rs\Omega_0^2)</math>

|}


====✏️ EXAMPLE E1-1.2: pendulum====
------
:
{|
:[[File:ExE1-2-1-eng.png|thumb|left|180px|link=]]
|A particle '''P''' of mass m hangs from an inextensible thread of length L attached to the ceiling. The assembly oscillates in a vertical plane fixed to the ground. We want to calculate the power of all the forces acting on the particle in two different reference frames: the ground and a platform rotating relative to the ground with a constant angular velocity <math>\Omega_0</math>. 

Ground reference frame: Since it is a Galilean reference frame, '''P''' is only subjected to interaction forces.
{|
:[[File:ExE1-2-2-neut.png|thumb|left|300px|link=]]

|<math>\dot{\Ws}_\Es^\Ts=(\nearrow\Ls\dot{\theta})\cdot(\nwarrow\Ts)=0</math>

<math>\dot{\Ws}_\Es^\text{weight}=(\nearrow\Ls\dot{\theta}\cdot(\downarrow \ms\gs)=-\ms\gs\Ls\dot{\theta}\sin\theta</math>

|}

Platform reference frame: It is a non Galilean reference frame, and the transportation force on '''P''' has to be taken into account.

{|
:[[File:ExE1-2-3-eng.png|thumb|left|400px|link=]]

|AB:ground, REL:platform

<math>\vel{P}{REL}=\vel{P}{AB}-\vel{P}{ar}=(\nearrow\Ls\dot{\theta})-(\otimes \Ls \Omega_0 \sin \theta)</math>

<math>\dot{\Ws}_\Es^\Ts\left[(\nearrow\Ls\dot{\theta})+(\odot\Ls\Omega_0\sin\theta)\right]\cdot(\nwarrow\Ts)=0</math>

<math>\dot{\Ws}_\Es^\text{weight}\left[(\nearrow\Ls\dot{\theta})+(\odot\Ls\Omega_0\sin\theta)\right]\cdot(\downarrow \ms\gs)=-\ms\gs\Ls\dot{\theta}\sin\theta</math>

<math>\dot{\Ws}_\Es^\text{ar}\left[(\nearrow\Ls\dot{\theta})+(\odot\Ls\Omega_0\sin\theta)\right]\cdot(\rightarrow \ms\Ls\Omega_0^2\sin\theta)=\ms\Ls^2\Omega_0^2\dot{\theta}\cos\theta</math>
|}


------
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==E1.2 Power of an action-reaction pair==

The total power associated with internal interactions in a constant-mass system is not zero in principle (since the two forces of each action-reaction pair are applied to particles that may have different velocities relative to the reference frame in which the calculation is made), and it is independent of the reference frame in which it is evaluated. It only depends on the separation velocity of the two particles. The power balance for the system is therefore written as: <math>\dot{\Ts}_\Rs^\text{syst}=\dot{\Ws}_\Rs^\text{ext} + \dot{\Ws}^\text{int} + \dot{\Ws}_\Rs^\text{ar} </math> .

<div>
=====💭 PROOF ➕=====

{|
:[[File:E1-2-neut.png|thumb|left|200px|link=]]
|<math>\dot{\Ws}^\text{ac-reac}=\overline{\Fs}_{\Ps \rightarrow \Qs}\cdot \vel{Q}{R} + \overline{\Fs}_{\Qs \rightarrow \Ps}\cdot \vel{P}{R}= \overline{\Fs}_{\Ps \rightarrow \Qs}\cdot \left[ \vel{Q}{R}-\vel{P}{R}\right]</math>

By the [[D1. Foundational laws of Newtonian dynamics#D1.6 Newton’s third law (action-reaction principle)|'''action-reaction principle''']], <math>\overline{\Fs}_{\Ps \rightarrow \Qs}=-\overline{\Fs}_{\Qs \rightarrow \Ps}</math>. Therefore: <math>\dot{\Ws}^\text{ac-reac}=\overline{\Fs}_{\Ps \rightarrow \Qs}\cdot \left[ \vel{Q}{R}-\vel{P}{R}\right]</math>

As <math>\overline{\Fs}_{\Ps \rightarrow \Qs}</math> has the <math>\PQvec</math> direction, only the components of the velocities in that direction have to be considered:

<math>\dot{\Ws}^\text{ac-reac}=\overline{\Fs}_{\Ps \rightarrow \Qs}\cdot \biggl( \vel{Q}{R} \Bigr]_{||\Ps\Qs}-\ \vel{P}{R} \Bigr]_{||\Ps\Qs}\biggl)</math>
|}

The difference <math>\biggl( \vel{Q}{R} \Bigr]_{\Ps\Qs}-\ \vel{P}{R} \Bigr]_{\Ps\Qs}\biggl)</math> is the separation velocity <math>\dot{\rho}</math> between '''P''' and '''Q''', and it does not dpend on the reference frame since it is the derivative of a scalar (the distance <math>\dot{\rho}</math> between the two particles). For the previous drawing, <math>\dot{\rho}> 0</math>. Since this is an attractive interaction, <math>\dot{\Ws}^\text{ac-reac} < 0</math>.

With a drawing, it is easy to see that:
* attractive interaction and particle approach <math>(\dot{\rho}<0)</math> : <math>\dot{\Ws}^\text{ac-reac} > 0</math> ;
* repulsive interaction and particle separation <math>(\dot{\rho}>0)</math> : <math>\dot{\Ws}^\text{ac-reac} > 0</math>
* repulsive interaction and particle approach <math>(\dot{\rho}<0)</math> : <math>\dot{\Ws}^\text{ac-reac} < 0</math>


</div>

There are two particular cases of interaction forces whose power always has the same sign:

* Friction action-reaction pair between '''P''' and '''Q''': since the friction force always opposes the relative velocity between the two points, its power is always negative:

<math>\hspace{1cm} \dot{\Ws}^\text{friction}=\dot{\Ws}^{\text{fric}\rightarrow\Ps}_\text{RTQ}+\dot{\Ws}^{\text{fric}\rightarrow\Qs}_\text{RTQ}=\overline{\Fs}^{\text{fric}\rightarrow\Ps} \cdot \vel{P}{RTQ} = - \abs{\overline{\Fs}^{\text{fric}\rightarrow\Ps}} \frac{\vel{P}{RTQ}}{\abs{\vel{P}{RTQ}}} \cdot \vel{P}{RTQ}<0;</math>

* Constraint action-reaction pair between '''P''' and '''Q''': from the [[D3. Interactions between rigid bodies#D3.4 Direct constraint interactions|'''analytical characterization equation''']] of the constraint forces between two points '''P''' and '''Q''' belonging to two rigid bodies <math>\mathrm{S}_\Ps</math> and <math>\mathrm{S}_\Qs</math>, it is easy to see that the associated power is zero:

<math>\hspace{1cm} \dot{\Ws}^\text{constraint}=\dot{\Ws}^{\text{constraint}\rightarrow\Ps}_{\mathrm{S}_\Ps}+\dot{\Ws}^{\text{constraint}\rightarrow\Qs}_{\mathrm{S}_\Ps}=\overline{\Fs}^{\text{constraint}\rightarrow\Ps} \cdot \overline{\mathbf{v}}_{\mathrm{S}_\Qs}(\Ps) = - 0;</math>

====✏️ EXAMPLE E1-2.1: power of an action-reaction friction pair====
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:
{|
:[[File:ExE1-2-1-2-eng.png|thumb|left|180px|link=]]
|A homogeneous wheel, with radius r and mass m, moves down an inclined plane without sliding until it joins a conveyor belt moving at a constant velocity <math>\vs_0</math> relative to the ground. At the moment it makes contact with the belt, the center of the wheel has a velocity <math>\vs_0</math> relative to the ground in the same direction as the belt. We want to calculate the power of the friction force between the belt and the wheel at this instant.
|}

At that momento, the <math>\mathbf{ICR}^\text{wheel}_\Es</math> is point <math>\Js_\text{wheel}</math> in contact with the belt. This point slides backwards with respect to the belt, therefore the friction force on the wheel is forwards:

<math>\left.\begin{array}{l}
\text { AB: ground } \\
\text { REL: belt }
\end{array}\right\} \quad \bar{\vs}_{\text {REL }}\left(\Js_{\text {wheel }}\right)=\bar{\vs}_{\text{AB}}\left(\Js_{\text {wheel }}\right)-\bar{\vs}_{\mathrm{ar}}\left(\Js_{\text {wheel }}\right)=-\bar{\vs}_{\mathrm{ar}}\left(\Js_{\text {wheel }}\right)=-\left(\rightarrow \mathrm{v}_0\right)=\left(\leftarrow \mathrm{v}_0\right) .</math>

[[File:ExE1-2-1-3-eng.png|thumb|center|500px|link=]]

Calculation of the friction power: <math>\dot{\Ws}^\text{friction}=\dot{\Ws}_\Rs^{\text{friction}\rightarrow \text{wheel}}+ \dot{\Ws}_\Rs^{\text{friction}\rightarrow \text{belt}}</math>, where R may be any reference frame. To quickly calculate this power, it is advisable to choose a reference frame R where one of the two points subjected to the action-reaction pair has zero speed, so that one of the two terms is directly zero:

<math>\dot{\Ws}^\text{friction}=\dot{\Ws}_\text{belt}^{\text{fric}\rightarrow \text{wheel}}+ \dot{\Ws}_\text{belt}^{\text{fric }\rightarrow \text{belt}}=\overline{\Fs}^{\text{fric}\rightarrow \text{wheel}} \cdot \bar{\vs}_{\text {belt}}\left(\Js_{\text {wheel }}\right)=(\rightarrow \mu\ms\gs)\cdot(\leftarrow \vs_0)=-\mu\ms\gs\vs_0.</math>

If the calculation is done in any other reference frame, the result is the same:

<math>\dot{\Ws}^\text{friction}=\dot{\Ws}_\text{E}^{\text{fric}\rightarrow \text{wheel}}+ \dot{\Ws}_\text{E}^{\text{fric} \rightarrow \text{belt}}=(\rightarrow \mu\ms\gs)\cdot 0 + (\leftarrow \mu\ms\gs)\cdot(\rightarrow \vs_0)=-\mu\ms\gs\vs_0.</math>

<math>\dot{\Ws}^\text{friction}=\dot{\Ws}_\text{vehicle}^{\text{fric}\rightarrow \text{wheel}}+ \dot{\Ws}_\text{vehicle}^{\text{fric} \rightarrow \text{belt}}=(\rightarrow \mu\ms\gs)\cdot (\rightarrow\vs_0) + (\leftarrow \mu\ms\gs) \cdot(\rightarrow 3\vs_0)=-\mu\ms\gs\vs_0.</math>

However, if only the power of one of the two forces is calculated (for example, the friction on the wheel), the result depends on the reference frame and can have any sign:

<math>\dot{\Ws}_\text{belt}^{\text{fric }\rightarrow \text{wheel}}=(\rightarrow \mu\ms\gs)\cdot(\leftarrow \vs_0)=-\mu\ms\gs\vs_0 ,</math>

<math>\dot{\Ws}_\text{E}^{\text{fric} \rightarrow \text{wheel}}=(\rightarrow \mu\ms\gs)\cdot 0 = 0 ,</math>

<math>\dot{\Ws}_\text{vehicle}^{\text{fric} \rightarrow \text{wheel}}=(\rightarrow \mu\ms\gs)\cdot (\rightarrow\vs_0)=\mu\ms\gs\vs_0.</math>


-----
-----

==E1.3 Power of a system of forces on a rigid body==

{|
|
:[[File:E1-3-eng.png|thumb|left|180px|link=]]
<center>'''Figure E1.3''': Kinematic torsor of a rigid body in a reference frame R, and forces on different points (dm) of that rigid body</center>

|When calculating the power of a system of forces on a rigid body S, the term associated with internal interactions is zero because those interactions are all constraint (cohesion) forces and, as seen in [[E1. Work-Energy Theorem: differential form#E1.2 Power of an action-reaction pair|'''section E1.2''']], their total power is zero.

The total power of a system of forces on S can be calculated from the torsor of the system at any point '''O''' of the rigid body, <math>(\overline{\Fs},\overline{\Ms}(\Os))</math>. Taking into account rigid body kinematics ('''Figure E1.3'''):

<math>\dot{\Ws}_\Rs^{\rightarrow \Ss} =\sum_\Ps\overline{\Fs}^{\rightarrow \Ps} \cdot \vel{P}{R}= \sum_\Ps \left[ \overline{\Fs}^{\rightarrow \Ps} \cdot \vel{O}{R} + \velang{S}{R} \times \OPvec \right] </math>

|}
<math>\dot{\Ws}_\Rs^{\rightarrow \Ss} =\left(\sum_\Ps\overline{\Fs}^{\rightarrow \Ps} \right)\cdot \vel{O}{R} + \sum_\Ps\overline{\Fs}^{\rightarrow \Ps} \cdot \left(\velang{S}{R} \times \OPvec \right) = \left( \sum_\Ps\overline{\Fs}^{\rightarrow \Ps} \right) \cdot \vel{O}{R} + \left[ \sum_\Ps \left( \OPvec \times \overline{\Fs}^{\rightarrow \Ps} \right)\right] \cdot \velang{S}{R} = \overline{\Fs}\cdot \vel{O}{R} + \overline{\Ms}(\Os)\cdot \velang{S}{R} </math>

This calculation also shows that the power of a moment (or a torque <math>\Gamma</math>) on a rigid body in a reference frame R is the product of the moment by the angular velocity of the rigid body: <math>\dot{\Ws}_\Rs^{\Gamma\rightarrow \mathrm{S}} = \overline{\Gamma} \cdot \velang{S}{R}</math>.

The power balance for a rigid body is: <math>\dot{\Ws}_\Rs^{ \rightarrow \mathrm{S}} = \dot{\Ws}_\Rs^{ \mathrm{ext}} + \dot{\Ws}_\Rs^{ \mathrm{ar}}.</math>

====✏️ EXAMPLE E1-3.1: roller sliding on a slope====
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:
{|
|
:[[File:ExE1-3-1-1-eng.png|thumb|left|180px|link=]]
|A homogeneous roller, with mass m and radius r, slides down a slope. The friction coefficient between the ground and the roller is <math>\mu</math>. We want to calculate the power of every force acting on the roller in the ground reference frame.
|}

:It is a 2D problem. The kinematic and dynamic descriptions are:
{|
|
[[File:ExE1-3-1-2-eng.png|thumb|right|250px|link=]]

|<math> \hspace{0.5cm}\vel{J}{E}=\vel{C}{T} + \velang{}{E} \times \CJvec= </math>

<math>\hspace{1.7cm}=\left( \rightarrow \vs_\Es \right) + \left( \otimes \Omega_\Es\right) \times \left( \downarrow \rs \right)</math>

<math> \hspace{0.5cm}\vel{J}{E} = \left[\rightarrow (\vs_\Es -\rs\Omega_\Es)\right]</math>

<math>\hspace{0.5cm} \dot{\Ws}_\Es^{\mathrm{ext}} = \dot{\Ws}_\Es^{\mathrm{weight}} + \dot{\Ws}_\Es^{\mathrm{friction}} + \dot{\Ws}_\Es^{\mathrm{constraint}}= \Bigr[\left(\downarrow \ms\gs\cos\beta\right)+\left(\rightarrow \ms\gs\sin\beta\right)\Bigr] \cdot (\rightarrow \vs_\Es) + \Bigr[(\leftarrow \mu \Ns) + (\uparrow \Ns ) \Bigr] \cdot \Bigr[\rightarrow (\vs_\Es -\rs\Omega_\Es)\Bigr]</math>

|}
:Since the roller center has no motion perpendicular to the plane, the AMT leads to <math>\Ns =\ms\gs\cos\beta</math>. Therefore:

{|
|
[[Archivo:ExE1-3-1-3-eng.png|thumb|right|140px|link=]]

|<math>\hspace{0.5cm} \dot{\Ws}_\Ts^{\mathrm{ext}} = \ms\gs\vs_\Ts \sin \beta - \ms\gs(\vs_\Ts -\rs\Omega_\Ts)\cos\beta=\ms\gs \Bigr[ \vs_\Ts (\sin\beta-\cos\beta)+\rs\Omega_\Ts \cos\beta \Bigr].</math>

<math>\hspace{0.5cm}</math>

If instead of considering the friction at '''J''' we consider the equivalent torsor at point '''C''', we must consider the force applied in '''C''' and the moment about '''C''', which is not zero:

<math>\hspace{0.5cm} \dot{\Ws}_\Es^{ \mathrm{friction}} = (\leftarrow \mu\Ns)\cdot(\rightarrow\vs_\Es)+(\otimes \mu\Ns\rs)\cdot(\otimes \Omega_\Es) =\mu\ms\gs\cos\beta(\rs\Omega_\Es-\vs_\Es)</math>,

<math>\hspace{0.5cm} \dot{\Ws}_\Es^{\mathrm{ext}} = \dot{\Ws}_\Es^{ \mathrm{weight}} + \dot{\Ws}_\Es^{ \mathrm{friction}}+ \dot{\Ws}_\Es^{ \mathrm{constraint}}= \ms\gs\vs_\Es\sin\beta+\mu\ms\gs\cos\beta (\rs\Omega_\Es-\vs_\Es)</math>.

<math>\hspace{0.5cm}</math>

The result is the same.
|}


====✏️ EXAMPLE E1-3.2: block sliding on a truck====
------
:
{|
|
:[[File:ExE1-3-2-1-eng.png|thumb|left|220px|link=]]
|A truck brakes with acceleration '''a''' relative to the ground at the instant in which its velocity (also relative to the ground) is v. At the same instant, the homogeneous block, of mass m, slides with velocity v' relative to the truck. We want to perform a power balance for the block in two different reference frames: the ground and the truck.
|}

:Ground reference frame: It is a 2D problem. The kinematic and dynamic descriptions are:
{|
|
[[File:ExE1-3-2-2-eng.png|thumb|right|200px|link=]]

| The constraint between the truck and the block allows only one translational DoF between the two. Therefore, in the 2D approximation, the associated torsor has two independent components. If characterized at the midpoint of the contact surface, these components are a normal force and a moment perpendicular to the plane. The AMT applied to the block leads to <math>\Ns=\ms\gs</math>.

|}
:The powers associated with all interactions on the block in the ground reference are:

:<math>\dot{\Ws}_\Es^{\rightarrow \mathrm{bloque}}=\dot{\Ws}_\Ts^{\mathrm{ext}}=\dot{\Ws}_\Ts^{\mathrm{peso}}+\dot{\Ws}_\Ts^{\mathrm{fricción}}+\dot{\Ws}_\Ts^{\mathrm{enlace}}= \Bigr[(\downarrow \ms\gs)+(\leftarrow \mu\ms\gs)+(\uparrow \Ns) \Bigr] \cdot \Bigr[\rightarrow(\vs+\vs')\Bigr] + (\odot \Ms_\es)\cdot 0 = -\mu\ms\gs(\vs+\vs')</math>

:Ya que el bloque se traslada respecto al suelo, su energía cinética total es inmediata, y se puede plantear fácilmente el balance de potencias:

:<math>\Ts_\Ts^\mathrm{bloque}=\frac{1}{2}\int_\mathrm{bloque}\ds\ms \vs_\Ts^2(\mathrm{bloque})=\frac{1}{2}\ms \vs_\Ts^2(\mathrm{bloc})\quad \Rightarrow \quad \dot{\Ts}_\Ts^\mathrm{bloque}=\ms\vs_\Ts( \mathrm{bloque})\dot{\vs}_\Ts( \mathrm{bloque}) </math>

:Como la única fuerza horizontal sobre el bloque es la fricción:

<math>\dot{\vs}_\Ts(\mathrm{bloque})=-\mu\gs\quad \Rightarrow \quad \dot{\Ts}_\Ts^\mathrm{bloque}=-\mu\ms\gs(\vs+\vs').</math>

:El balance de potencias es consistente:

:<math>\left.\begin{array}{l}
\dot{\mathrm{W}}_\Ts^{\rightarrow \text { bloque }}=\dot{\mathrm{W}}_{\mathrm{T}}^{\text {ext }}=-\mu \mathrm{mg}\left(\mathrm{v}+\mathrm{v}^{\prime}\right) \\
\dot{\mathrm{T}}_{\mathrm{T}}^{\text {bloque }}=-\mu \mathrm{mg}\left(\mathrm{v}+\mathrm{v}^{\prime}\right)
\end{array}\right\} \Rightarrow \dot{\mathrm{W}}_{\mathrm{T}}^{\rightarrow \text { bloque }}=\dot{\mathrm{T}}_{\mathrm{T}}^{\text {bloque }}</math>

:Referencia camión: Al estar el camión acelerado respecto al suelo, hay que tener en cuenta las fuerzas de arrastre sobre todos los diferenciales de masa (dm) del bloque:

{|
|

[[Archivo:ExE1-3-2-3-cat-esp.jpg|thumb|right|250px|link=]]
|<math>\cal{\Fs}^{\rightarrow \ds\ms}_{\mathrm{ar}}=-\ds\ms\overline{\as}_\mathrm{ar}(\ds\ms)=-\ds\ms (\leftarrow \as)=(\rightarrow \ds\ms \cdot \as)</math>.

El torsor resultante de este sistema de fuerzas en el centro de inercia '''G''' se reduce a una fuerza horizontal.

<math>\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{bloque}}=\dot{\Ws}_\mathrm{camión}^{\mathrm{ext}} + \dot{\Ws}_\mathrm{camión}^{\mathrm{ar}} =\dot{\Ws}_\mathrm{camión}^{\mathrm{peso}} + \dot{\Ws}_\mathrm{camión}^{\mathrm{fricción}} + \dot{\Ws}_\mathrm{camión}^{\mathrm{enlace}} + \dot{\Ws}_\mathrm{camión}^{\mathrm{ar}} , </math>

<math>\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{bloque}}= \Bigr[(\downarrow \ms\gs)+(\leftarrow \mu\ms\gs)+(\uparrow \Ns) + (\rightarrow \ms\as)\Bigr] \cdot(\rightarrow \vs')+(\odot\Ms_\es) \cdot0=\ms(\as-\mu\gs)\vs'.</math>
|}

:Se puede comprobar que el balance de potencias es consistente:

:<math>\Ts^\mathrm{bloque}_\mathrm{camión}=\frac{1}{2}\ms\vs^2_\mathrm{camión}(\mathrm{bloque}) \quad \Rightarrow \quad \dot{\Ts}^\mathrm{bloque}_\mathrm{camión} = \ms\vs_\mathrm{camión}(\mathrm{bloque})\dot{\vs}_\mathrm{camión}(\mathrm{bloque})</math>

:<math>\left.\begin{array}{l}
\mathrm{AB}:\mathrm{suelo} \\
\mathrm{REL}:\mathrm{camión}
\end{array}\right\} \quad \dot{\vs}_\mathrm{camión}(\mathrm{bloque})= \dot{\vs}_\mathrm{T}(\mathrm{bloque})-\dot{\vs}_\mathrm{ar}(\mathrm{bloque}) = (-\mu\gs)-(-\as)=\as-\mu\gs \quad \Rightarrow \quad \dot{\Ts}^\mathrm{bloque}_\mathrm{camión}=\ms\vs'(\as-\mu\gs)</math>

:<math>\left.\begin{array}{l}
\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{bloque}}=\dot{\Ws}_\mathrm{camión}^{\mathrm{ext}} + \dot{\Ws}_\mathrm{camión}^{\mathrm{ar}}= \ms(\as-\mu\gs)\vs'\\
\dot{\Ts}^\mathrm{bloque}_\mathrm{camión}=\ms\vs'(\as-\mu\gs)
\end{array}\right\} \Rightarrow\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{bloque}}= \dot{\Ts}^\mathrm{bloque}_\mathrm{camión}</math>


====✏️ EJEMPLO E1-3.3: camión frenando con carga====
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:
{|
|
:[[Archivo:ExE1-3-3-1-esp.png|thumb|left|220px|link=]]
|Un camión frena con aceleración '''a''' respecto al suelo en el instante en el que su velocidad (también respecto al suelo) es v. Los pares de frenada en el eje delantero y en el trasero son iguales y valen <math>\Gamma</math>, y las ruedas, de radio r y masa despreciable, no deslizan sobre el suelo. En el mismo instante, el bloque homogéneo, de masa m, desliz con velocidad v' relativa al camión. Se trata de hacer un balance de potencias para el sistema (bloque+camión) en la referencia del suelo y en la del camión.

|}

:Ya que la potencia de las interacciones internas no es nula en principio (no se trata de un único sólido rígido), es conveniente hacer un DGI del sistema para no olvidar ninguna:
{|
|
[[Archivo:ExE1-3-3-2-esp.png|thumb|right|250px|link=]]

|
|
c.p.s.d.: contacto puntual sin deslizamiento

c.m.c.d.: contacto multipuntual con deslizamiento

|}
:El sistema sobre el que se hace el balance de potencias es todo menos el suelo. Teniendo en cuenta que la potencia de los enlaces internos es siempre nula, las únicas interacciones que hay que tener en cuenta son el peso, los enlaces de las ruedas con el suelo, los pares de frenada y la fricción entre bloque y camión:

[[Archivo:ExE1-3-3-3-esp.png|thumb|center|300px|link=]]

:Referencia suelo:
:Las fuerzas de enlace del suelo sobre las ruedas están aplicadas a puntos de velocidad cero, y por tanto no hacen potencia. El peso es una fuerza vertical, y el movimiento del camión y del bloque es horizontal, por tanto su potencia también es cero.

:<math>\dot{\Ws}_\Ts^{\rightarrow \mathrm{sist}}=\dot{\Ws}_\Ts^{\mathrm{ext}}+\dot{\Ws}^{\mathrm{int}}=\dot{\Ws}_\Ts^{\mathrm{peso}}+\dot{\Ws}_\Ts^{\mathrm{enlace}}+\dot{\Ws}^{\mathrm{fricción}}+\dot{\Ws}^{\Gamma}=\dot{\Ws}^{\mathrm{fricción}}+\dot{\Ws}^{\Gamma}.</math>

:Las potencias de interacciones internas se pueden calcular en cualquier referencia. Por tanto:

:<math>\dot{\Ws}^{\mathrm{fricción}}= \dot{\Ws}_\mathrm{camión}^{\mathrm{fric}\rightarrow \mathrm{bloque}} +\dot{\Ws}_\mathrm{camió}^{\mathrm{fric}\rightarrow \mathrm{camión}} = \dot{\Ws}_\mathrm{camión}^{\mathrm{fric}\rightarrow \mathrm{bloque}}= (\leftarrow \mu\ms\gs)\cdot (\rightarrow \vs')=-\mu\ms\gs\vs'</math>

:<math>\dot{\Ws}^{\Gamma}= \dot{\Ws}_\mathrm{camión}^{\Gamma\rightarrow \mathrm{ruedas}} +\dot{\Ws}_\mathrm{camión}^{\Gamma\rightarrow \mathrm{camión}} = \dot{\Ws}_\mathrm{camión}^{\Gamma\rightarrow \mathrm{ruedas}}= 2 \left(\odot \Gamma \right)\cdot \left(\otimes \frac{\vs}{\rs}\right)=-2\Gamma\frac{\vs}{\rs}.</math>

:Por otro lado: <math>\Ts_\Ts^\mathrm{sist}=\frac{1}{2}\ms\vs_\Ts^2(\mathrm{bloque})+\frac{1}{2}\ms\vs_\Ts^2(\mathrm{camión})\quad \Rightarrow \quad \dot{\Ts}_\Ts^\mathrm{sist}=-\ms(\vs+\vs')\mu\gs-\Ms\vs\as. </math>

:Finalmente: <math>\dot{\Ws}_\Ts^{\rightarrow \mathrm{sist}}=\dot{\Ts}_\Ts^\mathrm{sist} \quad \Rightarrow \quad -\mu\ms\gs\vs'-2\Gamma\frac{\vs}{\rs}=-\ms(\vs+\vs')\mu\gs-\Ms\vs\as.</math>

:Esta ecuación permite calcular el par de frenada: <math>\Gamma=\frac{1}{2}(\mu\ms\gs+\Ms\as)\rs.</math>

:Referencia camión:

:En esta referencia hay que tener en cuenta las fuerzas de arrastre. Por otro lado, las fuerzas de enlace del suelo sobre las ruedas están aplicadas a puntos que ya no tienen velocidad cero, y por tanto hacen potencia. Como antes, la potencia del peso es cero.

:<math>\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{sist}}=\dot{\Ws}_\mathrm{camión}^{\mathrm{ext}}+\dot{\Ws}^{\mathrm{int}}+\dot{\Ws}_\mathrm{camión}^{\mathrm{ar}}=\dot{\Ws}_\mathrm{camión}^{\mathrm{enlace}}+\dot{\Ws}^{\mathrm{fricción}}+\dot{\Ws}^{\Gamma}+\dot{\Ws}_\mathrm{camión}^{\mathrm{ar}}.</math>

:Las fuerzas de arrastre sobre el camión no hacen potencia porque el camión está parado en esta referencia. La de las fuerzas de arrastre sobre el bloque se ha calculado en el [[E1. Teorema de la energía: versión diferencial#✏️ ✏️ EJEMPLO E1-3.2: bloque deslizando sobre un camión|'''ejemplo E1-3.2''']]. La potencia de la fricción y de los pares de frenada es la misma de antes ya que no depende de la referencia.

:<math>\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{sist}}=\dot{\Ws}_\mathrm{camión}^{\mathrm{enllaç}}+\dot{\Ws}^{\mathrm{fricción}}+\dot{\Ws}^{\Gamma}+\dot{\Ws}_\mathrm{camión}^{\mathrm{ar} \rightarrow \mathrm{bloque}}= \dot{\Ws}_\mathrm{camión}^{\mathrm{enlace}} -\mu\ms\gs\vs'-2\Gamma\frac{\vs}{\rs} +\ms\as\vs'.</math>

:Las descripciones cinemática y dinámica de la rueda son:

[[Archivo:ExE1-3-3-4-esp.png|thumb|center|430px|link=]]

:<math>\dot{\Ws}_\mathrm{camión}^{\mathrm{enlace}}=2\Bigr[(\leftarrow \Ts)+(\uparrow \Ns)\Bigr]\cdot(\leftarrow\vs)=2\Ts\vs.</math>

La fuerza tangencial de enlace se puede calcular a partir del TMC aplicado al centro '''C''' de la rueda. Como no tiene masa: <math>\sum\overline{\Ms}_\mathrm{ext}(\Cs)=\overline{0} \quad \Rightarrow \quad (\odot \Gamma)+(\otimes \Ts\rs)=0\quad \Rightarrow \quad \Ts=\Gamma/\rs.</math>

:Por tanto: <math>\dot{\Ws}_\mathrm{camió}^{\rightarrow \mathrm{sist}} =2\frac{\Gamma}{\rs}\vs-\mu\ms\gs\vs'-2\Gamma\frac{\vs}{\rs}+\ms\as\vs'=\ms(\as-\mu\gs)\vs'.</math>

:Por otro lado, como se ha visto en el [[E1. Teorema de la energía: versión diferencial#✏️ ✏️ EJEMPLO E1-3.2: bloque deslizando sobre un camión|'''ejemplo E1-3.2''']]:

:<math>\Ts_\mathrm{camión}^\mathrm{sist}= \Ts_\mathrm{camión}^\mathrm{bloque}=\frac{1}{2}\ms\vs_\mathrm{camión}^2(\mathrm{bloque}) \quad \Rightarrow \quad \dot{\Ts}_\mathrm{camión}^\mathrm{sist}= \dot{\Ts}_\mathrm{camión}^\mathrm{bloque}=\ms\vs'(\as-\mu\gs).</math>

:El balance de potencias es consistente:

:<math>\left.\begin{array}{l}
\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{sist}}= \ms(\as-\mu\gs)\vs'\\
\dot{\Ts}^\mathrm{sist}_\mathrm{camión}=\ms(\as-\mu\gs)\vs'
\end{array}\right\} \Rightarrow\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{sist}}= \dot{\Ts}^\mathrm{sist}_\mathrm{camión}</math>

-----
-----

==E1.4 Balance de potencias para un sistema multisólido: cálculo directo y cálculo indirecto==

El '''cálculo directo''' (a partir de su definición) de la potencia asociada a una interacción (una fuerza sobre un punto '''P''' o un momento sobre un sólido rígido S) exige el conocimiento de la interacción (fuerza o momento):

<math>\dot{\Ws}_\Rs^{\Fs\rightarrow \Ps}=\overline{\Fs}_{\rightarrow\Ps} \cdot \vel{P}{R} \quad , \quad \dot{\Ws}_\Rs^{\Ms\rightarrow \Ss}=\overline{\Ms}_{\rightarrow\Ss} \cdot \velang{S}{R} </math>

Cuando no se trata de una interacción formulable, hay que recurrir a los teoremas vectoriales para determinarla.

El '''cálculo indirecto''' es una alternativa que no necesita recurrir a estos teoremas. Se basa en la evaluación de un balance de potencias para un sistema y una referencia adecuados de manera que la potencia a determinar sea el único término desconocido de la ecuación (el único que no se puede calcular de manera directa). Cuando el balance incluye más de un término desconocido, se pueden combinar varios balances (para distintos sistemas y referencias) hasta tener un número suficiente de ecuaciones que permitan la determinación de todas las potencias desconocidas.

====✏️ EJEMPLO E1-4.1: polea frenada====
------
:
{|
|
:[[Archivo:ExE1-4-1-1-esp.png|thumb|left|150px|link=]]
|Un bloque de masa m cuelga de un hilo inextensible atado a un punto de la periferia de una polea de radio r. Entre soporte (fijo al suelo) y polea actúa un freno que garantiza que el bloque baja a velocidad constante respecto al suelo. Se trata de calcular la potencia del freno.

|}

:Càlculo directo: <math>\dot{\Ws}^{\mathrm{freno}}= \overline{\Gamma}^{\rightarrow \mathrm{polea}}\cdot\velang{polea}{T}=\left(\odot\Gamma\right)\cdot\left( \otimes \frac{\vs_0}{\rs}\right)=-\Gamma \frac{\vs_0}{\rs}.</math>

:El par de frenada no es un dato, y hay que recurrir al TMC para calcularlo.

:<math>\left.\begin{array}{l}
\text{SISTEMA: polea + bloque}\\
\text{TMC en } \Os
\end{array}\right\} \sum\overline{\Ms}_\mathrm{ext}(\Os)=\dot{\overline{\Hs}}_\mathrm{RTO}(\Os)</math>

:El momento cinético proviene de las dos piezas. Ya que '''O''' no es fijo al bloque, hay que hacer descomposición baricéntrica para calcular el momento cinético del bloque:

:<math>\overline{\Hs}_\mathrm{RTO=T} (\Os) =\overline{\Hs}_\Ts^\mathrm{politja}(\Os) + \overline{\Hs}_\Ts^\mathrm{bloc}(\Os)= \overline{\Hs}_\Ts^\mathrm{politja}(\Os) + \overline{\Hs}_\Ts^\mathrm{bloc}(\Gs_\mathrm{bloc})+ \overline{\Hs}_\Ts^{\oplus\mathrm{bloc}}(\Os) = </math>

:<math>\hspace{2.3cm}=\Is_\mathrm{O}^\mathrm{polea} \velang{polea}{T}+\Is_{\Gs_\mathrm{bloque}}^\mathrm{bloque}+\OGvec_\mathrm{bloque}\times \ms\overline{\vs}_\Ts(\Gs_\mathrm{bloque})=\left(\otimes \Is_\mathrm{O}\frac{\vs_0}{\rs}\right)+(\otimes \ms\vs_0\rs).</math>

:[[Archivo:ExE1-4-1-2-esp.png|thumb|right|180px|link=]]
:Se trata de un momento cinético constante en dirección y valor. Por tanto:

:<math>\dot{\overline{\Hs}}_\mathrm{RTO}(\Os)=0 \quad \Rightarrow \quad \sum{\overline{\Ms}}_\mathrm{ext}(\Os)=(\odot\Gamma)+(\otimes \ms\gs\rs) =0\quad \Rightarrow \quad \Gamma= \ms\gs\rs</math>.

:Finalmente: <math>\dot{\Ws}^{\mathrm{freno}}= -\Gamma \frac{\vs_0}{\rs} = -\ms\gs\vs_0.</math>

:Cálculo indirecto:

:<math>\left.\begin{array}{l}
\text{SISTEMA: polea + soporte + bloque}\\
\text{REF: suelo(Gal)}
\end{array}\right\} \quad \dot{\Ts}^\mathrm{sist}_\Ts=\dot{\Ws}^\mathrm{ext}_\Ts+ \dot{\Ws}^\mathrm{int}</math>

:<math>\left.\begin{array}{l}
\dot{\Ts}^\mathrm{sist}_\Ts=0\\
\dot{\Ws}^\mathrm{ext}_\Ts=\left(\downarrow \ms\gs\right)\cdot\left(\downarrow \vs_0\right)= \ms\gs\vs_0\\
\dot{\Ws}^\mathrm{int}=\dot{\Ws}^{\mathrm{freno}}
\end{array}\right\} \quad \Rightarrow \quad \dot{\Ws}^{\mathrm{freno}} =-\ms\gs\vs_0</math>



====✏️ EJEMPLO E1-4.2: mono y plátanos====
------
:
{|
|
:[[Archivo:ExE1-4-2-1-esp.png|thumb|left|180px|link=]]
|Unos plátanos de masa <math>(2/3)\ms</math> cuelgan de una cuerda inextensible que se apoya sin deslizar sobre una polea de radio r y masa despreciable, articulada respecto al suelo. En el otro extremo de la cuerda se agarra un mono de masa m. La polea gira con velocidad angular constante <math>\Omega_0</math> respecto al suelo bajo la acción de un motor. El mono se mueve respecto a la cuerda sin que sus manos se froten contra la cuerda, y de manera que la distancia de su centro de masas al suelo se mantiene constante. Se trata de calcular la potencia del motor.
|}

:Cálculo directo: <math>\dot{\Ws}^{\mathrm{freno}}= \overline{\Gamma}^{\rightarrow \mathrm{polea}}\cdot\velang{polea}{T}=\left(\odot\Gamma\right)\cdot\left( \odot \Omega_0\right)=\Gamma \Omega_0.</math>

:El par motor no es un dato, y hay que recurrir al TMC para calcularlo.

:<math>\left.\begin{array}{l}
\text{SISTEMA: polea + mono + plátanos}\\
\text{TMC en } \Os
\end{array}\right\} \sum\overline{\Ms}_\mathrm{ext}(\Os)=\dot{\overline{\Hs}}_\mathrm{RTO}(\Os)</math>

:El momento cinético proviene del mono y de los plátanos. En primera aproximación, se puede considerar la masa del mono concentrada en su tronco y su cabeza, de manera que no gira y su velocidad respecto al suelo es nula. Teniendo en cuenta que el punto '''O''' no se mueve respecto al mono pero sí respecto a los plátanos:

:<math>\overline{\Hs}_\mathrm{RTO=T} (\Os) =\overline{\Hs}_\Ts^\mathrm{mono}(\Os) + \overline{\Hs}_\Ts^\mathrm{plátanos}(\Os)= \overline{\Hs}_\Ts^\mathrm{mono}(\Os) + \overline{\Hs}_\Ts^\mathrm{plátanos}(\Gs_\mathrm{plát})+ \OGvec_\mathrm{plát} \times \frac{2}{3} \ms \overline{\vs}_\Ts(\Gs_\mathrm{plát})= </math>

:<math>\hspace{2.3cm}=\OGvec_\mathrm{plát}\times \frac{2}{3} \ms \overline{\vs}_\Ts(\Gs_\mathrm{plát}) = \Bigr[(\rightarrow \rs)+(\downarrow \xs)\Bigr] \times\frac{2}{3} \ms \left(\downarrow \rs\Omega_0\right)=\left(\otimes \frac{2}{3}\ms\rs^2\Omega_0\right).</math>

:Se trata de un momento cinético constante en dirección y valor. Por tanto:

:<math>\dot{\overline{\Hs}}_\mathrm{RTO}(\Os)=0 \quad \Rightarrow \quad \sum{\overline{\Ms}}_\mathrm{ext}(\Os)=(\odot\Gamma)+(\otimes \ms\gs\rs) + \left(\otimes \frac{2}{3}\ms\gs\rs\right)=0\quad \Rightarrow \quad \Gamma= \frac{5}{3}\ms\gs\rs</math>.

:Finalmente: <math>\dot{\Ws}^{\mathrm{motor}}= \Gamma \Omega_0 = \frac{5}{3}\ms\gs\rs\Omega_0.</math>

:Cálculo indirecto

:Hay que tener presente que es un problema con dos tipos de actuadores: el motor y la musculatura del mono. Hay diversos modelos biomecánicos para formular las fuerzas de los músculos, pero quedan fuera del ámbito de este curso. Ya que el mono se mueve, sus músculos desarrollan una potencia no nula.

:<math>\left.\begin{array}{l}
\text{SISTEMA: polea + mono + plátano}\\
\text{REF: suelo(Gal)}
\end{array}\right\} \quad \dot{\Ts}^\mathrm{sist}_\Ts=\dot{\Ws}^\mathrm{ext}_\Ts+ \dot{\Ws}^\mathrm{int}</math>

:<math>\left.\begin{array}{l}
\dot{\Ts}^\mathrm{sist}_\Ts=0\\
\dot{\Ws}^\mathrm{ext}_\Ts=\left(\downarrow \frac{2}{3}\ms\gs\right)\cdot\left(\downarrow \rs\Omega_0\right)= \frac{2}{3}\ms\gs\rs\Omega_0\\
\dot{\Ws}^\mathrm{int}=\dot{\Ws}^{\mathrm{musc}}+\dot{\Ws}^{\mathrm{motor}}
\end{array}\right\} \quad \Rightarrow \quad \dot{\Ws}^{\mathrm{musc}} + \dot{\Ws}^{\mathrm{motor}}=-\frac{2}{3}\ms\gs\rs\Omega_0</math>

:Este balance proporciona una ecuación con dos incógnitas. Se puede plantear un segundo balance para un sistema donde una de las dos no aparezca: un sistema sin el mono o un sistema sin el motor. Si se hace un balance de potencias para el sistema (mono + trozo de cuerda), las fuerzas externas a considerar son el peso y la tensión de la cuerda. Si se sigue trabajando en la referencia del suelo, esta tensión (desconocida) hará potencia no nula ya que está aplicada a un punto de velocidad ascendente.
:[[Archivo:ExE1-4-2-2-cat-esp.png|thumb|right|330px|link=]]

:Para evitar que aparezca un término asociado a la tensión de la cuerda, se puede trabajar en la referencia del trozo de cuerda que se ha considerado, que al tener velocidad ascendente constante es galileana. En esta referencia, el mono baja con velocidad constante:

:<math>\left.\begin{array}{l}
\text{REL: trozo de cuerda}\\
\text{AB: suelo}
\end{array}\right\}</math>

:<math>\overline{\vs}_\mathrm{REL}(\mathrm{mono})=\overline{\vs}_\mathrm{ar}(\mathrm{mono})=-\overline{\vs}_\mathrm{ar}(\mathrm{mono})=(\downarrow \rs\Omega_0)</math>

:<math>\left.\begin{array}{l}
\text{SISTEMA: mono+ trozo de cuerda}\\
\text{REF: trozo de cuerda(Gal)}
\end{array}\right\} \quad \dot{\Ts}^\mathrm{sist}_\mathrm{cuerda} =\dot{\Ws}^\mathrm{ext}_\mathrm{cuerda}+ \dot{\Ws}^\mathrm{int}</math>

:<math>\left.\begin{array}{l}
\dot{\Ts}^\mathrm{sist}_\mathrm{cuerda}=0\\
\dot{\Ws}^\mathrm{ext}_\mathrm{cuerda}=\left(\downarrow \ms\gs\right)\cdot\left(\downarrow \rs\Omega_0\right)= \ms\gs\rs\Omega_0\\
\dot{\Ws}^\mathrm{int}=\dot{\Ws}^{\mathrm{musc}}
\end{array}\right\} \quad \Rightarrow \quad \dot{\Ws}^{\mathrm{musc}} =-\ms\gs\rs\Omega_0</math>

:Combinando este resultado con el del balance anterior:

:<math>\dot{\Ws}^\mathrm{motor}=-\frac{2}{3}\ms\gs\rs\Omega_0-\dot{\Ws}^\mathrm{musc}=-\frac{5}{3}\ms\gs\rs\Omega_0</math>



© Universitat Politècnica de Catalunya. [[Mecánica:Derechos de autor |Todos los derechos reservados]]
-------------

<center>
[[D8. Conservaciones|<<< D8. Conservaciones]]

[[E2. Teorema de la energía: versión integrada|E2. Teorema de la energía: versión integrada>>>]]
</center>

E1. Work-Energy Theorem: differential form

2025-05-20T12:12:27Z

Bros: /* ✏️ EXAMPLE E1-1.2: pendulum */

<div class="noautonum">__TOC__</div>
<math>\newcommand{\uvec}{\overline{\textbf{u}}}
\newcommand{\vvec}{\overline{\textbf{v}}}
\newcommand{\evec}{\overline{\textbf{e}}}
\newcommand{\Omegavec}{\overline{\mathbf{\Omega}}}
\newcommand{\velang}[2]{\Omegavec^{\textrm{#1}}_{\textrm{#2}}}
\newcommand{\Alfavec}{\overline{\mathbf{\alpha}}}
\newcommand{\accang}[2]{\Alfavec^{\textrm{#1}}_{\textrm{#2}}}
\newcommand{\ds}{\textrm{d}}
\newcommand{\ps}{\textrm{p}}
\newcommand{\ns}{\textrm{n}}
\newcommand{\hs}{\textrm{h}}
\newcommand{\cs}{\textrm{c}}
\newcommand{\gs}{\textrm{g}}
\newcommand{\Ns}{\textrm{N}}
\newcommand{\Hs}{\textrm{H}}
\newcommand{\Fs}{\textrm{F}}
\newcommand{\ms}{\textrm{m}}
\newcommand{\ts}{\textrm{t}}
\newcommand{\us}{\textrm{u}}
\newcommand{\vs}{\textrm{v}}
\newcommand{\Rs}{\textrm{R}}
\newcommand{\Ws}{\textrm{W}}
\newcommand{\Ts}{\textrm{T}}
\newcommand{\Ls}{\textrm{L}}
\newcommand{\As}{\textrm{A}}
\newcommand{\Ds}{\textrm{D}}
\newcommand{\Bs}{\textrm{B}}
\newcommand{\es}{\textrm{e}}
\newcommand{\fs}{\textrm{f}}
\newcommand{\is}{\textrm{i}}
\newcommand{\Is}{\textrm{I}}
\newcommand{\ks}{\textrm{k}}
\newcommand{\js}{\textrm{j}}
\newcommand{\rs}{\textrm{r}}
\newcommand{\as}{\textrm{a}}
\newcommand{\ss}{\textrm{s}}
\newcommand{\Os}{\textbf{O}}
\newcommand{\Gs}{\textbf{G}}
\newcommand{\Cbf}{\textbf{C}}
\newcommand{\Or}{\Os_\Rs}
\newcommand{\Qs}{\textbf{Q}}
\newcommand{\Cs}{\textbf{C}}
\newcommand{\Ps}{\textbf{P}}
\newcommand{\Ss}{\textbf{S}}
\newcommand{\Js}{\textbf{J}}
\newcommand{\P}{\textrm{P}}
\newcommand{\Q}{\textrm{Q}}
\newcommand{\Ms}{\textrm{M}}
\newcommand{\deg}{^\textsf{o}}
\newcommand{\xs}{\textsf{x}}
\newcommand{\ys}{\textsf{y}}
\newcommand{\zs}{\textsf{z}}
\newcommand{\dert}[2]{\left.\frac{\ds{#1}}{\ds\ts}\right]_{\textrm{#2}}}
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\newcommand{\vec}[1]{\overline{#1}}
\newcommand{\vecbf}[1]{\overline{\textbf{#1}}}
\newcommand{\vecdot}[1]{\overline{\dot{#1}}}
\newcommand{\OQvec}{\vec{\Os\Qs}}
\newcommand{\GQvec}{\vec{\Gs\Qs}}
\newcommand{\GPvec}{\vec{\Gs\Ps}}
\newcommand{\PSvec}{\vec{\Ps\Ss}}
\newcommand{\QQvec}{\vec{\Qs\Qs}}
\newcommand{\QGvec}{\vec{\Qs\Gs}}
\newcommand{\QPvec}{\vec{\Qs\Ps}}
\newcommand{\JQvec}{\vec{\Js\Qs}}
\newcommand{\PQvec}{\vec{\Ps\Qs}}
\newcommand{\GJvec}{\vec{\Gs\Js}}
\newcommand{\OPvec}{\vec{\Os\textbf{P}}}
\newcommand{\OCvec}{\vec{\Os\Cs}}
\newcommand{\OGvec}{\vec{\Os\Gs}}
\newcommand{\CJvec}{\vec{\Cs\Js}}
\newcommand{\GCvec}{\vec{\Gs\Cs}}
\newcommand{\PGvec}{\vec{\Ps\Gs}}
\newcommand{\abs}[1]{\left|{#1}\right|}
\newcommand{\braq}[2]{\left\{{#1}\right\}_{\textrm{#2}}}
\newcommand{\matriz}[9]{
\begin{bmatrix}
{#1} & {#2} & {#3}\\
{#4} & {#5} & {#6}\\
{#7} & {#8} & {#9}
\end{bmatrix}}
\newcommand{\diag}[3]{
\begin{bmatrix}
{#1} & {0} & {0}\\
{0} & {#2} & {0}\\
{0} & {0} & {#3}
\end{bmatrix}}
\newcommand{\vector}[3]{
\begin{Bmatrix}
{#1}\\
{#2}\\
{#3}
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{#2}
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\newcommand{\agal}[1]{\vecbf{a}_{\textrm{Gal}} (#1)}
\newcommand{\angal}[1]{\vecbf{a}_{\textrm{NGal}} (#1)}
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\newcommand{\fvec}[2]{\overline{\mathbf{F}}_{\textrm{#1}\rightarrow\textrm{#2}}}
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\newcommand{\F}[1]{\vec{\Fs}_{#1}}
\newcommand{\Fcal}[2]{\vec{\cal{F}}^{\text{#1}}_{#2}}
</math>

The '''Work-Energy Theorem''' plays a prominent role in physics, as it bridges all branches of science (electromagnetism, thermodynamics, etc.): all of them deal with energy.

There is a certain parallel between the Vector Theorems and the '''Work-Energy Theorem''': they all allow us to understand the evolution of a magnitude that depends on the mechanical state of the system. In the case of the former, we are dealing with the instantaneous evolution of vector magnitudes (linear momentum, angular momentum), while the latter deals with the finite change (in a finite time interval) of a scalar magnitude ('''mechanical energy''').

This unit presents the differential form of this theorem (the '''power balance'''). The integrated form (which is the Work-Energy Theorem itself) is presented in [[E2. Teorema de la energía: versión integrada|'''unit E2''']].
 
------
------
==E1.1 Power balance in a system of particles==

:[[File:E1-1-eng.png|thumb|right|300px|link=]]
<center>'''Figure E1.1''' Forces on a particle of a system with constant matter</center>

Let us consider a system of constant matter, and let’s apply Newton's second law to each particle (or each mass differential) of this system in a general reference frame R (which may be non-Galilean) ('''Figure E1.1'''):

<math>\F{\rightarrow \Ps} + \Fcal{}{\text{ar}\rightarrow\Ps} + \Fcal{}{\text{Cor}\rightarrow\Ps} =\ms_\Ps\acc{P}{R}</math>

<math>\F{\rightarrow \Ps} </math> is the resultant interaction force on '''P''', and it comes from internal interactions (with particles of the same system) and external ones:

<math>\F{\rightarrow \Ps}=\F{\rightarrow \Ps}^\text{int}+\F{\rightarrow \Ps}^\text{ext}</math>

If both sides of the equation are multiplied by <math>\vel{P}{R}</math>, all forces (interaction and inertia forces) orthogonal to <math>\vel{P}{R}</math> are filtered out. Specifically, since <math>\acc{P}{R}=\accs{P}{R} + \accn{P}{R} </math>, where <math>\accn{P}{R} \perp \vel{P}{R}</math>, and since the Coriolis force is always orthogonal to <math>\vel{P}{R} (\Fcal{}{\text{Cor}\rightarrow\Ps} = -2\ms_\Ps \velang{R}{Gal} \times \vel{P}{R})</math>:

<math>\vel{P}{R} \cdot \left( \F{\rightarrow \Ps}^\text{int}+\F{\rightarrow \Ps}^\text{ext} + \Fcal{}{\text{ar}\rightarrow\Ps} \right) = \ms_\Ps\accs{P}{R} \cdot \vel{P}{R} = \frac{\ds}{\ds \ts} \left(\frac{1}{2}\ms_\Ps \vvec_\Rs^2(\Ps) \right) \equiv \frac{\ds\Ts_\Rs(\Ps)}{\ds\ts}=\dot{\Ts}_\Rs(\Ps)</math>

The scalar function <math>\Ts_\Rs(\Ps)</math> is the '''kinetic energy''' of the particle, and is a function of the [[D1. Foundational laws of Newtonian dynamics#D1.3 Newton’s Principle of Determinacy|'''mechanical state''']] of '''P'''. The scalar product of a force and the velocity of its application point is the '''force power''': <math>\vel{P}{R} \cdot \F{\rightarrow \Ps} \equiv \dot{\Ws}_\Rs^{\F{\rightarrow \Ps}} </math>. Both scalar quantities <math>(\Ts_\Rs(\Ps), \dot{\Ws}_\Rs^{\F{\rightarrow \Ps}})</math> depend on the reference frame. For the same set of interaction forces acting on '''P''', the reference frame determines which are filtered out and which are not.

The equation can be rewritten as <math>\dot{\Ws}_\Rs^{\text{ext}\rightarrow \Ps}+\dot{\Ws}_\Rs^{\text{int}\rightarrow \Ps}+\dot{\Ws}_\Rs^{\text{ar}\rightarrow \Ps} =\dot{\Ts}_\Rs(\Ps)</math>, and constitutes the '''power balance''' for particle '''P''' in the R reference frame. Summing for all particles in the system:

<math>\dot{\Ws}_\Rs^{\text{ext}}+\dot{\Ws}^{\text{int}}+\dot{\Ws}_\Rs^{\text{ar}} =\dot{\Ts}_\Rs^\text{syst}</math>

The calculation of kinetic energy is covered in [[E2. Teorema de la energía: versión integrada|'''unit E2''']]. This unit focuses on the calculation of powers. [[E1. Work-Energy Theorem: differential form#E1.2 Potencia de una pareja de acción-reacción|'''section E1-2''']] shows that the power of internal interactions does not depend on the reference frame (which is why the subscript R does not appear in the <math>\dot{\Ws}^{\text{int}}</math> term of the previous equation).

====✏️ EXAMPLE E1-1.1: free particle====
------
:
{|
:[[File:ExE1-1-1-1-eng.jpg|thumb|left|180px|link=]]
|A particle '''P''', with mass m, is released from the ring of a Ferris wheel when it is at its highest point. We want to calculate the power, at this instant, of all forces acting on the particle from three different reference frames: the ground, the ring, and the Ferris wheel cabin.
|}

:Ground reference frame: Since this is a Galilean reference frame, all forces on '''P''' are interaction forces.
{|
:[[File:ExE1-1-2- eng.png|thumb|right|230px|link=]]

|<math>\hspace{0.5cm}\sum\dot{\Ws}_\Ts=\dot{\Ws}_\Ts^\text{weight}= \left(\leftarrow \Rs\Omega_0\right) \cdot \left(\downarrow \ms\gs\right)=0</math>
|}

:Ring reference frame: It is a non-Galilean reference, and the transportation force on '''P''' must be taken into account.
{|
:[[File:ExE1-1-3-eng.png|thumb|right|250px|link=]]
|<math>\hspace{0.5cm}</math> AB:ground, REL:ring 

<math>\hspace{0.5cm}\vel{P}{REL}=\vel{P}{AB}-\vel{P}{ar}= \left(\leftarrow \Rs\Omega_0\right)-\left(\leftarrow \Rs\Omega_0\right)=\overline{0}</math> 

<math>\hspace{0.5cm}</math>

Since the velocity is zero, the total power is zero.
|}

:Cabin reference frame: It is also a non-Galilean reference, and the transportation force on '''P''' must be taken into account.

{|
:[[File:ExE1-1-4-eng.png|thumb|right|250px|link=]]
|<math>\hspace{0.5cm}</math>AB:ground, REL:cabin

<math>\hspace{0.5cm}\vel{P}{REL}=\vel{P}{AB}-\vel{P}{ar}= \left(\leftarrow \Rs\Omega_0\right)-\left(\uparrow \Rs\Omega_0\right)</math>

<math>\hspace{0.5cm}\dot{\Ws}_\text{cabin}^\text{weight}=\left[ \left(\leftarrow \Rs\Omega_0\right)+\left(\downarrow \Rs\Omega_0\right) \right] \cdot \left(\downarrow\ms\gs\right) =\ms\gs\Rs\Omega_0</math>

<math>\hspace{0.5cm}\dot{\Ws}_\text{cabin}^\text{ar}=\left[ \left(\leftarrow \Rs\Omega_0\right)+\left(\downarrow \Rs\Omega_0\right) \right] \cdot \left(\rightarrow\ms\Rs^2\Omega_0^2\right) =-\ms\gs\Rs^2 \Omega_0^3</math>

<math>\hspace{0.5cm}\sum\dot{\Ws}_\text{cabin}=\ms\Rs\Omega_0(\gs-\Rs\Omega_0^2)</math>

|}


====✏️ EXAMPLE E1-1.2: pendulum====
------
:
{|
:[[File:ExE1-2-1-eng.png|thumb|left|180px|link=]]
|A particle '''P''' of mass m hangs from an inextensible thread of length L attached to the ceiling. The assembly oscillates in a vertical plane fixed to the ground. We want to calculate the power of all the forces acting on the particle in two different reference frames: the ground and a platform rotating relative to the ground with a constant angular velocity <math>\Omega_0</math>. 

Ground reference frame: Since it is a Galilean reference frame, '''P''' is only subjected to interaction forces.
{|
:[[File:ExE1-2-2-neut.png|thumb|left|300px|link=]]

|<math>\dot{\Ws}_\Es^\Ts=(\nearrow\Ls\dot{\theta})\cdot(\nwarrow\Ts)=0</math>

<math>\dot{\Ws}_\Es^\text{weight}=(\nearrow\Ls\dot{\theta}\cdot(\downarrow \ms\gs)=-\ms\gs\Ls\dot{\theta}\sin\theta</math>

|}

Platform reference frame: It is a non Galilean reference frame, and the transportation force on '''P''' has to be taken into account.

{|
:[[File:ExE1-2-3-eng.png|thumb|left|400px|link=]]

|AB:ground, REL:platform

<math>\vel{P}{REL}=\vel{P}{AB}-\vel{P}{ar}=(\nearrow\Ls\dot{\theta})-(\otimes \Ls \Omega_0 \sin \theta)</math>

<math>\dot{\Ws}_\Es^\Ts\left[(\nearrow\Ls\dot{\theta})+(\odot\Ls\Omega_0\sin\theta)\right]\cdot(\nwarrow\Ts)=0</math>

<math>\dot{\Ws}_\Es^\text{weight}\left[(\nearrow\Ls\dot{\theta})+(\odot\Ls\Omega_0\sin\theta)\right]\cdot(\downarrow \ms\gs)=-\ms\gs\Ls\dot{\theta}\sin\theta</math>

<math>\dot{\Ws}_\Es^\text{ar}\left[(\nearrow\Ls\dot{\theta})+(\odot\Ls\Omega_0\sin\theta)\right]\cdot(\rightarrow \ms\Ls\Omega_0^2\sin\theta)=\ms\Ls^2\Omega_0^2\dot{\theta}\cos\theta</math>
|}


------
------

==E1.2 Power of an action-reaction pair==

The total power associated with internal interactions in a constant-mass system is not zero in principle (since the two forces of each action-reaction pair are applied to particles that may have different velocities relative to the reference frame in which the calculation is made), and it is independent of the reference frame in which it is evaluated. It only depends on the separation velocity of the two particles. The power balance for the system is therefore written as: <math>\dot{\Ts}_\Rs^\text{syst}=\dot{\Ws}_\Rs^\text{ext} + \dot{\Ws}^\text{int} + \dot{\Ws}_\Rs^\text{ar} </math> .

<div>
=====💭 PROOF ➕=====

{|
|
:[[File:E1-2-neut.png|thumb|left|200px|link=]]
|<math>\dot{\Ws}^\text{ac-reac}=\overline{\Fs}_{\Ps \rightarrow \Qs}\cdot \vel{Q}{R} + \overline{\Fs}_{\Qs \rightarrow \Ps}\cdot \vel{P}{R}= \overline{\Fs}_{\Ps \rightarrow \Qs}\cdot \left[ \vel{Q}{R}-\vel{P}{R}\right]</math>

By the [[D1. Foundational laws of Newtonian dynamics#D1.6 Newton’s third law (action-reaction principle)|'''action-reaction principle''']], <math>\overline{\Fs}_{\Ps \rightarrow \Qs}=-\overline{\Fs}_{\Qs \rightarrow \Ps}</math>. Therefore: <math>\dot{\Ws}^\text{ac-reac}=\overline{\Fs}_{\Ps \rightarrow \Qs}\cdot \left[ \vel{Q}{R}-\vel{P}{R}\right]</math>

As <math>\overline{\Fs}_{\Ps \rightarrow \Qs}</math> has the <math>\PQvec</math> direction, only the components of the velocities in that direction have to be considered:

<math>\dot{\Ws}^\text{ac-reac}=\overline{\Fs}_{\Ps \rightarrow \Qs}\cdot \biggl( \vel{Q}{R} \Bigr]_{||\Ps\Qs}-\ \vel{P}{R} \Bigr]_{||\Ps\Qs}\biggl)</math>
|}

The difference <math>\biggl( \vel{Q}{R} \Bigr]_{\Ps\Qs}-\ \vel{P}{R} \Bigr]_{\Ps\Qs}\biggl)</math> is the separation velocity <math>\dot{\rho}</math> between '''P''' and '''Q''', and it does not dpend on the reference frame since it is the derivative of a scalar (the distance <math>\dot{\rho}</math> between the two particles). For the previous drawing, <math>\dot{\rho}> 0</math>. Since this is an attractive interaction, <math>\dot{\Ws}^\text{ac-reac} < 0</math>.

With a drawing, it is easy to see that:
* attractive interaction and particle approach <math>(\dot{\rho}<0)</math> : <math>\dot{\Ws}^\text{ac-reac} > 0</math> ;
* repulsive interaction and particle separation <math>(\dot{\rho}>0)</math> : <math>\dot{\Ws}^\text{ac-reac} > 0</math>
* repulsive interaction and particle approach <math>(\dot{\rho}<0)</math> : <math>\dot{\Ws}^\text{ac-reac} < 0</math>


</div>

There are two particular cases of interaction forces whose power always has the same sign:

* Friction action-reaction pair between '''P''' and '''Q''': since the friction force always opposes the relative velocity between the two points, its power is always negative:

<math>\hspace{1cm} \dot{\Ws}^\text{friction}=\dot{\Ws}^{\text{fric}\rightarrow\Ps}_\text{RTQ}+\dot{\Ws}^{\text{fric}\rightarrow\Qs}_\text{RTQ}=\overline{\Fs}^{\text{fric}\rightarrow\Ps} \cdot \vel{P}{RTQ} = - \abs{\overline{\Fs}^{\text{fric}\rightarrow\Ps}} \frac{\vel{P}{RTQ}}{\abs{\vel{P}{RTQ}}} \cdot \vel{P}{RTQ}<0;</math>

* Constraint action-reaction pair between '''P''' and '''Q''': from the [[D3. Interactions between rigid bodies#D3.4 Direct constraint interactions|'''analytical characterization equation''']] of the constraint forces between two points '''P''' and '''Q''' belonging to two rigid bodies <math>\mathrm{S}_\Ps</math> and <math>\mathrm{S}_\Qs</math>, it is easy to see that the associated power is zero:

<math>\hspace{1cm} \dot{\Ws}^\text{constraint}=\dot{\Ws}^{\text{constraint}\rightarrow\Ps}_{\mathrm{S}_\Ps}+\dot{\Ws}^{\text{constraint}\rightarrow\Qs}_{\mathrm{S}_\Ps}=\overline{\Fs}^{\text{constraint}\rightarrow\Ps} \cdot \overline{\mathbf{v}}_{\mathrm{S}_\Qs}(\Ps) = - 0;</math>

====✏️ EXAMPLE E1-2.1: power of an action-reaction friction pair====
------
:
{|
|
:[[File:ExE1-2-1-2-eng.png|thumb|left|180px|link=]]
|A homogeneous wheel, with radius r and mass m, moves down an inclined plane without sliding until it joins a conveyor belt moving at a constant velocity <math>\vs_0</math> relative to the ground. At the moment it makes contact with the belt, the center of the wheel has a velocity <math>\vs_0</math> relative to the ground in the same direction as the belt. We want to calculate the power of the friction force between the belt and the wheel at this instant.
|}

At that momento, the <math>\mathbf{ICR}^\text{wheel}_\Es</math> is point <math>\Js_\text{wheel}</math> in contact with the belt. This point slides backwards with respect to the belt, therefore the friction force on the wheel is forwards:

<math>\left.\begin{array}{l}
\text { AB: ground } \\
\text { REL: belt }
\end{array}\right\} \quad \bar{\vs}_{\text {REL }}\left(\Js_{\text {wheel }}\right)=\bar{\vs}_{\text{AB}}\left(\Js_{\text {wheel }}\right)-\bar{\vs}_{\mathrm{ar}}\left(\Js_{\text {wheel }}\right)=-\bar{\vs}_{\mathrm{ar}}\left(\Js_{\text {wheel }}\right)=-\left(\rightarrow \mathrm{v}_0\right)=\left(\leftarrow \mathrm{v}_0\right) .</math>

[[File:ExE1-2-1-3-eng.png|thumb|center|500px|link=]]

Calculation of the friction power: <math>\dot{\Ws}^\text{friction}=\dot{\Ws}_\Rs^{\text{friction}\rightarrow \text{wheel}}+ \dot{\Ws}_\Rs^{\text{friction}\rightarrow \text{belt}}</math>, where R may be any reference frame. To quickly calculate this power, it is advisable to choose a reference frame R where one of the two points subjected to the action-reaction pair has zero speed, so that one of the two terms is directly zero:

<math>\dot{\Ws}^\text{friction}=\dot{\Ws}_\text{belt}^{\text{fric}\rightarrow \text{wheel}}+ \dot{\Ws}_\text{belt}^{\text{fric }\rightarrow \text{belt}}=\overline{\Fs}^{\text{fric}\rightarrow \text{wheel}} \cdot \bar{\vs}_{\text {belt}}\left(\Js_{\text {wheel }}\right)=(\rightarrow \mu\ms\gs)\cdot(\leftarrow \vs_0)=-\mu\ms\gs\vs_0.</math>

If the calculation is done in any other reference frame, the result is the same:

<math>\dot{\Ws}^\text{friction}=\dot{\Ws}_\text{E}^{\text{fric}\rightarrow \text{wheel}}+ \dot{\Ws}_\text{E}^{\text{fric} \rightarrow \text{belt}}=(\rightarrow \mu\ms\gs)\cdot 0 + (\leftarrow \mu\ms\gs)\cdot(\rightarrow \vs_0)=-\mu\ms\gs\vs_0.</math>

<math>\dot{\Ws}^\text{friction}=\dot{\Ws}_\text{vehicle}^{\text{fric}\rightarrow \text{wheel}}+ \dot{\Ws}_\text{vehicle}^{\text{fric} \rightarrow \text{belt}}=(\rightarrow \mu\ms\gs)\cdot (\rightarrow\vs_0) + (\leftarrow \mu\ms\gs) \cdot(\rightarrow 3\vs_0)=-\mu\ms\gs\vs_0.</math>

However, if only the power of one of the two forces is calculated (for example, the friction on the wheel), the result depends on the reference frame and can have any sign:

<math>\dot{\Ws}_\text{belt}^{\text{fric }\rightarrow \text{wheel}}=(\rightarrow \mu\ms\gs)\cdot(\leftarrow \vs_0)=-\mu\ms\gs\vs_0 ,</math>

<math>\dot{\Ws}_\text{E}^{\text{fric} \rightarrow \text{wheel}}=(\rightarrow \mu\ms\gs)\cdot 0 = 0 ,</math>

<math>\dot{\Ws}_\text{vehicle}^{\text{fric} \rightarrow \text{wheel}}=(\rightarrow \mu\ms\gs)\cdot (\rightarrow\vs_0)=\mu\ms\gs\vs_0.</math>


-----
-----

==E1.3 Power of a system of forces on a rigid body==

{|
|
:[[File:E1-3-eng.png|thumb|left|180px|link=]]
<center>'''Figure E1.3''': Kinematic torsor of a rigid body in a reference frame R, and forces on different points (dm) of that rigid body</center>

|When calculating the power of a system of forces on a rigid body S, the term associated with internal interactions is zero because those interactions are all constraint (cohesion) forces and, as seen in [[E1. Work-Energy Theorem: differential form#E1.2 Power of an action-reaction pair|'''section E1.2''']], their total power is zero.

The total power of a system of forces on S can be calculated from the torsor of the system at any point '''O''' of the rigid body, <math>(\overline{\Fs},\overline{\Ms}(\Os))</math>. Taking into account rigid body kinematics ('''Figure E1.3'''):

<math>\dot{\Ws}_\Rs^{\rightarrow \Ss} =\sum_\Ps\overline{\Fs}^{\rightarrow \Ps} \cdot \vel{P}{R}= \sum_\Ps \left[ \overline{\Fs}^{\rightarrow \Ps} \cdot \vel{O}{R} + \velang{S}{R} \times \OPvec \right] </math>

|}
<math>\dot{\Ws}_\Rs^{\rightarrow \Ss} =\left(\sum_\Ps\overline{\Fs}^{\rightarrow \Ps} \right)\cdot \vel{O}{R} + \sum_\Ps\overline{\Fs}^{\rightarrow \Ps} \cdot \left(\velang{S}{R} \times \OPvec \right) = \left( \sum_\Ps\overline{\Fs}^{\rightarrow \Ps} \right) \cdot \vel{O}{R} + \left[ \sum_\Ps \left( \OPvec \times \overline{\Fs}^{\rightarrow \Ps} \right)\right] \cdot \velang{S}{R} = \overline{\Fs}\cdot \vel{O}{R} + \overline{\Ms}(\Os)\cdot \velang{S}{R} </math>

This calculation also shows that the power of a moment (or a torque <math>\Gamma</math>) on a rigid body in a reference frame R is the product of the moment by the angular velocity of the rigid body: <math>\dot{\Ws}_\Rs^{\Gamma\rightarrow \mathrm{S}} = \overline{\Gamma} \cdot \velang{S}{R}</math>.

The power balance for a rigid body is: <math>\dot{\Ws}_\Rs^{ \rightarrow \mathrm{S}} = \dot{\Ws}_\Rs^{ \mathrm{ext}} + \dot{\Ws}_\Rs^{ \mathrm{ar}}.</math>

====✏️ EXAMPLE E1-3.1: roller sliding on a slope====
------
:
{|
|
:[[File:ExE1-3-1-1-eng.png|thumb|left|180px|link=]]
|A homogeneous roller, with mass m and radius r, slides down a slope. The friction coefficient between the ground and the roller is <math>\mu</math>. We want to calculate the power of every force acting on the roller in the ground reference frame.
|}

:It is a 2D problem. The kinematic and dynamic descriptions are:
{|
|
[[File:ExE1-3-1-2-eng.png|thumb|right|250px|link=]]

|<math> \hspace{0.5cm}\vel{J}{E}=\vel{C}{T} + \velang{}{E} \times \CJvec= </math>

<math>\hspace{1.7cm}=\left( \rightarrow \vs_\Es \right) + \left( \otimes \Omega_\Es\right) \times \left( \downarrow \rs \right)</math>

<math> \hspace{0.5cm}\vel{J}{E} = \left[\rightarrow (\vs_\Es -\rs\Omega_\Es)\right]</math>

<math>\hspace{0.5cm} \dot{\Ws}_\Es^{\mathrm{ext}} = \dot{\Ws}_\Es^{\mathrm{weight}} + \dot{\Ws}_\Es^{\mathrm{friction}} + \dot{\Ws}_\Es^{\mathrm{constraint}}= \Bigr[\left(\downarrow \ms\gs\cos\beta\right)+\left(\rightarrow \ms\gs\sin\beta\right)\Bigr] \cdot (\rightarrow \vs_\Es) + \Bigr[(\leftarrow \mu \Ns) + (\uparrow \Ns ) \Bigr] \cdot \Bigr[\rightarrow (\vs_\Es -\rs\Omega_\Es)\Bigr]</math>

|}
:Since the roller center has no motion perpendicular to the plane, the AMT leads to <math>\Ns =\ms\gs\cos\beta</math>. Therefore:

{|
|
[[Archivo:ExE1-3-1-3-eng.png|thumb|right|140px|link=]]

|<math>\hspace{0.5cm} \dot{\Ws}_\Ts^{\mathrm{ext}} = \ms\gs\vs_\Ts \sin \beta - \ms\gs(\vs_\Ts -\rs\Omega_\Ts)\cos\beta=\ms\gs \Bigr[ \vs_\Ts (\sin\beta-\cos\beta)+\rs\Omega_\Ts \cos\beta \Bigr].</math>

<math>\hspace{0.5cm}</math>

If instead of considering the friction at '''J''' we consider the equivalent torsor at point '''C''', we must consider the force applied in '''C''' and the moment about '''C''', which is not zero:

<math>\hspace{0.5cm} \dot{\Ws}_\Es^{ \mathrm{friction}} = (\leftarrow \mu\Ns)\cdot(\rightarrow\vs_\Es)+(\otimes \mu\Ns\rs)\cdot(\otimes \Omega_\Es) =\mu\ms\gs\cos\beta(\rs\Omega_\Es-\vs_\Es)</math>,

<math>\hspace{0.5cm} \dot{\Ws}_\Es^{\mathrm{ext}} = \dot{\Ws}_\Es^{ \mathrm{weight}} + \dot{\Ws}_\Es^{ \mathrm{friction}}+ \dot{\Ws}_\Es^{ \mathrm{constraint}}= \ms\gs\vs_\Es\sin\beta+\mu\ms\gs\cos\beta (\rs\Omega_\Es-\vs_\Es)</math>.

<math>\hspace{0.5cm}</math>

The result is the same.
|}


====✏️ EXAMPLE E1-3.2: block sliding on a truck====
------
:
{|
|
:[[File:ExE1-3-2-1-eng.png|thumb|left|220px|link=]]
|A truck brakes with acceleration '''a''' relative to the ground at the instant in which its velocity (also relative to the ground) is v. At the same instant, the homogeneous block, of mass m, slides with velocity v' relative to the truck. We want to perform a power balance for the block in two different reference frames: the ground and the truck.
|}

:Ground reference frame: It is a 2D problem. The kinematic and dynamic descriptions are:
{|
|
[[File:ExE1-3-2-2-eng.png|thumb|right|200px|link=]]

| The constraint between the truck and the block allows only one translational DoF between the two. Therefore, in the 2D approximation, the associated torsor has two independent components. If characterized at the midpoint of the contact surface, these components are a normal force and a moment perpendicular to the plane. The AMT applied to the block leads to <math>\Ns=\ms\gs</math>.

|}
:The powers associated with all interactions on the block in the ground reference are:

:<math>\dot{\Ws}_\Es^{\rightarrow \mathrm{bloque}}=\dot{\Ws}_\Ts^{\mathrm{ext}}=\dot{\Ws}_\Ts^{\mathrm{peso}}+\dot{\Ws}_\Ts^{\mathrm{fricción}}+\dot{\Ws}_\Ts^{\mathrm{enlace}}= \Bigr[(\downarrow \ms\gs)+(\leftarrow \mu\ms\gs)+(\uparrow \Ns) \Bigr] \cdot \Bigr[\rightarrow(\vs+\vs')\Bigr] + (\odot \Ms_\es)\cdot 0 = -\mu\ms\gs(\vs+\vs')</math>

:Ya que el bloque se traslada respecto al suelo, su energía cinética total es inmediata, y se puede plantear fácilmente el balance de potencias:

:<math>\Ts_\Ts^\mathrm{bloque}=\frac{1}{2}\int_\mathrm{bloque}\ds\ms \vs_\Ts^2(\mathrm{bloque})=\frac{1}{2}\ms \vs_\Ts^2(\mathrm{bloc})\quad \Rightarrow \quad \dot{\Ts}_\Ts^\mathrm{bloque}=\ms\vs_\Ts( \mathrm{bloque})\dot{\vs}_\Ts( \mathrm{bloque}) </math>

:Como la única fuerza horizontal sobre el bloque es la fricción:

<math>\dot{\vs}_\Ts(\mathrm{bloque})=-\mu\gs\quad \Rightarrow \quad \dot{\Ts}_\Ts^\mathrm{bloque}=-\mu\ms\gs(\vs+\vs').</math>

:El balance de potencias es consistente:

:<math>\left.\begin{array}{l}
\dot{\mathrm{W}}_\Ts^{\rightarrow \text { bloque }}=\dot{\mathrm{W}}_{\mathrm{T}}^{\text {ext }}=-\mu \mathrm{mg}\left(\mathrm{v}+\mathrm{v}^{\prime}\right) \\
\dot{\mathrm{T}}_{\mathrm{T}}^{\text {bloque }}=-\mu \mathrm{mg}\left(\mathrm{v}+\mathrm{v}^{\prime}\right)
\end{array}\right\} \Rightarrow \dot{\mathrm{W}}_{\mathrm{T}}^{\rightarrow \text { bloque }}=\dot{\mathrm{T}}_{\mathrm{T}}^{\text {bloque }}</math>

:Referencia camión: Al estar el camión acelerado respecto al suelo, hay que tener en cuenta las fuerzas de arrastre sobre todos los diferenciales de masa (dm) del bloque:

{|
|

[[Archivo:ExE1-3-2-3-cat-esp.jpg|thumb|right|250px|link=]]
|<math>\cal{\Fs}^{\rightarrow \ds\ms}_{\mathrm{ar}}=-\ds\ms\overline{\as}_\mathrm{ar}(\ds\ms)=-\ds\ms (\leftarrow \as)=(\rightarrow \ds\ms \cdot \as)</math>.

El torsor resultante de este sistema de fuerzas en el centro de inercia '''G''' se reduce a una fuerza horizontal.

<math>\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{bloque}}=\dot{\Ws}_\mathrm{camión}^{\mathrm{ext}} + \dot{\Ws}_\mathrm{camión}^{\mathrm{ar}} =\dot{\Ws}_\mathrm{camión}^{\mathrm{peso}} + \dot{\Ws}_\mathrm{camión}^{\mathrm{fricción}} + \dot{\Ws}_\mathrm{camión}^{\mathrm{enlace}} + \dot{\Ws}_\mathrm{camión}^{\mathrm{ar}} , </math>

<math>\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{bloque}}= \Bigr[(\downarrow \ms\gs)+(\leftarrow \mu\ms\gs)+(\uparrow \Ns) + (\rightarrow \ms\as)\Bigr] \cdot(\rightarrow \vs')+(\odot\Ms_\es) \cdot0=\ms(\as-\mu\gs)\vs'.</math>
|}

:Se puede comprobar que el balance de potencias es consistente:

:<math>\Ts^\mathrm{bloque}_\mathrm{camión}=\frac{1}{2}\ms\vs^2_\mathrm{camión}(\mathrm{bloque}) \quad \Rightarrow \quad \dot{\Ts}^\mathrm{bloque}_\mathrm{camión} = \ms\vs_\mathrm{camión}(\mathrm{bloque})\dot{\vs}_\mathrm{camión}(\mathrm{bloque})</math>

:<math>\left.\begin{array}{l}
\mathrm{AB}:\mathrm{suelo} \\
\mathrm{REL}:\mathrm{camión}
\end{array}\right\} \quad \dot{\vs}_\mathrm{camión}(\mathrm{bloque})= \dot{\vs}_\mathrm{T}(\mathrm{bloque})-\dot{\vs}_\mathrm{ar}(\mathrm{bloque}) = (-\mu\gs)-(-\as)=\as-\mu\gs \quad \Rightarrow \quad \dot{\Ts}^\mathrm{bloque}_\mathrm{camión}=\ms\vs'(\as-\mu\gs)</math>

:<math>\left.\begin{array}{l}
\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{bloque}}=\dot{\Ws}_\mathrm{camión}^{\mathrm{ext}} + \dot{\Ws}_\mathrm{camión}^{\mathrm{ar}}= \ms(\as-\mu\gs)\vs'\\
\dot{\Ts}^\mathrm{bloque}_\mathrm{camión}=\ms\vs'(\as-\mu\gs)
\end{array}\right\} \Rightarrow\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{bloque}}= \dot{\Ts}^\mathrm{bloque}_\mathrm{camión}</math>


====✏️ EJEMPLO E1-3.3: camión frenando con carga====
------
:
{|
|
:[[Archivo:ExE1-3-3-1-esp.png|thumb|left|220px|link=]]
|Un camión frena con aceleración '''a''' respecto al suelo en el instante en el que su velocidad (también respecto al suelo) es v. Los pares de frenada en el eje delantero y en el trasero son iguales y valen <math>\Gamma</math>, y las ruedas, de radio r y masa despreciable, no deslizan sobre el suelo. En el mismo instante, el bloque homogéneo, de masa m, desliz con velocidad v' relativa al camión. Se trata de hacer un balance de potencias para el sistema (bloque+camión) en la referencia del suelo y en la del camión.

|}

:Ya que la potencia de las interacciones internas no es nula en principio (no se trata de un único sólido rígido), es conveniente hacer un DGI del sistema para no olvidar ninguna:
{|
|
[[Archivo:ExE1-3-3-2-esp.png|thumb|right|250px|link=]]

|
|
c.p.s.d.: contacto puntual sin deslizamiento

c.m.c.d.: contacto multipuntual con deslizamiento

|}
:El sistema sobre el que se hace el balance de potencias es todo menos el suelo. Teniendo en cuenta que la potencia de los enlaces internos es siempre nula, las únicas interacciones que hay que tener en cuenta son el peso, los enlaces de las ruedas con el suelo, los pares de frenada y la fricción entre bloque y camión:

[[Archivo:ExE1-3-3-3-esp.png|thumb|center|300px|link=]]

:Referencia suelo:
:Las fuerzas de enlace del suelo sobre las ruedas están aplicadas a puntos de velocidad cero, y por tanto no hacen potencia. El peso es una fuerza vertical, y el movimiento del camión y del bloque es horizontal, por tanto su potencia también es cero.

:<math>\dot{\Ws}_\Ts^{\rightarrow \mathrm{sist}}=\dot{\Ws}_\Ts^{\mathrm{ext}}+\dot{\Ws}^{\mathrm{int}}=\dot{\Ws}_\Ts^{\mathrm{peso}}+\dot{\Ws}_\Ts^{\mathrm{enlace}}+\dot{\Ws}^{\mathrm{fricción}}+\dot{\Ws}^{\Gamma}=\dot{\Ws}^{\mathrm{fricción}}+\dot{\Ws}^{\Gamma}.</math>

:Las potencias de interacciones internas se pueden calcular en cualquier referencia. Por tanto:

:<math>\dot{\Ws}^{\mathrm{fricción}}= \dot{\Ws}_\mathrm{camión}^{\mathrm{fric}\rightarrow \mathrm{bloque}} +\dot{\Ws}_\mathrm{camió}^{\mathrm{fric}\rightarrow \mathrm{camión}} = \dot{\Ws}_\mathrm{camión}^{\mathrm{fric}\rightarrow \mathrm{bloque}}= (\leftarrow \mu\ms\gs)\cdot (\rightarrow \vs')=-\mu\ms\gs\vs'</math>

:<math>\dot{\Ws}^{\Gamma}= \dot{\Ws}_\mathrm{camión}^{\Gamma\rightarrow \mathrm{ruedas}} +\dot{\Ws}_\mathrm{camión}^{\Gamma\rightarrow \mathrm{camión}} = \dot{\Ws}_\mathrm{camión}^{\Gamma\rightarrow \mathrm{ruedas}}= 2 \left(\odot \Gamma \right)\cdot \left(\otimes \frac{\vs}{\rs}\right)=-2\Gamma\frac{\vs}{\rs}.</math>

:Por otro lado: <math>\Ts_\Ts^\mathrm{sist}=\frac{1}{2}\ms\vs_\Ts^2(\mathrm{bloque})+\frac{1}{2}\ms\vs_\Ts^2(\mathrm{camión})\quad \Rightarrow \quad \dot{\Ts}_\Ts^\mathrm{sist}=-\ms(\vs+\vs')\mu\gs-\Ms\vs\as. </math>

:Finalmente: <math>\dot{\Ws}_\Ts^{\rightarrow \mathrm{sist}}=\dot{\Ts}_\Ts^\mathrm{sist} \quad \Rightarrow \quad -\mu\ms\gs\vs'-2\Gamma\frac{\vs}{\rs}=-\ms(\vs+\vs')\mu\gs-\Ms\vs\as.</math>

:Esta ecuación permite calcular el par de frenada: <math>\Gamma=\frac{1}{2}(\mu\ms\gs+\Ms\as)\rs.</math>

:Referencia camión:

:En esta referencia hay que tener en cuenta las fuerzas de arrastre. Por otro lado, las fuerzas de enlace del suelo sobre las ruedas están aplicadas a puntos que ya no tienen velocidad cero, y por tanto hacen potencia. Como antes, la potencia del peso es cero.

:<math>\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{sist}}=\dot{\Ws}_\mathrm{camión}^{\mathrm{ext}}+\dot{\Ws}^{\mathrm{int}}+\dot{\Ws}_\mathrm{camión}^{\mathrm{ar}}=\dot{\Ws}_\mathrm{camión}^{\mathrm{enlace}}+\dot{\Ws}^{\mathrm{fricción}}+\dot{\Ws}^{\Gamma}+\dot{\Ws}_\mathrm{camión}^{\mathrm{ar}}.</math>

:Las fuerzas de arrastre sobre el camión no hacen potencia porque el camión está parado en esta referencia. La de las fuerzas de arrastre sobre el bloque se ha calculado en el [[E1. Teorema de la energía: versión diferencial#✏️ ✏️ EJEMPLO E1-3.2: bloque deslizando sobre un camión|'''ejemplo E1-3.2''']]. La potencia de la fricción y de los pares de frenada es la misma de antes ya que no depende de la referencia.

:<math>\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{sist}}=\dot{\Ws}_\mathrm{camión}^{\mathrm{enllaç}}+\dot{\Ws}^{\mathrm{fricción}}+\dot{\Ws}^{\Gamma}+\dot{\Ws}_\mathrm{camión}^{\mathrm{ar} \rightarrow \mathrm{bloque}}= \dot{\Ws}_\mathrm{camión}^{\mathrm{enlace}} -\mu\ms\gs\vs'-2\Gamma\frac{\vs}{\rs} +\ms\as\vs'.</math>

:Las descripciones cinemática y dinámica de la rueda son:

[[Archivo:ExE1-3-3-4-esp.png|thumb|center|430px|link=]]

:<math>\dot{\Ws}_\mathrm{camión}^{\mathrm{enlace}}=2\Bigr[(\leftarrow \Ts)+(\uparrow \Ns)\Bigr]\cdot(\leftarrow\vs)=2\Ts\vs.</math>

La fuerza tangencial de enlace se puede calcular a partir del TMC aplicado al centro '''C''' de la rueda. Como no tiene masa: <math>\sum\overline{\Ms}_\mathrm{ext}(\Cs)=\overline{0} \quad \Rightarrow \quad (\odot \Gamma)+(\otimes \Ts\rs)=0\quad \Rightarrow \quad \Ts=\Gamma/\rs.</math>

:Por tanto: <math>\dot{\Ws}_\mathrm{camió}^{\rightarrow \mathrm{sist}} =2\frac{\Gamma}{\rs}\vs-\mu\ms\gs\vs'-2\Gamma\frac{\vs}{\rs}+\ms\as\vs'=\ms(\as-\mu\gs)\vs'.</math>

:Por otro lado, como se ha visto en el [[E1. Teorema de la energía: versión diferencial#✏️ ✏️ EJEMPLO E1-3.2: bloque deslizando sobre un camión|'''ejemplo E1-3.2''']]:

:<math>\Ts_\mathrm{camión}^\mathrm{sist}= \Ts_\mathrm{camión}^\mathrm{bloque}=\frac{1}{2}\ms\vs_\mathrm{camión}^2(\mathrm{bloque}) \quad \Rightarrow \quad \dot{\Ts}_\mathrm{camión}^\mathrm{sist}= \dot{\Ts}_\mathrm{camión}^\mathrm{bloque}=\ms\vs'(\as-\mu\gs).</math>

:El balance de potencias es consistente:

:<math>\left.\begin{array}{l}
\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{sist}}= \ms(\as-\mu\gs)\vs'\\
\dot{\Ts}^\mathrm{sist}_\mathrm{camión}=\ms(\as-\mu\gs)\vs'
\end{array}\right\} \Rightarrow\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{sist}}= \dot{\Ts}^\mathrm{sist}_\mathrm{camión}</math>

-----
-----

==E1.4 Balance de potencias para un sistema multisólido: cálculo directo y cálculo indirecto==

El '''cálculo directo''' (a partir de su definición) de la potencia asociada a una interacción (una fuerza sobre un punto '''P''' o un momento sobre un sólido rígido S) exige el conocimiento de la interacción (fuerza o momento):

<math>\dot{\Ws}_\Rs^{\Fs\rightarrow \Ps}=\overline{\Fs}_{\rightarrow\Ps} \cdot \vel{P}{R} \quad , \quad \dot{\Ws}_\Rs^{\Ms\rightarrow \Ss}=\overline{\Ms}_{\rightarrow\Ss} \cdot \velang{S}{R} </math>

Cuando no se trata de una interacción formulable, hay que recurrir a los teoremas vectoriales para determinarla.

El '''cálculo indirecto''' es una alternativa que no necesita recurrir a estos teoremas. Se basa en la evaluación de un balance de potencias para un sistema y una referencia adecuados de manera que la potencia a determinar sea el único término desconocido de la ecuación (el único que no se puede calcular de manera directa). Cuando el balance incluye más de un término desconocido, se pueden combinar varios balances (para distintos sistemas y referencias) hasta tener un número suficiente de ecuaciones que permitan la determinación de todas las potencias desconocidas.

====✏️ EJEMPLO E1-4.1: polea frenada====
------
:
{|
|
:[[Archivo:ExE1-4-1-1-esp.png|thumb|left|150px|link=]]
|Un bloque de masa m cuelga de un hilo inextensible atado a un punto de la periferia de una polea de radio r. Entre soporte (fijo al suelo) y polea actúa un freno que garantiza que el bloque baja a velocidad constante respecto al suelo. Se trata de calcular la potencia del freno.

|}

:Càlculo directo: <math>\dot{\Ws}^{\mathrm{freno}}= \overline{\Gamma}^{\rightarrow \mathrm{polea}}\cdot\velang{polea}{T}=\left(\odot\Gamma\right)\cdot\left( \otimes \frac{\vs_0}{\rs}\right)=-\Gamma \frac{\vs_0}{\rs}.</math>

:El par de frenada no es un dato, y hay que recurrir al TMC para calcularlo.

:<math>\left.\begin{array}{l}
\text{SISTEMA: polea + bloque}\\
\text{TMC en } \Os
\end{array}\right\} \sum\overline{\Ms}_\mathrm{ext}(\Os)=\dot{\overline{\Hs}}_\mathrm{RTO}(\Os)</math>

:El momento cinético proviene de las dos piezas. Ya que '''O''' no es fijo al bloque, hay que hacer descomposición baricéntrica para calcular el momento cinético del bloque:

:<math>\overline{\Hs}_\mathrm{RTO=T} (\Os) =\overline{\Hs}_\Ts^\mathrm{politja}(\Os) + \overline{\Hs}_\Ts^\mathrm{bloc}(\Os)= \overline{\Hs}_\Ts^\mathrm{politja}(\Os) + \overline{\Hs}_\Ts^\mathrm{bloc}(\Gs_\mathrm{bloc})+ \overline{\Hs}_\Ts^{\oplus\mathrm{bloc}}(\Os) = </math>

:<math>\hspace{2.3cm}=\Is_\mathrm{O}^\mathrm{polea} \velang{polea}{T}+\Is_{\Gs_\mathrm{bloque}}^\mathrm{bloque}+\OGvec_\mathrm{bloque}\times \ms\overline{\vs}_\Ts(\Gs_\mathrm{bloque})=\left(\otimes \Is_\mathrm{O}\frac{\vs_0}{\rs}\right)+(\otimes \ms\vs_0\rs).</math>

:[[Archivo:ExE1-4-1-2-esp.png|thumb|right|180px|link=]]
:Se trata de un momento cinético constante en dirección y valor. Por tanto:

:<math>\dot{\overline{\Hs}}_\mathrm{RTO}(\Os)=0 \quad \Rightarrow \quad \sum{\overline{\Ms}}_\mathrm{ext}(\Os)=(\odot\Gamma)+(\otimes \ms\gs\rs) =0\quad \Rightarrow \quad \Gamma= \ms\gs\rs</math>.

:Finalmente: <math>\dot{\Ws}^{\mathrm{freno}}= -\Gamma \frac{\vs_0}{\rs} = -\ms\gs\vs_0.</math>

:Cálculo indirecto:

:<math>\left.\begin{array}{l}
\text{SISTEMA: polea + soporte + bloque}\\
\text{REF: suelo(Gal)}
\end{array}\right\} \quad \dot{\Ts}^\mathrm{sist}_\Ts=\dot{\Ws}^\mathrm{ext}_\Ts+ \dot{\Ws}^\mathrm{int}</math>

:<math>\left.\begin{array}{l}
\dot{\Ts}^\mathrm{sist}_\Ts=0\\
\dot{\Ws}^\mathrm{ext}_\Ts=\left(\downarrow \ms\gs\right)\cdot\left(\downarrow \vs_0\right)= \ms\gs\vs_0\\
\dot{\Ws}^\mathrm{int}=\dot{\Ws}^{\mathrm{freno}}
\end{array}\right\} \quad \Rightarrow \quad \dot{\Ws}^{\mathrm{freno}} =-\ms\gs\vs_0</math>



====✏️ EJEMPLO E1-4.2: mono y plátanos====
------
:
{|
|
:[[Archivo:ExE1-4-2-1-esp.png|thumb|left|180px|link=]]
|Unos plátanos de masa <math>(2/3)\ms</math> cuelgan de una cuerda inextensible que se apoya sin deslizar sobre una polea de radio r y masa despreciable, articulada respecto al suelo. En el otro extremo de la cuerda se agarra un mono de masa m. La polea gira con velocidad angular constante <math>\Omega_0</math> respecto al suelo bajo la acción de un motor. El mono se mueve respecto a la cuerda sin que sus manos se froten contra la cuerda, y de manera que la distancia de su centro de masas al suelo se mantiene constante. Se trata de calcular la potencia del motor.
|}

:Cálculo directo: <math>\dot{\Ws}^{\mathrm{freno}}= \overline{\Gamma}^{\rightarrow \mathrm{polea}}\cdot\velang{polea}{T}=\left(\odot\Gamma\right)\cdot\left( \odot \Omega_0\right)=\Gamma \Omega_0.</math>

:El par motor no es un dato, y hay que recurrir al TMC para calcularlo.

:<math>\left.\begin{array}{l}
\text{SISTEMA: polea + mono + plátanos}\\
\text{TMC en } \Os
\end{array}\right\} \sum\overline{\Ms}_\mathrm{ext}(\Os)=\dot{\overline{\Hs}}_\mathrm{RTO}(\Os)</math>

:El momento cinético proviene del mono y de los plátanos. En primera aproximación, se puede considerar la masa del mono concentrada en su tronco y su cabeza, de manera que no gira y su velocidad respecto al suelo es nula. Teniendo en cuenta que el punto '''O''' no se mueve respecto al mono pero sí respecto a los plátanos:

:<math>\overline{\Hs}_\mathrm{RTO=T} (\Os) =\overline{\Hs}_\Ts^\mathrm{mono}(\Os) + \overline{\Hs}_\Ts^\mathrm{plátanos}(\Os)= \overline{\Hs}_\Ts^\mathrm{mono}(\Os) + \overline{\Hs}_\Ts^\mathrm{plátanos}(\Gs_\mathrm{plát})+ \OGvec_\mathrm{plát} \times \frac{2}{3} \ms \overline{\vs}_\Ts(\Gs_\mathrm{plát})= </math>

:<math>\hspace{2.3cm}=\OGvec_\mathrm{plát}\times \frac{2}{3} \ms \overline{\vs}_\Ts(\Gs_\mathrm{plát}) = \Bigr[(\rightarrow \rs)+(\downarrow \xs)\Bigr] \times\frac{2}{3} \ms \left(\downarrow \rs\Omega_0\right)=\left(\otimes \frac{2}{3}\ms\rs^2\Omega_0\right).</math>

:Se trata de un momento cinético constante en dirección y valor. Por tanto:

:<math>\dot{\overline{\Hs}}_\mathrm{RTO}(\Os)=0 \quad \Rightarrow \quad \sum{\overline{\Ms}}_\mathrm{ext}(\Os)=(\odot\Gamma)+(\otimes \ms\gs\rs) + \left(\otimes \frac{2}{3}\ms\gs\rs\right)=0\quad \Rightarrow \quad \Gamma= \frac{5}{3}\ms\gs\rs</math>.

:Finalmente: <math>\dot{\Ws}^{\mathrm{motor}}= \Gamma \Omega_0 = \frac{5}{3}\ms\gs\rs\Omega_0.</math>

:Cálculo indirecto

:Hay que tener presente que es un problema con dos tipos de actuadores: el motor y la musculatura del mono. Hay diversos modelos biomecánicos para formular las fuerzas de los músculos, pero quedan fuera del ámbito de este curso. Ya que el mono se mueve, sus músculos desarrollan una potencia no nula.

:<math>\left.\begin{array}{l}
\text{SISTEMA: polea + mono + plátano}\\
\text{REF: suelo(Gal)}
\end{array}\right\} \quad \dot{\Ts}^\mathrm{sist}_\Ts=\dot{\Ws}^\mathrm{ext}_\Ts+ \dot{\Ws}^\mathrm{int}</math>

:<math>\left.\begin{array}{l}
\dot{\Ts}^\mathrm{sist}_\Ts=0\\
\dot{\Ws}^\mathrm{ext}_\Ts=\left(\downarrow \frac{2}{3}\ms\gs\right)\cdot\left(\downarrow \rs\Omega_0\right)= \frac{2}{3}\ms\gs\rs\Omega_0\\
\dot{\Ws}^\mathrm{int}=\dot{\Ws}^{\mathrm{musc}}+\dot{\Ws}^{\mathrm{motor}}
\end{array}\right\} \quad \Rightarrow \quad \dot{\Ws}^{\mathrm{musc}} + \dot{\Ws}^{\mathrm{motor}}=-\frac{2}{3}\ms\gs\rs\Omega_0</math>

:Este balance proporciona una ecuación con dos incógnitas. Se puede plantear un segundo balance para un sistema donde una de las dos no aparezca: un sistema sin el mono o un sistema sin el motor. Si se hace un balance de potencias para el sistema (mono + trozo de cuerda), las fuerzas externas a considerar son el peso y la tensión de la cuerda. Si se sigue trabajando en la referencia del suelo, esta tensión (desconocida) hará potencia no nula ya que está aplicada a un punto de velocidad ascendente.
:[[Archivo:ExE1-4-2-2-cat-esp.png|thumb|right|330px|link=]]

:Para evitar que aparezca un término asociado a la tensión de la cuerda, se puede trabajar en la referencia del trozo de cuerda que se ha considerado, que al tener velocidad ascendente constante es galileana. En esta referencia, el mono baja con velocidad constante:

:<math>\left.\begin{array}{l}
\text{REL: trozo de cuerda}\\
\text{AB: suelo}
\end{array}\right\}</math>

:<math>\overline{\vs}_\mathrm{REL}(\mathrm{mono})=\overline{\vs}_\mathrm{ar}(\mathrm{mono})=-\overline{\vs}_\mathrm{ar}(\mathrm{mono})=(\downarrow \rs\Omega_0)</math>

:<math>\left.\begin{array}{l}
\text{SISTEMA: mono+ trozo de cuerda}\\
\text{REF: trozo de cuerda(Gal)}
\end{array}\right\} \quad \dot{\Ts}^\mathrm{sist}_\mathrm{cuerda} =\dot{\Ws}^\mathrm{ext}_\mathrm{cuerda}+ \dot{\Ws}^\mathrm{int}</math>

:<math>\left.\begin{array}{l}
\dot{\Ts}^\mathrm{sist}_\mathrm{cuerda}=0\\
\dot{\Ws}^\mathrm{ext}_\mathrm{cuerda}=\left(\downarrow \ms\gs\right)\cdot\left(\downarrow \rs\Omega_0\right)= \ms\gs\rs\Omega_0\\
\dot{\Ws}^\mathrm{int}=\dot{\Ws}^{\mathrm{musc}}
\end{array}\right\} \quad \Rightarrow \quad \dot{\Ws}^{\mathrm{musc}} =-\ms\gs\rs\Omega_0</math>

:Combinando este resultado con el del balance anterior:

:<math>\dot{\Ws}^\mathrm{motor}=-\frac{2}{3}\ms\gs\rs\Omega_0-\dot{\Ws}^\mathrm{musc}=-\frac{5}{3}\ms\gs\rs\Omega_0</math>



© Universitat Politècnica de Catalunya. [[Mecánica:Derechos de autor |Todos los derechos reservados]]
-------------

<center>
[[D8. Conservaciones|<<< D8. Conservaciones]]

[[E2. Teorema de la energía: versión integrada|E2. Teorema de la energía: versión integrada>>>]]
</center>

E1. Work-Energy Theorem: differential form

2025-05-20T12:10:48Z

Bros:

<div class="noautonum">__TOC__</div>
<math>\newcommand{\uvec}{\overline{\textbf{u}}}
\newcommand{\vvec}{\overline{\textbf{v}}}
\newcommand{\evec}{\overline{\textbf{e}}}
\newcommand{\Omegavec}{\overline{\mathbf{\Omega}}}
\newcommand{\velang}[2]{\Omegavec^{\textrm{#1}}_{\textrm{#2}}}
\newcommand{\Alfavec}{\overline{\mathbf{\alpha}}}
\newcommand{\accang}[2]{\Alfavec^{\textrm{#1}}_{\textrm{#2}}}
\newcommand{\ds}{\textrm{d}}
\newcommand{\ps}{\textrm{p}}
\newcommand{\ns}{\textrm{n}}
\newcommand{\hs}{\textrm{h}}
\newcommand{\cs}{\textrm{c}}
\newcommand{\gs}{\textrm{g}}
\newcommand{\Ns}{\textrm{N}}
\newcommand{\Hs}{\textrm{H}}
\newcommand{\Fs}{\textrm{F}}
\newcommand{\ms}{\textrm{m}}
\newcommand{\ts}{\textrm{t}}
\newcommand{\us}{\textrm{u}}
\newcommand{\vs}{\textrm{v}}
\newcommand{\Rs}{\textrm{R}}
\newcommand{\Ws}{\textrm{W}}
\newcommand{\Ts}{\textrm{T}}
\newcommand{\Ls}{\textrm{L}}
\newcommand{\As}{\textrm{A}}
\newcommand{\Ds}{\textrm{D}}
\newcommand{\Bs}{\textrm{B}}
\newcommand{\es}{\textrm{e}}
\newcommand{\fs}{\textrm{f}}
\newcommand{\is}{\textrm{i}}
\newcommand{\Is}{\textrm{I}}
\newcommand{\ks}{\textrm{k}}
\newcommand{\js}{\textrm{j}}
\newcommand{\rs}{\textrm{r}}
\newcommand{\as}{\textrm{a}}
\newcommand{\ss}{\textrm{s}}
\newcommand{\Os}{\textbf{O}}
\newcommand{\Gs}{\textbf{G}}
\newcommand{\Cbf}{\textbf{C}}
\newcommand{\Or}{\Os_\Rs}
\newcommand{\Qs}{\textbf{Q}}
\newcommand{\Cs}{\textbf{C}}
\newcommand{\Ps}{\textbf{P}}
\newcommand{\Ss}{\textbf{S}}
\newcommand{\Js}{\textbf{J}}
\newcommand{\P}{\textrm{P}}
\newcommand{\Q}{\textrm{Q}}
\newcommand{\Ms}{\textrm{M}}
\newcommand{\deg}{^\textsf{o}}
\newcommand{\xs}{\textsf{x}}
\newcommand{\ys}{\textsf{y}}
\newcommand{\zs}{\textsf{z}}
\newcommand{\dert}[2]{\left.\frac{\ds{#1}}{\ds\ts}\right]_{\textrm{#2}}}
\newcommand{\ddert}[2]{\left.\frac{\ds^2{#1}}{\ds\ts^2}\right]_{\textrm{#2}}}
\newcommand{\vec}[1]{\overline{#1}}
\newcommand{\vecbf}[1]{\overline{\textbf{#1}}}
\newcommand{\vecdot}[1]{\overline{\dot{#1}}}
\newcommand{\OQvec}{\vec{\Os\Qs}}
\newcommand{\GQvec}{\vec{\Gs\Qs}}
\newcommand{\GPvec}{\vec{\Gs\Ps}}
\newcommand{\PSvec}{\vec{\Ps\Ss}}
\newcommand{\QQvec}{\vec{\Qs\Qs}}
\newcommand{\QGvec}{\vec{\Qs\Gs}}
\newcommand{\QPvec}{\vec{\Qs\Ps}}
\newcommand{\JQvec}{\vec{\Js\Qs}}
\newcommand{\PQvec}{\vec{\Ps\Qs}}
\newcommand{\GJvec}{\vec{\Gs\Js}}
\newcommand{\OPvec}{\vec{\Os\textbf{P}}}
\newcommand{\OCvec}{\vec{\Os\Cs}}
\newcommand{\OGvec}{\vec{\Os\Gs}}
\newcommand{\CJvec}{\vec{\Cs\Js}}
\newcommand{\GCvec}{\vec{\Gs\Cs}}
\newcommand{\PGvec}{\vec{\Ps\Gs}}
\newcommand{\abs}[1]{\left|{#1}\right|}
\newcommand{\braq}[2]{\left\{{#1}\right\}_{\textrm{#2}}}
\newcommand{\matriz}[9]{
\begin{bmatrix}
{#1} & {#2} & {#3}\\
{#4} & {#5} & {#6}\\
{#7} & {#8} & {#9}
\end{bmatrix}}
\newcommand{\diag}[3]{
\begin{bmatrix}
{#1} & {0} & {0}\\
{0} & {#2} & {0}\\
{0} & {0} & {#3}
\end{bmatrix}}
\newcommand{\vector}[3]{
\begin{Bmatrix}
{#1}\\
{#2}\\
{#3}
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{#1}\\
{#2}
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\newcommand{\vel}[2]{\vvec_{\textrm{#2}} (\textbf{#1})}
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\newcommand{\agal}[1]{\vecbf{a}_{\textrm{Gal}} (#1)}
\newcommand{\angal}[1]{\vecbf{a}_{\textrm{NGal}} (#1)}
\newcommand{\vgal}[1]{\vecbf{v}_{\textrm{Gal}} (#1)}
\newcommand{\fvec}[2]{\overline{\mathbf{F}}_{\textrm{#1}\rightarrow\textrm{#2}}}
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\newcommand{\F}[1]{\vec{\Fs}_{#1}}
\newcommand{\Fcal}[2]{\vec{\cal{F}}^{\text{#1}}_{#2}}
</math>

The '''Work-Energy Theorem''' plays a prominent role in physics, as it bridges all branches of science (electromagnetism, thermodynamics, etc.): all of them deal with energy.

There is a certain parallel between the Vector Theorems and the '''Work-Energy Theorem''': they all allow us to understand the evolution of a magnitude that depends on the mechanical state of the system. In the case of the former, we are dealing with the instantaneous evolution of vector magnitudes (linear momentum, angular momentum), while the latter deals with the finite change (in a finite time interval) of a scalar magnitude ('''mechanical energy''').

This unit presents the differential form of this theorem (the '''power balance'''). The integrated form (which is the Work-Energy Theorem itself) is presented in [[E2. Teorema de la energía: versión integrada|'''unit E2''']].
 
------
------
==E1.1 Power balance in a system of particles==

:[[File:E1-1-eng.png|thumb|right|300px|link=]]
<center>'''Figure E1.1''' Forces on a particle of a system with constant matter</center>

Let us consider a system of constant matter, and let’s apply Newton's second law to each particle (or each mass differential) of this system in a general reference frame R (which may be non-Galilean) ('''Figure E1.1'''):

<math>\F{\rightarrow \Ps} + \Fcal{}{\text{ar}\rightarrow\Ps} + \Fcal{}{\text{Cor}\rightarrow\Ps} =\ms_\Ps\acc{P}{R}</math>

<math>\F{\rightarrow \Ps} </math> is the resultant interaction force on '''P''', and it comes from internal interactions (with particles of the same system) and external ones:

<math>\F{\rightarrow \Ps}=\F{\rightarrow \Ps}^\text{int}+\F{\rightarrow \Ps}^\text{ext}</math>

If both sides of the equation are multiplied by <math>\vel{P}{R}</math>, all forces (interaction and inertia forces) orthogonal to <math>\vel{P}{R}</math> are filtered out. Specifically, since <math>\acc{P}{R}=\accs{P}{R} + \accn{P}{R} </math>, where <math>\accn{P}{R} \perp \vel{P}{R}</math>, and since the Coriolis force is always orthogonal to <math>\vel{P}{R} (\Fcal{}{\text{Cor}\rightarrow\Ps} = -2\ms_\Ps \velang{R}{Gal} \times \vel{P}{R})</math>:

<math>\vel{P}{R} \cdot \left( \F{\rightarrow \Ps}^\text{int}+\F{\rightarrow \Ps}^\text{ext} + \Fcal{}{\text{ar}\rightarrow\Ps} \right) = \ms_\Ps\accs{P}{R} \cdot \vel{P}{R} = \frac{\ds}{\ds \ts} \left(\frac{1}{2}\ms_\Ps \vvec_\Rs^2(\Ps) \right) \equiv \frac{\ds\Ts_\Rs(\Ps)}{\ds\ts}=\dot{\Ts}_\Rs(\Ps)</math>

The scalar function <math>\Ts_\Rs(\Ps)</math> is the '''kinetic energy''' of the particle, and is a function of the [[D1. Foundational laws of Newtonian dynamics#D1.3 Newton’s Principle of Determinacy|'''mechanical state''']] of '''P'''. The scalar product of a force and the velocity of its application point is the '''force power''': <math>\vel{P}{R} \cdot \F{\rightarrow \Ps} \equiv \dot{\Ws}_\Rs^{\F{\rightarrow \Ps}} </math>. Both scalar quantities <math>(\Ts_\Rs(\Ps), \dot{\Ws}_\Rs^{\F{\rightarrow \Ps}})</math> depend on the reference frame. For the same set of interaction forces acting on '''P''', the reference frame determines which are filtered out and which are not.

The equation can be rewritten as <math>\dot{\Ws}_\Rs^{\text{ext}\rightarrow \Ps}+\dot{\Ws}_\Rs^{\text{int}\rightarrow \Ps}+\dot{\Ws}_\Rs^{\text{ar}\rightarrow \Ps} =\dot{\Ts}_\Rs(\Ps)</math>, and constitutes the '''power balance''' for particle '''P''' in the R reference frame. Summing for all particles in the system:

<math>\dot{\Ws}_\Rs^{\text{ext}}+\dot{\Ws}^{\text{int}}+\dot{\Ws}_\Rs^{\text{ar}} =\dot{\Ts}_\Rs^\text{syst}</math>

The calculation of kinetic energy is covered in [[E2. Teorema de la energía: versión integrada|'''unit E2''']]. This unit focuses on the calculation of powers. [[E1. Work-Energy Theorem: differential form#E1.2 Potencia de una pareja de acción-reacción|'''section E1-2''']] shows that the power of internal interactions does not depend on the reference frame (which is why the subscript R does not appear in the <math>\dot{\Ws}^{\text{int}}</math> term of the previous equation).

====✏️ EXAMPLE E1-1.1: free particle====
------
:
{|
:[[File:ExE1-1-1-1-eng.jpg|thumb|left|180px|link=]]
|A particle '''P''', with mass m, is released from the ring of a Ferris wheel when it is at its highest point. We want to calculate the power, at this instant, of all forces acting on the particle from three different reference frames: the ground, the ring, and the Ferris wheel cabin.
|}

:Ground reference frame: Since this is a Galilean reference frame, all forces on '''P''' are interaction forces.
{|
:[[File:ExE1-1-2- eng.png|thumb|right|230px|link=]]

|<math>\hspace{0.5cm}\sum\dot{\Ws}_\Ts=\dot{\Ws}_\Ts^\text{weight}= \left(\leftarrow \Rs\Omega_0\right) \cdot \left(\downarrow \ms\gs\right)=0</math>
|}

:Ring reference frame: It is a non-Galilean reference, and the transportation force on '''P''' must be taken into account.
{|
:[[File:ExE1-1-3-eng.png|thumb|right|250px|link=]]
|<math>\hspace{0.5cm}</math> AB:ground, REL:ring 

<math>\hspace{0.5cm}\vel{P}{REL}=\vel{P}{AB}-\vel{P}{ar}= \left(\leftarrow \Rs\Omega_0\right)-\left(\leftarrow \Rs\Omega_0\right)=\overline{0}</math> 

<math>\hspace{0.5cm}</math>

Since the velocity is zero, the total power is zero.
|}

:Cabin reference frame: It is also a non-Galilean reference, and the transportation force on '''P''' must be taken into account.

{|
:[[File:ExE1-1-4-eng.png|thumb|right|250px|link=]]
|<math>\hspace{0.5cm}</math>AB:ground, REL:cabin

<math>\hspace{0.5cm}\vel{P}{REL}=\vel{P}{AB}-\vel{P}{ar}= \left(\leftarrow \Rs\Omega_0\right)-\left(\uparrow \Rs\Omega_0\right)</math>

<math>\hspace{0.5cm}\dot{\Ws}_\text{cabin}^\text{weight}=\left[ \left(\leftarrow \Rs\Omega_0\right)+\left(\downarrow \Rs\Omega_0\right) \right] \cdot \left(\downarrow\ms\gs\right) =\ms\gs\Rs\Omega_0</math>

<math>\hspace{0.5cm}\dot{\Ws}_\text{cabin}^\text{ar}=\left[ \left(\leftarrow \Rs\Omega_0\right)+\left(\downarrow \Rs\Omega_0\right) \right] \cdot \left(\rightarrow\ms\Rs^2\Omega_0^2\right) =-\ms\gs\Rs^2 \Omega_0^3</math>

<math>\hspace{0.5cm}\sum\dot{\Ws}_\text{cabin}=\ms\Rs\Omega_0(\gs-\Rs\Omega_0^2)</math>

|}


====✏️ EXAMPLE E1-1.2: pendulum====
------
:
{|
|
:[[File:ExE1-2-1-eng.png|thumb|left|180px|link=]]
|A particle '''P''' of mass m hangs from an inextensible thread of length L attached to the ceiling. The assembly oscillates in a vertical plane fixed to the ground. We want to calculate the power of all the forces acting on the particle in two different reference frames: the ground and a platform rotating relative to the ground with a constant angular velocity <math>\Omega_0</math>. 

Ground reference frame: Since it is a Galilean reference frame, '''P''' is only subjected to interaction forces.
{|
|
:[[File:ExE1-2-2-neut.png|thumb|left|300px|link=]]

|<math>\dot{\Ws}_\Es^\Ts=(\nearrow\Ls\dot{\theta})\cdot(\nwarrow\Ts)=0</math>

<math>\dot{\Ws}_\Es^\text{weight}=(\nearrow\Ls\dot{\theta}\cdot(\downarrow \ms\gs)=-\ms\gs\Ls\dot{\theta}\sin\theta</math>

|}

Platform reference frame: It is a non Galilean reference frame, and the transportation force on '''P''' has to be taken into account.

{|
|
:[[File:ExE1-2-3-eng.png|thumb|left|400px|link=]]

|AB:ground, REL:platform

<math>\vel{P}{REL}=\vel{P}{AB}-\vel{P}{ar}=(\nearrow\Ls\dot{\theta})-(\otimes \Ls \Omega_0 \sin \theta)</math>

<math>\dot{\Ws}_\Es^\Ts\left[(\nearrow\Ls\dot{\theta})+(\odot\Ls\Omega_0\sin\theta)\right]\cdot(\nwarrow\Ts)=0</math>

<math>\dot{\Ws}_\Es^\text{weight}\left[(\nearrow\Ls\dot{\theta})+(\odot\Ls\Omega_0\sin\theta)\right]\cdot(\downarrow \ms\gs)=-\ms\gs\Ls\dot{\theta}\sin\theta</math>

<math>\dot{\Ws}_\Es^\text{ar}\left[(\nearrow\Ls\dot{\theta})+(\odot\Ls\Omega_0\sin\theta)\right]\cdot(\rightarrow \ms\Ls\Omega_0^2\sin\theta)=\ms\Ls^2\Omega_0^2\dot{\theta}\cos\theta</math>
|}


------
------

==E1.2 Power of an action-reaction pair==

The total power associated with internal interactions in a constant-mass system is not zero in principle (since the two forces of each action-reaction pair are applied to particles that may have different velocities relative to the reference frame in which the calculation is made), and it is independent of the reference frame in which it is evaluated. It only depends on the separation velocity of the two particles. The power balance for the system is therefore written as: <math>\dot{\Ts}_\Rs^\text{syst}=\dot{\Ws}_\Rs^\text{ext} + \dot{\Ws}^\text{int} + \dot{\Ws}_\Rs^\text{ar} </math> .

<div>
=====💭 PROOF ➕=====

{|
|
:[[File:E1-2-neut.png|thumb|left|200px|link=]]
|<math>\dot{\Ws}^\text{ac-reac}=\overline{\Fs}_{\Ps \rightarrow \Qs}\cdot \vel{Q}{R} + \overline{\Fs}_{\Qs \rightarrow \Ps}\cdot \vel{P}{R}= \overline{\Fs}_{\Ps \rightarrow \Qs}\cdot \left[ \vel{Q}{R}-\vel{P}{R}\right]</math>

By the [[D1. Foundational laws of Newtonian dynamics#D1.6 Newton’s third law (action-reaction principle)|'''action-reaction principle''']], <math>\overline{\Fs}_{\Ps \rightarrow \Qs}=-\overline{\Fs}_{\Qs \rightarrow \Ps}</math>. Therefore: <math>\dot{\Ws}^\text{ac-reac}=\overline{\Fs}_{\Ps \rightarrow \Qs}\cdot \left[ \vel{Q}{R}-\vel{P}{R}\right]</math>

As <math>\overline{\Fs}_{\Ps \rightarrow \Qs}</math> has the <math>\PQvec</math> direction, only the components of the velocities in that direction have to be considered:

<math>\dot{\Ws}^\text{ac-reac}=\overline{\Fs}_{\Ps \rightarrow \Qs}\cdot \biggl( \vel{Q}{R} \Bigr]_{||\Ps\Qs}-\ \vel{P}{R} \Bigr]_{||\Ps\Qs}\biggl)</math>
|}

The difference <math>\biggl( \vel{Q}{R} \Bigr]_{\Ps\Qs}-\ \vel{P}{R} \Bigr]_{\Ps\Qs}\biggl)</math> is the separation velocity <math>\dot{\rho}</math> between '''P''' and '''Q''', and it does not dpend on the reference frame since it is the derivative of a scalar (the distance <math>\dot{\rho}</math> between the two particles). For the previous drawing, <math>\dot{\rho}> 0</math>. Since this is an attractive interaction, <math>\dot{\Ws}^\text{ac-reac} < 0</math>.

With a drawing, it is easy to see that:
* attractive interaction and particle approach <math>(\dot{\rho}<0)</math> : <math>\dot{\Ws}^\text{ac-reac} > 0</math> ;
* repulsive interaction and particle separation <math>(\dot{\rho}>0)</math> : <math>\dot{\Ws}^\text{ac-reac} > 0</math>
* repulsive interaction and particle approach <math>(\dot{\rho}<0)</math> : <math>\dot{\Ws}^\text{ac-reac} < 0</math>


</div>

There are two particular cases of interaction forces whose power always has the same sign:

* Friction action-reaction pair between '''P''' and '''Q''': since the friction force always opposes the relative velocity between the two points, its power is always negative:

<math>\hspace{1cm} \dot{\Ws}^\text{friction}=\dot{\Ws}^{\text{fric}\rightarrow\Ps}_\text{RTQ}+\dot{\Ws}^{\text{fric}\rightarrow\Qs}_\text{RTQ}=\overline{\Fs}^{\text{fric}\rightarrow\Ps} \cdot \vel{P}{RTQ} = - \abs{\overline{\Fs}^{\text{fric}\rightarrow\Ps}} \frac{\vel{P}{RTQ}}{\abs{\vel{P}{RTQ}}} \cdot \vel{P}{RTQ}<0;</math>

* Constraint action-reaction pair between '''P''' and '''Q''': from the [[D3. Interactions between rigid bodies#D3.4 Direct constraint interactions|'''analytical characterization equation''']] of the constraint forces between two points '''P''' and '''Q''' belonging to two rigid bodies <math>\mathrm{S}_\Ps</math> and <math>\mathrm{S}_\Qs</math>, it is easy to see that the associated power is zero:

<math>\hspace{1cm} \dot{\Ws}^\text{constraint}=\dot{\Ws}^{\text{constraint}\rightarrow\Ps}_{\mathrm{S}_\Ps}+\dot{\Ws}^{\text{constraint}\rightarrow\Qs}_{\mathrm{S}_\Ps}=\overline{\Fs}^{\text{constraint}\rightarrow\Ps} \cdot \overline{\mathbf{v}}_{\mathrm{S}_\Qs}(\Ps) = - 0;</math>

====✏️ EXAMPLE E1-2.1: power of an action-reaction friction pair====
------
:
{|
|
:[[File:ExE1-2-1-2-eng.png|thumb|left|180px|link=]]
|A homogeneous wheel, with radius r and mass m, moves down an inclined plane without sliding until it joins a conveyor belt moving at a constant velocity <math>\vs_0</math> relative to the ground. At the moment it makes contact with the belt, the center of the wheel has a velocity <math>\vs_0</math> relative to the ground in the same direction as the belt. We want to calculate the power of the friction force between the belt and the wheel at this instant.
|}

At that momento, the <math>\mathbf{ICR}^\text{wheel}_\Es</math> is point <math>\Js_\text{wheel}</math> in contact with the belt. This point slides backwards with respect to the belt, therefore the friction force on the wheel is forwards:

<math>\left.\begin{array}{l}
\text { AB: ground } \\
\text { REL: belt }
\end{array}\right\} \quad \bar{\vs}_{\text {REL }}\left(\Js_{\text {wheel }}\right)=\bar{\vs}_{\text{AB}}\left(\Js_{\text {wheel }}\right)-\bar{\vs}_{\mathrm{ar}}\left(\Js_{\text {wheel }}\right)=-\bar{\vs}_{\mathrm{ar}}\left(\Js_{\text {wheel }}\right)=-\left(\rightarrow \mathrm{v}_0\right)=\left(\leftarrow \mathrm{v}_0\right) .</math>

[[File:ExE1-2-1-3-eng.png|thumb|center|500px|link=]]

Calculation of the friction power: <math>\dot{\Ws}^\text{friction}=\dot{\Ws}_\Rs^{\text{friction}\rightarrow \text{wheel}}+ \dot{\Ws}_\Rs^{\text{friction}\rightarrow \text{belt}}</math>, where R may be any reference frame. To quickly calculate this power, it is advisable to choose a reference frame R where one of the two points subjected to the action-reaction pair has zero speed, so that one of the two terms is directly zero:

<math>\dot{\Ws}^\text{friction}=\dot{\Ws}_\text{belt}^{\text{fric}\rightarrow \text{wheel}}+ \dot{\Ws}_\text{belt}^{\text{fric }\rightarrow \text{belt}}=\overline{\Fs}^{\text{fric}\rightarrow \text{wheel}} \cdot \bar{\vs}_{\text {belt}}\left(\Js_{\text {wheel }}\right)=(\rightarrow \mu\ms\gs)\cdot(\leftarrow \vs_0)=-\mu\ms\gs\vs_0.</math>

If the calculation is done in any other reference frame, the result is the same:

<math>\dot{\Ws}^\text{friction}=\dot{\Ws}_\text{E}^{\text{fric}\rightarrow \text{wheel}}+ \dot{\Ws}_\text{E}^{\text{fric} \rightarrow \text{belt}}=(\rightarrow \mu\ms\gs)\cdot 0 + (\leftarrow \mu\ms\gs)\cdot(\rightarrow \vs_0)=-\mu\ms\gs\vs_0.</math>

<math>\dot{\Ws}^\text{friction}=\dot{\Ws}_\text{vehicle}^{\text{fric}\rightarrow \text{wheel}}+ \dot{\Ws}_\text{vehicle}^{\text{fric} \rightarrow \text{belt}}=(\rightarrow \mu\ms\gs)\cdot (\rightarrow\vs_0) + (\leftarrow \mu\ms\gs) \cdot(\rightarrow 3\vs_0)=-\mu\ms\gs\vs_0.</math>

However, if only the power of one of the two forces is calculated (for example, the friction on the wheel), the result depends on the reference frame and can have any sign:

<math>\dot{\Ws}_\text{belt}^{\text{fric }\rightarrow \text{wheel}}=(\rightarrow \mu\ms\gs)\cdot(\leftarrow \vs_0)=-\mu\ms\gs\vs_0 ,</math>

<math>\dot{\Ws}_\text{E}^{\text{fric} \rightarrow \text{wheel}}=(\rightarrow \mu\ms\gs)\cdot 0 = 0 ,</math>

<math>\dot{\Ws}_\text{vehicle}^{\text{fric} \rightarrow \text{wheel}}=(\rightarrow \mu\ms\gs)\cdot (\rightarrow\vs_0)=\mu\ms\gs\vs_0.</math>


-----
-----

==E1.3 Power of a system of forces on a rigid body==

{|
|
:[[File:E1-3-eng.png|thumb|left|180px|link=]]
<center>'''Figure E1.3''': Kinematic torsor of a rigid body in a reference frame R, and forces on different points (dm) of that rigid body</center>

|When calculating the power of a system of forces on a rigid body S, the term associated with internal interactions is zero because those interactions are all constraint (cohesion) forces and, as seen in [[E1. Work-Energy Theorem: differential form#E1.2 Power of an action-reaction pair|'''section E1.2''']], their total power is zero.

The total power of a system of forces on S can be calculated from the torsor of the system at any point '''O''' of the rigid body, <math>(\overline{\Fs},\overline{\Ms}(\Os))</math>. Taking into account rigid body kinematics ('''Figure E1.3'''):

<math>\dot{\Ws}_\Rs^{\rightarrow \Ss} =\sum_\Ps\overline{\Fs}^{\rightarrow \Ps} \cdot \vel{P}{R}= \sum_\Ps \left[ \overline{\Fs}^{\rightarrow \Ps} \cdot \vel{O}{R} + \velang{S}{R} \times \OPvec \right] </math>

|}
<math>\dot{\Ws}_\Rs^{\rightarrow \Ss} =\left(\sum_\Ps\overline{\Fs}^{\rightarrow \Ps} \right)\cdot \vel{O}{R} + \sum_\Ps\overline{\Fs}^{\rightarrow \Ps} \cdot \left(\velang{S}{R} \times \OPvec \right) = \left( \sum_\Ps\overline{\Fs}^{\rightarrow \Ps} \right) \cdot \vel{O}{R} + \left[ \sum_\Ps \left( \OPvec \times \overline{\Fs}^{\rightarrow \Ps} \right)\right] \cdot \velang{S}{R} = \overline{\Fs}\cdot \vel{O}{R} + \overline{\Ms}(\Os)\cdot \velang{S}{R} </math>

This calculation also shows that the power of a moment (or a torque <math>\Gamma</math>) on a rigid body in a reference frame R is the product of the moment by the angular velocity of the rigid body: <math>\dot{\Ws}_\Rs^{\Gamma\rightarrow \mathrm{S}} = \overline{\Gamma} \cdot \velang{S}{R}</math>.

The power balance for a rigid body is: <math>\dot{\Ws}_\Rs^{ \rightarrow \mathrm{S}} = \dot{\Ws}_\Rs^{ \mathrm{ext}} + \dot{\Ws}_\Rs^{ \mathrm{ar}}.</math>

====✏️ EXAMPLE E1-3.1: roller sliding on a slope====
------
:
{|
|
:[[File:ExE1-3-1-1-eng.png|thumb|left|180px|link=]]
|A homogeneous roller, with mass m and radius r, slides down a slope. The friction coefficient between the ground and the roller is <math>\mu</math>. We want to calculate the power of every force acting on the roller in the ground reference frame.
|}

:It is a 2D problem. The kinematic and dynamic descriptions are:
{|
|
[[File:ExE1-3-1-2-eng.png|thumb|right|250px|link=]]

|<math> \hspace{0.5cm}\vel{J}{E}=\vel{C}{T} + \velang{}{E} \times \CJvec= </math>

<math>\hspace{1.7cm}=\left( \rightarrow \vs_\Es \right) + \left( \otimes \Omega_\Es\right) \times \left( \downarrow \rs \right)</math>

<math> \hspace{0.5cm}\vel{J}{E} = \left[\rightarrow (\vs_\Es -\rs\Omega_\Es)\right]</math>

<math>\hspace{0.5cm} \dot{\Ws}_\Es^{\mathrm{ext}} = \dot{\Ws}_\Es^{\mathrm{weight}} + \dot{\Ws}_\Es^{\mathrm{friction}} + \dot{\Ws}_\Es^{\mathrm{constraint}}= \Bigr[\left(\downarrow \ms\gs\cos\beta\right)+\left(\rightarrow \ms\gs\sin\beta\right)\Bigr] \cdot (\rightarrow \vs_\Es) + \Bigr[(\leftarrow \mu \Ns) + (\uparrow \Ns ) \Bigr] \cdot \Bigr[\rightarrow (\vs_\Es -\rs\Omega_\Es)\Bigr]</math>

|}
:Since the roller center has no motion perpendicular to the plane, the AMT leads to <math>\Ns =\ms\gs\cos\beta</math>. Therefore:

{|
|
[[Archivo:ExE1-3-1-3-eng.png|thumb|right|140px|link=]]

|<math>\hspace{0.5cm} \dot{\Ws}_\Ts^{\mathrm{ext}} = \ms\gs\vs_\Ts \sin \beta - \ms\gs(\vs_\Ts -\rs\Omega_\Ts)\cos\beta=\ms\gs \Bigr[ \vs_\Ts (\sin\beta-\cos\beta)+\rs\Omega_\Ts \cos\beta \Bigr].</math>

<math>\hspace{0.5cm}</math>

If instead of considering the friction at '''J''' we consider the equivalent torsor at point '''C''', we must consider the force applied in '''C''' and the moment about '''C''', which is not zero:

<math>\hspace{0.5cm} \dot{\Ws}_\Es^{ \mathrm{friction}} = (\leftarrow \mu\Ns)\cdot(\rightarrow\vs_\Es)+(\otimes \mu\Ns\rs)\cdot(\otimes \Omega_\Es) =\mu\ms\gs\cos\beta(\rs\Omega_\Es-\vs_\Es)</math>,

<math>\hspace{0.5cm} \dot{\Ws}_\Es^{\mathrm{ext}} = \dot{\Ws}_\Es^{ \mathrm{weight}} + \dot{\Ws}_\Es^{ \mathrm{friction}}+ \dot{\Ws}_\Es^{ \mathrm{constraint}}= \ms\gs\vs_\Es\sin\beta+\mu\ms\gs\cos\beta (\rs\Omega_\Es-\vs_\Es)</math>.

<math>\hspace{0.5cm}</math>

The result is the same.
|}


====✏️ EXAMPLE E1-3.2: block sliding on a truck====
------
:
{|
|
:[[File:ExE1-3-2-1-eng.png|thumb|left|220px|link=]]
|A truck brakes with acceleration '''a''' relative to the ground at the instant in which its velocity (also relative to the ground) is v. At the same instant, the homogeneous block, of mass m, slides with velocity v' relative to the truck. We want to perform a power balance for the block in two different reference frames: the ground and the truck.
|}

:Ground reference frame: It is a 2D problem. The kinematic and dynamic descriptions are:
{|
|
[[File:ExE1-3-2-2-eng.png|thumb|right|200px|link=]]

| The constraint between the truck and the block allows only one translational DoF between the two. Therefore, in the 2D approximation, the associated torsor has two independent components. If characterized at the midpoint of the contact surface, these components are a normal force and a moment perpendicular to the plane. The AMT applied to the block leads to <math>\Ns=\ms\gs</math>.

|}
:The powers associated with all interactions on the block in the ground reference are:

:<math>\dot{\Ws}_\Es^{\rightarrow \mathrm{bloque}}=\dot{\Ws}_\Ts^{\mathrm{ext}}=\dot{\Ws}_\Ts^{\mathrm{peso}}+\dot{\Ws}_\Ts^{\mathrm{fricción}}+\dot{\Ws}_\Ts^{\mathrm{enlace}}= \Bigr[(\downarrow \ms\gs)+(\leftarrow \mu\ms\gs)+(\uparrow \Ns) \Bigr] \cdot \Bigr[\rightarrow(\vs+\vs')\Bigr] + (\odot \Ms_\es)\cdot 0 = -\mu\ms\gs(\vs+\vs')</math>

:Ya que el bloque se traslada respecto al suelo, su energía cinética total es inmediata, y se puede plantear fácilmente el balance de potencias:

:<math>\Ts_\Ts^\mathrm{bloque}=\frac{1}{2}\int_\mathrm{bloque}\ds\ms \vs_\Ts^2(\mathrm{bloque})=\frac{1}{2}\ms \vs_\Ts^2(\mathrm{bloc})\quad \Rightarrow \quad \dot{\Ts}_\Ts^\mathrm{bloque}=\ms\vs_\Ts( \mathrm{bloque})\dot{\vs}_\Ts( \mathrm{bloque}) </math>

:Como la única fuerza horizontal sobre el bloque es la fricción:

<math>\dot{\vs}_\Ts(\mathrm{bloque})=-\mu\gs\quad \Rightarrow \quad \dot{\Ts}_\Ts^\mathrm{bloque}=-\mu\ms\gs(\vs+\vs').</math>

:El balance de potencias es consistente:

:<math>\left.\begin{array}{l}
\dot{\mathrm{W}}_\Ts^{\rightarrow \text { bloque }}=\dot{\mathrm{W}}_{\mathrm{T}}^{\text {ext }}=-\mu \mathrm{mg}\left(\mathrm{v}+\mathrm{v}^{\prime}\right) \\
\dot{\mathrm{T}}_{\mathrm{T}}^{\text {bloque }}=-\mu \mathrm{mg}\left(\mathrm{v}+\mathrm{v}^{\prime}\right)
\end{array}\right\} \Rightarrow \dot{\mathrm{W}}_{\mathrm{T}}^{\rightarrow \text { bloque }}=\dot{\mathrm{T}}_{\mathrm{T}}^{\text {bloque }}</math>

:Referencia camión: Al estar el camión acelerado respecto al suelo, hay que tener en cuenta las fuerzas de arrastre sobre todos los diferenciales de masa (dm) del bloque:

{|
|

[[Archivo:ExE1-3-2-3-cat-esp.jpg|thumb|right|250px|link=]]
|<math>\cal{\Fs}^{\rightarrow \ds\ms}_{\mathrm{ar}}=-\ds\ms\overline{\as}_\mathrm{ar}(\ds\ms)=-\ds\ms (\leftarrow \as)=(\rightarrow \ds\ms \cdot \as)</math>.

El torsor resultante de este sistema de fuerzas en el centro de inercia '''G''' se reduce a una fuerza horizontal.

<math>\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{bloque}}=\dot{\Ws}_\mathrm{camión}^{\mathrm{ext}} + \dot{\Ws}_\mathrm{camión}^{\mathrm{ar}} =\dot{\Ws}_\mathrm{camión}^{\mathrm{peso}} + \dot{\Ws}_\mathrm{camión}^{\mathrm{fricción}} + \dot{\Ws}_\mathrm{camión}^{\mathrm{enlace}} + \dot{\Ws}_\mathrm{camión}^{\mathrm{ar}} , </math>

<math>\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{bloque}}= \Bigr[(\downarrow \ms\gs)+(\leftarrow \mu\ms\gs)+(\uparrow \Ns) + (\rightarrow \ms\as)\Bigr] \cdot(\rightarrow \vs')+(\odot\Ms_\es) \cdot0=\ms(\as-\mu\gs)\vs'.</math>
|}

:Se puede comprobar que el balance de potencias es consistente:

:<math>\Ts^\mathrm{bloque}_\mathrm{camión}=\frac{1}{2}\ms\vs^2_\mathrm{camión}(\mathrm{bloque}) \quad \Rightarrow \quad \dot{\Ts}^\mathrm{bloque}_\mathrm{camión} = \ms\vs_\mathrm{camión}(\mathrm{bloque})\dot{\vs}_\mathrm{camión}(\mathrm{bloque})</math>

:<math>\left.\begin{array}{l}
\mathrm{AB}:\mathrm{suelo} \\
\mathrm{REL}:\mathrm{camión}
\end{array}\right\} \quad \dot{\vs}_\mathrm{camión}(\mathrm{bloque})= \dot{\vs}_\mathrm{T}(\mathrm{bloque})-\dot{\vs}_\mathrm{ar}(\mathrm{bloque}) = (-\mu\gs)-(-\as)=\as-\mu\gs \quad \Rightarrow \quad \dot{\Ts}^\mathrm{bloque}_\mathrm{camión}=\ms\vs'(\as-\mu\gs)</math>

:<math>\left.\begin{array}{l}
\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{bloque}}=\dot{\Ws}_\mathrm{camión}^{\mathrm{ext}} + \dot{\Ws}_\mathrm{camión}^{\mathrm{ar}}= \ms(\as-\mu\gs)\vs'\\
\dot{\Ts}^\mathrm{bloque}_\mathrm{camión}=\ms\vs'(\as-\mu\gs)
\end{array}\right\} \Rightarrow\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{bloque}}= \dot{\Ts}^\mathrm{bloque}_\mathrm{camión}</math>


====✏️ EJEMPLO E1-3.3: camión frenando con carga====
------
:
{|
|
:[[Archivo:ExE1-3-3-1-esp.png|thumb|left|220px|link=]]
|Un camión frena con aceleración '''a''' respecto al suelo en el instante en el que su velocidad (también respecto al suelo) es v. Los pares de frenada en el eje delantero y en el trasero son iguales y valen <math>\Gamma</math>, y las ruedas, de radio r y masa despreciable, no deslizan sobre el suelo. En el mismo instante, el bloque homogéneo, de masa m, desliz con velocidad v' relativa al camión. Se trata de hacer un balance de potencias para el sistema (bloque+camión) en la referencia del suelo y en la del camión.

|}

:Ya que la potencia de las interacciones internas no es nula en principio (no se trata de un único sólido rígido), es conveniente hacer un DGI del sistema para no olvidar ninguna:
{|
|
[[Archivo:ExE1-3-3-2-esp.png|thumb|right|250px|link=]]

|
|
c.p.s.d.: contacto puntual sin deslizamiento

c.m.c.d.: contacto multipuntual con deslizamiento

|}
:El sistema sobre el que se hace el balance de potencias es todo menos el suelo. Teniendo en cuenta que la potencia de los enlaces internos es siempre nula, las únicas interacciones que hay que tener en cuenta son el peso, los enlaces de las ruedas con el suelo, los pares de frenada y la fricción entre bloque y camión:

[[Archivo:ExE1-3-3-3-esp.png|thumb|center|300px|link=]]

:Referencia suelo:
:Las fuerzas de enlace del suelo sobre las ruedas están aplicadas a puntos de velocidad cero, y por tanto no hacen potencia. El peso es una fuerza vertical, y el movimiento del camión y del bloque es horizontal, por tanto su potencia también es cero.

:<math>\dot{\Ws}_\Ts^{\rightarrow \mathrm{sist}}=\dot{\Ws}_\Ts^{\mathrm{ext}}+\dot{\Ws}^{\mathrm{int}}=\dot{\Ws}_\Ts^{\mathrm{peso}}+\dot{\Ws}_\Ts^{\mathrm{enlace}}+\dot{\Ws}^{\mathrm{fricción}}+\dot{\Ws}^{\Gamma}=\dot{\Ws}^{\mathrm{fricción}}+\dot{\Ws}^{\Gamma}.</math>

:Las potencias de interacciones internas se pueden calcular en cualquier referencia. Por tanto:

:<math>\dot{\Ws}^{\mathrm{fricción}}= \dot{\Ws}_\mathrm{camión}^{\mathrm{fric}\rightarrow \mathrm{bloque}} +\dot{\Ws}_\mathrm{camió}^{\mathrm{fric}\rightarrow \mathrm{camión}} = \dot{\Ws}_\mathrm{camión}^{\mathrm{fric}\rightarrow \mathrm{bloque}}= (\leftarrow \mu\ms\gs)\cdot (\rightarrow \vs')=-\mu\ms\gs\vs'</math>

:<math>\dot{\Ws}^{\Gamma}= \dot{\Ws}_\mathrm{camión}^{\Gamma\rightarrow \mathrm{ruedas}} +\dot{\Ws}_\mathrm{camión}^{\Gamma\rightarrow \mathrm{camión}} = \dot{\Ws}_\mathrm{camión}^{\Gamma\rightarrow \mathrm{ruedas}}= 2 \left(\odot \Gamma \right)\cdot \left(\otimes \frac{\vs}{\rs}\right)=-2\Gamma\frac{\vs}{\rs}.</math>

:Por otro lado: <math>\Ts_\Ts^\mathrm{sist}=\frac{1}{2}\ms\vs_\Ts^2(\mathrm{bloque})+\frac{1}{2}\ms\vs_\Ts^2(\mathrm{camión})\quad \Rightarrow \quad \dot{\Ts}_\Ts^\mathrm{sist}=-\ms(\vs+\vs')\mu\gs-\Ms\vs\as. </math>

:Finalmente: <math>\dot{\Ws}_\Ts^{\rightarrow \mathrm{sist}}=\dot{\Ts}_\Ts^\mathrm{sist} \quad \Rightarrow \quad -\mu\ms\gs\vs'-2\Gamma\frac{\vs}{\rs}=-\ms(\vs+\vs')\mu\gs-\Ms\vs\as.</math>

:Esta ecuación permite calcular el par de frenada: <math>\Gamma=\frac{1}{2}(\mu\ms\gs+\Ms\as)\rs.</math>

:Referencia camión:

:En esta referencia hay que tener en cuenta las fuerzas de arrastre. Por otro lado, las fuerzas de enlace del suelo sobre las ruedas están aplicadas a puntos que ya no tienen velocidad cero, y por tanto hacen potencia. Como antes, la potencia del peso es cero.

:<math>\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{sist}}=\dot{\Ws}_\mathrm{camión}^{\mathrm{ext}}+\dot{\Ws}^{\mathrm{int}}+\dot{\Ws}_\mathrm{camión}^{\mathrm{ar}}=\dot{\Ws}_\mathrm{camión}^{\mathrm{enlace}}+\dot{\Ws}^{\mathrm{fricción}}+\dot{\Ws}^{\Gamma}+\dot{\Ws}_\mathrm{camión}^{\mathrm{ar}}.</math>

:Las fuerzas de arrastre sobre el camión no hacen potencia porque el camión está parado en esta referencia. La de las fuerzas de arrastre sobre el bloque se ha calculado en el [[E1. Teorema de la energía: versión diferencial#✏️ ✏️ EJEMPLO E1-3.2: bloque deslizando sobre un camión|'''ejemplo E1-3.2''']]. La potencia de la fricción y de los pares de frenada es la misma de antes ya que no depende de la referencia.

:<math>\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{sist}}=\dot{\Ws}_\mathrm{camión}^{\mathrm{enllaç}}+\dot{\Ws}^{\mathrm{fricción}}+\dot{\Ws}^{\Gamma}+\dot{\Ws}_\mathrm{camión}^{\mathrm{ar} \rightarrow \mathrm{bloque}}= \dot{\Ws}_\mathrm{camión}^{\mathrm{enlace}} -\mu\ms\gs\vs'-2\Gamma\frac{\vs}{\rs} +\ms\as\vs'.</math>

:Las descripciones cinemática y dinámica de la rueda son:

[[Archivo:ExE1-3-3-4-esp.png|thumb|center|430px|link=]]

:<math>\dot{\Ws}_\mathrm{camión}^{\mathrm{enlace}}=2\Bigr[(\leftarrow \Ts)+(\uparrow \Ns)\Bigr]\cdot(\leftarrow\vs)=2\Ts\vs.</math>

La fuerza tangencial de enlace se puede calcular a partir del TMC aplicado al centro '''C''' de la rueda. Como no tiene masa: <math>\sum\overline{\Ms}_\mathrm{ext}(\Cs)=\overline{0} \quad \Rightarrow \quad (\odot \Gamma)+(\otimes \Ts\rs)=0\quad \Rightarrow \quad \Ts=\Gamma/\rs.</math>

:Por tanto: <math>\dot{\Ws}_\mathrm{camió}^{\rightarrow \mathrm{sist}} =2\frac{\Gamma}{\rs}\vs-\mu\ms\gs\vs'-2\Gamma\frac{\vs}{\rs}+\ms\as\vs'=\ms(\as-\mu\gs)\vs'.</math>

:Por otro lado, como se ha visto en el [[E1. Teorema de la energía: versión diferencial#✏️ ✏️ EJEMPLO E1-3.2: bloque deslizando sobre un camión|'''ejemplo E1-3.2''']]:

:<math>\Ts_\mathrm{camión}^\mathrm{sist}= \Ts_\mathrm{camión}^\mathrm{bloque}=\frac{1}{2}\ms\vs_\mathrm{camión}^2(\mathrm{bloque}) \quad \Rightarrow \quad \dot{\Ts}_\mathrm{camión}^\mathrm{sist}= \dot{\Ts}_\mathrm{camión}^\mathrm{bloque}=\ms\vs'(\as-\mu\gs).</math>

:El balance de potencias es consistente:

:<math>\left.\begin{array}{l}
\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{sist}}= \ms(\as-\mu\gs)\vs'\\
\dot{\Ts}^\mathrm{sist}_\mathrm{camión}=\ms(\as-\mu\gs)\vs'
\end{array}\right\} \Rightarrow\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{sist}}= \dot{\Ts}^\mathrm{sist}_\mathrm{camión}</math>

-----
-----

==E1.4 Balance de potencias para un sistema multisólido: cálculo directo y cálculo indirecto==

El '''cálculo directo''' (a partir de su definición) de la potencia asociada a una interacción (una fuerza sobre un punto '''P''' o un momento sobre un sólido rígido S) exige el conocimiento de la interacción (fuerza o momento):

<math>\dot{\Ws}_\Rs^{\Fs\rightarrow \Ps}=\overline{\Fs}_{\rightarrow\Ps} \cdot \vel{P}{R} \quad , \quad \dot{\Ws}_\Rs^{\Ms\rightarrow \Ss}=\overline{\Ms}_{\rightarrow\Ss} \cdot \velang{S}{R} </math>

Cuando no se trata de una interacción formulable, hay que recurrir a los teoremas vectoriales para determinarla.

El '''cálculo indirecto''' es una alternativa que no necesita recurrir a estos teoremas. Se basa en la evaluación de un balance de potencias para un sistema y una referencia adecuados de manera que la potencia a determinar sea el único término desconocido de la ecuación (el único que no se puede calcular de manera directa). Cuando el balance incluye más de un término desconocido, se pueden combinar varios balances (para distintos sistemas y referencias) hasta tener un número suficiente de ecuaciones que permitan la determinación de todas las potencias desconocidas.

====✏️ EJEMPLO E1-4.1: polea frenada====
------
:
{|
|
:[[Archivo:ExE1-4-1-1-esp.png|thumb|left|150px|link=]]
|Un bloque de masa m cuelga de un hilo inextensible atado a un punto de la periferia de una polea de radio r. Entre soporte (fijo al suelo) y polea actúa un freno que garantiza que el bloque baja a velocidad constante respecto al suelo. Se trata de calcular la potencia del freno.

|}

:Càlculo directo: <math>\dot{\Ws}^{\mathrm{freno}}= \overline{\Gamma}^{\rightarrow \mathrm{polea}}\cdot\velang{polea}{T}=\left(\odot\Gamma\right)\cdot\left( \otimes \frac{\vs_0}{\rs}\right)=-\Gamma \frac{\vs_0}{\rs}.</math>

:El par de frenada no es un dato, y hay que recurrir al TMC para calcularlo.

:<math>\left.\begin{array}{l}
\text{SISTEMA: polea + bloque}\\
\text{TMC en } \Os
\end{array}\right\} \sum\overline{\Ms}_\mathrm{ext}(\Os)=\dot{\overline{\Hs}}_\mathrm{RTO}(\Os)</math>

:El momento cinético proviene de las dos piezas. Ya que '''O''' no es fijo al bloque, hay que hacer descomposición baricéntrica para calcular el momento cinético del bloque:

:<math>\overline{\Hs}_\mathrm{RTO=T} (\Os) =\overline{\Hs}_\Ts^\mathrm{politja}(\Os) + \overline{\Hs}_\Ts^\mathrm{bloc}(\Os)= \overline{\Hs}_\Ts^\mathrm{politja}(\Os) + \overline{\Hs}_\Ts^\mathrm{bloc}(\Gs_\mathrm{bloc})+ \overline{\Hs}_\Ts^{\oplus\mathrm{bloc}}(\Os) = </math>

:<math>\hspace{2.3cm}=\Is_\mathrm{O}^\mathrm{polea} \velang{polea}{T}+\Is_{\Gs_\mathrm{bloque}}^\mathrm{bloque}+\OGvec_\mathrm{bloque}\times \ms\overline{\vs}_\Ts(\Gs_\mathrm{bloque})=\left(\otimes \Is_\mathrm{O}\frac{\vs_0}{\rs}\right)+(\otimes \ms\vs_0\rs).</math>

:[[Archivo:ExE1-4-1-2-esp.png|thumb|right|180px|link=]]
:Se trata de un momento cinético constante en dirección y valor. Por tanto:

:<math>\dot{\overline{\Hs}}_\mathrm{RTO}(\Os)=0 \quad \Rightarrow \quad \sum{\overline{\Ms}}_\mathrm{ext}(\Os)=(\odot\Gamma)+(\otimes \ms\gs\rs) =0\quad \Rightarrow \quad \Gamma= \ms\gs\rs</math>.

:Finalmente: <math>\dot{\Ws}^{\mathrm{freno}}= -\Gamma \frac{\vs_0}{\rs} = -\ms\gs\vs_0.</math>

:Cálculo indirecto:

:<math>\left.\begin{array}{l}
\text{SISTEMA: polea + soporte + bloque}\\
\text{REF: suelo(Gal)}
\end{array}\right\} \quad \dot{\Ts}^\mathrm{sist}_\Ts=\dot{\Ws}^\mathrm{ext}_\Ts+ \dot{\Ws}^\mathrm{int}</math>

:<math>\left.\begin{array}{l}
\dot{\Ts}^\mathrm{sist}_\Ts=0\\
\dot{\Ws}^\mathrm{ext}_\Ts=\left(\downarrow \ms\gs\right)\cdot\left(\downarrow \vs_0\right)= \ms\gs\vs_0\\
\dot{\Ws}^\mathrm{int}=\dot{\Ws}^{\mathrm{freno}}
\end{array}\right\} \quad \Rightarrow \quad \dot{\Ws}^{\mathrm{freno}} =-\ms\gs\vs_0</math>



====✏️ EJEMPLO E1-4.2: mono y plátanos====
------
:
{|
|
:[[Archivo:ExE1-4-2-1-esp.png|thumb|left|180px|link=]]
|Unos plátanos de masa <math>(2/3)\ms</math> cuelgan de una cuerda inextensible que se apoya sin deslizar sobre una polea de radio r y masa despreciable, articulada respecto al suelo. En el otro extremo de la cuerda se agarra un mono de masa m. La polea gira con velocidad angular constante <math>\Omega_0</math> respecto al suelo bajo la acción de un motor. El mono se mueve respecto a la cuerda sin que sus manos se froten contra la cuerda, y de manera que la distancia de su centro de masas al suelo se mantiene constante. Se trata de calcular la potencia del motor.
|}

:Cálculo directo: <math>\dot{\Ws}^{\mathrm{freno}}= \overline{\Gamma}^{\rightarrow \mathrm{polea}}\cdot\velang{polea}{T}=\left(\odot\Gamma\right)\cdot\left( \odot \Omega_0\right)=\Gamma \Omega_0.</math>

:El par motor no es un dato, y hay que recurrir al TMC para calcularlo.

:<math>\left.\begin{array}{l}
\text{SISTEMA: polea + mono + plátanos}\\
\text{TMC en } \Os
\end{array}\right\} \sum\overline{\Ms}_\mathrm{ext}(\Os)=\dot{\overline{\Hs}}_\mathrm{RTO}(\Os)</math>

:El momento cinético proviene del mono y de los plátanos. En primera aproximación, se puede considerar la masa del mono concentrada en su tronco y su cabeza, de manera que no gira y su velocidad respecto al suelo es nula. Teniendo en cuenta que el punto '''O''' no se mueve respecto al mono pero sí respecto a los plátanos:

:<math>\overline{\Hs}_\mathrm{RTO=T} (\Os) =\overline{\Hs}_\Ts^\mathrm{mono}(\Os) + \overline{\Hs}_\Ts^\mathrm{plátanos}(\Os)= \overline{\Hs}_\Ts^\mathrm{mono}(\Os) + \overline{\Hs}_\Ts^\mathrm{plátanos}(\Gs_\mathrm{plát})+ \OGvec_\mathrm{plát} \times \frac{2}{3} \ms \overline{\vs}_\Ts(\Gs_\mathrm{plát})= </math>

:<math>\hspace{2.3cm}=\OGvec_\mathrm{plát}\times \frac{2}{3} \ms \overline{\vs}_\Ts(\Gs_\mathrm{plát}) = \Bigr[(\rightarrow \rs)+(\downarrow \xs)\Bigr] \times\frac{2}{3} \ms \left(\downarrow \rs\Omega_0\right)=\left(\otimes \frac{2}{3}\ms\rs^2\Omega_0\right).</math>

:Se trata de un momento cinético constante en dirección y valor. Por tanto:

:<math>\dot{\overline{\Hs}}_\mathrm{RTO}(\Os)=0 \quad \Rightarrow \quad \sum{\overline{\Ms}}_\mathrm{ext}(\Os)=(\odot\Gamma)+(\otimes \ms\gs\rs) + \left(\otimes \frac{2}{3}\ms\gs\rs\right)=0\quad \Rightarrow \quad \Gamma= \frac{5}{3}\ms\gs\rs</math>.

:Finalmente: <math>\dot{\Ws}^{\mathrm{motor}}= \Gamma \Omega_0 = \frac{5}{3}\ms\gs\rs\Omega_0.</math>

:Cálculo indirecto

:Hay que tener presente que es un problema con dos tipos de actuadores: el motor y la musculatura del mono. Hay diversos modelos biomecánicos para formular las fuerzas de los músculos, pero quedan fuera del ámbito de este curso. Ya que el mono se mueve, sus músculos desarrollan una potencia no nula.

:<math>\left.\begin{array}{l}
\text{SISTEMA: polea + mono + plátano}\\
\text{REF: suelo(Gal)}
\end{array}\right\} \quad \dot{\Ts}^\mathrm{sist}_\Ts=\dot{\Ws}^\mathrm{ext}_\Ts+ \dot{\Ws}^\mathrm{int}</math>

:<math>\left.\begin{array}{l}
\dot{\Ts}^\mathrm{sist}_\Ts=0\\
\dot{\Ws}^\mathrm{ext}_\Ts=\left(\downarrow \frac{2}{3}\ms\gs\right)\cdot\left(\downarrow \rs\Omega_0\right)= \frac{2}{3}\ms\gs\rs\Omega_0\\
\dot{\Ws}^\mathrm{int}=\dot{\Ws}^{\mathrm{musc}}+\dot{\Ws}^{\mathrm{motor}}
\end{array}\right\} \quad \Rightarrow \quad \dot{\Ws}^{\mathrm{musc}} + \dot{\Ws}^{\mathrm{motor}}=-\frac{2}{3}\ms\gs\rs\Omega_0</math>

:Este balance proporciona una ecuación con dos incógnitas. Se puede plantear un segundo balance para un sistema donde una de las dos no aparezca: un sistema sin el mono o un sistema sin el motor. Si se hace un balance de potencias para el sistema (mono + trozo de cuerda), las fuerzas externas a considerar son el peso y la tensión de la cuerda. Si se sigue trabajando en la referencia del suelo, esta tensión (desconocida) hará potencia no nula ya que está aplicada a un punto de velocidad ascendente.
:[[Archivo:ExE1-4-2-2-cat-esp.png|thumb|right|330px|link=]]

:Para evitar que aparezca un término asociado a la tensión de la cuerda, se puede trabajar en la referencia del trozo de cuerda que se ha considerado, que al tener velocidad ascendente constante es galileana. En esta referencia, el mono baja con velocidad constante:

:<math>\left.\begin{array}{l}
\text{REL: trozo de cuerda}\\
\text{AB: suelo}
\end{array}\right\}</math>

:<math>\overline{\vs}_\mathrm{REL}(\mathrm{mono})=\overline{\vs}_\mathrm{ar}(\mathrm{mono})=-\overline{\vs}_\mathrm{ar}(\mathrm{mono})=(\downarrow \rs\Omega_0)</math>

:<math>\left.\begin{array}{l}
\text{SISTEMA: mono+ trozo de cuerda}\\
\text{REF: trozo de cuerda(Gal)}
\end{array}\right\} \quad \dot{\Ts}^\mathrm{sist}_\mathrm{cuerda} =\dot{\Ws}^\mathrm{ext}_\mathrm{cuerda}+ \dot{\Ws}^\mathrm{int}</math>

:<math>\left.\begin{array}{l}
\dot{\Ts}^\mathrm{sist}_\mathrm{cuerda}=0\\
\dot{\Ws}^\mathrm{ext}_\mathrm{cuerda}=\left(\downarrow \ms\gs\right)\cdot\left(\downarrow \rs\Omega_0\right)= \ms\gs\rs\Omega_0\\
\dot{\Ws}^\mathrm{int}=\dot{\Ws}^{\mathrm{musc}}
\end{array}\right\} \quad \Rightarrow \quad \dot{\Ws}^{\mathrm{musc}} =-\ms\gs\rs\Omega_0</math>

:Combinando este resultado con el del balance anterior:

:<math>\dot{\Ws}^\mathrm{motor}=-\frac{2}{3}\ms\gs\rs\Omega_0-\dot{\Ws}^\mathrm{musc}=-\frac{5}{3}\ms\gs\rs\Omega_0</math>



© Universitat Politècnica de Catalunya. [[Mecánica:Derechos de autor |Todos los derechos reservados]]
-------------

<center>
[[D8. Conservaciones|<<< D8. Conservaciones]]

[[E2. Teorema de la energía: versión integrada|E2. Teorema de la energía: versión integrada>>>]]
</center>

E1. Work-Energy Theorem: differential form

2025-05-20T12:10:02Z

Bros:

<div class="noautonum">__TOC__</div>
<math>\newcommand{\uvec}{\overline{\textbf{u}}}
\newcommand{\vvec}{\overline{\textbf{v}}}
\newcommand{\evec}{\overline{\textbf{e}}}
\newcommand{\Omegavec}{\overline{\mathbf{\Omega}}}
\newcommand{\velang}[2]{\Omegavec^{\textrm{#1}}_{\textrm{#2}}}
\newcommand{\Alfavec}{\overline{\mathbf{\alpha}}}
\newcommand{\accang}[2]{\Alfavec^{\textrm{#1}}_{\textrm{#2}}}
\newcommand{\ds}{\textrm{d}}
\newcommand{\ps}{\textrm{p}}
\newcommand{\ns}{\textrm{n}}
\newcommand{\hs}{\textrm{h}}
\newcommand{\cs}{\textrm{c}}
\newcommand{\gs}{\textrm{g}}
\newcommand{\Ns}{\textrm{N}}
\newcommand{\Hs}{\textrm{H}}
\newcommand{\Fs}{\textrm{F}}
\newcommand{\ms}{\textrm{m}}
\newcommand{\ts}{\textrm{t}}
\newcommand{\us}{\textrm{u}}
\newcommand{\vs}{\textrm{v}}
\newcommand{\Rs}{\textrm{R}}
\newcommand{\Ws}{\textrm{W}}
\newcommand{\Ts}{\textrm{T}}
\newcommand{\Ls}{\textrm{L}}
\newcommand{\As}{\textrm{A}}
\newcommand{\Ds}{\textrm{D}}
\newcommand{\Bs}{\textrm{B}}
\newcommand{\es}{\textrm{e}}
\newcommand{\fs}{\textrm{f}}
\newcommand{\is}{\textrm{i}}
\newcommand{\Is}{\textrm{I}}
\newcommand{\ks}{\textrm{k}}
\newcommand{\js}{\textrm{j}}
\newcommand{\rs}{\textrm{r}}
\newcommand{\as}{\textrm{a}}
\newcommand{\ss}{\textrm{s}}
\newcommand{\Os}{\textbf{O}}
\newcommand{\Gs}{\textbf{G}}
\newcommand{\Cbf}{\textbf{C}}
\newcommand{\Or}{\Os_\Rs}
\newcommand{\Qs}{\textbf{Q}}
\newcommand{\Cs}{\textbf{C}}
\newcommand{\Ps}{\textbf{P}}
\newcommand{\Ss}{\textbf{S}}
\newcommand{\Js}{\textbf{J}}
\newcommand{\P}{\textrm{P}}
\newcommand{\Q}{\textrm{Q}}
\newcommand{\Ms}{\textrm{M}}
\newcommand{\deg}{^\textsf{o}}
\newcommand{\xs}{\textsf{x}}
\newcommand{\ys}{\textsf{y}}
\newcommand{\zs}{\textsf{z}}
\newcommand{\dert}[2]{\left.\frac{\ds{#1}}{\ds\ts}\right]_{\textrm{#2}}}
\newcommand{\ddert}[2]{\left.\frac{\ds^2{#1}}{\ds\ts^2}\right]_{\textrm{#2}}}
\newcommand{\vec}[1]{\overline{#1}}
\newcommand{\vecbf}[1]{\overline{\textbf{#1}}}
\newcommand{\vecdot}[1]{\overline{\dot{#1}}}
\newcommand{\OQvec}{\vec{\Os\Qs}}
\newcommand{\GQvec}{\vec{\Gs\Qs}}
\newcommand{\GPvec}{\vec{\Gs\Ps}}
\newcommand{\PSvec}{\vec{\Ps\Ss}}
\newcommand{\QQvec}{\vec{\Qs\Qs}}
\newcommand{\QGvec}{\vec{\Qs\Gs}}
\newcommand{\QPvec}{\vec{\Qs\Ps}}
\newcommand{\JQvec}{\vec{\Js\Qs}}
\newcommand{\PQvec}{\vec{\Ps\Qs}}
\newcommand{\GJvec}{\vec{\Gs\Js}}
\newcommand{\OPvec}{\vec{\Os\textbf{P}}}
\newcommand{\OCvec}{\vec{\Os\Cs}}
\newcommand{\OGvec}{\vec{\Os\Gs}}
\newcommand{\CJvec}{\vec{\Cs\Js}}
\newcommand{\GCvec}{\vec{\Gs\Cs}}
\newcommand{\PGvec}{\vec{\Ps\Gs}}
\newcommand{\abs}[1]{\left|{#1}\right|}
\newcommand{\braq}[2]{\left\{{#1}\right\}_{\textrm{#2}}}
\newcommand{\matriz}[9]{
\begin{bmatrix}
{#1} & {#2} & {#3}\\
{#4} & {#5} & {#6}\\
{#7} & {#8} & {#9}
\end{bmatrix}}
\newcommand{\diag}[3]{
\begin{bmatrix}
{#1} & {0} & {0}\\
{0} & {#2} & {0}\\
{0} & {0} & {#3}
\end{bmatrix}}
\newcommand{\vector}[3]{
\begin{Bmatrix}
{#1}\\
{#2}\\
{#3}
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\newcommand{\vecdosd}[2]{
\begin{Bmatrix}
{#1}\\
{#2}
\end{Bmatrix}}
\newcommand{\vel}[2]{\vvec_{\textrm{#2}} (\textbf{#1})}
\newcommand{\acc}[2]{\vecbf{a}_{\textrm{#2}} (\textbf{#1})}
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\newcommand{\psio}{\dot{\psi}_0}
\newcommand{\Pll}{\textbf{P}_\textrm{lliure}}
\newcommand{\agal}[1]{\vecbf{a}_{\textrm{Gal}} (#1)}
\newcommand{\angal}[1]{\vecbf{a}_{\textrm{NGal}} (#1)}
\newcommand{\vgal}[1]{\vecbf{v}_{\textrm{Gal}} (#1)}
\newcommand{\fvec}[2]{\overline{\mathbf{F}}_{\textrm{#1}\rightarrow\textrm{#2}}}
\newcommand{\mvec}[2]{\overline{\mathbf{M}}_{\textrm{#1}\rightarrow\textrm{#2}}}
\newcommand{\F}[1]{\vec{\Fs}_{#1}}
\newcommand{\Fcal}[2]{\vec{\cal{F}}^{\text{#1}}_{#2}}
</math>

The '''Work-Energy Theorem''' plays a prominent role in physics, as it bridges all branches of science (electromagnetism, thermodynamics, etc.): all of them deal with energy.

There is a certain parallel between the Vector Theorems and the '''Work-Energy Theorem''': they all allow us to understand the evolution of a magnitude that depends on the mechanical state of the system. In the case of the former, we are dealing with the instantaneous evolution of vector magnitudes (linear momentum, angular momentum), while the latter deals with the finite change (in a finite time interval) of a scalar magnitude ('''mechanical energy''').

This unit presents the differential form of this theorem (the '''power balance'''). The integrated form (which is the Work-Energy Theorem itself) is presented in [[E2. Teorema de la energía: versión integrada|'''unit E2''']].
 
------
------
==E1.1 Power balance in a system of particles==

:[[File:E1-1-eng.png|thumb|right|300px|link=]]
<center>'''Figure E1.1''' Forces on a particle of a system with constant matter</center>

Let us consider a system of constant matter, and let’s apply Newton's second law to each particle (or each mass differential) of this system in a general reference frame R (which may be non-Galilean) ('''Figure E1.1'''):

<math>\F{\rightarrow \Ps} + \Fcal{}{\text{ar}\rightarrow\Ps} + \Fcal{}{\text{Cor}\rightarrow\Ps} =\ms_\Ps\acc{P}{R}</math>

<math>\F{\rightarrow \Ps} </math> is the resultant interaction force on '''P''', and it comes from internal interactions (with particles of the same system) and external ones:

<math>\F{\rightarrow \Ps}=\F{\rightarrow \Ps}^\text{int}+\F{\rightarrow \Ps}^\text{ext}</math>

If both sides of the equation are multiplied by <math>\vel{P}{R}</math>, all forces (interaction and inertia forces) orthogonal to <math>\vel{P}{R}</math> are filtered out. Specifically, since <math>\acc{P}{R}=\accs{P}{R} + \accn{P}{R} </math>, where <math>\accn{P}{R} \perp \vel{P}{R}</math>, and since the Coriolis force is always orthogonal to <math>\vel{P}{R} (\Fcal{}{\text{Cor}\rightarrow\Ps} = -2\ms_\Ps \velang{R}{Gal} \times \vel{P}{R})</math>:

<math>\vel{P}{R} \cdot \left( \F{\rightarrow \Ps}^\text{int}+\F{\rightarrow \Ps}^\text{ext} + \Fcal{}{\text{ar}\rightarrow\Ps} \right) = \ms_\Ps\accs{P}{R} \cdot \vel{P}{R} = \frac{\ds}{\ds \ts} \left(\frac{1}{2}\ms_\Ps \vvec_\Rs^2(\Ps) \right) \equiv \frac{\ds\Ts_\Rs(\Ps)}{\ds\ts}=\dot{\Ts}_\Rs(\Ps)</math>

The scalar function <math>\Ts_\Rs(\Ps)</math> is the '''kinetic energy''' of the particle, and is a function of the [[D1. Foundational laws of Newtonian dynamics#D1.3 Newton’s Principle of Determinacy|'''mechanical state''']] of '''P'''. The scalar product of a force and the velocity of its application point is the '''force power''': <math>\vel{P}{R} \cdot \F{\rightarrow \Ps} \equiv \dot{\Ws}_\Rs^{\F{\rightarrow \Ps}} </math>. Both scalar quantities <math>(\Ts_\Rs(\Ps), \dot{\Ws}_\Rs^{\F{\rightarrow \Ps}})</math> depend on the reference frame. For the same set of interaction forces acting on '''P''', the reference frame determines which are filtered out and which are not.

The equation can be rewritten as <math>\dot{\Ws}_\Rs^{\text{ext}\rightarrow \Ps}+\dot{\Ws}_\Rs^{\text{int}\rightarrow \Ps}+\dot{\Ws}_\Rs^{\text{ar}\rightarrow \Ps} =\dot{\Ts}_\Rs(\Ps)</math>, and constitutes the '''power balance''' for particle '''P''' in the R reference frame. Summing for all particles in the system:

<math>\dot{\Ws}_\Rs^{\text{ext}}+\dot{\Ws}^{\text{int}}+\dot{\Ws}_\Rs^{\text{ar}} =\dot{\Ts}_\Rs^\text{syst}</math>

The calculation of kinetic energy is covered in [[E2. Teorema de la energía: versión integrada|'''unit E2''']]. This unit focuses on the calculation of powers. [[E1. Work-Energy Theorem: differential form#E1.2 Potencia de una pareja de acción-reacción|'''section E1-2''']] shows that the power of internal interactions does not depend on the reference frame (which is why the subscript R does not appear in the <math>\dot{\Ws}^{\text{int}}</math> term of the previous equation).

====✏️ EXAMPLE E1-1.1: free particle====
------
:
{|
:[[File:ExE1-1-1-1-eng.jpg|thumb|left|180px|link=]]
|A particle '''P''', with mass m, is released from the ring of a Ferris wheel when it is at its highest point. We want to calculate the power, at this instant, of all forces acting on the particle from three different reference frames: the ground, the ring, and the Ferris wheel cabin.
|}

:Ground reference frame: Since this is a Galilean reference frame, all forces on '''P''' are interaction forces.
{|
|
:[[File:ExE1-1-2- eng.png|thumb|right|230px|link=]]

|<math>\hspace{0.5cm}\sum\dot{\Ws}_\Ts=\dot{\Ws}_\Ts^\text{weight}= \left(\leftarrow \Rs\Omega_0\right) \cdot \left(\downarrow \ms\gs\right)=0</math>
|}

:Ring reference frame: It is a non-Galilean reference, and the transportation force on '''P''' must be taken into account.
{|
|
:[[File:ExE1-1-3-eng.png|thumb|right|250px|link=]]
|<math>\hspace{0.5cm}</math> AB:ground, REL:ring 

<math>\hspace{0.5cm}\vel{P}{REL}=\vel{P}{AB}-\vel{P}{ar}= \left(\leftarrow \Rs\Omega_0\right)-\left(\leftarrow \Rs\Omega_0\right)=\overline{0}</math> 

<math>\hspace{0.5cm}</math>

Since the velocity is zero, the total power is zero.
|}

:Cabin reference frame: It is also a non-Galilean reference, and the transportation force on '''P''' must be taken into account.

{|
|
:[[File:ExE1-1-4-eng.png|thumb|right|250px|link=]]
|<math>\hspace{0.5cm}</math>AB:ground, REL:cabin

<math>\hspace{0.5cm}\vel{P}{REL}=\vel{P}{AB}-\vel{P}{ar}= \left(\leftarrow \Rs\Omega_0\right)-\left(\uparrow \Rs\Omega_0\right)</math>

<math>\hspace{0.5cm}\dot{\Ws}_\text{cabin}^\text{weight}=\left[ \left(\leftarrow \Rs\Omega_0\right)+\left(\downarrow \Rs\Omega_0\right) \right] \cdot \left(\downarrow\ms\gs\right) =\ms\gs\Rs\Omega_0</math>

<math>\hspace{0.5cm}\dot{\Ws}_\text{cabin}^\text{ar}=\left[ \left(\leftarrow \Rs\Omega_0\right)+\left(\downarrow \Rs\Omega_0\right) \right] \cdot \left(\rightarrow\ms\Rs^2\Omega_0^2\right) =-\ms\gs\Rs^2 \Omega_0^3</math>

<math>\hspace{0.5cm}\sum\dot{\Ws}_\text{cabin}=\ms\Rs\Omega_0(\gs-\Rs\Omega_0^2)</math>

|}


====✏️ EXAMPLE E1-1.2: pendulum====
------
:
{|
|
:[[File:ExE1-2-1-eng.png|thumb|left|180px|link=]]
|A particle '''P''' of mass m hangs from an inextensible thread of length L attached to the ceiling. The assembly oscillates in a vertical plane fixed to the ground. We want to calculate the power of all the forces acting on the particle in two different reference frames: the ground and a platform rotating relative to the ground with a constant angular velocity <math>\Omega_0</math>. 

Ground reference frame: Since it is a Galilean reference frame, '''P''' is only subjected to interaction forces.
{|
|
:[[File:ExE1-2-2-neut.png|thumb|left|300px|link=]]

|<math>\dot{\Ws}_\Es^\Ts=(\nearrow\Ls\dot{\theta})\cdot(\nwarrow\Ts)=0</math>

<math>\dot{\Ws}_\Es^\text{weight}=(\nearrow\Ls\dot{\theta}\cdot(\downarrow \ms\gs)=-\ms\gs\Ls\dot{\theta}\sin\theta</math>

|}

Platform reference frame: It is a non Galilean reference frame, and the transportation force on '''P''' has to be taken into account.

{|
|
:[[File:ExE1-2-3-eng.png|thumb|left|400px|link=]]

|AB:ground, REL:platform

<math>\vel{P}{REL}=\vel{P}{AB}-\vel{P}{ar}=(\nearrow\Ls\dot{\theta})-(\otimes \Ls \Omega_0 \sin \theta)</math>

<math>\dot{\Ws}_\Es^\Ts\left[(\nearrow\Ls\dot{\theta})+(\odot\Ls\Omega_0\sin\theta)\right]\cdot(\nwarrow\Ts)=0</math>

<math>\dot{\Ws}_\Es^\text{weight}\left[(\nearrow\Ls\dot{\theta})+(\odot\Ls\Omega_0\sin\theta)\right]\cdot(\downarrow \ms\gs)=-\ms\gs\Ls\dot{\theta}\sin\theta</math>

<math>\dot{\Ws}_\Es^\text{ar}\left[(\nearrow\Ls\dot{\theta})+(\odot\Ls\Omega_0\sin\theta)\right]\cdot(\rightarrow \ms\Ls\Omega_0^2\sin\theta)=\ms\Ls^2\Omega_0^2\dot{\theta}\cos\theta</math>
|}


------
------

==E1.2 Power of an action-reaction pair==

The total power associated with internal interactions in a constant-mass system is not zero in principle (since the two forces of each action-reaction pair are applied to particles that may have different velocities relative to the reference frame in which the calculation is made), and it is independent of the reference frame in which it is evaluated. It only depends on the separation velocity of the two particles. The power balance for the system is therefore written as: <math>\dot{\Ts}_\Rs^\text{syst}=\dot{\Ws}_\Rs^\text{ext} + \dot{\Ws}^\text{int} + \dot{\Ws}_\Rs^\text{ar} </math> .

<div>
=====💭 PROOF ➕=====

{|
|
:[[File:E1-2-neut.png|thumb|left|200px|link=]]
|<math>\dot{\Ws}^\text{ac-reac}=\overline{\Fs}_{\Ps \rightarrow \Qs}\cdot \vel{Q}{R} + \overline{\Fs}_{\Qs \rightarrow \Ps}\cdot \vel{P}{R}= \overline{\Fs}_{\Ps \rightarrow \Qs}\cdot \left[ \vel{Q}{R}-\vel{P}{R}\right]</math>

By the [[D1. Foundational laws of Newtonian dynamics#D1.6 Newton’s third law (action-reaction principle)|'''action-reaction principle''']], <math>\overline{\Fs}_{\Ps \rightarrow \Qs}=-\overline{\Fs}_{\Qs \rightarrow \Ps}</math>. Therefore: <math>\dot{\Ws}^\text{ac-reac}=\overline{\Fs}_{\Ps \rightarrow \Qs}\cdot \left[ \vel{Q}{R}-\vel{P}{R}\right]</math>

As <math>\overline{\Fs}_{\Ps \rightarrow \Qs}</math> has the <math>\PQvec</math> direction, only the components of the velocities in that direction have to be considered:

<math>\dot{\Ws}^\text{ac-reac}=\overline{\Fs}_{\Ps \rightarrow \Qs}\cdot \biggl( \vel{Q}{R} \Bigr]_{||\Ps\Qs}-\ \vel{P}{R} \Bigr]_{||\Ps\Qs}\biggl)</math>
|}

The difference <math>\biggl( \vel{Q}{R} \Bigr]_{\Ps\Qs}-\ \vel{P}{R} \Bigr]_{\Ps\Qs}\biggl)</math> is the separation velocity <math>\dot{\rho}</math> between '''P''' and '''Q''', and it does not dpend on the reference frame since it is the derivative of a scalar (the distance <math>\dot{\rho}</math> between the two particles). For the previous drawing, <math>\dot{\rho}> 0</math>. Since this is an attractive interaction, <math>\dot{\Ws}^\text{ac-reac} < 0</math>.

With a drawing, it is easy to see that:
* attractive interaction and particle approach <math>(\dot{\rho}<0)</math> : <math>\dot{\Ws}^\text{ac-reac} > 0</math> ;
* repulsive interaction and particle separation <math>(\dot{\rho}>0)</math> : <math>\dot{\Ws}^\text{ac-reac} > 0</math>
* repulsive interaction and particle approach <math>(\dot{\rho}<0)</math> : <math>\dot{\Ws}^\text{ac-reac} < 0</math>


</div>

There are two particular cases of interaction forces whose power always has the same sign:

* Friction action-reaction pair between '''P''' and '''Q''': since the friction force always opposes the relative velocity between the two points, its power is always negative:

<math>\hspace{1cm} \dot{\Ws}^\text{friction}=\dot{\Ws}^{\text{fric}\rightarrow\Ps}_\text{RTQ}+\dot{\Ws}^{\text{fric}\rightarrow\Qs}_\text{RTQ}=\overline{\Fs}^{\text{fric}\rightarrow\Ps} \cdot \vel{P}{RTQ} = - \abs{\overline{\Fs}^{\text{fric}\rightarrow\Ps}} \frac{\vel{P}{RTQ}}{\abs{\vel{P}{RTQ}}} \cdot \vel{P}{RTQ}<0;</math>

* Constraint action-reaction pair between '''P''' and '''Q''': from the [[D3. Interactions between rigid bodies#D3.4 Direct constraint interactions|'''analytical characterization equation''']] of the constraint forces between two points '''P''' and '''Q''' belonging to two rigid bodies <math>\mathrm{S}_\Ps</math> and <math>\mathrm{S}_\Qs</math>, it is easy to see that the associated power is zero:

<math>\hspace{1cm} \dot{\Ws}^\text{constraint}=\dot{\Ws}^{\text{constraint}\rightarrow\Ps}_{\mathrm{S}_\Ps}+\dot{\Ws}^{\text{constraint}\rightarrow\Qs}_{\mathrm{S}_\Ps}=\overline{\Fs}^{\text{constraint}\rightarrow\Ps} \cdot \overline{\mathbf{v}}_{\mathrm{S}_\Qs}(\Ps) = - 0;</math>

====✏️ EXAMPLE E1-2.1: power of an action-reaction friction pair====
------
:
{|
|
:[[File:ExE1-2-1-2-eng.png|thumb|left|180px|link=]]
|A homogeneous wheel, with radius r and mass m, moves down an inclined plane without sliding until it joins a conveyor belt moving at a constant velocity <math>\vs_0</math> relative to the ground. At the moment it makes contact with the belt, the center of the wheel has a velocity <math>\vs_0</math> relative to the ground in the same direction as the belt. We want to calculate the power of the friction force between the belt and the wheel at this instant.
|}

At that momento, the <math>\mathbf{ICR}^\text{wheel}_\Es</math> is point <math>\Js_\text{wheel}</math> in contact with the belt. This point slides backwards with respect to the belt, therefore the friction force on the wheel is forwards:

<math>\left.\begin{array}{l}
\text { AB: ground } \\
\text { REL: belt }
\end{array}\right\} \quad \bar{\vs}_{\text {REL }}\left(\Js_{\text {wheel }}\right)=\bar{\vs}_{\text{AB}}\left(\Js_{\text {wheel }}\right)-\bar{\vs}_{\mathrm{ar}}\left(\Js_{\text {wheel }}\right)=-\bar{\vs}_{\mathrm{ar}}\left(\Js_{\text {wheel }}\right)=-\left(\rightarrow \mathrm{v}_0\right)=\left(\leftarrow \mathrm{v}_0\right) .</math>

[[File:ExE1-2-1-3-eng.png|thumb|center|500px|link=]]

Calculation of the friction power: <math>\dot{\Ws}^\text{friction}=\dot{\Ws}_\Rs^{\text{friction}\rightarrow \text{wheel}}+ \dot{\Ws}_\Rs^{\text{friction}\rightarrow \text{belt}}</math>, where R may be any reference frame. To quickly calculate this power, it is advisable to choose a reference frame R where one of the two points subjected to the action-reaction pair has zero speed, so that one of the two terms is directly zero:

<math>\dot{\Ws}^\text{friction}=\dot{\Ws}_\text{belt}^{\text{fric}\rightarrow \text{wheel}}+ \dot{\Ws}_\text{belt}^{\text{fric }\rightarrow \text{belt}}=\overline{\Fs}^{\text{fric}\rightarrow \text{wheel}} \cdot \bar{\vs}_{\text {belt}}\left(\Js_{\text {wheel }}\right)=(\rightarrow \mu\ms\gs)\cdot(\leftarrow \vs_0)=-\mu\ms\gs\vs_0.</math>

If the calculation is done in any other reference frame, the result is the same:

<math>\dot{\Ws}^\text{friction}=\dot{\Ws}_\text{E}^{\text{fric}\rightarrow \text{wheel}}+ \dot{\Ws}_\text{E}^{\text{fric} \rightarrow \text{belt}}=(\rightarrow \mu\ms\gs)\cdot 0 + (\leftarrow \mu\ms\gs)\cdot(\rightarrow \vs_0)=-\mu\ms\gs\vs_0.</math>

<math>\dot{\Ws}^\text{friction}=\dot{\Ws}_\text{vehicle}^{\text{fric}\rightarrow \text{wheel}}+ \dot{\Ws}_\text{vehicle}^{\text{fric} \rightarrow \text{belt}}=(\rightarrow \mu\ms\gs)\cdot (\rightarrow\vs_0) + (\leftarrow \mu\ms\gs) \cdot(\rightarrow 3\vs_0)=-\mu\ms\gs\vs_0.</math>

However, if only the power of one of the two forces is calculated (for example, the friction on the wheel), the result depends on the reference frame and can have any sign:

<math>\dot{\Ws}_\text{belt}^{\text{fric }\rightarrow \text{wheel}}=(\rightarrow \mu\ms\gs)\cdot(\leftarrow \vs_0)=-\mu\ms\gs\vs_0 ,</math>

<math>\dot{\Ws}_\text{E}^{\text{fric} \rightarrow \text{wheel}}=(\rightarrow \mu\ms\gs)\cdot 0 = 0 ,</math>

<math>\dot{\Ws}_\text{vehicle}^{\text{fric} \rightarrow \text{wheel}}=(\rightarrow \mu\ms\gs)\cdot (\rightarrow\vs_0)=\mu\ms\gs\vs_0.</math>


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==E1.3 Power of a system of forces on a rigid body==

{|
|
:[[File:E1-3-eng.png|thumb|left|180px|link=]]
<center>'''Figure E1.3''': Kinematic torsor of a rigid body in a reference frame R, and forces on different points (dm) of that rigid body</center>

|When calculating the power of a system of forces on a rigid body S, the term associated with internal interactions is zero because those interactions are all constraint (cohesion) forces and, as seen in [[E1. Work-Energy Theorem: differential form#E1.2 Power of an action-reaction pair|'''section E1.2''']], their total power is zero.

The total power of a system of forces on S can be calculated from the torsor of the system at any point '''O''' of the rigid body, <math>(\overline{\Fs},\overline{\Ms}(\Os))</math>. Taking into account rigid body kinematics ('''Figure E1.3'''):

<math>\dot{\Ws}_\Rs^{\rightarrow \Ss} =\sum_\Ps\overline{\Fs}^{\rightarrow \Ps} \cdot \vel{P}{R}= \sum_\Ps \left[ \overline{\Fs}^{\rightarrow \Ps} \cdot \vel{O}{R} + \velang{S}{R} \times \OPvec \right] </math>

|}
<math>\dot{\Ws}_\Rs^{\rightarrow \Ss} =\left(\sum_\Ps\overline{\Fs}^{\rightarrow \Ps} \right)\cdot \vel{O}{R} + \sum_\Ps\overline{\Fs}^{\rightarrow \Ps} \cdot \left(\velang{S}{R} \times \OPvec \right) = \left( \sum_\Ps\overline{\Fs}^{\rightarrow \Ps} \right) \cdot \vel{O}{R} + \left[ \sum_\Ps \left( \OPvec \times \overline{\Fs}^{\rightarrow \Ps} \right)\right] \cdot \velang{S}{R} = \overline{\Fs}\cdot \vel{O}{R} + \overline{\Ms}(\Os)\cdot \velang{S}{R} </math>

This calculation also shows that the power of a moment (or a torque <math>\Gamma</math>) on a rigid body in a reference frame R is the product of the moment by the angular velocity of the rigid body: <math>\dot{\Ws}_\Rs^{\Gamma\rightarrow \mathrm{S}} = \overline{\Gamma} \cdot \velang{S}{R}</math>.

The power balance for a rigid body is: <math>\dot{\Ws}_\Rs^{ \rightarrow \mathrm{S}} = \dot{\Ws}_\Rs^{ \mathrm{ext}} + \dot{\Ws}_\Rs^{ \mathrm{ar}}.</math>

====✏️ EXAMPLE E1-3.1: roller sliding on a slope====
------
:
{|
|
:[[File:ExE1-3-1-1-eng.png|thumb|left|180px|link=]]
|A homogeneous roller, with mass m and radius r, slides down a slope. The friction coefficient between the ground and the roller is <math>\mu</math>. We want to calculate the power of every force acting on the roller in the ground reference frame.
|}

:It is a 2D problem. The kinematic and dynamic descriptions are:
{|
|
[[File:ExE1-3-1-2-eng.png|thumb|right|250px|link=]]

|<math> \hspace{0.5cm}\vel{J}{E}=\vel{C}{T} + \velang{}{E} \times \CJvec= </math>

<math>\hspace{1.7cm}=\left( \rightarrow \vs_\Es \right) + \left( \otimes \Omega_\Es\right) \times \left( \downarrow \rs \right)</math>

<math> \hspace{0.5cm}\vel{J}{E} = \left[\rightarrow (\vs_\Es -\rs\Omega_\Es)\right]</math>

<math>\hspace{0.5cm} \dot{\Ws}_\Es^{\mathrm{ext}} = \dot{\Ws}_\Es^{\mathrm{weight}} + \dot{\Ws}_\Es^{\mathrm{friction}} + \dot{\Ws}_\Es^{\mathrm{constraint}}= \Bigr[\left(\downarrow \ms\gs\cos\beta\right)+\left(\rightarrow \ms\gs\sin\beta\right)\Bigr] \cdot (\rightarrow \vs_\Es) + \Bigr[(\leftarrow \mu \Ns) + (\uparrow \Ns ) \Bigr] \cdot \Bigr[\rightarrow (\vs_\Es -\rs\Omega_\Es)\Bigr]</math>

|}
:Since the roller center has no motion perpendicular to the plane, the AMT leads to <math>\Ns =\ms\gs\cos\beta</math>. Therefore:

{|
|
[[Archivo:ExE1-3-1-3-eng.png|thumb|right|140px|link=]]

|<math>\hspace{0.5cm} \dot{\Ws}_\Ts^{\mathrm{ext}} = \ms\gs\vs_\Ts \sin \beta - \ms\gs(\vs_\Ts -\rs\Omega_\Ts)\cos\beta=\ms\gs \Bigr[ \vs_\Ts (\sin\beta-\cos\beta)+\rs\Omega_\Ts \cos\beta \Bigr].</math>

<math>\hspace{0.5cm}</math>

If instead of considering the friction at '''J''' we consider the equivalent torsor at point '''C''', we must consider the force applied in '''C''' and the moment about '''C''', which is not zero:

<math>\hspace{0.5cm} \dot{\Ws}_\Es^{ \mathrm{friction}} = (\leftarrow \mu\Ns)\cdot(\rightarrow\vs_\Es)+(\otimes \mu\Ns\rs)\cdot(\otimes \Omega_\Es) =\mu\ms\gs\cos\beta(\rs\Omega_\Es-\vs_\Es)</math>,

<math>\hspace{0.5cm} \dot{\Ws}_\Es^{\mathrm{ext}} = \dot{\Ws}_\Es^{ \mathrm{weight}} + \dot{\Ws}_\Es^{ \mathrm{friction}}+ \dot{\Ws}_\Es^{ \mathrm{constraint}}= \ms\gs\vs_\Es\sin\beta+\mu\ms\gs\cos\beta (\rs\Omega_\Es-\vs_\Es)</math>.

<math>\hspace{0.5cm}</math>

The result is the same.
|}


====✏️ EXAMPLE E1-3.2: block sliding on a truck====
------
:
{|
|
:[[File:ExE1-3-2-1-eng.png|thumb|left|220px|link=]]
|A truck brakes with acceleration '''a''' relative to the ground at the instant in which its velocity (also relative to the ground) is v. At the same instant, the homogeneous block, of mass m, slides with velocity v' relative to the truck. We want to perform a power balance for the block in two different reference frames: the ground and the truck.
|}

:Ground reference frame: It is a 2D problem. The kinematic and dynamic descriptions are:
{|
|
[[File:ExE1-3-2-2-eng.png|thumb|right|200px|link=]]

| The constraint between the truck and the block allows only one translational DoF between the two. Therefore, in the 2D approximation, the associated torsor has two independent components. If characterized at the midpoint of the contact surface, these components are a normal force and a moment perpendicular to the plane. The AMT applied to the block leads to <math>\Ns=\ms\gs</math>.

|}
:The powers associated with all interactions on the block in the ground reference are:

:<math>\dot{\Ws}_\Es^{\rightarrow \mathrm{bloque}}=\dot{\Ws}_\Ts^{\mathrm{ext}}=\dot{\Ws}_\Ts^{\mathrm{peso}}+\dot{\Ws}_\Ts^{\mathrm{fricción}}+\dot{\Ws}_\Ts^{\mathrm{enlace}}= \Bigr[(\downarrow \ms\gs)+(\leftarrow \mu\ms\gs)+(\uparrow \Ns) \Bigr] \cdot \Bigr[\rightarrow(\vs+\vs')\Bigr] + (\odot \Ms_\es)\cdot 0 = -\mu\ms\gs(\vs+\vs')</math>

:Ya que el bloque se traslada respecto al suelo, su energía cinética total es inmediata, y se puede plantear fácilmente el balance de potencias:

:<math>\Ts_\Ts^\mathrm{bloque}=\frac{1}{2}\int_\mathrm{bloque}\ds\ms \vs_\Ts^2(\mathrm{bloque})=\frac{1}{2}\ms \vs_\Ts^2(\mathrm{bloc})\quad \Rightarrow \quad \dot{\Ts}_\Ts^\mathrm{bloque}=\ms\vs_\Ts( \mathrm{bloque})\dot{\vs}_\Ts( \mathrm{bloque}) </math>

:Como la única fuerza horizontal sobre el bloque es la fricción:

<math>\dot{\vs}_\Ts(\mathrm{bloque})=-\mu\gs\quad \Rightarrow \quad \dot{\Ts}_\Ts^\mathrm{bloque}=-\mu\ms\gs(\vs+\vs').</math>

:El balance de potencias es consistente:

:<math>\left.\begin{array}{l}
\dot{\mathrm{W}}_\Ts^{\rightarrow \text { bloque }}=\dot{\mathrm{W}}_{\mathrm{T}}^{\text {ext }}=-\mu \mathrm{mg}\left(\mathrm{v}+\mathrm{v}^{\prime}\right) \\
\dot{\mathrm{T}}_{\mathrm{T}}^{\text {bloque }}=-\mu \mathrm{mg}\left(\mathrm{v}+\mathrm{v}^{\prime}\right)
\end{array}\right\} \Rightarrow \dot{\mathrm{W}}_{\mathrm{T}}^{\rightarrow \text { bloque }}=\dot{\mathrm{T}}_{\mathrm{T}}^{\text {bloque }}</math>

:Referencia camión: Al estar el camión acelerado respecto al suelo, hay que tener en cuenta las fuerzas de arrastre sobre todos los diferenciales de masa (dm) del bloque:

{|
|

[[Archivo:ExE1-3-2-3-cat-esp.jpg|thumb|right|250px|link=]]
|<math>\cal{\Fs}^{\rightarrow \ds\ms}_{\mathrm{ar}}=-\ds\ms\overline{\as}_\mathrm{ar}(\ds\ms)=-\ds\ms (\leftarrow \as)=(\rightarrow \ds\ms \cdot \as)</math>.

El torsor resultante de este sistema de fuerzas en el centro de inercia '''G''' se reduce a una fuerza horizontal.

<math>\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{bloque}}=\dot{\Ws}_\mathrm{camión}^{\mathrm{ext}} + \dot{\Ws}_\mathrm{camión}^{\mathrm{ar}} =\dot{\Ws}_\mathrm{camión}^{\mathrm{peso}} + \dot{\Ws}_\mathrm{camión}^{\mathrm{fricción}} + \dot{\Ws}_\mathrm{camión}^{\mathrm{enlace}} + \dot{\Ws}_\mathrm{camión}^{\mathrm{ar}} , </math>

<math>\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{bloque}}= \Bigr[(\downarrow \ms\gs)+(\leftarrow \mu\ms\gs)+(\uparrow \Ns) + (\rightarrow \ms\as)\Bigr] \cdot(\rightarrow \vs')+(\odot\Ms_\es) \cdot0=\ms(\as-\mu\gs)\vs'.</math>
|}

:Se puede comprobar que el balance de potencias es consistente:

:<math>\Ts^\mathrm{bloque}_\mathrm{camión}=\frac{1}{2}\ms\vs^2_\mathrm{camión}(\mathrm{bloque}) \quad \Rightarrow \quad \dot{\Ts}^\mathrm{bloque}_\mathrm{camión} = \ms\vs_\mathrm{camión}(\mathrm{bloque})\dot{\vs}_\mathrm{camión}(\mathrm{bloque})</math>

:<math>\left.\begin{array}{l}
\mathrm{AB}:\mathrm{suelo} \\
\mathrm{REL}:\mathrm{camión}
\end{array}\right\} \quad \dot{\vs}_\mathrm{camión}(\mathrm{bloque})= \dot{\vs}_\mathrm{T}(\mathrm{bloque})-\dot{\vs}_\mathrm{ar}(\mathrm{bloque}) = (-\mu\gs)-(-\as)=\as-\mu\gs \quad \Rightarrow \quad \dot{\Ts}^\mathrm{bloque}_\mathrm{camión}=\ms\vs'(\as-\mu\gs)</math>

:<math>\left.\begin{array}{l}
\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{bloque}}=\dot{\Ws}_\mathrm{camión}^{\mathrm{ext}} + \dot{\Ws}_\mathrm{camión}^{\mathrm{ar}}= \ms(\as-\mu\gs)\vs'\\
\dot{\Ts}^\mathrm{bloque}_\mathrm{camión}=\ms\vs'(\as-\mu\gs)
\end{array}\right\} \Rightarrow\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{bloque}}= \dot{\Ts}^\mathrm{bloque}_\mathrm{camión}</math>


====✏️ EJEMPLO E1-3.3: camión frenando con carga====
------
:
{|
|
:[[Archivo:ExE1-3-3-1-esp.png|thumb|left|220px|link=]]
|Un camión frena con aceleración '''a''' respecto al suelo en el instante en el que su velocidad (también respecto al suelo) es v. Los pares de frenada en el eje delantero y en el trasero son iguales y valen <math>\Gamma</math>, y las ruedas, de radio r y masa despreciable, no deslizan sobre el suelo. En el mismo instante, el bloque homogéneo, de masa m, desliz con velocidad v' relativa al camión. Se trata de hacer un balance de potencias para el sistema (bloque+camión) en la referencia del suelo y en la del camión.

|}

:Ya que la potencia de las interacciones internas no es nula en principio (no se trata de un único sólido rígido), es conveniente hacer un DGI del sistema para no olvidar ninguna:
{|
|
[[Archivo:ExE1-3-3-2-esp.png|thumb|right|250px|link=]]

|
|
c.p.s.d.: contacto puntual sin deslizamiento

c.m.c.d.: contacto multipuntual con deslizamiento

|}
:El sistema sobre el que se hace el balance de potencias es todo menos el suelo. Teniendo en cuenta que la potencia de los enlaces internos es siempre nula, las únicas interacciones que hay que tener en cuenta son el peso, los enlaces de las ruedas con el suelo, los pares de frenada y la fricción entre bloque y camión:

[[Archivo:ExE1-3-3-3-esp.png|thumb|center|300px|link=]]

:Referencia suelo:
:Las fuerzas de enlace del suelo sobre las ruedas están aplicadas a puntos de velocidad cero, y por tanto no hacen potencia. El peso es una fuerza vertical, y el movimiento del camión y del bloque es horizontal, por tanto su potencia también es cero.

:<math>\dot{\Ws}_\Ts^{\rightarrow \mathrm{sist}}=\dot{\Ws}_\Ts^{\mathrm{ext}}+\dot{\Ws}^{\mathrm{int}}=\dot{\Ws}_\Ts^{\mathrm{peso}}+\dot{\Ws}_\Ts^{\mathrm{enlace}}+\dot{\Ws}^{\mathrm{fricción}}+\dot{\Ws}^{\Gamma}=\dot{\Ws}^{\mathrm{fricción}}+\dot{\Ws}^{\Gamma}.</math>

:Las potencias de interacciones internas se pueden calcular en cualquier referencia. Por tanto:

:<math>\dot{\Ws}^{\mathrm{fricción}}= \dot{\Ws}_\mathrm{camión}^{\mathrm{fric}\rightarrow \mathrm{bloque}} +\dot{\Ws}_\mathrm{camió}^{\mathrm{fric}\rightarrow \mathrm{camión}} = \dot{\Ws}_\mathrm{camión}^{\mathrm{fric}\rightarrow \mathrm{bloque}}= (\leftarrow \mu\ms\gs)\cdot (\rightarrow \vs')=-\mu\ms\gs\vs'</math>

:<math>\dot{\Ws}^{\Gamma}= \dot{\Ws}_\mathrm{camión}^{\Gamma\rightarrow \mathrm{ruedas}} +\dot{\Ws}_\mathrm{camión}^{\Gamma\rightarrow \mathrm{camión}} = \dot{\Ws}_\mathrm{camión}^{\Gamma\rightarrow \mathrm{ruedas}}= 2 \left(\odot \Gamma \right)\cdot \left(\otimes \frac{\vs}{\rs}\right)=-2\Gamma\frac{\vs}{\rs}.</math>

:Por otro lado: <math>\Ts_\Ts^\mathrm{sist}=\frac{1}{2}\ms\vs_\Ts^2(\mathrm{bloque})+\frac{1}{2}\ms\vs_\Ts^2(\mathrm{camión})\quad \Rightarrow \quad \dot{\Ts}_\Ts^\mathrm{sist}=-\ms(\vs+\vs')\mu\gs-\Ms\vs\as. </math>

:Finalmente: <math>\dot{\Ws}_\Ts^{\rightarrow \mathrm{sist}}=\dot{\Ts}_\Ts^\mathrm{sist} \quad \Rightarrow \quad -\mu\ms\gs\vs'-2\Gamma\frac{\vs}{\rs}=-\ms(\vs+\vs')\mu\gs-\Ms\vs\as.</math>

:Esta ecuación permite calcular el par de frenada: <math>\Gamma=\frac{1}{2}(\mu\ms\gs+\Ms\as)\rs.</math>

:Referencia camión:

:En esta referencia hay que tener en cuenta las fuerzas de arrastre. Por otro lado, las fuerzas de enlace del suelo sobre las ruedas están aplicadas a puntos que ya no tienen velocidad cero, y por tanto hacen potencia. Como antes, la potencia del peso es cero.

:<math>\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{sist}}=\dot{\Ws}_\mathrm{camión}^{\mathrm{ext}}+\dot{\Ws}^{\mathrm{int}}+\dot{\Ws}_\mathrm{camión}^{\mathrm{ar}}=\dot{\Ws}_\mathrm{camión}^{\mathrm{enlace}}+\dot{\Ws}^{\mathrm{fricción}}+\dot{\Ws}^{\Gamma}+\dot{\Ws}_\mathrm{camión}^{\mathrm{ar}}.</math>

:Las fuerzas de arrastre sobre el camión no hacen potencia porque el camión está parado en esta referencia. La de las fuerzas de arrastre sobre el bloque se ha calculado en el [[E1. Teorema de la energía: versión diferencial#✏️ ✏️ EJEMPLO E1-3.2: bloque deslizando sobre un camión|'''ejemplo E1-3.2''']]. La potencia de la fricción y de los pares de frenada es la misma de antes ya que no depende de la referencia.

:<math>\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{sist}}=\dot{\Ws}_\mathrm{camión}^{\mathrm{enllaç}}+\dot{\Ws}^{\mathrm{fricción}}+\dot{\Ws}^{\Gamma}+\dot{\Ws}_\mathrm{camión}^{\mathrm{ar} \rightarrow \mathrm{bloque}}= \dot{\Ws}_\mathrm{camión}^{\mathrm{enlace}} -\mu\ms\gs\vs'-2\Gamma\frac{\vs}{\rs} +\ms\as\vs'.</math>

:Las descripciones cinemática y dinámica de la rueda son:

[[Archivo:ExE1-3-3-4-esp.png|thumb|center|430px|link=]]

:<math>\dot{\Ws}_\mathrm{camión}^{\mathrm{enlace}}=2\Bigr[(\leftarrow \Ts)+(\uparrow \Ns)\Bigr]\cdot(\leftarrow\vs)=2\Ts\vs.</math>

La fuerza tangencial de enlace se puede calcular a partir del TMC aplicado al centro '''C''' de la rueda. Como no tiene masa: <math>\sum\overline{\Ms}_\mathrm{ext}(\Cs)=\overline{0} \quad \Rightarrow \quad (\odot \Gamma)+(\otimes \Ts\rs)=0\quad \Rightarrow \quad \Ts=\Gamma/\rs.</math>

:Por tanto: <math>\dot{\Ws}_\mathrm{camió}^{\rightarrow \mathrm{sist}} =2\frac{\Gamma}{\rs}\vs-\mu\ms\gs\vs'-2\Gamma\frac{\vs}{\rs}+\ms\as\vs'=\ms(\as-\mu\gs)\vs'.</math>

:Por otro lado, como se ha visto en el [[E1. Teorema de la energía: versión diferencial#✏️ ✏️ EJEMPLO E1-3.2: bloque deslizando sobre un camión|'''ejemplo E1-3.2''']]:

:<math>\Ts_\mathrm{camión}^\mathrm{sist}= \Ts_\mathrm{camión}^\mathrm{bloque}=\frac{1}{2}\ms\vs_\mathrm{camión}^2(\mathrm{bloque}) \quad \Rightarrow \quad \dot{\Ts}_\mathrm{camión}^\mathrm{sist}= \dot{\Ts}_\mathrm{camión}^\mathrm{bloque}=\ms\vs'(\as-\mu\gs).</math>

:El balance de potencias es consistente:

:<math>\left.\begin{array}{l}
\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{sist}}= \ms(\as-\mu\gs)\vs'\\
\dot{\Ts}^\mathrm{sist}_\mathrm{camión}=\ms(\as-\mu\gs)\vs'
\end{array}\right\} \Rightarrow\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{sist}}= \dot{\Ts}^\mathrm{sist}_\mathrm{camión}</math>

-----
-----

==E1.4 Balance de potencias para un sistema multisólido: cálculo directo y cálculo indirecto==

El '''cálculo directo''' (a partir de su definición) de la potencia asociada a una interacción (una fuerza sobre un punto '''P''' o un momento sobre un sólido rígido S) exige el conocimiento de la interacción (fuerza o momento):

<math>\dot{\Ws}_\Rs^{\Fs\rightarrow \Ps}=\overline{\Fs}_{\rightarrow\Ps} \cdot \vel{P}{R} \quad , \quad \dot{\Ws}_\Rs^{\Ms\rightarrow \Ss}=\overline{\Ms}_{\rightarrow\Ss} \cdot \velang{S}{R} </math>

Cuando no se trata de una interacción formulable, hay que recurrir a los teoremas vectoriales para determinarla.

El '''cálculo indirecto''' es una alternativa que no necesita recurrir a estos teoremas. Se basa en la evaluación de un balance de potencias para un sistema y una referencia adecuados de manera que la potencia a determinar sea el único término desconocido de la ecuación (el único que no se puede calcular de manera directa). Cuando el balance incluye más de un término desconocido, se pueden combinar varios balances (para distintos sistemas y referencias) hasta tener un número suficiente de ecuaciones que permitan la determinación de todas las potencias desconocidas.

====✏️ EJEMPLO E1-4.1: polea frenada====
------
:
{|
|
:[[Archivo:ExE1-4-1-1-esp.png|thumb|left|150px|link=]]
|Un bloque de masa m cuelga de un hilo inextensible atado a un punto de la periferia de una polea de radio r. Entre soporte (fijo al suelo) y polea actúa un freno que garantiza que el bloque baja a velocidad constante respecto al suelo. Se trata de calcular la potencia del freno.

|}

:Càlculo directo: <math>\dot{\Ws}^{\mathrm{freno}}= \overline{\Gamma}^{\rightarrow \mathrm{polea}}\cdot\velang{polea}{T}=\left(\odot\Gamma\right)\cdot\left( \otimes \frac{\vs_0}{\rs}\right)=-\Gamma \frac{\vs_0}{\rs}.</math>

:El par de frenada no es un dato, y hay que recurrir al TMC para calcularlo.

:<math>\left.\begin{array}{l}
\text{SISTEMA: polea + bloque}\\
\text{TMC en } \Os
\end{array}\right\} \sum\overline{\Ms}_\mathrm{ext}(\Os)=\dot{\overline{\Hs}}_\mathrm{RTO}(\Os)</math>

:El momento cinético proviene de las dos piezas. Ya que '''O''' no es fijo al bloque, hay que hacer descomposición baricéntrica para calcular el momento cinético del bloque:

:<math>\overline{\Hs}_\mathrm{RTO=T} (\Os) =\overline{\Hs}_\Ts^\mathrm{politja}(\Os) + \overline{\Hs}_\Ts^\mathrm{bloc}(\Os)= \overline{\Hs}_\Ts^\mathrm{politja}(\Os) + \overline{\Hs}_\Ts^\mathrm{bloc}(\Gs_\mathrm{bloc})+ \overline{\Hs}_\Ts^{\oplus\mathrm{bloc}}(\Os) = </math>

:<math>\hspace{2.3cm}=\Is_\mathrm{O}^\mathrm{polea} \velang{polea}{T}+\Is_{\Gs_\mathrm{bloque}}^\mathrm{bloque}+\OGvec_\mathrm{bloque}\times \ms\overline{\vs}_\Ts(\Gs_\mathrm{bloque})=\left(\otimes \Is_\mathrm{O}\frac{\vs_0}{\rs}\right)+(\otimes \ms\vs_0\rs).</math>

:[[Archivo:ExE1-4-1-2-esp.png|thumb|right|180px|link=]]
:Se trata de un momento cinético constante en dirección y valor. Por tanto:

:<math>\dot{\overline{\Hs}}_\mathrm{RTO}(\Os)=0 \quad \Rightarrow \quad \sum{\overline{\Ms}}_\mathrm{ext}(\Os)=(\odot\Gamma)+(\otimes \ms\gs\rs) =0\quad \Rightarrow \quad \Gamma= \ms\gs\rs</math>.

:Finalmente: <math>\dot{\Ws}^{\mathrm{freno}}= -\Gamma \frac{\vs_0}{\rs} = -\ms\gs\vs_0.</math>

:Cálculo indirecto:

:<math>\left.\begin{array}{l}
\text{SISTEMA: polea + soporte + bloque}\\
\text{REF: suelo(Gal)}
\end{array}\right\} \quad \dot{\Ts}^\mathrm{sist}_\Ts=\dot{\Ws}^\mathrm{ext}_\Ts+ \dot{\Ws}^\mathrm{int}</math>

:<math>\left.\begin{array}{l}
\dot{\Ts}^\mathrm{sist}_\Ts=0\\
\dot{\Ws}^\mathrm{ext}_\Ts=\left(\downarrow \ms\gs\right)\cdot\left(\downarrow \vs_0\right)= \ms\gs\vs_0\\
\dot{\Ws}^\mathrm{int}=\dot{\Ws}^{\mathrm{freno}}
\end{array}\right\} \quad \Rightarrow \quad \dot{\Ws}^{\mathrm{freno}} =-\ms\gs\vs_0</math>



====✏️ EJEMPLO E1-4.2: mono y plátanos====
------
:
{|
|
:[[Archivo:ExE1-4-2-1-esp.png|thumb|left|180px|link=]]
|Unos plátanos de masa <math>(2/3)\ms</math> cuelgan de una cuerda inextensible que se apoya sin deslizar sobre una polea de radio r y masa despreciable, articulada respecto al suelo. En el otro extremo de la cuerda se agarra un mono de masa m. La polea gira con velocidad angular constante <math>\Omega_0</math> respecto al suelo bajo la acción de un motor. El mono se mueve respecto a la cuerda sin que sus manos se froten contra la cuerda, y de manera que la distancia de su centro de masas al suelo se mantiene constante. Se trata de calcular la potencia del motor.
|}

:Cálculo directo: <math>\dot{\Ws}^{\mathrm{freno}}= \overline{\Gamma}^{\rightarrow \mathrm{polea}}\cdot\velang{polea}{T}=\left(\odot\Gamma\right)\cdot\left( \odot \Omega_0\right)=\Gamma \Omega_0.</math>

:El par motor no es un dato, y hay que recurrir al TMC para calcularlo.

:<math>\left.\begin{array}{l}
\text{SISTEMA: polea + mono + plátanos}\\
\text{TMC en } \Os
\end{array}\right\} \sum\overline{\Ms}_\mathrm{ext}(\Os)=\dot{\overline{\Hs}}_\mathrm{RTO}(\Os)</math>

:El momento cinético proviene del mono y de los plátanos. En primera aproximación, se puede considerar la masa del mono concentrada en su tronco y su cabeza, de manera que no gira y su velocidad respecto al suelo es nula. Teniendo en cuenta que el punto '''O''' no se mueve respecto al mono pero sí respecto a los plátanos:

:<math>\overline{\Hs}_\mathrm{RTO=T} (\Os) =\overline{\Hs}_\Ts^\mathrm{mono}(\Os) + \overline{\Hs}_\Ts^\mathrm{plátanos}(\Os)= \overline{\Hs}_\Ts^\mathrm{mono}(\Os) + \overline{\Hs}_\Ts^\mathrm{plátanos}(\Gs_\mathrm{plát})+ \OGvec_\mathrm{plát} \times \frac{2}{3} \ms \overline{\vs}_\Ts(\Gs_\mathrm{plát})= </math>

:<math>\hspace{2.3cm}=\OGvec_\mathrm{plát}\times \frac{2}{3} \ms \overline{\vs}_\Ts(\Gs_\mathrm{plát}) = \Bigr[(\rightarrow \rs)+(\downarrow \xs)\Bigr] \times\frac{2}{3} \ms \left(\downarrow \rs\Omega_0\right)=\left(\otimes \frac{2}{3}\ms\rs^2\Omega_0\right).</math>

:Se trata de un momento cinético constante en dirección y valor. Por tanto:

:<math>\dot{\overline{\Hs}}_\mathrm{RTO}(\Os)=0 \quad \Rightarrow \quad \sum{\overline{\Ms}}_\mathrm{ext}(\Os)=(\odot\Gamma)+(\otimes \ms\gs\rs) + \left(\otimes \frac{2}{3}\ms\gs\rs\right)=0\quad \Rightarrow \quad \Gamma= \frac{5}{3}\ms\gs\rs</math>.

:Finalmente: <math>\dot{\Ws}^{\mathrm{motor}}= \Gamma \Omega_0 = \frac{5}{3}\ms\gs\rs\Omega_0.</math>

:Cálculo indirecto

:Hay que tener presente que es un problema con dos tipos de actuadores: el motor y la musculatura del mono. Hay diversos modelos biomecánicos para formular las fuerzas de los músculos, pero quedan fuera del ámbito de este curso. Ya que el mono se mueve, sus músculos desarrollan una potencia no nula.

:<math>\left.\begin{array}{l}
\text{SISTEMA: polea + mono + plátano}\\
\text{REF: suelo(Gal)}
\end{array}\right\} \quad \dot{\Ts}^\mathrm{sist}_\Ts=\dot{\Ws}^\mathrm{ext}_\Ts+ \dot{\Ws}^\mathrm{int}</math>

:<math>\left.\begin{array}{l}
\dot{\Ts}^\mathrm{sist}_\Ts=0\\
\dot{\Ws}^\mathrm{ext}_\Ts=\left(\downarrow \frac{2}{3}\ms\gs\right)\cdot\left(\downarrow \rs\Omega_0\right)= \frac{2}{3}\ms\gs\rs\Omega_0\\
\dot{\Ws}^\mathrm{int}=\dot{\Ws}^{\mathrm{musc}}+\dot{\Ws}^{\mathrm{motor}}
\end{array}\right\} \quad \Rightarrow \quad \dot{\Ws}^{\mathrm{musc}} + \dot{\Ws}^{\mathrm{motor}}=-\frac{2}{3}\ms\gs\rs\Omega_0</math>

:Este balance proporciona una ecuación con dos incógnitas. Se puede plantear un segundo balance para un sistema donde una de las dos no aparezca: un sistema sin el mono o un sistema sin el motor. Si se hace un balance de potencias para el sistema (mono + trozo de cuerda), las fuerzas externas a considerar son el peso y la tensión de la cuerda. Si se sigue trabajando en la referencia del suelo, esta tensión (desconocida) hará potencia no nula ya que está aplicada a un punto de velocidad ascendente.
:[[Archivo:ExE1-4-2-2-cat-esp.png|thumb|right|330px|link=]]

:Para evitar que aparezca un término asociado a la tensión de la cuerda, se puede trabajar en la referencia del trozo de cuerda que se ha considerado, que al tener velocidad ascendente constante es galileana. En esta referencia, el mono baja con velocidad constante:

:<math>\left.\begin{array}{l}
\text{REL: trozo de cuerda}\\
\text{AB: suelo}
\end{array}\right\}</math>

:<math>\overline{\vs}_\mathrm{REL}(\mathrm{mono})=\overline{\vs}_\mathrm{ar}(\mathrm{mono})=-\overline{\vs}_\mathrm{ar}(\mathrm{mono})=(\downarrow \rs\Omega_0)</math>

:<math>\left.\begin{array}{l}
\text{SISTEMA: mono+ trozo de cuerda}\\
\text{REF: trozo de cuerda(Gal)}
\end{array}\right\} \quad \dot{\Ts}^\mathrm{sist}_\mathrm{cuerda} =\dot{\Ws}^\mathrm{ext}_\mathrm{cuerda}+ \dot{\Ws}^\mathrm{int}</math>

:<math>\left.\begin{array}{l}
\dot{\Ts}^\mathrm{sist}_\mathrm{cuerda}=0\\
\dot{\Ws}^\mathrm{ext}_\mathrm{cuerda}=\left(\downarrow \ms\gs\right)\cdot\left(\downarrow \rs\Omega_0\right)= \ms\gs\rs\Omega_0\\
\dot{\Ws}^\mathrm{int}=\dot{\Ws}^{\mathrm{musc}}
\end{array}\right\} \quad \Rightarrow \quad \dot{\Ws}^{\mathrm{musc}} =-\ms\gs\rs\Omega_0</math>

:Combinando este resultado con el del balance anterior:

:<math>\dot{\Ws}^\mathrm{motor}=-\frac{2}{3}\ms\gs\rs\Omega_0-\dot{\Ws}^\mathrm{musc}=-\frac{5}{3}\ms\gs\rs\Omega_0</math>



© Universitat Politècnica de Catalunya. [[Mecánica:Derechos de autor |Todos los derechos reservados]]
-------------

<center>
[[D8. Conservaciones|<<< D8. Conservaciones]]

[[E2. Teorema de la energía: versión integrada|E2. Teorema de la energía: versión integrada>>>]]
</center>

E1. Work-Energy Theorem: differential form

2025-05-20T11:31:55Z

Bros:

<div class="noautonum">__TOC__</div>
<math>\newcommand{\uvec}{\overline{\textbf{u}}}
\newcommand{\vvec}{\overline{\textbf{v}}}
\newcommand{\evec}{\overline{\textbf{e}}}
\newcommand{\Omegavec}{\overline{\mathbf{\Omega}}}
\newcommand{\velang}[2]{\Omegavec^{\textrm{#1}}_{\textrm{#2}}}
\newcommand{\Alfavec}{\overline{\mathbf{\alpha}}}
\newcommand{\accang}[2]{\Alfavec^{\textrm{#1}}_{\textrm{#2}}}
\newcommand{\ds}{\textrm{d}}
\newcommand{\ps}{\textrm{p}}
\newcommand{\ns}{\textrm{n}}
\newcommand{\hs}{\textrm{h}}
\newcommand{\cs}{\textrm{c}}
\newcommand{\gs}{\textrm{g}}
\newcommand{\Ns}{\textrm{N}}
\newcommand{\Hs}{\textrm{H}}
\newcommand{\Fs}{\textrm{F}}
\newcommand{\ms}{\textrm{m}}
\newcommand{\ts}{\textrm{t}}
\newcommand{\us}{\textrm{u}}
\newcommand{\vs}{\textrm{v}}
\newcommand{\Rs}{\textrm{R}}
\newcommand{\Ws}{\textrm{W}}
\newcommand{\Ts}{\textrm{T}}
\newcommand{\Ls}{\textrm{L}}
\newcommand{\As}{\textrm{A}}
\newcommand{\Ds}{\textrm{D}}
\newcommand{\Bs}{\textrm{B}}
\newcommand{\es}{\textrm{e}}
\newcommand{\fs}{\textrm{f}}
\newcommand{\is}{\textrm{i}}
\newcommand{\Is}{\textrm{I}}
\newcommand{\ks}{\textrm{k}}
\newcommand{\js}{\textrm{j}}
\newcommand{\rs}{\textrm{r}}
\newcommand{\as}{\textrm{a}}
\newcommand{\ss}{\textrm{s}}
\newcommand{\Os}{\textbf{O}}
\newcommand{\Gs}{\textbf{G}}
\newcommand{\Cbf}{\textbf{C}}
\newcommand{\Or}{\Os_\Rs}
\newcommand{\Qs}{\textbf{Q}}
\newcommand{\Cs}{\textbf{C}}
\newcommand{\Ps}{\textbf{P}}
\newcommand{\Ss}{\textbf{S}}
\newcommand{\Js}{\textbf{J}}
\newcommand{\P}{\textrm{P}}
\newcommand{\Q}{\textrm{Q}}
\newcommand{\Ms}{\textrm{M}}
\newcommand{\deg}{^\textsf{o}}
\newcommand{\xs}{\textsf{x}}
\newcommand{\ys}{\textsf{y}}
\newcommand{\zs}{\textsf{z}}
\newcommand{\dert}[2]{\left.\frac{\ds{#1}}{\ds\ts}\right]_{\textrm{#2}}}
\newcommand{\ddert}[2]{\left.\frac{\ds^2{#1}}{\ds\ts^2}\right]_{\textrm{#2}}}
\newcommand{\vec}[1]{\overline{#1}}
\newcommand{\vecbf}[1]{\overline{\textbf{#1}}}
\newcommand{\vecdot}[1]{\overline{\dot{#1}}}
\newcommand{\OQvec}{\vec{\Os\Qs}}
\newcommand{\GQvec}{\vec{\Gs\Qs}}
\newcommand{\GPvec}{\vec{\Gs\Ps}}
\newcommand{\PSvec}{\vec{\Ps\Ss}}
\newcommand{\QQvec}{\vec{\Qs\Qs}}
\newcommand{\QGvec}{\vec{\Qs\Gs}}
\newcommand{\QPvec}{\vec{\Qs\Ps}}
\newcommand{\JQvec}{\vec{\Js\Qs}}
\newcommand{\PQvec}{\vec{\Ps\Qs}}
\newcommand{\GJvec}{\vec{\Gs\Js}}
\newcommand{\OPvec}{\vec{\Os\textbf{P}}}
\newcommand{\OCvec}{\vec{\Os\Cs}}
\newcommand{\OGvec}{\vec{\Os\Gs}}
\newcommand{\CJvec}{\vec{\Cs\Js}}
\newcommand{\GCvec}{\vec{\Gs\Cs}}
\newcommand{\PGvec}{\vec{\Ps\Gs}}
\newcommand{\abs}[1]{\left|{#1}\right|}
\newcommand{\braq}[2]{\left\{{#1}\right\}_{\textrm{#2}}}
\newcommand{\matriz}[9]{
\begin{bmatrix}
{#1} & {#2} & {#3}\\
{#4} & {#5} & {#6}\\
{#7} & {#8} & {#9}
\end{bmatrix}}
\newcommand{\diag}[3]{
\begin{bmatrix}
{#1} & {0} & {0}\\
{0} & {#2} & {0}\\
{0} & {0} & {#3}
\end{bmatrix}}
\newcommand{\vector}[3]{
\begin{Bmatrix}
{#1}\\
{#2}\\
{#3}
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\newcommand{\vecdosd}[2]{
\begin{Bmatrix}
{#1}\\
{#2}
\end{Bmatrix}}
\newcommand{\vel}[2]{\vvec_{\textrm{#2}} (\textbf{#1})}
\newcommand{\acc}[2]{\vecbf{a}_{\textrm{#2}} (\textbf{#1})}
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\newcommand{\psio}{\dot{\psi}_0}
\newcommand{\Pll}{\textbf{P}_\textrm{lliure}}
\newcommand{\agal}[1]{\vecbf{a}_{\textrm{Gal}} (#1)}
\newcommand{\angal}[1]{\vecbf{a}_{\textrm{NGal}} (#1)}
\newcommand{\vgal}[1]{\vecbf{v}_{\textrm{Gal}} (#1)}
\newcommand{\fvec}[2]{\overline{\mathbf{F}}_{\textrm{#1}\rightarrow\textrm{#2}}}
\newcommand{\mvec}[2]{\overline{\mathbf{M}}_{\textrm{#1}\rightarrow\textrm{#2}}}
\newcommand{\F}[1]{\vec{\Fs}_{#1}}
\newcommand{\Fcal}[2]{\vec{\cal{F}}^{\text{#1}}_{#2}}
</math>

The '''Work-Energy Theorem''' plays a prominent role in physics, as it bridges all branches of science (electromagnetism, thermodynamics, etc.): all of them deal with energy.

There is a certain parallel between the Vector Theorems and the '''Work-Energy Theorem''': they all allow us to understand the evolution of a magnitude that depends on the mechanical state of the system. In the case of the former, we are dealing with the instantaneous evolution of vector magnitudes (linear momentum, angular momentum), while the latter deals with the finite change (in a finite time interval) of a scalar magnitude ('''mechanical energy''').

This unit presents the differential form of this theorem (the '''power balance'''). The integrated form (which is the Work-Energy Theorem itself) is presented in [[E2. Teorema de la energía: versión integrada|'''unit E2''']].
 
------
------
==E1.1 Power balance in a system of particles==

:[[File:E1-1-eng.png|thumb|right|300px|link=]]
<center>'''Figure E1.1''' Forces on a particle of a system with constant matter</center>

Let us consider a system of constant matter, and let’s apply Newton's second law to each particle (or each mass differential) of this system in a general reference frame R (which may be non-Galilean) ('''Figure E1.1'''):

<math>\F{\rightarrow \Ps} + \Fcal{}{\text{ar}\rightarrow\Ps} + \Fcal{}{\text{Cor}\rightarrow\Ps} =\ms_\Ps\acc{P}{R}</math>

<math>\F{\rightarrow \Ps} </math> is the resultant interaction force on '''P''', and it comes from internal interactions (with particles of the same system) and external ones:

<math>\F{\rightarrow \Ps}=\F{\rightarrow \Ps}^\text{int}+\F{\rightarrow \Ps}^\text{ext}</math>

If both sides of the equation are multiplied by <math>\vel{P}{R}</math>, all forces (interaction and inertia forces) orthogonal to <math>\vel{P}{R}</math> are filtered out. Specifically, since <math>\acc{P}{R}=\accs{P}{R} + \accn{P}{R} </math>, where <math>\accn{P}{R} \perp \vel{P}{R}</math>, and since the Coriolis force is always orthogonal to <math>\vel{P}{R} (\Fcal{}{\text{Cor}\rightarrow\Ps} = -2\ms_\Ps \velang{R}{Gal} \times \vel{P}{R})</math>:

<math>\vel{P}{R} \cdot \left( \F{\rightarrow \Ps}^\text{int}+\F{\rightarrow \Ps}^\text{ext} + \Fcal{}{\text{ar}\rightarrow\Ps} \right) = \ms_\Ps\accs{P}{R} \cdot \vel{P}{R} = \frac{\ds}{\ds \ts} \left(\frac{1}{2}\ms_\Ps \vvec_\Rs^2(\Ps) \right) \equiv \frac{\ds\Ts_\Rs(\Ps)}{\ds\ts}=\dot{\Ts}_\Rs(\Ps)</math>

The scalar function <math>\Ts_\Rs(\Ps)</math> is the '''kinetic energy''' of the particle, and is a function of the [[D1. Foundational laws of Newtonian dynamics#D1.3 Newton’s Principle of Determinacy|'''mechanical state''']] of '''P'''. The scalar product of a force and the velocity of its application point is the '''force power''': <math>\vel{P}{R} \cdot \F{\rightarrow \Ps} \equiv \dot{\Ws}_\Rs^{\F{\rightarrow \Ps}} </math>. Both scalar quantities <math>(\Ts_\Rs(\Ps), \dot{\Ws}_\Rs^{\F{\rightarrow \Ps}})</math> depend on the reference frame. For the same set of interaction forces acting on '''P''', the reference frame determines which are filtered out and which are not.

The equation can be rewritten as <math>\dot{\Ws}_\Rs^{\text{ext}\rightarrow \Ps}+\dot{\Ws}_\Rs^{\text{int}\rightarrow \Ps}+\dot{\Ws}_\Rs^{\text{ar}\rightarrow \Ps} =\dot{\Ts}_\Rs(\Ps)</math>, and constitutes the '''power balance''' for particle '''P''' in the R reference frame. Summing for all particles in the system:

<math>\dot{\Ws}_\Rs^{\text{ext}}+\dot{\Ws}^{\text{int}}+\dot{\Ws}_\Rs^{\text{ar}} =\dot{\Ts}_\Rs^\text{syst}</math>

The calculation of kinetic energy is covered in [[E2. Teorema de la energía: versión integrada|'''unit E2''']]. This unit focuses on the calculation of powers. [[E1. Work-Energy Theorem: differential form#E1.2 Potencia de una pareja de acción-reacción|'''section E1-2''']] shows that the power of internal interactions does not depend on the reference frame (which is why the subscript R does not appear in the <math>\dot{\Ws}^{\text{int}}</math> term of the previous equation).

====✏️ EXAMPLE E1-1.1: free particle====
------
:
{|
|
:[[File:ExE1-1-1-1-eng.jpg|thumb|left|180px|link=]]
|A particle '''P''', with mass m, is released from the ring of a Ferris wheel when it is at its highest point. We want to calculate the power, at this instant, of all forces acting on the particle from three different reference frames: the ground, the ring, and the Ferris wheel cabin.
|}

:Ground reference frame: Since this is a Galilean reference frame, all forces on '''P''' are interaction forces.
{|
|
:[[File:ExE1-1-2- eng.png|thumb|right|230px|link=]]

|<math>\hspace{0.5cm}\sum\dot{\Ws}_\Ts=\dot{\Ws}_\Ts^\text{weight}= \left(\leftarrow \Rs\Omega_0\right) \cdot \left(\downarrow \ms\gs\right)=0</math>
|}

:Ring reference frame: It is a non-Galilean reference, and the transportation force on '''P''' must be taken into account.
{|
|
:[[File:ExE1-1-3-eng.png|thumb|right|250px|link=]]
|<math>\hspace{0.5cm}</math> AB:ground, REL:ring 

<math>\hspace{0.5cm}\vel{P}{REL}=\vel{P}{AB}-\vel{P}{ar}= \left(\leftarrow \Rs\Omega_0\right)-\left(\leftarrow \Rs\Omega_0\right)=\overline{0}</math> 

<math>\hspace{0.5cm}</math>

Since the velocity is zero, the total power is zero.
|}

:Cabin reference frame: It is also a non-Galilean reference, and the transportation force on '''P''' must be taken into account.

{|
|
:[[File:ExE1-1-4-eng.png|thumb|right|250px|link=]]
|<math>\hspace{0.5cm}</math>AB:ground, REL:cabin

<math>\hspace{0.5cm}\vel{P}{REL}=\vel{P}{AB}-\vel{P}{ar}= \left(\leftarrow \Rs\Omega_0\right)-\left(\uparrow \Rs\Omega_0\right)</math>

<math>\hspace{0.5cm}\dot{\Ws}_\text{cabin}^\text{weight}=\left[ \left(\leftarrow \Rs\Omega_0\right)+\left(\downarrow \Rs\Omega_0\right) \right] \cdot \left(\downarrow\ms\gs\right) =\ms\gs\Rs\Omega_0</math>

<math>\hspace{0.5cm}\dot{\Ws}_\text{cabin}^\text{ar}=\left[ \left(\leftarrow \Rs\Omega_0\right)+\left(\downarrow \Rs\Omega_0\right) \right] \cdot \left(\rightarrow\ms\Rs^2\Omega_0^2\right) =-\ms\gs\Rs^2 \Omega_0^3</math>

<math>\hspace{0.5cm}\sum\dot{\Ws}_\text{cabin}=\ms\Rs\Omega_0(\gs-\Rs\Omega_0^2)</math>

|}


====✏️ EXAMPLE E1-1.2: pendulum====
------
:
{|
|
:[[File:ExE1-2-1-eng.png|thumb|left|180px|link=]]
|A particle '''P''' of mass m hangs from an inextensible thread of length L attached to the ceiling. The assembly oscillates in a vertical plane fixed to the ground. We want to calculate the power of all the forces acting on the particle in two different reference frames: the ground and a platform rotating relative to the ground with a constant angular velocity <math>\Omega_0</math>. 

Ground reference frame: Since it is a Galilean reference frame, '''P''' is only subjected to interaction forces.
{|
|
:[[File:ExE1-2-2-neut.png|thumb|left|300px|link=]]

|<math>\dot{\Ws}_\Es^\Ts=(\nearrow\Ls\dot{\theta})\cdot(\nwarrow\Ts)=0</math>

<math>\dot{\Ws}_\Es^\text{weight}=(\nearrow\Ls\dot{\theta}\cdot(\downarrow \ms\gs)=-\ms\gs\Ls\dot{\theta}\sin\theta</math>

|}

Platform reference frame: It is a non Galilean reference frame, and the transportation force on '''P''' has to be taken into account.

{|
|
:[[File:ExE1-2-3-eng.png|thumb|left|400px|link=]]

|AB:ground, REL:platform

<math>\vel{P}{REL}=\vel{P}{AB}-\vel{P}{ar}=(\nearrow\Ls\dot{\theta})-(\otimes \Ls \Omega_0 \sin \theta)</math>

<math>\dot{\Ws}_\Es^\Ts\left[(\nearrow\Ls\dot{\theta})+(\odot\Ls\Omega_0\sin\theta)\right]\cdot(\nwarrow\Ts)=0</math>

<math>\dot{\Ws}_\Es^\text{weight}\left[(\nearrow\Ls\dot{\theta})+(\odot\Ls\Omega_0\sin\theta)\right]\cdot(\downarrow \ms\gs)=-\ms\gs\Ls\dot{\theta}\sin\theta</math>

<math>\dot{\Ws}_\Es^\text{ar}\left[(\nearrow\Ls\dot{\theta})+(\odot\Ls\Omega_0\sin\theta)\right]\cdot(\rightarrow \ms\Ls\Omega_0^2\sin\theta)=\ms\Ls^2\Omega_0^2\dot{\theta}\cos\theta</math>
|}


------
------

==E1.2 Power of an action-reaction pair==

The total power associated with internal interactions in a constant-mass system is not zero in principle (since the two forces of each action-reaction pair are applied to particles that may have different velocities relative to the reference frame in which the calculation is made), and it is independent of the reference frame in which it is evaluated. It only depends on the separation velocity of the two particles. The power balance for the system is therefore written as: <math>\dot{\Ts}_\Rs^\text{syst}=\dot{\Ws}_\Rs^\text{ext} + \dot{\Ws}^\text{int} + \dot{\Ws}_\Rs^\text{ar} </math> .

<div>
=====💭 PROOF ➕=====

{|
|
:[[File:E1-2-neut.png|thumb|left|200px|link=]]
|<math>\dot{\Ws}^\text{ac-reac}=\overline{\Fs}_{\Ps \rightarrow \Qs}\cdot \vel{Q}{R} + \overline{\Fs}_{\Qs \rightarrow \Ps}\cdot \vel{P}{R}= \overline{\Fs}_{\Ps \rightarrow \Qs}\cdot \left[ \vel{Q}{R}-\vel{P}{R}\right]</math>

By the [[D1. Foundational laws of Newtonian dynamics#D1.6 Newton’s third law (action-reaction principle)|'''action-reaction principle''']], <math>\overline{\Fs}_{\Ps \rightarrow \Qs}=-\overline{\Fs}_{\Qs \rightarrow \Ps}</math>. Therefore: <math>\dot{\Ws}^\text{ac-reac}=\overline{\Fs}_{\Ps \rightarrow \Qs}\cdot \left[ \vel{Q}{R}-\vel{P}{R}\right]</math>

As <math>\overline{\Fs}_{\Ps \rightarrow \Qs}</math> has the <math>\PQvec</math> direction, only the components of the velocities in that direction have to be considered:

<math>\dot{\Ws}^\text{ac-reac}=\overline{\Fs}_{\Ps \rightarrow \Qs}\cdot \biggl( \vel{Q}{R} \Bigr]_{||\Ps\Qs}-\ \vel{P}{R} \Bigr]_{||\Ps\Qs}\biggl)</math>
|}

The difference <math>\biggl( \vel{Q}{R} \Bigr]_{\Ps\Qs}-\ \vel{P}{R} \Bigr]_{\Ps\Qs}\biggl)</math> is the separation velocity <math>\dot{\rho}</math> between '''P''' and '''Q''', and it does not dpend on the reference frame since it is the derivative of a scalar (the distance <math>\dot{\rho}</math> between the two particles). For the previous drawing, <math>\dot{\rho}> 0</math>. Since this is an attractive interaction, <math>\dot{\Ws}^\text{ac-reac} < 0</math>.

With a drawing, it is easy to see that:
* attractive interaction and particle approach <math>(\dot{\rho}<0)</math> : <math>\dot{\Ws}^\text{ac-reac} > 0</math> ;
* repulsive interaction and particle separation <math>(\dot{\rho}>0)</math> : <math>\dot{\Ws}^\text{ac-reac} > 0</math>
* repulsive interaction and particle approach <math>(\dot{\rho}<0)</math> : <math>\dot{\Ws}^\text{ac-reac} < 0</math>


</div>

There are two particular cases of interaction forces whose power always has the same sign:

* Friction action-reaction pair between '''P''' and '''Q''': since the friction force always opposes the relative velocity between the two points, its power is always negative:

<math>\hspace{1cm} \dot{\Ws}^\text{friction}=\dot{\Ws}^{\text{fric}\rightarrow\Ps}_\text{RTQ}+\dot{\Ws}^{\text{fric}\rightarrow\Qs}_\text{RTQ}=\overline{\Fs}^{\text{fric}\rightarrow\Ps} \cdot \vel{P}{RTQ} = - \abs{\overline{\Fs}^{\text{fric}\rightarrow\Ps}} \frac{\vel{P}{RTQ}}{\abs{\vel{P}{RTQ}}} \cdot \vel{P}{RTQ}<0;</math>

* Constraint action-reaction pair between '''P''' and '''Q''': from the [[D3. Interactions between rigid bodies#D3.4 Direct constraint interactions|'''analytical characterization equation''']] of the constraint forces between two points '''P''' and '''Q''' belonging to two rigid bodies <math>\mathrm{S}_\Ps</math> and <math>\mathrm{S}_\Qs</math>, it is easy to see that the associated power is zero:

<math>\hspace{1cm} \dot{\Ws}^\text{constraint}=\dot{\Ws}^{\text{constraint}\rightarrow\Ps}_{\mathrm{S}_\Ps}+\dot{\Ws}^{\text{constraint}\rightarrow\Qs}_{\mathrm{S}_\Ps}=\overline{\Fs}^{\text{constraint}\rightarrow\Ps} \cdot \overline{\mathbf{v}}_{\mathrm{S}_\Qs}(\Ps) = - 0;</math>

====✏️ EXAMPLE E1-2.1: power of an action-reaction friction pair====
------
:
{|
|
:[[File:ExE1-2-1-2-eng.png|thumb|left|180px|link=]]
|A homogeneous wheel, with radius r and mass m, moves down an inclined plane without sliding until it joins a conveyor belt moving at a constant velocity <math>\vs_0</math> relative to the ground. At the moment it makes contact with the belt, the center of the wheel has a velocity <math>\vs_0</math> relative to the ground in the same direction as the belt. We want to calculate the power of the friction force between the belt and the wheel at this instant.
|}

At that momento, the <math>\mathbf{ICR}^\text{wheel}_\Es</math> is point <math>\Js_\text{wheel}</math> in contact with the belt. This point slides backwards with respect to the belt, therefore the friction force on the wheel is forwards:

<math>\left.\begin{array}{l}
\text { AB: ground } \\
\text { REL: belt }
\end{array}\right\} \quad \bar{\vs}_{\text {REL }}\left(\Js_{\text {wheel }}\right)=\bar{\vs}_{\text{AB}}\left(\Js_{\text {wheel }}\right)-\bar{\vs}_{\mathrm{ar}}\left(\Js_{\text {wheel }}\right)=-\bar{\vs}_{\mathrm{ar}}\left(\Js_{\text {wheel }}\right)=-\left(\rightarrow \mathrm{v}_0\right)=\left(\leftarrow \mathrm{v}_0\right) .</math>

[[File:ExE1-2-1-3-eng.png|thumb|center|500px|link=]]

Calculation of the friction power: <math>\dot{\Ws}^\text{friction}=\dot{\Ws}_\Rs^{\text{friction}\rightarrow \text{wheel}}+ \dot{\Ws}_\Rs^{\text{friction}\rightarrow \text{belt}}</math>, where R may be any reference frame. To quickly calculate this power, it is advisable to choose a reference frame R where one of the two points subjected to the action-reaction pair has zero speed, so that one of the two terms is directly zero:

<math>\dot{\Ws}^\text{friction}=\dot{\Ws}_\text{belt}^{\text{fric}\rightarrow \text{wheel}}+ \dot{\Ws}_\text{belt}^{\text{fric }\rightarrow \text{belt}}=\overline{\Fs}^{\text{fric}\rightarrow \text{wheel}} \cdot \bar{\vs}_{\text {belt}}\left(\Js_{\text {wheel }}\right)=(\rightarrow \mu\ms\gs)\cdot(\leftarrow \vs_0)=-\mu\ms\gs\vs_0.</math>

If the calculation is done in any other reference frame, the result is the same:

<math>\dot{\Ws}^\text{friction}=\dot{\Ws}_\text{E}^{\text{fric}\rightarrow \text{wheel}}+ \dot{\Ws}_\text{E}^{\text{fric} \rightarrow \text{belt}}=(\rightarrow \mu\ms\gs)\cdot 0 + (\leftarrow \mu\ms\gs)\cdot(\rightarrow \vs_0)=-\mu\ms\gs\vs_0.</math>

<math>\dot{\Ws}^\text{friction}=\dot{\Ws}_\text{vehicle}^{\text{fric}\rightarrow \text{wheel}}+ \dot{\Ws}_\text{vehicle}^{\text{fric} \rightarrow \text{belt}}=(\rightarrow \mu\ms\gs)\cdot (\rightarrow\vs_0) + (\leftarrow \mu\ms\gs) \cdot(\rightarrow 3\vs_0)=-\mu\ms\gs\vs_0.</math>

However, if only the power of one of the two forces is calculated (for example, the friction on the wheel), the result depends on the reference frame and can have any sign:

<math>\dot{\Ws}_\text{belt}^{\text{fric }\rightarrow \text{wheel}}=(\rightarrow \mu\ms\gs)\cdot(\leftarrow \vs_0)=-\mu\ms\gs\vs_0 ,</math>

<math>\dot{\Ws}_\text{E}^{\text{fric} \rightarrow \text{wheel}}=(\rightarrow \mu\ms\gs)\cdot 0 = 0 ,</math>

<math>\dot{\Ws}_\text{vehicle}^{\text{fric} \rightarrow \text{wheel}}=(\rightarrow \mu\ms\gs)\cdot (\rightarrow\vs_0)=\mu\ms\gs\vs_0.</math>


-----
-----

==E1.3 Power of a system of forces on a rigid body==

{|
|
:[[File:E1-3-eng.png|thumb|left|180px|link=]]
<center>'''Figure E1.3''': Kinematic torsor of a rigid body in a reference frame R, and forces on different points (dm) of that rigid body</center>

|When calculating the power of a system of forces on a rigid body S, the term associated with internal interactions is zero because those interactions are all constraint (cohesion) forces and, as seen in [[E1. Work-Energy Theorem: differential form#E1.2 Power of an action-reaction pair|'''section E1.2''']], their total power is zero.

The total power of a system of forces on S can be calculated from the torsor of the system at any point '''O''' of the rigid body, <math>(\overline{\Fs},\overline{\Ms}(\Os))</math>. Taking into account rigid body kinematics ('''Figure E1.3'''):

<math>\dot{\Ws}_\Rs^{\rightarrow \Ss} =\sum_\Ps\overline{\Fs}^{\rightarrow \Ps} \cdot \vel{P}{R}= \sum_\Ps \left[ \overline{\Fs}^{\rightarrow \Ps} \cdot \vel{O}{R} + \velang{S}{R} \times \OPvec \right] </math>

|}
<math>\dot{\Ws}_\Rs^{\rightarrow \Ss} =\left(\sum_\Ps\overline{\Fs}^{\rightarrow \Ps} \right)\cdot \vel{O}{R} + \sum_\Ps\overline{\Fs}^{\rightarrow \Ps} \cdot \left(\velang{S}{R} \times \OPvec \right) = \left( \sum_\Ps\overline{\Fs}^{\rightarrow \Ps} \right) \cdot \vel{O}{R} + \left[ \sum_\Ps \left( \OPvec \times \overline{\Fs}^{\rightarrow \Ps} \right)\right] \cdot \velang{S}{R} = \overline{\Fs}\cdot \vel{O}{R} + \overline{\Ms}(\Os)\cdot \velang{S}{R} </math>

This calculation also shows that the power of a moment (or a torque <math>\Gamma</math>) on a rigid body in a reference frame R is the product of the moment by the angular velocity of the rigid body: <math>\dot{\Ws}_\Rs^{\Gamma\rightarrow \mathrm{S}} = \overline{\Gamma} \cdot \velang{S}{R}</math>.

The power balance for a rigid body is: <math>\dot{\Ws}_\Rs^{ \rightarrow \mathrm{S}} = \dot{\Ws}_\Rs^{ \mathrm{ext}} + \dot{\Ws}_\Rs^{ \mathrm{ar}}.</math>

====✏️ EXAMPLE E1-3.1: roller sliding on a slope====
------
:
{|
|
:[[File:ExE1-3-1-1-eng.png|thumb|left|180px|link=]]
|A homogeneous roller, with mass m and radius r, slides down a slope. The friction coefficient between the ground and the roller is <math>\mu</math>. We want to calculate the power of every force acting on the roller in the ground reference frame.
|}

:It is a 2D problem. The kinematic and dynamic descriptions are:
{|
|
[[File:ExE1-3-1-2-eng.png|thumb|right|250px|link=]]

|<math> \hspace{0.5cm}\vel{J}{E}=\vel{C}{T} + \velang{}{E} \times \CJvec= </math>

<math>\hspace{1.7cm}=\left( \rightarrow \vs_\Es \right) + \left( \otimes \Omega_\Es\right) \times \left( \downarrow \rs \right)</math>

<math> \hspace{0.5cm}\vel{J}{E} = \left[\rightarrow (\vs_\Es -\rs\Omega_\Es)\right]</math>

<math>\hspace{0.5cm} \dot{\Ws}_\Es^{\mathrm{ext}} = \dot{\Ws}_\Es^{\mathrm{weight}} + \dot{\Ws}_\Es^{\mathrm{friction}} + \dot{\Ws}_\Es^{\mathrm{constraint}}= \Bigr[\left(\downarrow \ms\gs\cos\beta\right)+\left(\rightarrow \ms\gs\sin\beta\right)\Bigr] \cdot (\rightarrow \vs_\Es) + \Bigr[(\leftarrow \mu \Ns) + (\uparrow \Ns ) \Bigr] \cdot \Bigr[\rightarrow (\vs_\Es -\rs\Omega_\Es)\Bigr]</math>

|}
:Since the roller center has no motion perpendicular to the plane, the AMT leads to <math>\Ns =\ms\gs\cos\beta</math>. Therefore:

{|
|
[[Archivo:ExE1-3-1-3-eng.png|thumb|right|140px|link=]]

|<math>\hspace{0.5cm} \dot{\Ws}_\Ts^{\mathrm{ext}} = \ms\gs\vs_\Ts \sin \beta - \ms\gs(\vs_\Ts -\rs\Omega_\Ts)\cos\beta=\ms\gs \Bigr[ \vs_\Ts (\sin\beta-\cos\beta)+\rs\Omega_\Ts \cos\beta \Bigr].</math>

<math>\hspace{0.5cm}</math>

If instead of considering the friction at '''J''' we consider the equivalent torsor at point '''C''', we must consider the force applied in '''C''' and the moment about '''C''', which is not zero:

<math>\hspace{0.5cm} \dot{\Ws}_\Es^{ \mathrm{friction}} = (\leftarrow \mu\Ns)\cdot(\rightarrow\vs_\Es)+(\otimes \mu\Ns\rs)\cdot(\otimes \Omega_\Es) =\mu\ms\gs\cos\beta(\rs\Omega_\Es-\vs_\Es)</math>,

<math>\hspace{0.5cm} \dot{\Ws}_\Es^{\mathrm{ext}} = \dot{\Ws}_\Es^{ \mathrm{weight}} + \dot{\Ws}_\Es^{ \mathrm{friction}}+ \dot{\Ws}_\Es^{ \mathrm{constraint}}= \ms\gs\vs_\Es\sin\beta+\mu\ms\gs\cos\beta (\rs\Omega_\Es-\vs_\Es)</math>.

<math>\hspace{0.5cm}</math>

The result is the same.
|}


====✏️ EXAMPLE E1-3.2: block sliding on a truck====
------
:
{|
|
:[[File:ExE1-3-2-1-eng.png|thumb|left|220px|link=]]
|A truck brakes with acceleration '''a''' relative to the ground at the instant in which its velocity (also relative to the ground) is v. At the same instant, the homogeneous block, of mass m, slides with velocity v' relative to the truck. We want to perform a power balance for the block in two different reference frames: the ground and the truck.
|}

:Ground reference frame: It is a 2D problem. The kinematic and dynamic descriptions are:
{|
|
[[File:ExE1-3-2-2-eng.png|thumb|right|200px|link=]]

| The constraint between the truck and the block allows only one translational DoF between the two. Therefore, in the 2D approximation, the associated torsor has two independent components. If characterized at the midpoint of the contact surface, these components are a normal force and a moment perpendicular to the plane. The AMT applied to the block leads to <math>\Ns=\ms\gs</math>.

|}
:The powers associated with all interactions on the block in the ground reference are:

:<math>\dot{\Ws}_\Es^{\rightarrow \mathrm{bloque}}=\dot{\Ws}_\Ts^{\mathrm{ext}}=\dot{\Ws}_\Ts^{\mathrm{peso}}+\dot{\Ws}_\Ts^{\mathrm{fricción}}+\dot{\Ws}_\Ts^{\mathrm{enlace}}= \Bigr[(\downarrow \ms\gs)+(\leftarrow \mu\ms\gs)+(\uparrow \Ns) \Bigr] \cdot \Bigr[\rightarrow(\vs+\vs')\Bigr] + (\odot \Ms_\es)\cdot 0 = -\mu\ms\gs(\vs+\vs')</math>

:Ya que el bloque se traslada respecto al suelo, su energía cinética total es inmediata, y se puede plantear fácilmente el balance de potencias:

:<math>\Ts_\Ts^\mathrm{bloque}=\frac{1}{2}\int_\mathrm{bloque}\ds\ms \vs_\Ts^2(\mathrm{bloque})=\frac{1}{2}\ms \vs_\Ts^2(\mathrm{bloc})\quad \Rightarrow \quad \dot{\Ts}_\Ts^\mathrm{bloque}=\ms\vs_\Ts( \mathrm{bloque})\dot{\vs}_\Ts( \mathrm{bloque}) </math>

:Como la única fuerza horizontal sobre el bloque es la fricción:

<math>\dot{\vs}_\Ts(\mathrm{bloque})=-\mu\gs\quad \Rightarrow \quad \dot{\Ts}_\Ts^\mathrm{bloque}=-\mu\ms\gs(\vs+\vs').</math>

:El balance de potencias es consistente:

:<math>\left.\begin{array}{l}
\dot{\mathrm{W}}_\Ts^{\rightarrow \text { bloque }}=\dot{\mathrm{W}}_{\mathrm{T}}^{\text {ext }}=-\mu \mathrm{mg}\left(\mathrm{v}+\mathrm{v}^{\prime}\right) \\
\dot{\mathrm{T}}_{\mathrm{T}}^{\text {bloque }}=-\mu \mathrm{mg}\left(\mathrm{v}+\mathrm{v}^{\prime}\right)
\end{array}\right\} \Rightarrow \dot{\mathrm{W}}_{\mathrm{T}}^{\rightarrow \text { bloque }}=\dot{\mathrm{T}}_{\mathrm{T}}^{\text {bloque }}</math>

:Referencia camión: Al estar el camión acelerado respecto al suelo, hay que tener en cuenta las fuerzas de arrastre sobre todos los diferenciales de masa (dm) del bloque:

{|
|

[[Archivo:ExE1-3-2-3-cat-esp.jpg|thumb|right|250px|link=]]
|<math>\cal{\Fs}^{\rightarrow \ds\ms}_{\mathrm{ar}}=-\ds\ms\overline{\as}_\mathrm{ar}(\ds\ms)=-\ds\ms (\leftarrow \as)=(\rightarrow \ds\ms \cdot \as)</math>.

El torsor resultante de este sistema de fuerzas en el centro de inercia '''G''' se reduce a una fuerza horizontal.

<math>\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{bloque}}=\dot{\Ws}_\mathrm{camión}^{\mathrm{ext}} + \dot{\Ws}_\mathrm{camión}^{\mathrm{ar}} =\dot{\Ws}_\mathrm{camión}^{\mathrm{peso}} + \dot{\Ws}_\mathrm{camión}^{\mathrm{fricción}} + \dot{\Ws}_\mathrm{camión}^{\mathrm{enlace}} + \dot{\Ws}_\mathrm{camión}^{\mathrm{ar}} , </math>

<math>\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{bloque}}= \Bigr[(\downarrow \ms\gs)+(\leftarrow \mu\ms\gs)+(\uparrow \Ns) + (\rightarrow \ms\as)\Bigr] \cdot(\rightarrow \vs')+(\odot\Ms_\es) \cdot0=\ms(\as-\mu\gs)\vs'.</math>
|}

:Se puede comprobar que el balance de potencias es consistente:

:<math>\Ts^\mathrm{bloque}_\mathrm{camión}=\frac{1}{2}\ms\vs^2_\mathrm{camión}(\mathrm{bloque}) \quad \Rightarrow \quad \dot{\Ts}^\mathrm{bloque}_\mathrm{camión} = \ms\vs_\mathrm{camión}(\mathrm{bloque})\dot{\vs}_\mathrm{camión}(\mathrm{bloque})</math>

:<math>\left.\begin{array}{l}
\mathrm{AB}:\mathrm{suelo} \\
\mathrm{REL}:\mathrm{camión}
\end{array}\right\} \quad \dot{\vs}_\mathrm{camión}(\mathrm{bloque})= \dot{\vs}_\mathrm{T}(\mathrm{bloque})-\dot{\vs}_\mathrm{ar}(\mathrm{bloque}) = (-\mu\gs)-(-\as)=\as-\mu\gs \quad \Rightarrow \quad \dot{\Ts}^\mathrm{bloque}_\mathrm{camión}=\ms\vs'(\as-\mu\gs)</math>

:<math>\left.\begin{array}{l}
\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{bloque}}=\dot{\Ws}_\mathrm{camión}^{\mathrm{ext}} + \dot{\Ws}_\mathrm{camión}^{\mathrm{ar}}= \ms(\as-\mu\gs)\vs'\\
\dot{\Ts}^\mathrm{bloque}_\mathrm{camión}=\ms\vs'(\as-\mu\gs)
\end{array}\right\} \Rightarrow\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{bloque}}= \dot{\Ts}^\mathrm{bloque}_\mathrm{camión}</math>


====✏️ EJEMPLO E1-3.3: camión frenando con carga====
------
:
{|
|
:[[Archivo:ExE1-3-3-1-esp.png|thumb|left|220px|link=]]
|Un camión frena con aceleración '''a''' respecto al suelo en el instante en el que su velocidad (también respecto al suelo) es v. Los pares de frenada en el eje delantero y en el trasero son iguales y valen <math>\Gamma</math>, y las ruedas, de radio r y masa despreciable, no deslizan sobre el suelo. En el mismo instante, el bloque homogéneo, de masa m, desliz con velocidad v' relativa al camión. Se trata de hacer un balance de potencias para el sistema (bloque+camión) en la referencia del suelo y en la del camión.

|}

:Ya que la potencia de las interacciones internas no es nula en principio (no se trata de un único sólido rígido), es conveniente hacer un DGI del sistema para no olvidar ninguna:
{|
|
[[Archivo:ExE1-3-3-2-esp.png|thumb|right|250px|link=]]

|
|
c.p.s.d.: contacto puntual sin deslizamiento

c.m.c.d.: contacto multipuntual con deslizamiento

|}
:El sistema sobre el que se hace el balance de potencias es todo menos el suelo. Teniendo en cuenta que la potencia de los enlaces internos es siempre nula, las únicas interacciones que hay que tener en cuenta son el peso, los enlaces de las ruedas con el suelo, los pares de frenada y la fricción entre bloque y camión:

[[Archivo:ExE1-3-3-3-esp.png|thumb|center|300px|link=]]

:Referencia suelo:
:Las fuerzas de enlace del suelo sobre las ruedas están aplicadas a puntos de velocidad cero, y por tanto no hacen potencia. El peso es una fuerza vertical, y el movimiento del camión y del bloque es horizontal, por tanto su potencia también es cero.

:<math>\dot{\Ws}_\Ts^{\rightarrow \mathrm{sist}}=\dot{\Ws}_\Ts^{\mathrm{ext}}+\dot{\Ws}^{\mathrm{int}}=\dot{\Ws}_\Ts^{\mathrm{peso}}+\dot{\Ws}_\Ts^{\mathrm{enlace}}+\dot{\Ws}^{\mathrm{fricción}}+\dot{\Ws}^{\Gamma}=\dot{\Ws}^{\mathrm{fricción}}+\dot{\Ws}^{\Gamma}.</math>

:Las potencias de interacciones internas se pueden calcular en cualquier referencia. Por tanto:

:<math>\dot{\Ws}^{\mathrm{fricción}}= \dot{\Ws}_\mathrm{camión}^{\mathrm{fric}\rightarrow \mathrm{bloque}} +\dot{\Ws}_\mathrm{camió}^{\mathrm{fric}\rightarrow \mathrm{camión}} = \dot{\Ws}_\mathrm{camión}^{\mathrm{fric}\rightarrow \mathrm{bloque}}= (\leftarrow \mu\ms\gs)\cdot (\rightarrow \vs')=-\mu\ms\gs\vs'</math>

:<math>\dot{\Ws}^{\Gamma}= \dot{\Ws}_\mathrm{camión}^{\Gamma\rightarrow \mathrm{ruedas}} +\dot{\Ws}_\mathrm{camión}^{\Gamma\rightarrow \mathrm{camión}} = \dot{\Ws}_\mathrm{camión}^{\Gamma\rightarrow \mathrm{ruedas}}= 2 \left(\odot \Gamma \right)\cdot \left(\otimes \frac{\vs}{\rs}\right)=-2\Gamma\frac{\vs}{\rs}.</math>

:Por otro lado: <math>\Ts_\Ts^\mathrm{sist}=\frac{1}{2}\ms\vs_\Ts^2(\mathrm{bloque})+\frac{1}{2}\ms\vs_\Ts^2(\mathrm{camión})\quad \Rightarrow \quad \dot{\Ts}_\Ts^\mathrm{sist}=-\ms(\vs+\vs')\mu\gs-\Ms\vs\as. </math>

:Finalmente: <math>\dot{\Ws}_\Ts^{\rightarrow \mathrm{sist}}=\dot{\Ts}_\Ts^\mathrm{sist} \quad \Rightarrow \quad -\mu\ms\gs\vs'-2\Gamma\frac{\vs}{\rs}=-\ms(\vs+\vs')\mu\gs-\Ms\vs\as.</math>

:Esta ecuación permite calcular el par de frenada: <math>\Gamma=\frac{1}{2}(\mu\ms\gs+\Ms\as)\rs.</math>

:Referencia camión:

:En esta referencia hay que tener en cuenta las fuerzas de arrastre. Por otro lado, las fuerzas de enlace del suelo sobre las ruedas están aplicadas a puntos que ya no tienen velocidad cero, y por tanto hacen potencia. Como antes, la potencia del peso es cero.

:<math>\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{sist}}=\dot{\Ws}_\mathrm{camión}^{\mathrm{ext}}+\dot{\Ws}^{\mathrm{int}}+\dot{\Ws}_\mathrm{camión}^{\mathrm{ar}}=\dot{\Ws}_\mathrm{camión}^{\mathrm{enlace}}+\dot{\Ws}^{\mathrm{fricción}}+\dot{\Ws}^{\Gamma}+\dot{\Ws}_\mathrm{camión}^{\mathrm{ar}}.</math>

:Las fuerzas de arrastre sobre el camión no hacen potencia porque el camión está parado en esta referencia. La de las fuerzas de arrastre sobre el bloque se ha calculado en el [[E1. Teorema de la energía: versión diferencial#✏️ ✏️ EJEMPLO E1-3.2: bloque deslizando sobre un camión|'''ejemplo E1-3.2''']]. La potencia de la fricción y de los pares de frenada es la misma de antes ya que no depende de la referencia.

:<math>\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{sist}}=\dot{\Ws}_\mathrm{camión}^{\mathrm{enllaç}}+\dot{\Ws}^{\mathrm{fricción}}+\dot{\Ws}^{\Gamma}+\dot{\Ws}_\mathrm{camión}^{\mathrm{ar} \rightarrow \mathrm{bloque}}= \dot{\Ws}_\mathrm{camión}^{\mathrm{enlace}} -\mu\ms\gs\vs'-2\Gamma\frac{\vs}{\rs} +\ms\as\vs'.</math>

:Las descripciones cinemática y dinámica de la rueda son:

[[Archivo:ExE1-3-3-4-esp.png|thumb|center|430px|link=]]

:<math>\dot{\Ws}_\mathrm{camión}^{\mathrm{enlace}}=2\Bigr[(\leftarrow \Ts)+(\uparrow \Ns)\Bigr]\cdot(\leftarrow\vs)=2\Ts\vs.</math>

La fuerza tangencial de enlace se puede calcular a partir del TMC aplicado al centro '''C''' de la rueda. Como no tiene masa: <math>\sum\overline{\Ms}_\mathrm{ext}(\Cs)=\overline{0} \quad \Rightarrow \quad (\odot \Gamma)+(\otimes \Ts\rs)=0\quad \Rightarrow \quad \Ts=\Gamma/\rs.</math>

:Por tanto: <math>\dot{\Ws}_\mathrm{camió}^{\rightarrow \mathrm{sist}} =2\frac{\Gamma}{\rs}\vs-\mu\ms\gs\vs'-2\Gamma\frac{\vs}{\rs}+\ms\as\vs'=\ms(\as-\mu\gs)\vs'.</math>

:Por otro lado, como se ha visto en el [[E1. Teorema de la energía: versión diferencial#✏️ ✏️ EJEMPLO E1-3.2: bloque deslizando sobre un camión|'''ejemplo E1-3.2''']]:

:<math>\Ts_\mathrm{camión}^\mathrm{sist}= \Ts_\mathrm{camión}^\mathrm{bloque}=\frac{1}{2}\ms\vs_\mathrm{camión}^2(\mathrm{bloque}) \quad \Rightarrow \quad \dot{\Ts}_\mathrm{camión}^\mathrm{sist}= \dot{\Ts}_\mathrm{camión}^\mathrm{bloque}=\ms\vs'(\as-\mu\gs).</math>

:El balance de potencias es consistente:

:<math>\left.\begin{array}{l}
\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{sist}}= \ms(\as-\mu\gs)\vs'\\
\dot{\Ts}^\mathrm{sist}_\mathrm{camión}=\ms(\as-\mu\gs)\vs'
\end{array}\right\} \Rightarrow\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{sist}}= \dot{\Ts}^\mathrm{sist}_\mathrm{camión}</math>

-----
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==E1.4 Balance de potencias para un sistema multisólido: cálculo directo y cálculo indirecto==

El '''cálculo directo''' (a partir de su definición) de la potencia asociada a una interacción (una fuerza sobre un punto '''P''' o un momento sobre un sólido rígido S) exige el conocimiento de la interacción (fuerza o momento):

<math>\dot{\Ws}_\Rs^{\Fs\rightarrow \Ps}=\overline{\Fs}_{\rightarrow\Ps} \cdot \vel{P}{R} \quad , \quad \dot{\Ws}_\Rs^{\Ms\rightarrow \Ss}=\overline{\Ms}_{\rightarrow\Ss} \cdot \velang{S}{R} </math>

Cuando no se trata de una interacción formulable, hay que recurrir a los teoremas vectoriales para determinarla.

El '''cálculo indirecto''' es una alternativa que no necesita recurrir a estos teoremas. Se basa en la evaluación de un balance de potencias para un sistema y una referencia adecuados de manera que la potencia a determinar sea el único término desconocido de la ecuación (el único que no se puede calcular de manera directa). Cuando el balance incluye más de un término desconocido, se pueden combinar varios balances (para distintos sistemas y referencias) hasta tener un número suficiente de ecuaciones que permitan la determinación de todas las potencias desconocidas.

====✏️ EJEMPLO E1-4.1: polea frenada====
------
:
{|
|
:[[Archivo:ExE1-4-1-1-esp.png|thumb|left|150px|link=]]
|Un bloque de masa m cuelga de un hilo inextensible atado a un punto de la periferia de una polea de radio r. Entre soporte (fijo al suelo) y polea actúa un freno que garantiza que el bloque baja a velocidad constante respecto al suelo. Se trata de calcular la potencia del freno.

|}

:Càlculo directo: <math>\dot{\Ws}^{\mathrm{freno}}= \overline{\Gamma}^{\rightarrow \mathrm{polea}}\cdot\velang{polea}{T}=\left(\odot\Gamma\right)\cdot\left( \otimes \frac{\vs_0}{\rs}\right)=-\Gamma \frac{\vs_0}{\rs}.</math>

:El par de frenada no es un dato, y hay que recurrir al TMC para calcularlo.

:<math>\left.\begin{array}{l}
\text{SISTEMA: polea + bloque}\\
\text{TMC en } \Os
\end{array}\right\} \sum\overline{\Ms}_\mathrm{ext}(\Os)=\dot{\overline{\Hs}}_\mathrm{RTO}(\Os)</math>

:El momento cinético proviene de las dos piezas. Ya que '''O''' no es fijo al bloque, hay que hacer descomposición baricéntrica para calcular el momento cinético del bloque:

:<math>\overline{\Hs}_\mathrm{RTO=T} (\Os) =\overline{\Hs}_\Ts^\mathrm{politja}(\Os) + \overline{\Hs}_\Ts^\mathrm{bloc}(\Os)= \overline{\Hs}_\Ts^\mathrm{politja}(\Os) + \overline{\Hs}_\Ts^\mathrm{bloc}(\Gs_\mathrm{bloc})+ \overline{\Hs}_\Ts^{\oplus\mathrm{bloc}}(\Os) = </math>

:<math>\hspace{2.3cm}=\Is_\mathrm{O}^\mathrm{polea} \velang{polea}{T}+\Is_{\Gs_\mathrm{bloque}}^\mathrm{bloque}+\OGvec_\mathrm{bloque}\times \ms\overline{\vs}_\Ts(\Gs_\mathrm{bloque})=\left(\otimes \Is_\mathrm{O}\frac{\vs_0}{\rs}\right)+(\otimes \ms\vs_0\rs).</math>

:[[Archivo:ExE1-4-1-2-esp.png|thumb|right|180px|link=]]
:Se trata de un momento cinético constante en dirección y valor. Por tanto:

:<math>\dot{\overline{\Hs}}_\mathrm{RTO}(\Os)=0 \quad \Rightarrow \quad \sum{\overline{\Ms}}_\mathrm{ext}(\Os)=(\odot\Gamma)+(\otimes \ms\gs\rs) =0\quad \Rightarrow \quad \Gamma= \ms\gs\rs</math>.

:Finalmente: <math>\dot{\Ws}^{\mathrm{freno}}= -\Gamma \frac{\vs_0}{\rs} = -\ms\gs\vs_0.</math>

:Cálculo indirecto:

:<math>\left.\begin{array}{l}
\text{SISTEMA: polea + soporte + bloque}\\
\text{REF: suelo(Gal)}
\end{array}\right\} \quad \dot{\Ts}^\mathrm{sist}_\Ts=\dot{\Ws}^\mathrm{ext}_\Ts+ \dot{\Ws}^\mathrm{int}</math>

:<math>\left.\begin{array}{l}
\dot{\Ts}^\mathrm{sist}_\Ts=0\\
\dot{\Ws}^\mathrm{ext}_\Ts=\left(\downarrow \ms\gs\right)\cdot\left(\downarrow \vs_0\right)= \ms\gs\vs_0\\
\dot{\Ws}^\mathrm{int}=\dot{\Ws}^{\mathrm{freno}}
\end{array}\right\} \quad \Rightarrow \quad \dot{\Ws}^{\mathrm{freno}} =-\ms\gs\vs_0</math>



====✏️ EJEMPLO E1-4.2: mono y plátanos====
------
:
{|
|
:[[Archivo:ExE1-4-2-1-esp.png|thumb|left|180px|link=]]
|Unos plátanos de masa <math>(2/3)\ms</math> cuelgan de una cuerda inextensible que se apoya sin deslizar sobre una polea de radio r y masa despreciable, articulada respecto al suelo. En el otro extremo de la cuerda se agarra un mono de masa m. La polea gira con velocidad angular constante <math>\Omega_0</math> respecto al suelo bajo la acción de un motor. El mono se mueve respecto a la cuerda sin que sus manos se froten contra la cuerda, y de manera que la distancia de su centro de masas al suelo se mantiene constante. Se trata de calcular la potencia del motor.
|}

:Cálculo directo: <math>\dot{\Ws}^{\mathrm{freno}}= \overline{\Gamma}^{\rightarrow \mathrm{polea}}\cdot\velang{polea}{T}=\left(\odot\Gamma\right)\cdot\left( \odot \Omega_0\right)=\Gamma \Omega_0.</math>

:El par motor no es un dato, y hay que recurrir al TMC para calcularlo.

:<math>\left.\begin{array}{l}
\text{SISTEMA: polea + mono + plátanos}\\
\text{TMC en } \Os
\end{array}\right\} \sum\overline{\Ms}_\mathrm{ext}(\Os)=\dot{\overline{\Hs}}_\mathrm{RTO}(\Os)</math>

:El momento cinético proviene del mono y de los plátanos. En primera aproximación, se puede considerar la masa del mono concentrada en su tronco y su cabeza, de manera que no gira y su velocidad respecto al suelo es nula. Teniendo en cuenta que el punto '''O''' no se mueve respecto al mono pero sí respecto a los plátanos:

:<math>\overline{\Hs}_\mathrm{RTO=T} (\Os) =\overline{\Hs}_\Ts^\mathrm{mono}(\Os) + \overline{\Hs}_\Ts^\mathrm{plátanos}(\Os)= \overline{\Hs}_\Ts^\mathrm{mono}(\Os) + \overline{\Hs}_\Ts^\mathrm{plátanos}(\Gs_\mathrm{plát})+ \OGvec_\mathrm{plát} \times \frac{2}{3} \ms \overline{\vs}_\Ts(\Gs_\mathrm{plát})= </math>

:<math>\hspace{2.3cm}=\OGvec_\mathrm{plát}\times \frac{2}{3} \ms \overline{\vs}_\Ts(\Gs_\mathrm{plát}) = \Bigr[(\rightarrow \rs)+(\downarrow \xs)\Bigr] \times\frac{2}{3} \ms \left(\downarrow \rs\Omega_0\right)=\left(\otimes \frac{2}{3}\ms\rs^2\Omega_0\right).</math>

:Se trata de un momento cinético constante en dirección y valor. Por tanto:

:<math>\dot{\overline{\Hs}}_\mathrm{RTO}(\Os)=0 \quad \Rightarrow \quad \sum{\overline{\Ms}}_\mathrm{ext}(\Os)=(\odot\Gamma)+(\otimes \ms\gs\rs) + \left(\otimes \frac{2}{3}\ms\gs\rs\right)=0\quad \Rightarrow \quad \Gamma= \frac{5}{3}\ms\gs\rs</math>.

:Finalmente: <math>\dot{\Ws}^{\mathrm{motor}}= \Gamma \Omega_0 = \frac{5}{3}\ms\gs\rs\Omega_0.</math>

:Cálculo indirecto

:Hay que tener presente que es un problema con dos tipos de actuadores: el motor y la musculatura del mono. Hay diversos modelos biomecánicos para formular las fuerzas de los músculos, pero quedan fuera del ámbito de este curso. Ya que el mono se mueve, sus músculos desarrollan una potencia no nula.

:<math>\left.\begin{array}{l}
\text{SISTEMA: polea + mono + plátano}\\
\text{REF: suelo(Gal)}
\end{array}\right\} \quad \dot{\Ts}^\mathrm{sist}_\Ts=\dot{\Ws}^\mathrm{ext}_\Ts+ \dot{\Ws}^\mathrm{int}</math>

:<math>\left.\begin{array}{l}
\dot{\Ts}^\mathrm{sist}_\Ts=0\\
\dot{\Ws}^\mathrm{ext}_\Ts=\left(\downarrow \frac{2}{3}\ms\gs\right)\cdot\left(\downarrow \rs\Omega_0\right)= \frac{2}{3}\ms\gs\rs\Omega_0\\
\dot{\Ws}^\mathrm{int}=\dot{\Ws}^{\mathrm{musc}}+\dot{\Ws}^{\mathrm{motor}}
\end{array}\right\} \quad \Rightarrow \quad \dot{\Ws}^{\mathrm{musc}} + \dot{\Ws}^{\mathrm{motor}}=-\frac{2}{3}\ms\gs\rs\Omega_0</math>

:Este balance proporciona una ecuación con dos incógnitas. Se puede plantear un segundo balance para un sistema donde una de las dos no aparezca: un sistema sin el mono o un sistema sin el motor. Si se hace un balance de potencias para el sistema (mono + trozo de cuerda), las fuerzas externas a considerar son el peso y la tensión de la cuerda. Si se sigue trabajando en la referencia del suelo, esta tensión (desconocida) hará potencia no nula ya que está aplicada a un punto de velocidad ascendente.
:[[Archivo:ExE1-4-2-2-cat-esp.png|thumb|right|330px|link=]]

:Para evitar que aparezca un término asociado a la tensión de la cuerda, se puede trabajar en la referencia del trozo de cuerda que se ha considerado, que al tener velocidad ascendente constante es galileana. En esta referencia, el mono baja con velocidad constante:

:<math>\left.\begin{array}{l}
\text{REL: trozo de cuerda}\\
\text{AB: suelo}
\end{array}\right\}</math>

:<math>\overline{\vs}_\mathrm{REL}(\mathrm{mono})=\overline{\vs}_\mathrm{ar}(\mathrm{mono})=-\overline{\vs}_\mathrm{ar}(\mathrm{mono})=(\downarrow \rs\Omega_0)</math>

:<math>\left.\begin{array}{l}
\text{SISTEMA: mono+ trozo de cuerda}\\
\text{REF: trozo de cuerda(Gal)}
\end{array}\right\} \quad \dot{\Ts}^\mathrm{sist}_\mathrm{cuerda} =\dot{\Ws}^\mathrm{ext}_\mathrm{cuerda}+ \dot{\Ws}^\mathrm{int}</math>

:<math>\left.\begin{array}{l}
\dot{\Ts}^\mathrm{sist}_\mathrm{cuerda}=0\\
\dot{\Ws}^\mathrm{ext}_\mathrm{cuerda}=\left(\downarrow \ms\gs\right)\cdot\left(\downarrow \rs\Omega_0\right)= \ms\gs\rs\Omega_0\\
\dot{\Ws}^\mathrm{int}=\dot{\Ws}^{\mathrm{musc}}
\end{array}\right\} \quad \Rightarrow \quad \dot{\Ws}^{\mathrm{musc}} =-\ms\gs\rs\Omega_0</math>

:Combinando este resultado con el del balance anterior:

:<math>\dot{\Ws}^\mathrm{motor}=-\frac{2}{3}\ms\gs\rs\Omega_0-\dot{\Ws}^\mathrm{musc}=-\frac{5}{3}\ms\gs\rs\Omega_0</math>



© Universitat Politècnica de Catalunya. [[Mecánica:Derechos de autor |Todos los derechos reservados]]
-------------

<center>
[[D8. Conservaciones|<<< D8. Conservaciones]]

[[E2. Teorema de la energía: versión integrada|E2. Teorema de la energía: versión integrada>>>]]
</center>

File:ExE1-3-2-2-eng.png

2025-05-20T11:31:29Z

Bros:

ExE1-3-2-2-eng

File:ExE1-3-2-1-eng.png

2025-05-20T11:30:47Z

Bros:

ExE1-3-2-1-eng

E1. Work-Energy Theorem: differential form

2025-05-14T07:39:11Z

Bros:

<div class="noautonum">__TOC__</div>
<math>\newcommand{\uvec}{\overline{\textbf{u}}}
\newcommand{\vvec}{\overline{\textbf{v}}}
\newcommand{\evec}{\overline{\textbf{e}}}
\newcommand{\Omegavec}{\overline{\mathbf{\Omega}}}
\newcommand{\velang}[2]{\Omegavec^{\textrm{#1}}_{\textrm{#2}}}
\newcommand{\Alfavec}{\overline{\mathbf{\alpha}}}
\newcommand{\accang}[2]{\Alfavec^{\textrm{#1}}_{\textrm{#2}}}
\newcommand{\ds}{\textrm{d}}
\newcommand{\ps}{\textrm{p}}
\newcommand{\ns}{\textrm{n}}
\newcommand{\hs}{\textrm{h}}
\newcommand{\cs}{\textrm{c}}
\newcommand{\gs}{\textrm{g}}
\newcommand{\Ns}{\textrm{N}}
\newcommand{\Hs}{\textrm{H}}
\newcommand{\Fs}{\textrm{F}}
\newcommand{\ms}{\textrm{m}}
\newcommand{\ts}{\textrm{t}}
\newcommand{\us}{\textrm{u}}
\newcommand{\vs}{\textrm{v}}
\newcommand{\Rs}{\textrm{R}}
\newcommand{\Ws}{\textrm{W}}
\newcommand{\Ts}{\textrm{T}}
\newcommand{\Ls}{\textrm{L}}
\newcommand{\As}{\textrm{A}}
\newcommand{\Ds}{\textrm{D}}
\newcommand{\Bs}{\textrm{B}}
\newcommand{\es}{\textrm{e}}
\newcommand{\fs}{\textrm{f}}
\newcommand{\is}{\textrm{i}}
\newcommand{\Is}{\textrm{I}}
\newcommand{\ks}{\textrm{k}}
\newcommand{\js}{\textrm{j}}
\newcommand{\rs}{\textrm{r}}
\newcommand{\as}{\textrm{a}}
\newcommand{\ss}{\textrm{s}}
\newcommand{\Os}{\textbf{O}}
\newcommand{\Gs}{\textbf{G}}
\newcommand{\Cbf}{\textbf{C}}
\newcommand{\Or}{\Os_\Rs}
\newcommand{\Qs}{\textbf{Q}}
\newcommand{\Cs}{\textbf{C}}
\newcommand{\Ps}{\textbf{P}}
\newcommand{\Ss}{\textbf{S}}
\newcommand{\Js}{\textbf{J}}
\newcommand{\P}{\textrm{P}}
\newcommand{\Q}{\textrm{Q}}
\newcommand{\Ms}{\textrm{M}}
\newcommand{\deg}{^\textsf{o}}
\newcommand{\xs}{\textsf{x}}
\newcommand{\ys}{\textsf{y}}
\newcommand{\zs}{\textsf{z}}
\newcommand{\dert}[2]{\left.\frac{\ds{#1}}{\ds\ts}\right]_{\textrm{#2}}}
\newcommand{\ddert}[2]{\left.\frac{\ds^2{#1}}{\ds\ts^2}\right]_{\textrm{#2}}}
\newcommand{\vec}[1]{\overline{#1}}
\newcommand{\vecbf}[1]{\overline{\textbf{#1}}}
\newcommand{\vecdot}[1]{\overline{\dot{#1}}}
\newcommand{\OQvec}{\vec{\Os\Qs}}
\newcommand{\GQvec}{\vec{\Gs\Qs}}
\newcommand{\GPvec}{\vec{\Gs\Ps}}
\newcommand{\PSvec}{\vec{\Ps\Ss}}
\newcommand{\QQvec}{\vec{\Qs\Qs}}
\newcommand{\QGvec}{\vec{\Qs\Gs}}
\newcommand{\QPvec}{\vec{\Qs\Ps}}
\newcommand{\JQvec}{\vec{\Js\Qs}}
\newcommand{\PQvec}{\vec{\Ps\Qs}}
\newcommand{\GJvec}{\vec{\Gs\Js}}
\newcommand{\OPvec}{\vec{\Os\textbf{P}}}
\newcommand{\OCvec}{\vec{\Os\Cs}}
\newcommand{\OGvec}{\vec{\Os\Gs}}
\newcommand{\CJvec}{\vec{\Cs\Js}}
\newcommand{\GCvec}{\vec{\Gs\Cs}}
\newcommand{\PGvec}{\vec{\Ps\Gs}}
\newcommand{\abs}[1]{\left|{#1}\right|}
\newcommand{\braq}[2]{\left\{{#1}\right\}_{\textrm{#2}}}
\newcommand{\matriz}[9]{
\begin{bmatrix}
{#1} & {#2} & {#3}\\
{#4} & {#5} & {#6}\\
{#7} & {#8} & {#9}
\end{bmatrix}}
\newcommand{\diag}[3]{
\begin{bmatrix}
{#1} & {0} & {0}\\
{0} & {#2} & {0}\\
{0} & {0} & {#3}
\end{bmatrix}}
\newcommand{\vector}[3]{
\begin{Bmatrix}
{#1}\\
{#2}\\
{#3}
\end{Bmatrix}}
\newcommand{\vecdosd}[2]{
\begin{Bmatrix}
{#1}\\
{#2}
\end{Bmatrix}}
\newcommand{\vel}[2]{\vvec_{\textrm{#2}} (\textbf{#1})}
\newcommand{\acc}[2]{\vecbf{a}_{\textrm{#2}} (\textbf{#1})}
\newcommand{\accs}[2]{\vecbf{a}_{\textrm{#2}}^{\textrm{s}} (\textbf{#1})}
\newcommand{\accn}[2]{\vecbf{a}_{\textrm{#2}}^{\textrm{n}} (\textbf{#1})}
\newcommand{\velo}[1]{\vvec_{\textrm{#1}}}
\newcommand{\accso}[1]{\vecbf{a}_{\textrm{#1}}^{\textrm{s}}}
\newcommand{\accno}[1]{\vecbf{a}_{\textrm{#1}}^{\textrm{n}}}
\newcommand{\re}[2]{\Re_{\textrm{#2}}(\textbf{#1})}
\newcommand{\psio}{\dot{\psi}_0}
\newcommand{\Pll}{\textbf{P}_\textrm{lliure}}
\newcommand{\agal}[1]{\vecbf{a}_{\textrm{Gal}} (#1)}
\newcommand{\angal}[1]{\vecbf{a}_{\textrm{NGal}} (#1)}
\newcommand{\vgal}[1]{\vecbf{v}_{\textrm{Gal}} (#1)}
\newcommand{\fvec}[2]{\overline{\mathbf{F}}_{\textrm{#1}\rightarrow\textrm{#2}}}
\newcommand{\mvec}[2]{\overline{\mathbf{M}}_{\textrm{#1}\rightarrow\textrm{#2}}}
\newcommand{\F}[1]{\vec{\Fs}_{#1}}
\newcommand{\Fcal}[2]{\vec{\cal{F}}^{\text{#1}}_{#2}}
</math>

The '''Work-Energy Theorem''' plays a prominent role in physics, as it bridges all branches of science (electromagnetism, thermodynamics, etc.): all of them deal with energy.

There is a certain parallel between the Vector Theorems and the '''Work-Energy Theorem''': they all allow us to understand the evolution of a magnitude that depends on the mechanical state of the system. In the case of the former, we are dealing with the instantaneous evolution of vector magnitudes (linear momentum, angular momentum), while the latter deals with the finite change (in a finite time interval) of a scalar magnitude ('''mechanical energy''').

This unit presents the differential form of this theorem (the '''power balance'''). The integrated form (which is the Work-Energy Theorem itself) is presented in [[E2. Teorema de la energía: versión integrada|'''unit E2''']].
 
------
------
==E1.1 Power balance in a system of particles==

:[[File:E1-1-eng.png|thumb|right|300px|link=]]
<center>'''Figure E1.1''' Forces on a particle of a system with constant matter</center>

Let us consider a system of constant matter, and let’s apply Newton's second law to each particle (or each mass differential) of this system in a general reference frame R (which may be non-Galilean) ('''Figure E1.1'''):

<math>\F{\rightarrow \Ps} + \Fcal{}{\text{ar}\rightarrow\Ps} + \Fcal{}{\text{Cor}\rightarrow\Ps} =\ms_\Ps\acc{P}{R}</math>

<math>\F{\rightarrow \Ps} </math> is the resultant interaction force on '''P''', and it comes from internal interactions (with particles of the same system) and external ones:

<math>\F{\rightarrow \Ps}=\F{\rightarrow \Ps}^\text{int}+\F{\rightarrow \Ps}^\text{ext}</math>

If both sides of the equation are multiplied by <math>\vel{P}{R}</math>, all forces (interaction and inertia forces) orthogonal to <math>\vel{P}{R}</math> are filtered out. Specifically, since <math>\acc{P}{R}=\accs{P}{R} + \accn{P}{R} </math>, where <math>\accn{P}{R} \perp \vel{P}{R}</math>, and since the Coriolis force is always orthogonal to <math>\vel{P}{R} (\Fcal{}{\text{Cor}\rightarrow\Ps} = -2\ms_\Ps \velang{R}{Gal} \times \vel{P}{R})</math>:

<math>\vel{P}{R} \cdot \left( \F{\rightarrow \Ps}^\text{int}+\F{\rightarrow \Ps}^\text{ext} + \Fcal{}{\text{ar}\rightarrow\Ps} \right) = \ms_\Ps\accs{P}{R} \cdot \vel{P}{R} = \frac{\ds}{\ds \ts} \left(\frac{1}{2}\ms_\Ps \vvec_\Rs^2(\Ps) \right) \equiv \frac{\ds\Ts_\Rs(\Ps)}{\ds\ts}=\dot{\Ts}_\Rs(\Ps)</math>

The scalar function <math>\Ts_\Rs(\Ps)</math> is the '''kinetic energy''' of the particle, and is a function of the [[D1. Foundational laws of Newtonian dynamics#D1.3 Newton’s Principle of Determinacy|'''mechanical state''']] of '''P'''. The scalar product of a force and the velocity of its application point is the '''force power''': <math>\vel{P}{R} \cdot \F{\rightarrow \Ps} \equiv \dot{\Ws}_\Rs^{\F{\rightarrow \Ps}} </math>. Both scalar quantities <math>(\Ts_\Rs(\Ps), \dot{\Ws}_\Rs^{\F{\rightarrow \Ps}})</math> depend on the reference frame. For the same set of interaction forces acting on '''P''', the reference frame determines which are filtered out and which are not.

The equation can be rewritten as <math>\dot{\Ws}_\Rs^{\text{ext}\rightarrow \Ps}+\dot{\Ws}_\Rs^{\text{int}\rightarrow \Ps}+\dot{\Ws}_\Rs^{\text{ar}\rightarrow \Ps} =\dot{\Ts}_\Rs(\Ps)</math>, and constitutes the '''power balance''' for particle '''P''' in the R reference frame. Summing for all particles in the system:

<math>\dot{\Ws}_\Rs^{\text{ext}}+\dot{\Ws}^{\text{int}}+\dot{\Ws}_\Rs^{\text{ar}} =\dot{\Ts}_\Rs^\text{syst}</math>

The calculation of kinetic energy is covered in [[E2. Teorema de la energía: versión integrada|'''unit E2''']]. This unit focuses on the calculation of powers. [[E1. Work-Energy Theorem: differential form#E1.2 Potencia de una pareja de acción-reacción|'''section E1-2''']] shows that the power of internal interactions does not depend on the reference frame (which is why the subscript R does not appear in the <math>\dot{\Ws}^{\text{int}}</math> term of the previous equation).

====✏️ EXAMPLE E1-1.1: free particle====
------
:
{|
|
:[[File:ExE1-1-1-1-eng.jpg|thumb|left|180px|link=]]
|A particle '''P''', with mass m, is released from the ring of a Ferris wheel when it is at its highest point. We want to calculate the power, at this instant, of all forces acting on the particle from three different reference frames: the ground, the ring, and the Ferris wheel cabin.
|}

:Ground reference frame: Since this is a Galilean reference frame, all forces on '''P''' are interaction forces.
{|
|
:[[File:ExE1-1-2- eng.png|thumb|right|230px|link=]]

|<math>\hspace{0.5cm}\sum\dot{\Ws}_\Ts=\dot{\Ws}_\Ts^\text{weight}= \left(\leftarrow \Rs\Omega_0\right) \cdot \left(\downarrow \ms\gs\right)=0</math>
|}

:Ring reference frame: It is a non-Galilean reference, and the transportation force on '''P''' must be taken into account.
{|
|
:[[File:ExE1-1-3-eng.png|thumb|right|250px|link=]]
|<math>\hspace{0.5cm}</math> AB:ground, REL:ring 

<math>\hspace{0.5cm}\vel{P}{REL}=\vel{P}{AB}-\vel{P}{ar}= \left(\leftarrow \Rs\Omega_0\right)-\left(\leftarrow \Rs\Omega_0\right)=\overline{0}</math> 

<math>\hspace{0.5cm}</math>

Since the velocity is zero, the total power is zero.
|}

:Cabin reference frame: It is also a non-Galilean reference, and the transportation force on '''P''' must be taken into account.

{|
|
:[[File:ExE1-1-4-eng.png|thumb|right|250px|link=]]
|<math>\hspace{0.5cm}</math>AB:ground, REL:cabin

<math>\hspace{0.5cm}\vel{P}{REL}=\vel{P}{AB}-\vel{P}{ar}= \left(\leftarrow \Rs\Omega_0\right)-\left(\uparrow \Rs\Omega_0\right)</math>

<math>\hspace{0.5cm}\dot{\Ws}_\text{cabin}^\text{weight}=\left[ \left(\leftarrow \Rs\Omega_0\right)+\left(\downarrow \Rs\Omega_0\right) \right] \cdot \left(\downarrow\ms\gs\right) =\ms\gs\Rs\Omega_0</math>

<math>\hspace{0.5cm}\dot{\Ws}_\text{cabin}^\text{ar}=\left[ \left(\leftarrow \Rs\Omega_0\right)+\left(\downarrow \Rs\Omega_0\right) \right] \cdot \left(\rightarrow\ms\Rs^2\Omega_0^2\right) =-\ms\gs\Rs^2 \Omega_0^3</math>

<math>\hspace{0.5cm}\sum\dot{\Ws}_\text{cabin}=\ms\Rs\Omega_0(\gs-\Rs\Omega_0^2)</math>

|}


====✏️ EXAMPLE E1-1.2: pendulum====
------
:
{|
|
:[[File:ExE1-2-1-eng.png|thumb|left|180px|link=]]
|A particle '''P''' of mass m hangs from an inextensible thread of length L attached to the ceiling. The assembly oscillates in a vertical plane fixed to the ground. We want to calculate the power of all the forces acting on the particle in two different reference frames: the ground and a platform rotating relative to the ground with a constant angular velocity <math>\Omega_0</math>. 

Ground reference frame: Since it is a Galilean reference frame, '''P''' is only subjected to interaction forces.
{|
|
:[[File:ExE1-2-2-neut.png|thumb|left|300px|link=]]

|<math>\dot{\Ws}_\Es^\Ts=(\nearrow\Ls\dot{\theta})\cdot(\nwarrow\Ts)=0</math>

<math>\dot{\Ws}_\Es^\text{weight}=(\nearrow\Ls\dot{\theta}\cdot(\downarrow \ms\gs)=-\ms\gs\Ls\dot{\theta}\sin\theta</math>

|}

Platform reference frame: It is a non Galilean reference frame, and the transportation force on '''P''' has to be taken into account.

{|
|
:[[File:ExE1-2-3-eng.png|thumb|left|400px|link=]]

|AB:ground, REL:platform

<math>\vel{P}{REL}=\vel{P}{AB}-\vel{P}{ar}=(\nearrow\Ls\dot{\theta})-(\otimes \Ls \Omega_0 \sin \theta)</math>

<math>\dot{\Ws}_\Es^\Ts\left[(\nearrow\Ls\dot{\theta})+(\odot\Ls\Omega_0\sin\theta)\right]\cdot(\nwarrow\Ts)=0</math>

<math>\dot{\Ws}_\Es^\text{weight}\left[(\nearrow\Ls\dot{\theta})+(\odot\Ls\Omega_0\sin\theta)\right]\cdot(\downarrow \ms\gs)=-\ms\gs\Ls\dot{\theta}\sin\theta</math>

<math>\dot{\Ws}_\Es^\text{ar}\left[(\nearrow\Ls\dot{\theta})+(\odot\Ls\Omega_0\sin\theta)\right]\cdot(\rightarrow \ms\Ls\Omega_0^2\sin\theta)=\ms\Ls^2\Omega_0^2\dot{\theta}\cos\theta</math>
|}


------
------

==E1.2 Power of an action-reaction pair==

The total power associated with internal interactions in a constant-mass system is not zero in principle (since the two forces of each action-reaction pair are applied to particles that may have different velocities relative to the reference frame in which the calculation is made), and it is independent of the reference frame in which it is evaluated. It only depends on the separation velocity of the two particles. The power balance for the system is therefore written as: <math>\dot{\Ts}_\Rs^\text{syst}=\dot{\Ws}_\Rs^\text{ext} + \dot{\Ws}^\text{int} + \dot{\Ws}_\Rs^\text{ar} </math> .

<div>
=====💭 PROOF ➕=====

{|
|
:[[File:E1-2-neut.png|thumb|left|200px|link=]]
|<math>\dot{\Ws}^\text{ac-reac}=\overline{\Fs}_{\Ps \rightarrow \Qs}\cdot \vel{Q}{R} + \overline{\Fs}_{\Qs \rightarrow \Ps}\cdot \vel{P}{R}= \overline{\Fs}_{\Ps \rightarrow \Qs}\cdot \left[ \vel{Q}{R}-\vel{P}{R}\right]</math>

By the [[D1. Foundational laws of Newtonian dynamics#D1.6 Newton’s third law (action-reaction principle)|'''action-reaction principle''']], <math>\overline{\Fs}_{\Ps \rightarrow \Qs}=-\overline{\Fs}_{\Qs \rightarrow \Ps}</math>. Therefore: <math>\dot{\Ws}^\text{ac-reac}=\overline{\Fs}_{\Ps \rightarrow \Qs}\cdot \left[ \vel{Q}{R}-\vel{P}{R}\right]</math>

As <math>\overline{\Fs}_{\Ps \rightarrow \Qs}</math> has the <math>\PQvec</math> direction, only the components of the velocities in that direction have to be considered:

<math>\dot{\Ws}^\text{ac-reac}=\overline{\Fs}_{\Ps \rightarrow \Qs}\cdot \biggl( \vel{Q}{R} \Bigr]_{||\Ps\Qs}-\ \vel{P}{R} \Bigr]_{||\Ps\Qs}\biggl)</math>
|}

The difference <math>\biggl( \vel{Q}{R} \Bigr]_{\Ps\Qs}-\ \vel{P}{R} \Bigr]_{\Ps\Qs}\biggl)</math> is the separation velocity <math>\dot{\rho}</math> between '''P''' and '''Q''', and it does not dpend on the reference frame since it is the derivative of a scalar (the distance <math>\dot{\rho}</math> between the two particles). For the previous drawing, <math>\dot{\rho}> 0</math>. Since this is an attractive interaction, <math>\dot{\Ws}^\text{ac-reac} < 0</math>.

With a drawing, it is easy to see that:
* attractive interaction and particle approach <math>(\dot{\rho}<0)</math> : <math>\dot{\Ws}^\text{ac-reac} > 0</math> ;
* repulsive interaction and particle separation <math>(\dot{\rho}>0)</math> : <math>\dot{\Ws}^\text{ac-reac} > 0</math>
* repulsive interaction and particle approach <math>(\dot{\rho}<0)</math> : <math>\dot{\Ws}^\text{ac-reac} < 0</math>


</div>

There are two particular cases of interaction forces whose power always has the same sign:

* Friction action-reaction pair between '''P''' and '''Q''': since the friction force always opposes the relative velocity between the two points, its power is always negative:

<math>\hspace{1cm} \dot{\Ws}^\text{friction}=\dot{\Ws}^{\text{fric}\rightarrow\Ps}_\text{RTQ}+\dot{\Ws}^{\text{fric}\rightarrow\Qs}_\text{RTQ}=\overline{\Fs}^{\text{fric}\rightarrow\Ps} \cdot \vel{P}{RTQ} = - \abs{\overline{\Fs}^{\text{fric}\rightarrow\Ps}} \frac{\vel{P}{RTQ}}{\abs{\vel{P}{RTQ}}} \cdot \vel{P}{RTQ}<0;</math>

* Constraint action-reaction pair between '''P''' and '''Q''': from the [[D3. Interactions between rigid bodies#D3.4 Direct constraint interactions|'''analytical characterization equation''']] of the constraint forces between two points '''P''' and '''Q''' belonging to two rigid bodies <math>\mathrm{S}_\Ps</math> and <math>\mathrm{S}_\Qs</math>, it is easy to see that the associated power is zero:

<math>\hspace{1cm} \dot{\Ws}^\text{constraint}=\dot{\Ws}^{\text{constraint}\rightarrow\Ps}_{\mathrm{S}_\Ps}+\dot{\Ws}^{\text{constraint}\rightarrow\Qs}_{\mathrm{S}_\Ps}=\overline{\Fs}^{\text{constraint}\rightarrow\Ps} \cdot \overline{\mathbf{v}}_{\mathrm{S}_\Qs}(\Ps) = - 0;</math>

====✏️ EXAMPLE E1-2.1: power of an action-reaction friction pair====
------
:
{|
|
:[[File:ExE1-2-1-2-eng.png|thumb|left|180px|link=]]
|A homogeneous wheel, with radius r and mass m, moves down an inclined plane without sliding until it joins a conveyor belt moving at a constant velocity <math>\vs_0</math> relative to the ground. At the moment it makes contact with the belt, the center of the wheel has a velocity <math>\vs_0</math> relative to the ground in the same direction as the belt. We want to calculate the power of the friction force between the belt and the wheel at this instant.
|}

At that momento, the <math>\mathbf{ICR}^\text{wheel}_\Es</math> is point <math>\Js_\text{wheel}</math> in contact with the belt. This point slides backwards with respect to the belt, therefore the friction force on the wheel is forwards:

<math>\left.\begin{array}{l}
\text { AB: ground } \\
\text { REL: belt }
\end{array}\right\} \quad \bar{\vs}_{\text {REL }}\left(\Js_{\text {wheel }}\right)=\bar{\vs}_{\text{AB}}\left(\Js_{\text {wheel }}\right)-\bar{\vs}_{\mathrm{ar}}\left(\Js_{\text {wheel }}\right)=-\bar{\vs}_{\mathrm{ar}}\left(\Js_{\text {wheel }}\right)=-\left(\rightarrow \mathrm{v}_0\right)=\left(\leftarrow \mathrm{v}_0\right) .</math>

[[File:ExE1-2-1-3-eng.png|thumb|center|500px|link=]]

Calculation of the friction power: <math>\dot{\Ws}^\text{friction}=\dot{\Ws}_\Rs^{\text{friction}\rightarrow \text{wheel}}+ \dot{\Ws}_\Rs^{\text{friction}\rightarrow \text{belt}}</math>, where R may be any reference frame. To quickly calculate this power, it is advisable to choose a reference frame R where one of the two points subjected to the action-reaction pair has zero speed, so that one of the two terms is directly zero:

<math>\dot{\Ws}^\text{friction}=\dot{\Ws}_\text{belt}^{\text{fric}\rightarrow \text{wheel}}+ \dot{\Ws}_\text{belt}^{\text{fric }\rightarrow \text{belt}}=\overline{\Fs}^{\text{fric}\rightarrow \text{wheel}} \cdot \bar{\vs}_{\text {belt}}\left(\Js_{\text {wheel }}\right)=(\rightarrow \mu\ms\gs)\cdot(\leftarrow \vs_0)=-\mu\ms\gs\vs_0.</math>

If the calculation is done in any other reference frame, the result is the same:

<math>\dot{\Ws}^\text{friction}=\dot{\Ws}_\text{E}^{\text{fric}\rightarrow \text{wheel}}+ \dot{\Ws}_\text{E}^{\text{fric} \rightarrow \text{belt}}=(\rightarrow \mu\ms\gs)\cdot 0 + (\leftarrow \mu\ms\gs)\cdot(\rightarrow \vs_0)=-\mu\ms\gs\vs_0.</math>

<math>\dot{\Ws}^\text{friction}=\dot{\Ws}_\text{vehicle}^{\text{fric}\rightarrow \text{wheel}}+ \dot{\Ws}_\text{vehicle}^{\text{fric} \rightarrow \text{belt}}=(\rightarrow \mu\ms\gs)\cdot (\rightarrow\vs_0) + (\leftarrow \mu\ms\gs) \cdot(\rightarrow 3\vs_0)=-\mu\ms\gs\vs_0.</math>

However, if only the power of one of the two forces is calculated (for example, the friction on the wheel), the result depends on the reference frame and can have any sign:

<math>\dot{\Ws}_\text{belt}^{\text{fric }\rightarrow \text{wheel}}=(\rightarrow \mu\ms\gs)\cdot(\leftarrow \vs_0)=-\mu\ms\gs\vs_0 ,</math>

<math>\dot{\Ws}_\text{E}^{\text{fric} \rightarrow \text{wheel}}=(\rightarrow \mu\ms\gs)\cdot 0 = 0 ,</math>

<math>\dot{\Ws}_\text{vehicle}^{\text{fric} \rightarrow \text{wheel}}=(\rightarrow \mu\ms\gs)\cdot (\rightarrow\vs_0)=\mu\ms\gs\vs_0.</math>


-----
-----

==E1.3 Power of a system of forces on a rigid body==

{|
|
:[[File:E1-3-eng.png|thumb|left|180px|link=]]
<center>'''Figure E1.3''': Kinematic torsor of a rigid body in a reference frame R, and forces on different points (dm) of that rigid body</center>

|When calculating the power of a system of forces on a rigid body S, the term associated with internal interactions is zero because those interactions are all constraint (cohesion) forces and, as seen in [[E1. Work-Energy Theorem: differential form#E1.2 Power of an action-reaction pair|'''section E1.2''']], their total power is zero.

The total power of a system of forces on S can be calculated from the torsor of the system at any point '''O''' of the rigid body, <math>(\overline{\Fs},\overline{\Ms}(\Os))</math>. Taking into account rigid body kinematics ('''Figure E1.3'''):

<math>\dot{\Ws}_\Rs^{\rightarrow \Ss} =\sum_\Ps\overline{\Fs}^{\rightarrow \Ps} \cdot \vel{P}{R}= \sum_\Ps \left[ \overline{\Fs}^{\rightarrow \Ps} \cdot \vel{O}{R} + \velang{S}{R} \times \OPvec \right] </math>

|}
<math>\dot{\Ws}_\Rs^{\rightarrow \Ss} =\left(\sum_\Ps\overline{\Fs}^{\rightarrow \Ps} \right)\cdot \vel{O}{R} + \sum_\Ps\overline{\Fs}^{\rightarrow \Ps} \cdot \left(\velang{S}{R} \times \OPvec \right) = \left( \sum_\Ps\overline{\Fs}^{\rightarrow \Ps} \right) \cdot \vel{O}{R} + \left[ \sum_\Ps \left( \OPvec \times \overline{\Fs}^{\rightarrow \Ps} \right)\right] \cdot \velang{S}{R} = \overline{\Fs}\cdot \vel{O}{R} + \overline{\Ms}(\Os)\cdot \velang{S}{R} </math>

This calculation also shows that the power of a moment (or a torque <math>\Gamma</math>) on a rigid body in a reference frame R is the product of the moment by the angular velocity of the rigid body: <math>\dot{\Ws}_\Rs^{\Gamma\rightarrow \mathrm{S}} = \overline{\Gamma} \cdot \velang{S}{R}</math>.

The power balance for a rigid body is: <math>\dot{\Ws}_\Rs^{ \rightarrow \mathrm{S}} = \dot{\Ws}_\Rs^{ \mathrm{ext}} + \dot{\Ws}_\Rs^{ \mathrm{ar}}.</math>

====✏️ EXAMPLE E1-3.1: roller sliding on a slope====
------
:
{|
|
:[[File:ExE1-3-1-1-eng.png|thumb|left|180px|link=]]
|A homogeneous roller, with mass m and radius r, slides down a slope. The friction coefficient between the ground and the roller is <math>\mu</math>. We want to calculate the power of every force acting on the roller in the ground reference frame.
|}

:It is a 2D problem. The kinematic and dynamic descriptions are:
{|
|
[[File:ExE1-3-1-2-eng.png|thumb|right|250px|link=]]

|<math> \hspace{0.5cm}\vel{J}{E}=\vel{C}{T} + \velang{}{E} \times \CJvec= </math>

<math>\hspace{1.7cm}=\left( \rightarrow \vs_\Es \right) + \left( \otimes \Omega_\Es\right) \times \left( \downarrow \rs \right)</math>

<math> \hspace{0.5cm}\vel{J}{E} = \left[\rightarrow (\vs_\Es -\rs\Omega_\Es)\right]</math>

<math>\hspace{0.5cm} \dot{\Ws}_\Es^{\mathrm{ext}} = \dot{\Ws}_\Es^{\mathrm{weight}} + \dot{\Ws}_\Es^{\mathrm{friction}} + \dot{\Ws}_\Es^{\mathrm{constraint}}= \Bigr[\left(\downarrow \ms\gs\cos\beta\right)+\left(\rightarrow \ms\gs\sin\beta\right)\Bigr] \cdot (\rightarrow \vs_\Es) + \Bigr[(\leftarrow \mu \Ns) + (\uparrow \Ns ) \Bigr] \cdot \Bigr[\rightarrow (\vs_\Es -\rs\Omega_\Es)\Bigr]</math>

|}
:Since the roller center has no motion perpendicular to the plane, the AMT leads to <math>\Ns =\ms\gs\cos\beta</math>. Therefore:

{|
|
[[Archivo:ExE1-3-1-3-eng.png|thumb|right|140px|link=]]

|<math>\hspace{0.5cm} \dot{\Ws}_\Ts^{\mathrm{ext}} = \ms\gs\vs_\Ts \sin \beta - \ms\gs(\vs_\Ts -\rs\Omega_\Ts)\cos\beta=\ms\gs \Bigr[ \vs_\Ts (\sin\beta-\cos\beta)+\rs\Omega_\Ts \cos\beta \Bigr].</math>

<math>\hspace{0.5cm}</math>

Si en lugar de considerar la fricción en '''J''' se considera el torsor equivalente en '''C''', hay que considerar la fuerza aplicada en '''C''' y el momento respecto a '''C''', que no es nulo:

<math>\hspace{0.5cm} \dot{\Ws}_\Ts^{ \mathrm{fricción}} = (\leftarrow \mu\Ns)\cdot(\rightarrow\vs_\Ts)+(\otimes \mu\Ns\rs)\cdot(\otimes \Omega_\Ts) =\mu\ms\gs\cos\beta(\rs\Omega_\Ts-\vs_\Ts)</math>,

<math>\hspace{0.5cm} \dot{\Ws}_\Ts^{\mathrm{ext}} = \dot{\Ws}_\Ts^{ \mathrm{peso}} + \dot{\Ws}_\Ts^{ \mathrm{fricción}}+ \dot{\Ws}_\Ts^{ \mathrm{enlace}}= \ms\gs\vs_\Ts\sin\beta+\mu\ms\gs\cos\beta (\rs\Omega_\Ts-\vs_\Ts)</math>.

<math>\hspace{0.5cm}</math>El resultado es el mismo.
|}


====✏️ EJEMPLO E1-3.2: bloque deslizando sobre un camión====
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:
{|
|
:[[Archivo:ExE1-3-2-1-esp.png|thumb|left|220px|link=]]
|Un camión frena con aceleración '''a''' respecto al suelo en el instante en el que su velocidad (también respecto al suelo) es v. En el mismo instante, el bloque homogéneo, de masa m, desliza con velocidad v' relativa al camión. Se trata de hacer un balance de potencias para el bloque en la referencia del suelo y en la del camión.

|}

:Referencia suelo: Se trata de un problema plano. Las descripciones cinemática y dinámica son:
{|
|
[[Archivo:ExE1-3-2-2-cat-esp.png|thumb|right|200px|link=]]

| El enlace entre camión y bloque permite sólo un GL de translación entre los dos. Por tanto, en la aproximación 2D, el torsor asociado tiene dos componentes independientes. Si se caracteriza en el punto medio del contacto entre los dos elementos, son una fuerza normal y un momento perpendicular al plano. El TCM aplicado al bloque conduce a <math>\Ns=\ms\gs</math>.

|}
:Las potencias asociadas a todas las interacciones sobre el bloque en la referencia del suelo son:

:<math>\dot{\Ws}_\Ts^{\rightarrow \mathrm{bloque}}=\dot{\Ws}_\Ts^{\mathrm{ext}}=\dot{\Ws}_\Ts^{\mathrm{peso}}+\dot{\Ws}_\Ts^{\mathrm{fricción}}+\dot{\Ws}_\Ts^{\mathrm{enlace}}= \Bigr[(\downarrow \ms\gs)+(\leftarrow \mu\ms\gs)+(\uparrow \Ns) \Bigr] \cdot \Bigr[\rightarrow(\vs+\vs')\Bigr] + (\odot \Ms_\es)\cdot 0 = -\mu\ms\gs(\vs+\vs')</math>

:Ya que el bloque se traslada respecto al suelo, su energía cinética total es inmediata, y se puede plantear fácilmente el balance de potencias:

:<math>\Ts_\Ts^\mathrm{bloque}=\frac{1}{2}\int_\mathrm{bloque}\ds\ms \vs_\Ts^2(\mathrm{bloque})=\frac{1}{2}\ms \vs_\Ts^2(\mathrm{bloc})\quad \Rightarrow \quad \dot{\Ts}_\Ts^\mathrm{bloque}=\ms\vs_\Ts( \mathrm{bloque})\dot{\vs}_\Ts( \mathrm{bloque}) </math>

:Como la única fuerza horizontal sobre el bloque es la fricción:

<math>\dot{\vs}_\Ts(\mathrm{bloque})=-\mu\gs\quad \Rightarrow \quad \dot{\Ts}_\Ts^\mathrm{bloque}=-\mu\ms\gs(\vs+\vs').</math>

:El balance de potencias es consistente:

:<math>\left.\begin{array}{l}
\dot{\mathrm{W}}_\Ts^{\rightarrow \text { bloque }}=\dot{\mathrm{W}}_{\mathrm{T}}^{\text {ext }}=-\mu \mathrm{mg}\left(\mathrm{v}+\mathrm{v}^{\prime}\right) \\
\dot{\mathrm{T}}_{\mathrm{T}}^{\text {bloque }}=-\mu \mathrm{mg}\left(\mathrm{v}+\mathrm{v}^{\prime}\right)
\end{array}\right\} \Rightarrow \dot{\mathrm{W}}_{\mathrm{T}}^{\rightarrow \text { bloque }}=\dot{\mathrm{T}}_{\mathrm{T}}^{\text {bloque }}</math>

:Referencia camión: Al estar el camión acelerado respecto al suelo, hay que tener en cuenta las fuerzas de arrastre sobre todos los diferenciales de masa (dm) del bloque:

{|
|

[[Archivo:ExE1-3-2-3-cat-esp.jpg|thumb|right|250px|link=]]
|<math>\cal{\Fs}^{\rightarrow \ds\ms}_{\mathrm{ar}}=-\ds\ms\overline{\as}_\mathrm{ar}(\ds\ms)=-\ds\ms (\leftarrow \as)=(\rightarrow \ds\ms \cdot \as)</math>.

El torsor resultante de este sistema de fuerzas en el centro de inercia '''G''' se reduce a una fuerza horizontal.

<math>\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{bloque}}=\dot{\Ws}_\mathrm{camión}^{\mathrm{ext}} + \dot{\Ws}_\mathrm{camión}^{\mathrm{ar}} =\dot{\Ws}_\mathrm{camión}^{\mathrm{peso}} + \dot{\Ws}_\mathrm{camión}^{\mathrm{fricción}} + \dot{\Ws}_\mathrm{camión}^{\mathrm{enlace}} + \dot{\Ws}_\mathrm{camión}^{\mathrm{ar}} , </math>

<math>\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{bloque}}= \Bigr[(\downarrow \ms\gs)+(\leftarrow \mu\ms\gs)+(\uparrow \Ns) + (\rightarrow \ms\as)\Bigr] \cdot(\rightarrow \vs')+(\odot\Ms_\es) \cdot0=\ms(\as-\mu\gs)\vs'.</math>
|}

:Se puede comprobar que el balance de potencias es consistente:

:<math>\Ts^\mathrm{bloque}_\mathrm{camión}=\frac{1}{2}\ms\vs^2_\mathrm{camión}(\mathrm{bloque}) \quad \Rightarrow \quad \dot{\Ts}^\mathrm{bloque}_\mathrm{camión} = \ms\vs_\mathrm{camión}(\mathrm{bloque})\dot{\vs}_\mathrm{camión}(\mathrm{bloque})</math>

:<math>\left.\begin{array}{l}
\mathrm{AB}:\mathrm{suelo} \\
\mathrm{REL}:\mathrm{camión}
\end{array}\right\} \quad \dot{\vs}_\mathrm{camión}(\mathrm{bloque})= \dot{\vs}_\mathrm{T}(\mathrm{bloque})-\dot{\vs}_\mathrm{ar}(\mathrm{bloque}) = (-\mu\gs)-(-\as)=\as-\mu\gs \quad \Rightarrow \quad \dot{\Ts}^\mathrm{bloque}_\mathrm{camión}=\ms\vs'(\as-\mu\gs)</math>

:<math>\left.\begin{array}{l}
\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{bloque}}=\dot{\Ws}_\mathrm{camión}^{\mathrm{ext}} + \dot{\Ws}_\mathrm{camión}^{\mathrm{ar}}= \ms(\as-\mu\gs)\vs'\\
\dot{\Ts}^\mathrm{bloque}_\mathrm{camión}=\ms\vs'(\as-\mu\gs)
\end{array}\right\} \Rightarrow\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{bloque}}= \dot{\Ts}^\mathrm{bloque}_\mathrm{camión}</math>


====✏️ EJEMPLO E1-3.3: camión frenando con carga====
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:
{|
|
:[[Archivo:ExE1-3-3-1-esp.png|thumb|left|220px|link=]]
|Un camión frena con aceleración '''a''' respecto al suelo en el instante en el que su velocidad (también respecto al suelo) es v. Los pares de frenada en el eje delantero y en el trasero son iguales y valen <math>\Gamma</math>, y las ruedas, de radio r y masa despreciable, no deslizan sobre el suelo. En el mismo instante, el bloque homogéneo, de masa m, desliz con velocidad v' relativa al camión. Se trata de hacer un balance de potencias para el sistema (bloque+camión) en la referencia del suelo y en la del camión.

|}

:Ya que la potencia de las interacciones internas no es nula en principio (no se trata de un único sólido rígido), es conveniente hacer un DGI del sistema para no olvidar ninguna:
{|
|
[[Archivo:ExE1-3-3-2-esp.png|thumb|right|250px|link=]]

|
|
c.p.s.d.: contacto puntual sin deslizamiento

c.m.c.d.: contacto multipuntual con deslizamiento

|}
:El sistema sobre el que se hace el balance de potencias es todo menos el suelo. Teniendo en cuenta que la potencia de los enlaces internos es siempre nula, las únicas interacciones que hay que tener en cuenta son el peso, los enlaces de las ruedas con el suelo, los pares de frenada y la fricción entre bloque y camión:

[[Archivo:ExE1-3-3-3-esp.png|thumb|center|300px|link=]]

:Referencia suelo:
:Las fuerzas de enlace del suelo sobre las ruedas están aplicadas a puntos de velocidad cero, y por tanto no hacen potencia. El peso es una fuerza vertical, y el movimiento del camión y del bloque es horizontal, por tanto su potencia también es cero.

:<math>\dot{\Ws}_\Ts^{\rightarrow \mathrm{sist}}=\dot{\Ws}_\Ts^{\mathrm{ext}}+\dot{\Ws}^{\mathrm{int}}=\dot{\Ws}_\Ts^{\mathrm{peso}}+\dot{\Ws}_\Ts^{\mathrm{enlace}}+\dot{\Ws}^{\mathrm{fricción}}+\dot{\Ws}^{\Gamma}=\dot{\Ws}^{\mathrm{fricción}}+\dot{\Ws}^{\Gamma}.</math>

:Las potencias de interacciones internas se pueden calcular en cualquier referencia. Por tanto:

:<math>\dot{\Ws}^{\mathrm{fricción}}= \dot{\Ws}_\mathrm{camión}^{\mathrm{fric}\rightarrow \mathrm{bloque}} +\dot{\Ws}_\mathrm{camió}^{\mathrm{fric}\rightarrow \mathrm{camión}} = \dot{\Ws}_\mathrm{camión}^{\mathrm{fric}\rightarrow \mathrm{bloque}}= (\leftarrow \mu\ms\gs)\cdot (\rightarrow \vs')=-\mu\ms\gs\vs'</math>

:<math>\dot{\Ws}^{\Gamma}= \dot{\Ws}_\mathrm{camión}^{\Gamma\rightarrow \mathrm{ruedas}} +\dot{\Ws}_\mathrm{camión}^{\Gamma\rightarrow \mathrm{camión}} = \dot{\Ws}_\mathrm{camión}^{\Gamma\rightarrow \mathrm{ruedas}}= 2 \left(\odot \Gamma \right)\cdot \left(\otimes \frac{\vs}{\rs}\right)=-2\Gamma\frac{\vs}{\rs}.</math>

:Por otro lado: <math>\Ts_\Ts^\mathrm{sist}=\frac{1}{2}\ms\vs_\Ts^2(\mathrm{bloque})+\frac{1}{2}\ms\vs_\Ts^2(\mathrm{camión})\quad \Rightarrow \quad \dot{\Ts}_\Ts^\mathrm{sist}=-\ms(\vs+\vs')\mu\gs-\Ms\vs\as. </math>

:Finalmente: <math>\dot{\Ws}_\Ts^{\rightarrow \mathrm{sist}}=\dot{\Ts}_\Ts^\mathrm{sist} \quad \Rightarrow \quad -\mu\ms\gs\vs'-2\Gamma\frac{\vs}{\rs}=-\ms(\vs+\vs')\mu\gs-\Ms\vs\as.</math>

:Esta ecuación permite calcular el par de frenada: <math>\Gamma=\frac{1}{2}(\mu\ms\gs+\Ms\as)\rs.</math>

:Referencia camión:

:En esta referencia hay que tener en cuenta las fuerzas de arrastre. Por otro lado, las fuerzas de enlace del suelo sobre las ruedas están aplicadas a puntos que ya no tienen velocidad cero, y por tanto hacen potencia. Como antes, la potencia del peso es cero.

:<math>\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{sist}}=\dot{\Ws}_\mathrm{camión}^{\mathrm{ext}}+\dot{\Ws}^{\mathrm{int}}+\dot{\Ws}_\mathrm{camión}^{\mathrm{ar}}=\dot{\Ws}_\mathrm{camión}^{\mathrm{enlace}}+\dot{\Ws}^{\mathrm{fricción}}+\dot{\Ws}^{\Gamma}+\dot{\Ws}_\mathrm{camión}^{\mathrm{ar}}.</math>

:Las fuerzas de arrastre sobre el camión no hacen potencia porque el camión está parado en esta referencia. La de las fuerzas de arrastre sobre el bloque se ha calculado en el [[E1. Teorema de la energía: versión diferencial#✏️ ✏️ EJEMPLO E1-3.2: bloque deslizando sobre un camión|'''ejemplo E1-3.2''']]. La potencia de la fricción y de los pares de frenada es la misma de antes ya que no depende de la referencia.

:<math>\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{sist}}=\dot{\Ws}_\mathrm{camión}^{\mathrm{enllaç}}+\dot{\Ws}^{\mathrm{fricción}}+\dot{\Ws}^{\Gamma}+\dot{\Ws}_\mathrm{camión}^{\mathrm{ar} \rightarrow \mathrm{bloque}}= \dot{\Ws}_\mathrm{camión}^{\mathrm{enlace}} -\mu\ms\gs\vs'-2\Gamma\frac{\vs}{\rs} +\ms\as\vs'.</math>

:Las descripciones cinemática y dinámica de la rueda son:

[[Archivo:ExE1-3-3-4-esp.png|thumb|center|430px|link=]]

:<math>\dot{\Ws}_\mathrm{camión}^{\mathrm{enlace}}=2\Bigr[(\leftarrow \Ts)+(\uparrow \Ns)\Bigr]\cdot(\leftarrow\vs)=2\Ts\vs.</math>

La fuerza tangencial de enlace se puede calcular a partir del TMC aplicado al centro '''C''' de la rueda. Como no tiene masa: <math>\sum\overline{\Ms}_\mathrm{ext}(\Cs)=\overline{0} \quad \Rightarrow \quad (\odot \Gamma)+(\otimes \Ts\rs)=0\quad \Rightarrow \quad \Ts=\Gamma/\rs.</math>

:Por tanto: <math>\dot{\Ws}_\mathrm{camió}^{\rightarrow \mathrm{sist}} =2\frac{\Gamma}{\rs}\vs-\mu\ms\gs\vs'-2\Gamma\frac{\vs}{\rs}+\ms\as\vs'=\ms(\as-\mu\gs)\vs'.</math>

:Por otro lado, como se ha visto en el [[E1. Teorema de la energía: versión diferencial#✏️ ✏️ EJEMPLO E1-3.2: bloque deslizando sobre un camión|'''ejemplo E1-3.2''']]:

:<math>\Ts_\mathrm{camión}^\mathrm{sist}= \Ts_\mathrm{camión}^\mathrm{bloque}=\frac{1}{2}\ms\vs_\mathrm{camión}^2(\mathrm{bloque}) \quad \Rightarrow \quad \dot{\Ts}_\mathrm{camión}^\mathrm{sist}= \dot{\Ts}_\mathrm{camión}^\mathrm{bloque}=\ms\vs'(\as-\mu\gs).</math>

:El balance de potencias es consistente:

:<math>\left.\begin{array}{l}
\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{sist}}= \ms(\as-\mu\gs)\vs'\\
\dot{\Ts}^\mathrm{sist}_\mathrm{camión}=\ms(\as-\mu\gs)\vs'
\end{array}\right\} \Rightarrow\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{sist}}= \dot{\Ts}^\mathrm{sist}_\mathrm{camión}</math>

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==E1.4 Balance de potencias para un sistema multisólido: cálculo directo y cálculo indirecto==

El '''cálculo directo''' (a partir de su definición) de la potencia asociada a una interacción (una fuerza sobre un punto '''P''' o un momento sobre un sólido rígido S) exige el conocimiento de la interacción (fuerza o momento):

<math>\dot{\Ws}_\Rs^{\Fs\rightarrow \Ps}=\overline{\Fs}_{\rightarrow\Ps} \cdot \vel{P}{R} \quad , \quad \dot{\Ws}_\Rs^{\Ms\rightarrow \Ss}=\overline{\Ms}_{\rightarrow\Ss} \cdot \velang{S}{R} </math>

Cuando no se trata de una interacción formulable, hay que recurrir a los teoremas vectoriales para determinarla.

El '''cálculo indirecto''' es una alternativa que no necesita recurrir a estos teoremas. Se basa en la evaluación de un balance de potencias para un sistema y una referencia adecuados de manera que la potencia a determinar sea el único término desconocido de la ecuación (el único que no se puede calcular de manera directa). Cuando el balance incluye más de un término desconocido, se pueden combinar varios balances (para distintos sistemas y referencias) hasta tener un número suficiente de ecuaciones que permitan la determinación de todas las potencias desconocidas.

====✏️ EJEMPLO E1-4.1: polea frenada====
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:
{|
|
:[[Archivo:ExE1-4-1-1-esp.png|thumb|left|150px|link=]]
|Un bloque de masa m cuelga de un hilo inextensible atado a un punto de la periferia de una polea de radio r. Entre soporte (fijo al suelo) y polea actúa un freno que garantiza que el bloque baja a velocidad constante respecto al suelo. Se trata de calcular la potencia del freno.

|}

:Càlculo directo: <math>\dot{\Ws}^{\mathrm{freno}}= \overline{\Gamma}^{\rightarrow \mathrm{polea}}\cdot\velang{polea}{T}=\left(\odot\Gamma\right)\cdot\left( \otimes \frac{\vs_0}{\rs}\right)=-\Gamma \frac{\vs_0}{\rs}.</math>

:El par de frenada no es un dato, y hay que recurrir al TMC para calcularlo.

:<math>\left.\begin{array}{l}
\text{SISTEMA: polea + bloque}\\
\text{TMC en } \Os
\end{array}\right\} \sum\overline{\Ms}_\mathrm{ext}(\Os)=\dot{\overline{\Hs}}_\mathrm{RTO}(\Os)</math>

:El momento cinético proviene de las dos piezas. Ya que '''O''' no es fijo al bloque, hay que hacer descomposición baricéntrica para calcular el momento cinético del bloque:

:<math>\overline{\Hs}_\mathrm{RTO=T} (\Os) =\overline{\Hs}_\Ts^\mathrm{politja}(\Os) + \overline{\Hs}_\Ts^\mathrm{bloc}(\Os)= \overline{\Hs}_\Ts^\mathrm{politja}(\Os) + \overline{\Hs}_\Ts^\mathrm{bloc}(\Gs_\mathrm{bloc})+ \overline{\Hs}_\Ts^{\oplus\mathrm{bloc}}(\Os) = </math>

:<math>\hspace{2.3cm}=\Is_\mathrm{O}^\mathrm{polea} \velang{polea}{T}+\Is_{\Gs_\mathrm{bloque}}^\mathrm{bloque}+\OGvec_\mathrm{bloque}\times \ms\overline{\vs}_\Ts(\Gs_\mathrm{bloque})=\left(\otimes \Is_\mathrm{O}\frac{\vs_0}{\rs}\right)+(\otimes \ms\vs_0\rs).</math>

:[[Archivo:ExE1-4-1-2-esp.png|thumb|right|180px|link=]]
:Se trata de un momento cinético constante en dirección y valor. Por tanto:

:<math>\dot{\overline{\Hs}}_\mathrm{RTO}(\Os)=0 \quad \Rightarrow \quad \sum{\overline{\Ms}}_\mathrm{ext}(\Os)=(\odot\Gamma)+(\otimes \ms\gs\rs) =0\quad \Rightarrow \quad \Gamma= \ms\gs\rs</math>.

:Finalmente: <math>\dot{\Ws}^{\mathrm{freno}}= -\Gamma \frac{\vs_0}{\rs} = -\ms\gs\vs_0.</math>

:Cálculo indirecto:

:<math>\left.\begin{array}{l}
\text{SISTEMA: polea + soporte + bloque}\\
\text{REF: suelo(Gal)}
\end{array}\right\} \quad \dot{\Ts}^\mathrm{sist}_\Ts=\dot{\Ws}^\mathrm{ext}_\Ts+ \dot{\Ws}^\mathrm{int}</math>

:<math>\left.\begin{array}{l}
\dot{\Ts}^\mathrm{sist}_\Ts=0\\
\dot{\Ws}^\mathrm{ext}_\Ts=\left(\downarrow \ms\gs\right)\cdot\left(\downarrow \vs_0\right)= \ms\gs\vs_0\\
\dot{\Ws}^\mathrm{int}=\dot{\Ws}^{\mathrm{freno}}
\end{array}\right\} \quad \Rightarrow \quad \dot{\Ws}^{\mathrm{freno}} =-\ms\gs\vs_0</math>



====✏️ EJEMPLO E1-4.2: mono y plátanos====
------
:
{|
|
:[[Archivo:ExE1-4-2-1-esp.png|thumb|left|180px|link=]]
|Unos plátanos de masa <math>(2/3)\ms</math> cuelgan de una cuerda inextensible que se apoya sin deslizar sobre una polea de radio r y masa despreciable, articulada respecto al suelo. En el otro extremo de la cuerda se agarra un mono de masa m. La polea gira con velocidad angular constante <math>\Omega_0</math> respecto al suelo bajo la acción de un motor. El mono se mueve respecto a la cuerda sin que sus manos se froten contra la cuerda, y de manera que la distancia de su centro de masas al suelo se mantiene constante. Se trata de calcular la potencia del motor.
|}

:Cálculo directo: <math>\dot{\Ws}^{\mathrm{freno}}= \overline{\Gamma}^{\rightarrow \mathrm{polea}}\cdot\velang{polea}{T}=\left(\odot\Gamma\right)\cdot\left( \odot \Omega_0\right)=\Gamma \Omega_0.</math>

:El par motor no es un dato, y hay que recurrir al TMC para calcularlo.

:<math>\left.\begin{array}{l}
\text{SISTEMA: polea + mono + plátanos}\\
\text{TMC en } \Os
\end{array}\right\} \sum\overline{\Ms}_\mathrm{ext}(\Os)=\dot{\overline{\Hs}}_\mathrm{RTO}(\Os)</math>

:El momento cinético proviene del mono y de los plátanos. En primera aproximación, se puede considerar la masa del mono concentrada en su tronco y su cabeza, de manera que no gira y su velocidad respecto al suelo es nula. Teniendo en cuenta que el punto '''O''' no se mueve respecto al mono pero sí respecto a los plátanos:

:<math>\overline{\Hs}_\mathrm{RTO=T} (\Os) =\overline{\Hs}_\Ts^\mathrm{mono}(\Os) + \overline{\Hs}_\Ts^\mathrm{plátanos}(\Os)= \overline{\Hs}_\Ts^\mathrm{mono}(\Os) + \overline{\Hs}_\Ts^\mathrm{plátanos}(\Gs_\mathrm{plát})+ \OGvec_\mathrm{plát} \times \frac{2}{3} \ms \overline{\vs}_\Ts(\Gs_\mathrm{plát})= </math>

:<math>\hspace{2.3cm}=\OGvec_\mathrm{plát}\times \frac{2}{3} \ms \overline{\vs}_\Ts(\Gs_\mathrm{plát}) = \Bigr[(\rightarrow \rs)+(\downarrow \xs)\Bigr] \times\frac{2}{3} \ms \left(\downarrow \rs\Omega_0\right)=\left(\otimes \frac{2}{3}\ms\rs^2\Omega_0\right).</math>

:Se trata de un momento cinético constante en dirección y valor. Por tanto:

:<math>\dot{\overline{\Hs}}_\mathrm{RTO}(\Os)=0 \quad \Rightarrow \quad \sum{\overline{\Ms}}_\mathrm{ext}(\Os)=(\odot\Gamma)+(\otimes \ms\gs\rs) + \left(\otimes \frac{2}{3}\ms\gs\rs\right)=0\quad \Rightarrow \quad \Gamma= \frac{5}{3}\ms\gs\rs</math>.

:Finalmente: <math>\dot{\Ws}^{\mathrm{motor}}= \Gamma \Omega_0 = \frac{5}{3}\ms\gs\rs\Omega_0.</math>

:Cálculo indirecto

:Hay que tener presente que es un problema con dos tipos de actuadores: el motor y la musculatura del mono. Hay diversos modelos biomecánicos para formular las fuerzas de los músculos, pero quedan fuera del ámbito de este curso. Ya que el mono se mueve, sus músculos desarrollan una potencia no nula.

:<math>\left.\begin{array}{l}
\text{SISTEMA: polea + mono + plátano}\\
\text{REF: suelo(Gal)}
\end{array}\right\} \quad \dot{\Ts}^\mathrm{sist}_\Ts=\dot{\Ws}^\mathrm{ext}_\Ts+ \dot{\Ws}^\mathrm{int}</math>

:<math>\left.\begin{array}{l}
\dot{\Ts}^\mathrm{sist}_\Ts=0\\
\dot{\Ws}^\mathrm{ext}_\Ts=\left(\downarrow \frac{2}{3}\ms\gs\right)\cdot\left(\downarrow \rs\Omega_0\right)= \frac{2}{3}\ms\gs\rs\Omega_0\\
\dot{\Ws}^\mathrm{int}=\dot{\Ws}^{\mathrm{musc}}+\dot{\Ws}^{\mathrm{motor}}
\end{array}\right\} \quad \Rightarrow \quad \dot{\Ws}^{\mathrm{musc}} + \dot{\Ws}^{\mathrm{motor}}=-\frac{2}{3}\ms\gs\rs\Omega_0</math>

:Este balance proporciona una ecuación con dos incógnitas. Se puede plantear un segundo balance para un sistema donde una de las dos no aparezca: un sistema sin el mono o un sistema sin el motor. Si se hace un balance de potencias para el sistema (mono + trozo de cuerda), las fuerzas externas a considerar son el peso y la tensión de la cuerda. Si se sigue trabajando en la referencia del suelo, esta tensión (desconocida) hará potencia no nula ya que está aplicada a un punto de velocidad ascendente.
:[[Archivo:ExE1-4-2-2-cat-esp.png|thumb|right|330px|link=]]

:Para evitar que aparezca un término asociado a la tensión de la cuerda, se puede trabajar en la referencia del trozo de cuerda que se ha considerado, que al tener velocidad ascendente constante es galileana. En esta referencia, el mono baja con velocidad constante:

:<math>\left.\begin{array}{l}
\text{REL: trozo de cuerda}\\
\text{AB: suelo}
\end{array}\right\}</math>

:<math>\overline{\vs}_\mathrm{REL}(\mathrm{mono})=\overline{\vs}_\mathrm{ar}(\mathrm{mono})=-\overline{\vs}_\mathrm{ar}(\mathrm{mono})=(\downarrow \rs\Omega_0)</math>

:<math>\left.\begin{array}{l}
\text{SISTEMA: mono+ trozo de cuerda}\\
\text{REF: trozo de cuerda(Gal)}
\end{array}\right\} \quad \dot{\Ts}^\mathrm{sist}_\mathrm{cuerda} =\dot{\Ws}^\mathrm{ext}_\mathrm{cuerda}+ \dot{\Ws}^\mathrm{int}</math>

:<math>\left.\begin{array}{l}
\dot{\Ts}^\mathrm{sist}_\mathrm{cuerda}=0\\
\dot{\Ws}^\mathrm{ext}_\mathrm{cuerda}=\left(\downarrow \ms\gs\right)\cdot\left(\downarrow \rs\Omega_0\right)= \ms\gs\rs\Omega_0\\
\dot{\Ws}^\mathrm{int}=\dot{\Ws}^{\mathrm{musc}}
\end{array}\right\} \quad \Rightarrow \quad \dot{\Ws}^{\mathrm{musc}} =-\ms\gs\rs\Omega_0</math>

:Combinando este resultado con el del balance anterior:

:<math>\dot{\Ws}^\mathrm{motor}=-\frac{2}{3}\ms\gs\rs\Omega_0-\dot{\Ws}^\mathrm{musc}=-\frac{5}{3}\ms\gs\rs\Omega_0</math>



© Universitat Politècnica de Catalunya. [[Mecánica:Derechos de autor |Todos los derechos reservados]]
-------------

<center>
[[D8. Conservaciones|<<< D8. Conservaciones]]

[[E2. Teorema de la energía: versión integrada|E2. Teorema de la energía: versión integrada>>>]]
</center>

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E1. Work-Energy Theorem: differential form

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<div class="noautonum">__TOC__</div>
<math>\newcommand{\uvec}{\overline{\textbf{u}}}
\newcommand{\vvec}{\overline{\textbf{v}}}
\newcommand{\evec}{\overline{\textbf{e}}}
\newcommand{\Omegavec}{\overline{\mathbf{\Omega}}}
\newcommand{\velang}[2]{\Omegavec^{\textrm{#1}}_{\textrm{#2}}}
\newcommand{\Alfavec}{\overline{\mathbf{\alpha}}}
\newcommand{\accang}[2]{\Alfavec^{\textrm{#1}}_{\textrm{#2}}}
\newcommand{\ds}{\textrm{d}}
\newcommand{\ps}{\textrm{p}}
\newcommand{\ns}{\textrm{n}}
\newcommand{\hs}{\textrm{h}}
\newcommand{\cs}{\textrm{c}}
\newcommand{\gs}{\textrm{g}}
\newcommand{\Ns}{\textrm{N}}
\newcommand{\Hs}{\textrm{H}}
\newcommand{\Fs}{\textrm{F}}
\newcommand{\ms}{\textrm{m}}
\newcommand{\ts}{\textrm{t}}
\newcommand{\us}{\textrm{u}}
\newcommand{\vs}{\textrm{v}}
\newcommand{\Rs}{\textrm{R}}
\newcommand{\Ws}{\textrm{W}}
\newcommand{\Ts}{\textrm{T}}
\newcommand{\Ls}{\textrm{L}}
\newcommand{\As}{\textrm{A}}
\newcommand{\Ds}{\textrm{D}}
\newcommand{\Bs}{\textrm{B}}
\newcommand{\es}{\textrm{e}}
\newcommand{\fs}{\textrm{f}}
\newcommand{\is}{\textrm{i}}
\newcommand{\Is}{\textrm{I}}
\newcommand{\ks}{\textrm{k}}
\newcommand{\js}{\textrm{j}}
\newcommand{\rs}{\textrm{r}}
\newcommand{\as}{\textrm{a}}
\newcommand{\ss}{\textrm{s}}
\newcommand{\Os}{\textbf{O}}
\newcommand{\Gs}{\textbf{G}}
\newcommand{\Cbf}{\textbf{C}}
\newcommand{\Or}{\Os_\Rs}
\newcommand{\Qs}{\textbf{Q}}
\newcommand{\Cs}{\textbf{C}}
\newcommand{\Ps}{\textbf{P}}
\newcommand{\Ss}{\textbf{S}}
\newcommand{\Js}{\textbf{J}}
\newcommand{\P}{\textrm{P}}
\newcommand{\Q}{\textrm{Q}}
\newcommand{\Ms}{\textrm{M}}
\newcommand{\deg}{^\textsf{o}}
\newcommand{\xs}{\textsf{x}}
\newcommand{\ys}{\textsf{y}}
\newcommand{\zs}{\textsf{z}}
\newcommand{\dert}[2]{\left.\frac{\ds{#1}}{\ds\ts}\right]_{\textrm{#2}}}
\newcommand{\ddert}[2]{\left.\frac{\ds^2{#1}}{\ds\ts^2}\right]_{\textrm{#2}}}
\newcommand{\vec}[1]{\overline{#1}}
\newcommand{\vecbf}[1]{\overline{\textbf{#1}}}
\newcommand{\vecdot}[1]{\overline{\dot{#1}}}
\newcommand{\OQvec}{\vec{\Os\Qs}}
\newcommand{\GQvec}{\vec{\Gs\Qs}}
\newcommand{\GPvec}{\vec{\Gs\Ps}}
\newcommand{\PSvec}{\vec{\Ps\Ss}}
\newcommand{\QQvec}{\vec{\Qs\Qs}}
\newcommand{\QGvec}{\vec{\Qs\Gs}}
\newcommand{\QPvec}{\vec{\Qs\Ps}}
\newcommand{\JQvec}{\vec{\Js\Qs}}
\newcommand{\PQvec}{\vec{\Ps\Qs}}
\newcommand{\GJvec}{\vec{\Gs\Js}}
\newcommand{\OPvec}{\vec{\Os\textbf{P}}}
\newcommand{\OCvec}{\vec{\Os\Cs}}
\newcommand{\OGvec}{\vec{\Os\Gs}}
\newcommand{\CJvec}{\vec{\Cs\Js}}
\newcommand{\GCvec}{\vec{\Gs\Cs}}
\newcommand{\PGvec}{\vec{\Ps\Gs}}
\newcommand{\abs}[1]{\left|{#1}\right|}
\newcommand{\braq}[2]{\left\{{#1}\right\}_{\textrm{#2}}}
\newcommand{\matriz}[9]{
\begin{bmatrix}
{#1} & {#2} & {#3}\\
{#4} & {#5} & {#6}\\
{#7} & {#8} & {#9}
\end{bmatrix}}
\newcommand{\diag}[3]{
\begin{bmatrix}
{#1} & {0} & {0}\\
{0} & {#2} & {0}\\
{0} & {0} & {#3}
\end{bmatrix}}
\newcommand{\vector}[3]{
\begin{Bmatrix}
{#1}\\
{#2}\\
{#3}
\end{Bmatrix}}
\newcommand{\vecdosd}[2]{
\begin{Bmatrix}
{#1}\\
{#2}
\end{Bmatrix}}
\newcommand{\vel}[2]{\vvec_{\textrm{#2}} (\textbf{#1})}
\newcommand{\acc}[2]{\vecbf{a}_{\textrm{#2}} (\textbf{#1})}
\newcommand{\accs}[2]{\vecbf{a}_{\textrm{#2}}^{\textrm{s}} (\textbf{#1})}
\newcommand{\accn}[2]{\vecbf{a}_{\textrm{#2}}^{\textrm{n}} (\textbf{#1})}
\newcommand{\velo}[1]{\vvec_{\textrm{#1}}}
\newcommand{\accso}[1]{\vecbf{a}_{\textrm{#1}}^{\textrm{s}}}
\newcommand{\accno}[1]{\vecbf{a}_{\textrm{#1}}^{\textrm{n}}}
\newcommand{\re}[2]{\Re_{\textrm{#2}}(\textbf{#1})}
\newcommand{\psio}{\dot{\psi}_0}
\newcommand{\Pll}{\textbf{P}_\textrm{lliure}}
\newcommand{\agal}[1]{\vecbf{a}_{\textrm{Gal}} (#1)}
\newcommand{\angal}[1]{\vecbf{a}_{\textrm{NGal}} (#1)}
\newcommand{\vgal}[1]{\vecbf{v}_{\textrm{Gal}} (#1)}
\newcommand{\fvec}[2]{\overline{\mathbf{F}}_{\textrm{#1}\rightarrow\textrm{#2}}}
\newcommand{\mvec}[2]{\overline{\mathbf{M}}_{\textrm{#1}\rightarrow\textrm{#2}}}
\newcommand{\F}[1]{\vec{\Fs}_{#1}}
\newcommand{\Fcal}[2]{\vec{\cal{F}}^{\text{#1}}_{#2}}
</math>

The '''Work-Energy Theorem''' plays a prominent role in physics, as it bridges all branches of science (electromagnetism, thermodynamics, etc.): all of them deal with energy.

There is a certain parallel between the Vector Theorems and the '''Work-Energy Theorem''': they all allow us to understand the evolution of a magnitude that depends on the mechanical state of the system. In the case of the former, we are dealing with the instantaneous evolution of vector magnitudes (linear momentum, angular momentum), while the latter deals with the finite change (in a finite time interval) of a scalar magnitude ('''mechanical energy''').

This unit presents the differential form of this theorem (the '''power balance'''). The integrated form (which is the Work-Energy Theorem itself) is presented in [[E2. Teorema de la energía: versión integrada|'''unit E2''']].
 
------
------
==E1.1 Power balance in a system of particles==

:[[File:E1-1-eng.png|thumb|right|300px|link=]]
<center>'''Figure E1.1''' Forces on a particle of a system with constant matter</center>

Let us consider a system of constant matter, and let’s apply Newton's second law to each particle (or each mass differential) of this system in a general reference frame R (which may be non-Galilean) ('''Figure E1.1'''):

<math>\F{\rightarrow \Ps} + \Fcal{}{\text{ar}\rightarrow\Ps} + \Fcal{}{\text{Cor}\rightarrow\Ps} =\ms_\Ps\acc{P}{R}</math>

<math>\F{\rightarrow \Ps} </math> is the resultant interaction force on '''P''', and it comes from internal interactions (with particles of the same system) and external ones:

<math>\F{\rightarrow \Ps}=\F{\rightarrow \Ps}^\text{int}+\F{\rightarrow \Ps}^\text{ext}</math>

If both sides of the equation are multiplied by <math>\vel{P}{R}</math>, all forces (interaction and inertia forces) orthogonal to <math>\vel{P}{R}</math> are filtered out. Specifically, since <math>\acc{P}{R}=\accs{P}{R} + \accn{P}{R} </math>, where <math>\accn{P}{R} \perp \vel{P}{R}</math>, and since the Coriolis force is always orthogonal to <math>\vel{P}{R} (\Fcal{}{\text{Cor}\rightarrow\Ps} = -2\ms_\Ps \velang{R}{Gal} \times \vel{P}{R})</math>:

<math>\vel{P}{R} \cdot \left( \F{\rightarrow \Ps}^\text{int}+\F{\rightarrow \Ps}^\text{ext} + \Fcal{}{\text{ar}\rightarrow\Ps} \right) = \ms_\Ps\accs{P}{R} \cdot \vel{P}{R} = \frac{\ds}{\ds \ts} \left(\frac{1}{2}\ms_\Ps \vvec_\Rs^2(\Ps) \right) \equiv \frac{\ds\Ts_\Rs(\Ps)}{\ds\ts}=\dot{\Ts}_\Rs(\Ps)</math>

The scalar function <math>\Ts_\Rs(\Ps)</math> is the '''kinetic energy''' of the particle, and is a function of the [[D1. Foundational laws of Newtonian dynamics#D1.3 Newton’s Principle of Determinacy|'''mechanical state''']] of '''P'''. The scalar product of a force and the velocity of its application point is the '''force power''': <math>\vel{P}{R} \cdot \F{\rightarrow \Ps} \equiv \dot{\Ws}_\Rs^{\F{\rightarrow \Ps}} </math>. Both scalar quantities <math>(\Ts_\Rs(\Ps), \dot{\Ws}_\Rs^{\F{\rightarrow \Ps}})</math> depend on the reference frame. For the same set of interaction forces acting on '''P''', the reference frame determines which are filtered out and which are not.

The equation can be rewritten as <math>\dot{\Ws}_\Rs^{\text{ext}\rightarrow \Ps}+\dot{\Ws}_\Rs^{\text{int}\rightarrow \Ps}+\dot{\Ws}_\Rs^{\text{ar}\rightarrow \Ps} =\dot{\Ts}_\Rs(\Ps)</math>, and constitutes the '''power balance''' for particle '''P''' in the R reference frame. Summing for all particles in the system:

<math>\dot{\Ws}_\Rs^{\text{ext}}+\dot{\Ws}^{\text{int}}+\dot{\Ws}_\Rs^{\text{ar}} =\dot{\Ts}_\Rs^\text{syst}</math>

The calculation of kinetic energy is covered in [[E2. Teorema de la energía: versión integrada|'''unit E2''']]. This unit focuses on the calculation of powers. [[E1. Work-Energy Theorem: differential form#E1.2 Potencia de una pareja de acción-reacción|'''section E1-2''']] shows that the power of internal interactions does not depend on the reference frame (which is why the subscript R does not appear in the <math>\dot{\Ws}^{\text{int}}</math> term of the previous equation).

====✏️ EXAMPLE E1-1.1: free particle====
------
:
{|
|
:[[File:ExE1-1-1-1-eng.jpg|thumb|left|180px|link=]]
|A particle '''P''', with mass m, is released from the ring of a Ferris wheel when it is at its highest point. We want to calculate the power, at this instant, of all forces acting on the particle from three different reference frames: the ground, the ring, and the Ferris wheel cabin.
|}

:Ground reference frame: Since this is a Galilean reference frame, all forces on '''P''' are interaction forces.
{|
|
:[[File:ExE1-1-2- eng.png|thumb|right|230px|link=]]

|<math>\hspace{0.5cm}\sum\dot{\Ws}_\Ts=\dot{\Ws}_\Ts^\text{weight}= \left(\leftarrow \Rs\Omega_0\right) \cdot \left(\downarrow \ms\gs\right)=0</math>
|}

:Ring reference frame: It is a non-Galilean reference, and the transportation force on '''P''' must be taken into account.
{|
|
:[[File:ExE1-1-3-eng.png|thumb|right|250px|link=]]
|<math>\hspace{0.5cm}</math> AB:ground, REL:ring 

<math>\hspace{0.5cm}\vel{P}{REL}=\vel{P}{AB}-\vel{P}{ar}= \left(\leftarrow \Rs\Omega_0\right)-\left(\leftarrow \Rs\Omega_0\right)=\overline{0}</math> 

<math>\hspace{0.5cm}</math>

Since the velocity is zero, the total power is zero.
|}

:Cabin reference frame: It is also a non-Galilean reference, and the transportation force on '''P''' must be taken into account.

{|
|
:[[File:ExE1-1-4-eng.png|thumb|right|250px|link=]]
|<math>\hspace{0.5cm}</math>AB:ground, REL:cabin

<math>\hspace{0.5cm}\vel{P}{REL}=\vel{P}{AB}-\vel{P}{ar}= \left(\leftarrow \Rs\Omega_0\right)-\left(\uparrow \Rs\Omega_0\right)</math>

<math>\hspace{0.5cm}\dot{\Ws}_\text{cabin}^\text{weight}=\left[ \left(\leftarrow \Rs\Omega_0\right)+\left(\downarrow \Rs\Omega_0\right) \right] \cdot \left(\downarrow\ms\gs\right) =\ms\gs\Rs\Omega_0</math>

<math>\hspace{0.5cm}\dot{\Ws}_\text{cabin}^\text{ar}=\left[ \left(\leftarrow \Rs\Omega_0\right)+\left(\downarrow \Rs\Omega_0\right) \right] \cdot \left(\rightarrow\ms\Rs^2\Omega_0^2\right) =-\ms\gs\Rs^2 \Omega_0^3</math>

<math>\hspace{0.5cm}\sum\dot{\Ws}_\text{cabin}=\ms\Rs\Omega_0(\gs-\Rs\Omega_0^2)</math>

|}


====✏️ EXAMPLE E1-1.2: pendulum====
------
:
{|
|
:[[File:ExE1-2-1-eng.png|thumb|left|180px|link=]]
|A particle '''P''' of mass m hangs from an inextensible thread of length L attached to the ceiling. The assembly oscillates in a vertical plane fixed to the ground. We want to calculate the power of all the forces acting on the particle in two different reference frames: the ground and a platform rotating relative to the ground with a constant angular velocity <math>\Omega_0</math>. 

Ground reference frame: Since it is a Galilean reference frame, '''P''' is only subjected to interaction forces.
{|
|
:[[File:ExE1-2-2-neut.png|thumb|left|300px|link=]]

|<math>\dot{\Ws}_\Es^\Ts=(\nearrow\Ls\dot{\theta})\cdot(\nwarrow\Ts)=0</math>

<math>\dot{\Ws}_\Es^\text{weight}=(\nearrow\Ls\dot{\theta}\cdot(\downarrow \ms\gs)=-\ms\gs\Ls\dot{\theta}\sin\theta</math>

|}

Platform reference frame: It is a non Galilean reference frame, and the transportation force on '''P''' has to be taken into account.

{|
|
:[[File:ExE1-2-3-eng.png|thumb|left|400px|link=]]

|AB:ground, REL:platform

<math>\vel{P}{REL}=\vel{P}{AB}-\vel{P}{ar}=(\nearrow\Ls\dot{\theta})-(\otimes \Ls \Omega_0 \sin \theta)</math>

<math>\dot{\Ws}_\Es^\Ts\left[(\nearrow\Ls\dot{\theta})+(\odot\Ls\Omega_0\sin\theta)\right]\cdot(\nwarrow\Ts)=0</math>

<math>\dot{\Ws}_\Es^\text{weight}\left[(\nearrow\Ls\dot{\theta})+(\odot\Ls\Omega_0\sin\theta)\right]\cdot(\downarrow \ms\gs)=-\ms\gs\Ls\dot{\theta}\sin\theta</math>

<math>\dot{\Ws}_\Es^\text{ar}\left[(\nearrow\Ls\dot{\theta})+(\odot\Ls\Omega_0\sin\theta)\right]\cdot(\rightarrow \ms\Ls\Omega_0^2\sin\theta)=\ms\Ls^2\Omega_0^2\dot{\theta}\cos\theta</math>
|}


------
------

==E1.2 Power of an action-reaction pair==

The total power associated with internal interactions in a constant-mass system is not zero in principle (since the two forces of each action-reaction pair are applied to particles that may have different velocities relative to the reference frame in which the calculation is made), and it is independent of the reference frame in which it is evaluated. It only depends on the separation velocity of the two particles. The power balance for the system is therefore written as: <math>\dot{\Ts}_\Rs^\text{syst}=\dot{\Ws}_\Rs^\text{ext} + \dot{\Ws}^\text{int} + \dot{\Ws}_\Rs^\text{ar} </math> .

<div>
=====💭 PROOF ➕=====

{|
|
:[[File:E1-2-neut.png|thumb|left|200px|link=]]
|<math>\dot{\Ws}^\text{ac-reac}=\overline{\Fs}_{\Ps \rightarrow \Qs}\cdot \vel{Q}{R} + \overline{\Fs}_{\Qs \rightarrow \Ps}\cdot \vel{P}{R}= \overline{\Fs}_{\Ps \rightarrow \Qs}\cdot \left[ \vel{Q}{R}-\vel{P}{R}\right]</math>

By the [[D1. Foundational laws of Newtonian dynamics#D1.6 Newton’s third law (action-reaction principle)|'''action-reaction principle''']], <math>\overline{\Fs}_{\Ps \rightarrow \Qs}=-\overline{\Fs}_{\Qs \rightarrow \Ps}</math>. Therefore: <math>\dot{\Ws}^\text{ac-reac}=\overline{\Fs}_{\Ps \rightarrow \Qs}\cdot \left[ \vel{Q}{R}-\vel{P}{R}\right]</math>

As <math>\overline{\Fs}_{\Ps \rightarrow \Qs}</math> has the <math>\PQvec</math> direction, only the components of the velocities in that direction have to be considered:

<math>\dot{\Ws}^\text{ac-reac}=\overline{\Fs}_{\Ps \rightarrow \Qs}\cdot \biggl( \vel{Q}{R} \Bigr]_{||\Ps\Qs}-\ \vel{P}{R} \Bigr]_{||\Ps\Qs}\biggl)</math>
|}

The difference <math>\biggl( \vel{Q}{R} \Bigr]_{\Ps\Qs}-\ \vel{P}{R} \Bigr]_{\Ps\Qs}\biggl)</math> is the separation velocity <math>\dot{\rho}</math> between '''P''' and '''Q''', and it does not dpend on the reference frame since it is the derivative of a scalar (the distance <math>\dot{\rho}</math> between the two particles). For the previous drawing, <math>\dot{\rho}> 0</math>. Since this is an attractive interaction, <math>\dot{\Ws}^\text{ac-reac} < 0</math>.

With a drawing, it is easy to see that:
* attractive interaction and particle approach <math>(\dot{\rho}<0)</math> : <math>\dot{\Ws}^\text{ac-reac} > 0</math> ;
* repulsive interaction and particle separation <math>(\dot{\rho}>0)</math> : <math>\dot{\Ws}^\text{ac-reac} > 0</math>
* repulsive interaction and particle approach <math>(\dot{\rho}<0)</math> : <math>\dot{\Ws}^\text{ac-reac} < 0</math>


</div>

There are two particular cases of interaction forces whose power always has the same sign:

* Friction action-reaction pair between '''P''' and '''Q''': since the friction force always opposes the relative velocity between the two points, its power is always negative:

<math>\hspace{1cm} \dot{\Ws}^\text{friction}=\dot{\Ws}^{\text{fric}\rightarrow\Ps}_\text{RTQ}+\dot{\Ws}^{\text{fric}\rightarrow\Qs}_\text{RTQ}=\overline{\Fs}^{\text{fric}\rightarrow\Ps} \cdot \vel{P}{RTQ} = - \abs{\overline{\Fs}^{\text{fric}\rightarrow\Ps}} \frac{\vel{P}{RTQ}}{\abs{\vel{P}{RTQ}}} \cdot \vel{P}{RTQ}<0;</math>

* Constraint action-reaction pair between '''P''' and '''Q''': from the [[D3. Interactions between rigid bodies#D3.4 Direct constraint interactions|'''analytical characterization equation''']] of the constraint forces between two points '''P''' and '''Q''' belonging to two rigid bodies <math>\mathrm{S}_\Ps</math> and <math>\mathrm{S}_\Qs</math>, it is easy to see that the associated power is zero:

<math>\hspace{1cm} \dot{\Ws}^\text{constraint}=\dot{\Ws}^{\text{constraint}\rightarrow\Ps}_{\mathrm{S}_\Ps}+\dot{\Ws}^{\text{constraint}\rightarrow\Qs}_{\mathrm{S}_\Ps}=\overline{\Fs}^{\text{constraint}\rightarrow\Ps} \cdot \overline{\mathbf{v}}_{\mathrm{S}_\Qs}(\Ps) = - 0;</math>

====✏️ EXAMPLE E1-2.1: power of an action-reaction friction pair====
------
:
{|
|
:[[File:ExE1-2-1-2-eng.png|thumb|left|180px|link=]]
|A homogeneous wheel, with radius r and mass m, moves down an inclined plane without sliding until it joins a conveyor belt moving at a constant velocity <math>\vs_0</math> relative to the ground. At the moment it makes contact with the belt, the center of the wheel has a velocity <math>\vs_0</math> relative to the ground in the same direction as the belt. We want to calculate the power of the friction force between the belt and the wheel at this instant.
|}

At that momento, the <math>\mathbf{ICR}^\text{wheel}_\Es</math> is point <math>\Js_\text{wheel}</math> in contact with the belt. This point slides backwards with respect to the belt, therefore the friction force on the wheel is forwards:

<math>\left.\begin{array}{l}
\text { AB: ground } \\
\text { REL: belt }
\end{array}\right\} \quad \bar{\vs}_{\text {REL }}\left(\Js_{\text {wheel }}\right)=\bar{\vs}_{\text{AB}}\left(\Js_{\text {wheel }}\right)-\bar{\vs}_{\mathrm{ar}}\left(\Js_{\text {wheel }}\right)=-\bar{\vs}_{\mathrm{ar}}\left(\Js_{\text {wheel }}\right)=-\left(\rightarrow \mathrm{v}_0\right)=\left(\leftarrow \mathrm{v}_0\right) .</math>

[[File:ExE1-2-1-3-eng.png|thumb|center|500px|link=]]

Calculation of the friction power: <math>\dot{\Ws}^\text{friction}=\dot{\Ws}_\Rs^{\text{friction}\rightarrow \text{wheel}}+ \dot{\Ws}_\Rs^{\text{friction}\rightarrow \text{belt}}</math>, where R may be any reference frame. To quickly calculate this power, it is advisable to choose a reference frame R where one of the two points subjected to the action-reaction pair has zero speed, so that one of the two terms is directly zero:

<math>\dot{\Ws}^\text{friction}=\dot{\Ws}_\text{belt}^{\text{fric}\rightarrow \text{wheel}}+ \dot{\Ws}_\text{belt}^{\text{fric }\rightarrow \text{belt}}=\overline{\Fs}^{\text{fric}\rightarrow \text{wheel}} \cdot \bar{\vs}_{\text {belt}}\left(\Js_{\text {wheel }}\right)=(\rightarrow \mu\ms\gs)\cdot(\leftarrow \vs_0)=-\mu\ms\gs\vs_0.</math>

Si el cálculo se hace en cualquier otra referencia, el resultado es el mismo:

<math>\dot{\Ws}^\text{fricción}=\dot{\Ws}_\text{T}^{\text{fric}\rightarrow \text{rueda}}+ \dot{\Ws}_\text{T}^{\text{fric} \rightarrow \text{cinta}}=(\rightarrow \mu\ms\gs)\cdot 0 + (\leftarrow \mu\ms\gs)\cdot(\rightarrow \vs_0)=-\mu\ms\gs\vs_0.</math>

<math>\dot{\Ws}^\text{fricción}=\dot{\Ws}_\text{vehículo}^{\text{fric}\rightarrow \text{rueda}}+ \dot{\Ws}_\text{vehículo}^{\text{fric} \rightarrow \text{cinta}}=(\rightarrow \mu\ms\gs)\cdot (\rightarrow\vs_0) + (\leftarrow \mu\ms\gs) \cdot(\rightarrow 3\vs_0)=-\mu\ms\gs\vs_0.</math>

:[[Archivo:ExE1-2-1-3-esp.png|thumb|center|450px|link=]]

En cambio, si sólo se calcula la potencia de una de las dos fuerzas (por ejemplo, la fricción sobre la rueda), el resultado depende de la referencia y puede tener cualquier signo:

<math>\dot{\Ws}_\text{rueda}^{\text{fric }\rightarrow \text{roda}}=(\rightarrow \mu\ms\gs)\cdot(\leftarrow \vs_0)=-\mu\ms\gs\vs_0 ,</math>

<math>\dot{\Ws}_\text{T}^{\text{fric} \rightarrow \text{rueda}}=(\rightarrow \mu\ms\gs)\cdot 0 = 0 ,</math>

<math>\dot{\Ws}_\text{vehicle}^{\text{fric} \rightarrow \text{rueda}}=(\rightarrow \mu\ms\gs)\cdot (\rightarrow\vs_0)=\mu\ms\gs\vs_0.</math>


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==E1.3 Potencia de un sistema de fuerzas sobre un sólido rígido==

{|
|
:[[Archivo:E1-3-esp.png|thumb|left|180px|link=]]
<center>'''Figura E1.3''': Torsor cinemático de un sólido rígido en una referenciay forces sobre los distintos puntos (dm) de este sóldio</center>

|Cuando el sistema sobre el que se calcula el balance de potencias es un sólido rígido S, no hay término asociado a interacciones internas porque todas son de enlace (fuerzas de cohesión) y, como se ha visto en la [[E1. Teorema de la energía: versión diferencial#E1.2 Potencia de un par de acción reacción|'''secciónE1.2''']], su potencia total es nula.

La potencia total de un sistema de fuerzas sobre S se puede calcular a partir del torsor del sistema en cualquier punto '''O''' del sólido <math>(\overline{\Fs},\overline{\Ms}(\Os))</math>. Teniendo en cuenta la cinemática de sólido rígido ('''Figura E1.3'''):

<math>\dot{\Ws}_\Rs^{\rightarrow \Ps} =\sum_\Ps\overline{\Fs}^{\rightarrow \Ps} \cdot \vel{P}{R}= \sum_\Ps \left[ \overline{\Fs}^{\rightarrow \Ps} \cdot \vel{O}{R} + \velang{S}{R} \times \OPvec \right] </math>

|}
<math>\dot{\Ws}_\Rs^{\rightarrow \Ps} =\left(\sum_\Ps\overline{\Fs}^{\rightarrow \Ps} \right)\cdot \vel{O}{R} + \sum_\Ps\overline{\Fs}^{\rightarrow \Ps} \cdot \left(\velang{S}{R} \times \OPvec \right) = \left( \sum_\Ps\overline{\Fs}^{\rightarrow \Ps} \right) \cdot \vel{O}{R} + \left[ \sum_\Ps \left( \OPvec \times \overline{\Fs}^{\rightarrow \Ps} \right)\right] \cdot \velang{S}{R} = \overline{\Fs}\cdot \vel{O}{R} + \overline{\Ms}(\Os)\cdot \velang{S}{R} </math>

Este cálculo demuestra también que la potencia de un momento (o un par <math>\Gamma</math>) sobre un sólido en una referencia R es el producto del momento por la velocidad angular del sólido: <math>\dot{\Ws}_\Rs^{\Gamma\rightarrow \mathrm{S}} = \overline{\Gamma} \cdot \velang{S}{R}</math>.

El balance de potencias para un sólido rígido es: <math>\dot{\Ws}_\Rs^{ \rightarrow \mathrm{S}} = \dot{\Ws}_\Rs^{ \mathrm{ext}} + \dot{\Ws}_\Rs^{ \mathrm{ar}}.</math>

====✏️ EJEMPLO E1-3.1: rodillo deslizando sobre un plano inclinado====
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:
{|
|
:[[Archivo:ExE1-3-1-1-esp.png|thumb|left|180px|link=]]
|Un rodillo homogéneo, de masa m y radi r, baja deslizando por una pendiente. El coeficiente de fricción entre suelo y rodillo es <math>\mu</math>. trata de calcular las potencias de todas las fuerzas que actúan sobre el rodillo en la referencia del suelo.
|}

:Se trata de un problema plano. Las descripciones cinemática y dinámica son:
{|
|
[[Archivo:ExE1-3-1-2-cat-esp.png|thumb|right|250px|link=]]

|<math> \hspace{0.5cm}\vel{J}{T}=\vel{C}{T} + \velang{}{T} \times \CJvec= </math>

<math>\hspace{1.7cm}=\left( \rightarrow \vs_\Ts \right) + \left( \otimes \Omega_\Ts\right) \times \left( \downarrow \rs \right)</math>

<math> \hspace{0.5cm}\vel{J}{T} = \left[\rightarrow (\vs_\Ts -\rs\Omega_\Ts)\right]</math>

<math>\hspace{0.5cm} \dot{\Ws}_\Ts^{\mathrm{ext}} = \dot{\Ws}_\Ts^{\mathrm{peso}} + \dot{\Ws}_\Ts^{\mathrm{fricción}} + \dot{\Ws}_\Ts^{\mathrm{enlace}}= \Bigr[\left(\downarrow \ms\gs\cos\beta\right)+\left(\rightarrow \ms\gs\sin\beta\right)\Bigr] \cdot (\rightarrow \vs_\Ts) + \Bigr[(\leftarrow \mu \Ns) + (\uparrow \Ns ) \Bigr] \cdot \Bigr[\rightarrow (\vs_\Ts -\rs\Omega_\Ts)\Bigr]</math>

|}
:Ya que el centro del rodillo no tiene movimiento perpendicular al plano, el TCM conduce a <math>\Ns =\ms\gs\cos\beta</math>. Por tanto:
{|
|
[[Archivo:ExE1-3-1-3-eng.png|thumb|right|140px|link=]]

|<math>\hspace{0.5cm} \dot{\Ws}_\Ts^{\mathrm{ext}} = \ms\gs\vs_\Ts \sin \beta - \ms\gs(\vs_\Ts -\rs\Omega_\Ts)\cos\beta=\ms\gs \Bigr[ \vs_\Ts (\sin\beta-\cos\beta)+\rs\Omega_\Ts \cos\beta \Bigr].</math>

<math>\hspace{0.5cm}</math>

Si en lugar de considerar la fricción en '''J''' se considera el torsor equivalente en '''C''', hay que considerar la fuerza aplicada en '''C''' y el momento respecto a '''C''', que no es nulo:

<math>\hspace{0.5cm} \dot{\Ws}_\Ts^{ \mathrm{fricción}} = (\leftarrow \mu\Ns)\cdot(\rightarrow\vs_\Ts)+(\otimes \mu\Ns\rs)\cdot(\otimes \Omega_\Ts) =\mu\ms\gs\cos\beta(\rs\Omega_\Ts-\vs_\Ts)</math>,

<math>\hspace{0.5cm} \dot{\Ws}_\Ts^{\mathrm{ext}} = \dot{\Ws}_\Ts^{ \mathrm{peso}} + \dot{\Ws}_\Ts^{ \mathrm{fricción}}+ \dot{\Ws}_\Ts^{ \mathrm{enlace}}= \ms\gs\vs_\Ts\sin\beta+\mu\ms\gs\cos\beta (\rs\Omega_\Ts-\vs_\Ts)</math>.

<math>\hspace{0.5cm}</math>El resultado es el mismo.
|}


====✏️ EJEMPLO E1-3.2: bloque deslizando sobre un camión====
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:
{|
|
:[[Archivo:ExE1-3-2-1-esp.png|thumb|left|220px|link=]]
|Un camión frena con aceleración '''a''' respecto al suelo en el instante en el que su velocidad (también respecto al suelo) es v. En el mismo instante, el bloque homogéneo, de masa m, desliza con velocidad v' relativa al camión. Se trata de hacer un balance de potencias para el bloque en la referencia del suelo y en la del camión.

|}

:Referencia suelo: Se trata de un problema plano. Las descripciones cinemática y dinámica son:
{|
|
[[Archivo:ExE1-3-2-2-cat-esp.png|thumb|right|200px|link=]]

| El enlace entre camión y bloque permite sólo un GL de translación entre los dos. Por tanto, en la aproximación 2D, el torsor asociado tiene dos componentes independientes. Si se caracteriza en el punto medio del contacto entre los dos elementos, son una fuerza normal y un momento perpendicular al plano. El TCM aplicado al bloque conduce a <math>\Ns=\ms\gs</math>.

|}
:Las potencias asociadas a todas las interacciones sobre el bloque en la referencia del suelo son:

:<math>\dot{\Ws}_\Ts^{\rightarrow \mathrm{bloque}}=\dot{\Ws}_\Ts^{\mathrm{ext}}=\dot{\Ws}_\Ts^{\mathrm{peso}}+\dot{\Ws}_\Ts^{\mathrm{fricción}}+\dot{\Ws}_\Ts^{\mathrm{enlace}}= \Bigr[(\downarrow \ms\gs)+(\leftarrow \mu\ms\gs)+(\uparrow \Ns) \Bigr] \cdot \Bigr[\rightarrow(\vs+\vs')\Bigr] + (\odot \Ms_\es)\cdot 0 = -\mu\ms\gs(\vs+\vs')</math>

:Ya que el bloque se traslada respecto al suelo, su energía cinética total es inmediata, y se puede plantear fácilmente el balance de potencias:

:<math>\Ts_\Ts^\mathrm{bloque}=\frac{1}{2}\int_\mathrm{bloque}\ds\ms \vs_\Ts^2(\mathrm{bloque})=\frac{1}{2}\ms \vs_\Ts^2(\mathrm{bloc})\quad \Rightarrow \quad \dot{\Ts}_\Ts^\mathrm{bloque}=\ms\vs_\Ts( \mathrm{bloque})\dot{\vs}_\Ts( \mathrm{bloque}) </math>

:Como la única fuerza horizontal sobre el bloque es la fricción:

<math>\dot{\vs}_\Ts(\mathrm{bloque})=-\mu\gs\quad \Rightarrow \quad \dot{\Ts}_\Ts^\mathrm{bloque}=-\mu\ms\gs(\vs+\vs').</math>

:El balance de potencias es consistente:

:<math>\left.\begin{array}{l}
\dot{\mathrm{W}}_\Ts^{\rightarrow \text { bloque }}=\dot{\mathrm{W}}_{\mathrm{T}}^{\text {ext }}=-\mu \mathrm{mg}\left(\mathrm{v}+\mathrm{v}^{\prime}\right) \\
\dot{\mathrm{T}}_{\mathrm{T}}^{\text {bloque }}=-\mu \mathrm{mg}\left(\mathrm{v}+\mathrm{v}^{\prime}\right)
\end{array}\right\} \Rightarrow \dot{\mathrm{W}}_{\mathrm{T}}^{\rightarrow \text { bloque }}=\dot{\mathrm{T}}_{\mathrm{T}}^{\text {bloque }}</math>

:Referencia camión: Al estar el camión acelerado respecto al suelo, hay que tener en cuenta las fuerzas de arrastre sobre todos los diferenciales de masa (dm) del bloque:

{|
|

[[Archivo:ExE1-3-2-3-cat-esp.jpg|thumb|right|250px|link=]]
|<math>\cal{\Fs}^{\rightarrow \ds\ms}_{\mathrm{ar}}=-\ds\ms\overline{\as}_\mathrm{ar}(\ds\ms)=-\ds\ms (\leftarrow \as)=(\rightarrow \ds\ms \cdot \as)</math>.

El torsor resultante de este sistema de fuerzas en el centro de inercia '''G''' se reduce a una fuerza horizontal.

<math>\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{bloque}}=\dot{\Ws}_\mathrm{camión}^{\mathrm{ext}} + \dot{\Ws}_\mathrm{camión}^{\mathrm{ar}} =\dot{\Ws}_\mathrm{camión}^{\mathrm{peso}} + \dot{\Ws}_\mathrm{camión}^{\mathrm{fricción}} + \dot{\Ws}_\mathrm{camión}^{\mathrm{enlace}} + \dot{\Ws}_\mathrm{camión}^{\mathrm{ar}} , </math>

<math>\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{bloque}}= \Bigr[(\downarrow \ms\gs)+(\leftarrow \mu\ms\gs)+(\uparrow \Ns) + (\rightarrow \ms\as)\Bigr] \cdot(\rightarrow \vs')+(\odot\Ms_\es) \cdot0=\ms(\as-\mu\gs)\vs'.</math>
|}

:Se puede comprobar que el balance de potencias es consistente:

:<math>\Ts^\mathrm{bloque}_\mathrm{camión}=\frac{1}{2}\ms\vs^2_\mathrm{camión}(\mathrm{bloque}) \quad \Rightarrow \quad \dot{\Ts}^\mathrm{bloque}_\mathrm{camión} = \ms\vs_\mathrm{camión}(\mathrm{bloque})\dot{\vs}_\mathrm{camión}(\mathrm{bloque})</math>

:<math>\left.\begin{array}{l}
\mathrm{AB}:\mathrm{suelo} \\
\mathrm{REL}:\mathrm{camión}
\end{array}\right\} \quad \dot{\vs}_\mathrm{camión}(\mathrm{bloque})= \dot{\vs}_\mathrm{T}(\mathrm{bloque})-\dot{\vs}_\mathrm{ar}(\mathrm{bloque}) = (-\mu\gs)-(-\as)=\as-\mu\gs \quad \Rightarrow \quad \dot{\Ts}^\mathrm{bloque}_\mathrm{camión}=\ms\vs'(\as-\mu\gs)</math>

:<math>\left.\begin{array}{l}
\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{bloque}}=\dot{\Ws}_\mathrm{camión}^{\mathrm{ext}} + \dot{\Ws}_\mathrm{camión}^{\mathrm{ar}}= \ms(\as-\mu\gs)\vs'\\
\dot{\Ts}^\mathrm{bloque}_\mathrm{camión}=\ms\vs'(\as-\mu\gs)
\end{array}\right\} \Rightarrow\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{bloque}}= \dot{\Ts}^\mathrm{bloque}_\mathrm{camión}</math>


====✏️ EJEMPLO E1-3.3: camión frenando con carga====
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:
{|
|
:[[Archivo:ExE1-3-3-1-esp.png|thumb|left|220px|link=]]
|Un camión frena con aceleración '''a''' respecto al suelo en el instante en el que su velocidad (también respecto al suelo) es v. Los pares de frenada en el eje delantero y en el trasero son iguales y valen <math>\Gamma</math>, y las ruedas, de radio r y masa despreciable, no deslizan sobre el suelo. En el mismo instante, el bloque homogéneo, de masa m, desliz con velocidad v' relativa al camión. Se trata de hacer un balance de potencias para el sistema (bloque+camión) en la referencia del suelo y en la del camión.

|}

:Ya que la potencia de las interacciones internas no es nula en principio (no se trata de un único sólido rígido), es conveniente hacer un DGI del sistema para no olvidar ninguna:
{|
|
[[Archivo:ExE1-3-3-2-esp.png|thumb|right|250px|link=]]

|
|
c.p.s.d.: contacto puntual sin deslizamiento

c.m.c.d.: contacto multipuntual con deslizamiento

|}
:El sistema sobre el que se hace el balance de potencias es todo menos el suelo. Teniendo en cuenta que la potencia de los enlaces internos es siempre nula, las únicas interacciones que hay que tener en cuenta son el peso, los enlaces de las ruedas con el suelo, los pares de frenada y la fricción entre bloque y camión:

[[Archivo:ExE1-3-3-3-esp.png|thumb|center|300px|link=]]

:Referencia suelo:
:Las fuerzas de enlace del suelo sobre las ruedas están aplicadas a puntos de velocidad cero, y por tanto no hacen potencia. El peso es una fuerza vertical, y el movimiento del camión y del bloque es horizontal, por tanto su potencia también es cero.

:<math>\dot{\Ws}_\Ts^{\rightarrow \mathrm{sist}}=\dot{\Ws}_\Ts^{\mathrm{ext}}+\dot{\Ws}^{\mathrm{int}}=\dot{\Ws}_\Ts^{\mathrm{peso}}+\dot{\Ws}_\Ts^{\mathrm{enlace}}+\dot{\Ws}^{\mathrm{fricción}}+\dot{\Ws}^{\Gamma}=\dot{\Ws}^{\mathrm{fricción}}+\dot{\Ws}^{\Gamma}.</math>

:Las potencias de interacciones internas se pueden calcular en cualquier referencia. Por tanto:

:<math>\dot{\Ws}^{\mathrm{fricción}}= \dot{\Ws}_\mathrm{camión}^{\mathrm{fric}\rightarrow \mathrm{bloque}} +\dot{\Ws}_\mathrm{camió}^{\mathrm{fric}\rightarrow \mathrm{camión}} = \dot{\Ws}_\mathrm{camión}^{\mathrm{fric}\rightarrow \mathrm{bloque}}= (\leftarrow \mu\ms\gs)\cdot (\rightarrow \vs')=-\mu\ms\gs\vs'</math>

:<math>\dot{\Ws}^{\Gamma}= \dot{\Ws}_\mathrm{camión}^{\Gamma\rightarrow \mathrm{ruedas}} +\dot{\Ws}_\mathrm{camión}^{\Gamma\rightarrow \mathrm{camión}} = \dot{\Ws}_\mathrm{camión}^{\Gamma\rightarrow \mathrm{ruedas}}= 2 \left(\odot \Gamma \right)\cdot \left(\otimes \frac{\vs}{\rs}\right)=-2\Gamma\frac{\vs}{\rs}.</math>

:Por otro lado: <math>\Ts_\Ts^\mathrm{sist}=\frac{1}{2}\ms\vs_\Ts^2(\mathrm{bloque})+\frac{1}{2}\ms\vs_\Ts^2(\mathrm{camión})\quad \Rightarrow \quad \dot{\Ts}_\Ts^\mathrm{sist}=-\ms(\vs+\vs')\mu\gs-\Ms\vs\as. </math>

:Finalmente: <math>\dot{\Ws}_\Ts^{\rightarrow \mathrm{sist}}=\dot{\Ts}_\Ts^\mathrm{sist} \quad \Rightarrow \quad -\mu\ms\gs\vs'-2\Gamma\frac{\vs}{\rs}=-\ms(\vs+\vs')\mu\gs-\Ms\vs\as.</math>

:Esta ecuación permite calcular el par de frenada: <math>\Gamma=\frac{1}{2}(\mu\ms\gs+\Ms\as)\rs.</math>

:Referencia camión:

:En esta referencia hay que tener en cuenta las fuerzas de arrastre. Por otro lado, las fuerzas de enlace del suelo sobre las ruedas están aplicadas a puntos que ya no tienen velocidad cero, y por tanto hacen potencia. Como antes, la potencia del peso es cero.

:<math>\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{sist}}=\dot{\Ws}_\mathrm{camión}^{\mathrm{ext}}+\dot{\Ws}^{\mathrm{int}}+\dot{\Ws}_\mathrm{camión}^{\mathrm{ar}}=\dot{\Ws}_\mathrm{camión}^{\mathrm{enlace}}+\dot{\Ws}^{\mathrm{fricción}}+\dot{\Ws}^{\Gamma}+\dot{\Ws}_\mathrm{camión}^{\mathrm{ar}}.</math>

:Las fuerzas de arrastre sobre el camión no hacen potencia porque el camión está parado en esta referencia. La de las fuerzas de arrastre sobre el bloque se ha calculado en el [[E1. Teorema de la energía: versión diferencial#✏️ ✏️ EJEMPLO E1-3.2: bloque deslizando sobre un camión|'''ejemplo E1-3.2''']]. La potencia de la fricción y de los pares de frenada es la misma de antes ya que no depende de la referencia.

:<math>\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{sist}}=\dot{\Ws}_\mathrm{camión}^{\mathrm{enllaç}}+\dot{\Ws}^{\mathrm{fricción}}+\dot{\Ws}^{\Gamma}+\dot{\Ws}_\mathrm{camión}^{\mathrm{ar} \rightarrow \mathrm{bloque}}= \dot{\Ws}_\mathrm{camión}^{\mathrm{enlace}} -\mu\ms\gs\vs'-2\Gamma\frac{\vs}{\rs} +\ms\as\vs'.</math>

:Las descripciones cinemática y dinámica de la rueda son:

[[Archivo:ExE1-3-3-4-esp.png|thumb|center|430px|link=]]

:<math>\dot{\Ws}_\mathrm{camión}^{\mathrm{enlace}}=2\Bigr[(\leftarrow \Ts)+(\uparrow \Ns)\Bigr]\cdot(\leftarrow\vs)=2\Ts\vs.</math>

La fuerza tangencial de enlace se puede calcular a partir del TMC aplicado al centro '''C''' de la rueda. Como no tiene masa: <math>\sum\overline{\Ms}_\mathrm{ext}(\Cs)=\overline{0} \quad \Rightarrow \quad (\odot \Gamma)+(\otimes \Ts\rs)=0\quad \Rightarrow \quad \Ts=\Gamma/\rs.</math>

:Por tanto: <math>\dot{\Ws}_\mathrm{camió}^{\rightarrow \mathrm{sist}} =2\frac{\Gamma}{\rs}\vs-\mu\ms\gs\vs'-2\Gamma\frac{\vs}{\rs}+\ms\as\vs'=\ms(\as-\mu\gs)\vs'.</math>

:Por otro lado, como se ha visto en el [[E1. Teorema de la energía: versión diferencial#✏️ ✏️ EJEMPLO E1-3.2: bloque deslizando sobre un camión|'''ejemplo E1-3.2''']]:

:<math>\Ts_\mathrm{camión}^\mathrm{sist}= \Ts_\mathrm{camión}^\mathrm{bloque}=\frac{1}{2}\ms\vs_\mathrm{camión}^2(\mathrm{bloque}) \quad \Rightarrow \quad \dot{\Ts}_\mathrm{camión}^\mathrm{sist}= \dot{\Ts}_\mathrm{camión}^\mathrm{bloque}=\ms\vs'(\as-\mu\gs).</math>

:El balance de potencias es consistente:

:<math>\left.\begin{array}{l}
\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{sist}}= \ms(\as-\mu\gs)\vs'\\
\dot{\Ts}^\mathrm{sist}_\mathrm{camión}=\ms(\as-\mu\gs)\vs'
\end{array}\right\} \Rightarrow\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{sist}}= \dot{\Ts}^\mathrm{sist}_\mathrm{camión}</math>

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==E1.4 Balance de potencias para un sistema multisólido: cálculo directo y cálculo indirecto==

El '''cálculo directo''' (a partir de su definición) de la potencia asociada a una interacción (una fuerza sobre un punto '''P''' o un momento sobre un sólido rígido S) exige el conocimiento de la interacción (fuerza o momento):

<math>\dot{\Ws}_\Rs^{\Fs\rightarrow \Ps}=\overline{\Fs}_{\rightarrow\Ps} \cdot \vel{P}{R} \quad , \quad \dot{\Ws}_\Rs^{\Ms\rightarrow \Ss}=\overline{\Ms}_{\rightarrow\Ss} \cdot \velang{S}{R} </math>

Cuando no se trata de una interacción formulable, hay que recurrir a los teoremas vectoriales para determinarla.

El '''cálculo indirecto''' es una alternativa que no necesita recurrir a estos teoremas. Se basa en la evaluación de un balance de potencias para un sistema y una referencia adecuados de manera que la potencia a determinar sea el único término desconocido de la ecuación (el único que no se puede calcular de manera directa). Cuando el balance incluye más de un término desconocido, se pueden combinar varios balances (para distintos sistemas y referencias) hasta tener un número suficiente de ecuaciones que permitan la determinación de todas las potencias desconocidas.

====✏️ EJEMPLO E1-4.1: polea frenada====
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:
{|
|
:[[Archivo:ExE1-4-1-1-esp.png|thumb|left|150px|link=]]
|Un bloque de masa m cuelga de un hilo inextensible atado a un punto de la periferia de una polea de radio r. Entre soporte (fijo al suelo) y polea actúa un freno que garantiza que el bloque baja a velocidad constante respecto al suelo. Se trata de calcular la potencia del freno.

|}

:Càlculo directo: <math>\dot{\Ws}^{\mathrm{freno}}= \overline{\Gamma}^{\rightarrow \mathrm{polea}}\cdot\velang{polea}{T}=\left(\odot\Gamma\right)\cdot\left( \otimes \frac{\vs_0}{\rs}\right)=-\Gamma \frac{\vs_0}{\rs}.</math>

:El par de frenada no es un dato, y hay que recurrir al TMC para calcularlo.

:<math>\left.\begin{array}{l}
\text{SISTEMA: polea + bloque}\\
\text{TMC en } \Os
\end{array}\right\} \sum\overline{\Ms}_\mathrm{ext}(\Os)=\dot{\overline{\Hs}}_\mathrm{RTO}(\Os)</math>

:El momento cinético proviene de las dos piezas. Ya que '''O''' no es fijo al bloque, hay que hacer descomposición baricéntrica para calcular el momento cinético del bloque:

:<math>\overline{\Hs}_\mathrm{RTO=T} (\Os) =\overline{\Hs}_\Ts^\mathrm{politja}(\Os) + \overline{\Hs}_\Ts^\mathrm{bloc}(\Os)= \overline{\Hs}_\Ts^\mathrm{politja}(\Os) + \overline{\Hs}_\Ts^\mathrm{bloc}(\Gs_\mathrm{bloc})+ \overline{\Hs}_\Ts^{\oplus\mathrm{bloc}}(\Os) = </math>

:<math>\hspace{2.3cm}=\Is_\mathrm{O}^\mathrm{polea} \velang{polea}{T}+\Is_{\Gs_\mathrm{bloque}}^\mathrm{bloque}+\OGvec_\mathrm{bloque}\times \ms\overline{\vs}_\Ts(\Gs_\mathrm{bloque})=\left(\otimes \Is_\mathrm{O}\frac{\vs_0}{\rs}\right)+(\otimes \ms\vs_0\rs).</math>

:[[Archivo:ExE1-4-1-2-esp.png|thumb|right|180px|link=]]
:Se trata de un momento cinético constante en dirección y valor. Por tanto:

:<math>\dot{\overline{\Hs}}_\mathrm{RTO}(\Os)=0 \quad \Rightarrow \quad \sum{\overline{\Ms}}_\mathrm{ext}(\Os)=(\odot\Gamma)+(\otimes \ms\gs\rs) =0\quad \Rightarrow \quad \Gamma= \ms\gs\rs</math>.

:Finalmente: <math>\dot{\Ws}^{\mathrm{freno}}= -\Gamma \frac{\vs_0}{\rs} = -\ms\gs\vs_0.</math>

:Cálculo indirecto:

:<math>\left.\begin{array}{l}
\text{SISTEMA: polea + soporte + bloque}\\
\text{REF: suelo(Gal)}
\end{array}\right\} \quad \dot{\Ts}^\mathrm{sist}_\Ts=\dot{\Ws}^\mathrm{ext}_\Ts+ \dot{\Ws}^\mathrm{int}</math>

:<math>\left.\begin{array}{l}
\dot{\Ts}^\mathrm{sist}_\Ts=0\\
\dot{\Ws}^\mathrm{ext}_\Ts=\left(\downarrow \ms\gs\right)\cdot\left(\downarrow \vs_0\right)= \ms\gs\vs_0\\
\dot{\Ws}^\mathrm{int}=\dot{\Ws}^{\mathrm{freno}}
\end{array}\right\} \quad \Rightarrow \quad \dot{\Ws}^{\mathrm{freno}} =-\ms\gs\vs_0</math>



====✏️ EJEMPLO E1-4.2: mono y plátanos====
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:
{|
|
:[[Archivo:ExE1-4-2-1-esp.png|thumb|left|180px|link=]]
|Unos plátanos de masa <math>(2/3)\ms</math> cuelgan de una cuerda inextensible que se apoya sin deslizar sobre una polea de radio r y masa despreciable, articulada respecto al suelo. En el otro extremo de la cuerda se agarra un mono de masa m. La polea gira con velocidad angular constante <math>\Omega_0</math> respecto al suelo bajo la acción de un motor. El mono se mueve respecto a la cuerda sin que sus manos se froten contra la cuerda, y de manera que la distancia de su centro de masas al suelo se mantiene constante. Se trata de calcular la potencia del motor.
|}

:Cálculo directo: <math>\dot{\Ws}^{\mathrm{freno}}= \overline{\Gamma}^{\rightarrow \mathrm{polea}}\cdot\velang{polea}{T}=\left(\odot\Gamma\right)\cdot\left( \odot \Omega_0\right)=\Gamma \Omega_0.</math>

:El par motor no es un dato, y hay que recurrir al TMC para calcularlo.

:<math>\left.\begin{array}{l}
\text{SISTEMA: polea + mono + plátanos}\\
\text{TMC en } \Os
\end{array}\right\} \sum\overline{\Ms}_\mathrm{ext}(\Os)=\dot{\overline{\Hs}}_\mathrm{RTO}(\Os)</math>

:El momento cinético proviene del mono y de los plátanos. En primera aproximación, se puede considerar la masa del mono concentrada en su tronco y su cabeza, de manera que no gira y su velocidad respecto al suelo es nula. Teniendo en cuenta que el punto '''O''' no se mueve respecto al mono pero sí respecto a los plátanos:

:<math>\overline{\Hs}_\mathrm{RTO=T} (\Os) =\overline{\Hs}_\Ts^\mathrm{mono}(\Os) + \overline{\Hs}_\Ts^\mathrm{plátanos}(\Os)= \overline{\Hs}_\Ts^\mathrm{mono}(\Os) + \overline{\Hs}_\Ts^\mathrm{plátanos}(\Gs_\mathrm{plát})+ \OGvec_\mathrm{plát} \times \frac{2}{3} \ms \overline{\vs}_\Ts(\Gs_\mathrm{plát})= </math>

:<math>\hspace{2.3cm}=\OGvec_\mathrm{plát}\times \frac{2}{3} \ms \overline{\vs}_\Ts(\Gs_\mathrm{plát}) = \Bigr[(\rightarrow \rs)+(\downarrow \xs)\Bigr] \times\frac{2}{3} \ms \left(\downarrow \rs\Omega_0\right)=\left(\otimes \frac{2}{3}\ms\rs^2\Omega_0\right).</math>

:Se trata de un momento cinético constante en dirección y valor. Por tanto:

:<math>\dot{\overline{\Hs}}_\mathrm{RTO}(\Os)=0 \quad \Rightarrow \quad \sum{\overline{\Ms}}_\mathrm{ext}(\Os)=(\odot\Gamma)+(\otimes \ms\gs\rs) + \left(\otimes \frac{2}{3}\ms\gs\rs\right)=0\quad \Rightarrow \quad \Gamma= \frac{5}{3}\ms\gs\rs</math>.

:Finalmente: <math>\dot{\Ws}^{\mathrm{motor}}= \Gamma \Omega_0 = \frac{5}{3}\ms\gs\rs\Omega_0.</math>

:Cálculo indirecto

:Hay que tener presente que es un problema con dos tipos de actuadores: el motor y la musculatura del mono. Hay diversos modelos biomecánicos para formular las fuerzas de los músculos, pero quedan fuera del ámbito de este curso. Ya que el mono se mueve, sus músculos desarrollan una potencia no nula.

:<math>\left.\begin{array}{l}
\text{SISTEMA: polea + mono + plátano}\\
\text{REF: suelo(Gal)}
\end{array}\right\} \quad \dot{\Ts}^\mathrm{sist}_\Ts=\dot{\Ws}^\mathrm{ext}_\Ts+ \dot{\Ws}^\mathrm{int}</math>

:<math>\left.\begin{array}{l}
\dot{\Ts}^\mathrm{sist}_\Ts=0\\
\dot{\Ws}^\mathrm{ext}_\Ts=\left(\downarrow \frac{2}{3}\ms\gs\right)\cdot\left(\downarrow \rs\Omega_0\right)= \frac{2}{3}\ms\gs\rs\Omega_0\\
\dot{\Ws}^\mathrm{int}=\dot{\Ws}^{\mathrm{musc}}+\dot{\Ws}^{\mathrm{motor}}
\end{array}\right\} \quad \Rightarrow \quad \dot{\Ws}^{\mathrm{musc}} + \dot{\Ws}^{\mathrm{motor}}=-\frac{2}{3}\ms\gs\rs\Omega_0</math>

:Este balance proporciona una ecuación con dos incógnitas. Se puede plantear un segundo balance para un sistema donde una de las dos no aparezca: un sistema sin el mono o un sistema sin el motor. Si se hace un balance de potencias para el sistema (mono + trozo de cuerda), las fuerzas externas a considerar son el peso y la tensión de la cuerda. Si se sigue trabajando en la referencia del suelo, esta tensión (desconocida) hará potencia no nula ya que está aplicada a un punto de velocidad ascendente.
:[[Archivo:ExE1-4-2-2-cat-esp.png|thumb|right|330px|link=]]

:Para evitar que aparezca un término asociado a la tensión de la cuerda, se puede trabajar en la referencia del trozo de cuerda que se ha considerado, que al tener velocidad ascendente constante es galileana. En esta referencia, el mono baja con velocidad constante:

:<math>\left.\begin{array}{l}
\text{REL: trozo de cuerda}\\
\text{AB: suelo}
\end{array}\right\}</math>

:<math>\overline{\vs}_\mathrm{REL}(\mathrm{mono})=\overline{\vs}_\mathrm{ar}(\mathrm{mono})=-\overline{\vs}_\mathrm{ar}(\mathrm{mono})=(\downarrow \rs\Omega_0)</math>

:<math>\left.\begin{array}{l}
\text{SISTEMA: mono+ trozo de cuerda}\\
\text{REF: trozo de cuerda(Gal)}
\end{array}\right\} \quad \dot{\Ts}^\mathrm{sist}_\mathrm{cuerda} =\dot{\Ws}^\mathrm{ext}_\mathrm{cuerda}+ \dot{\Ws}^\mathrm{int}</math>

:<math>\left.\begin{array}{l}
\dot{\Ts}^\mathrm{sist}_\mathrm{cuerda}=0\\
\dot{\Ws}^\mathrm{ext}_\mathrm{cuerda}=\left(\downarrow \ms\gs\right)\cdot\left(\downarrow \rs\Omega_0\right)= \ms\gs\rs\Omega_0\\
\dot{\Ws}^\mathrm{int}=\dot{\Ws}^{\mathrm{musc}}
\end{array}\right\} \quad \Rightarrow \quad \dot{\Ws}^{\mathrm{musc}} =-\ms\gs\rs\Omega_0</math>

:Combinando este resultado con el del balance anterior:

:<math>\dot{\Ws}^\mathrm{motor}=-\frac{2}{3}\ms\gs\rs\Omega_0-\dot{\Ws}^\mathrm{musc}=-\frac{5}{3}\ms\gs\rs\Omega_0</math>



© Universitat Politècnica de Catalunya. [[Mecánica:Derechos de autor |Todos los derechos reservados]]
-------------

<center>
[[D8. Conservaciones|<<< D8. Conservaciones]]

[[E2. Teorema de la energía: versión integrada|E2. Teorema de la energía: versión integrada>>>]]
</center>

File:ExE1-2-1-3-eng.png

2025-05-14T07:22:33Z

Bros:

ExE1-2-1-3-eng

File:ExE1-2-1-2-eng.png

2025-05-14T07:20:09Z

Bros:

ExE1-2-1-2-eng

D4. Vector theorems

2025-05-14T07:14:54Z

Bros: /* D4.8 Barycentric decomposition of the angular momentum */

<div class="noautonum">__TOC__</div>

<math>
\newcommand{\uvec}{\overline{\textbf{u}}}
\newcommand{\vvec}{\overline{\textbf{v}}}
\newcommand{\evec}{\overline{\textbf{e}}}
\newcommand{\Omegavec}{\overline{\mathbf{\Omega}}}
\newcommand{\velang}[2]{\Omegavec^{\textrm{#1}}_{\textrm{#2}}}
\newcommand{\Alfavec}{\overline{\mathbf{\alpha}}}
\newcommand{\accang}[2]{\Alfavec^{\textrm{#1}}_{\textrm{#2}}}
\newcommand{\as}{\textrm{a}}
\newcommand{\ds}{\textrm{d}}
\newcommand{\hs}{\textrm{h}}
\newcommand{\ks}{\textrm{k}}
\newcommand{\Ms}{\textrm{M}}
\newcommand{\Ns}{\textrm{N}}
\newcommand{\Fs}{\textrm{F}}
\newcommand{\ms}{\textrm{m}}
\newcommand{\ns}{\textrm{n}}
\newcommand{\ts}{\textrm{t}}
\newcommand{\Is}{\textrm{I}}
\newcommand{\us}{\textrm{u}}
\newcommand{\qs}{\textrm{q}}
\newcommand{\vs}{\textrm{v}}
\newcommand{\Rs}{\textrm{R}}
\newcommand{\Ts}{\textrm{T}}
\newcommand{\Ls}{\textrm{L}}
\newcommand{\Bs}{\textrm{B}}
\newcommand{\es}{\textrm{e}}
\newcommand{\fs}{\textrm{f}}
\newcommand{\gs}{\textrm{g}}
\newcommand{\is}{\textrm{i}}
\newcommand{\js}{\textrm{j}}
\newcommand{\rs}{\textrm{r}}
\newcommand{\Os}{\textbf{O}}
\newcommand{\Gs}{\textbf{G}}
\newcommand{\Cbf}{\textbf{C}}
\newcommand{\Or}{\Os_\Rs}
\newcommand{\Qs}{\textbf{Q}}
\newcommand{\Js}{\textbf{J}}
\newcommand{\Cs}{\textbf{C}}
\newcommand{\Ps}{\textbf{P}}
\newcommand{\Ss}{\textbf{S}}
\newcommand{\ss}{\textsf{s}}
\newcommand{\P}{\textrm{P}}
\newcommand{\Q}{\textrm{Q}}
\newcommand{\deg}{^\textsf{o}}
\newcommand{\xs}{\textsf{x}}
\newcommand{\ys}{\textsf{y}}
\newcommand{\zs}{\textsf{z}}
\newcommand{\dert}[2]{\left.\frac{\ds{#1}}{\ds\ts}\right]_{\textrm{#2}}}
\newcommand{\ddert}[2]{\left.\frac{\ds^2{#1}}{\ds\ts^2}\right]_{\textrm{#2}}}
\newcommand{\vec}[1]{\overline{#1}}
\newcommand{\vecbf}[1]{\overline{\textbf{#1}}}
\newcommand{\vecdot}[1]{\overline{\dot{#1}}}
\newcommand{\GPvec}{\vec{\Gs\Ps}}
\newcommand{\OQvec}{\vec{\Os\Qs}}
\newcommand{\JGvec}{\vec{\Js\Gs}}
\newcommand{\QGvec}{\vec{\Qs\Gs}}
\newcommand{\QPvec}{\vec{\Qs\Ps}}
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\newcommand{\OCvec}{\vec{\Os\Cs}}
\newcommand{\OGvec}{\vec{\Os\Gs}}
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\begin{Bmatrix}
{#1}\\
{#2}\\
{#3}
\end{Bmatrix}}
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\begin{Bmatrix}
{#1}\\
{#2}
\end{Bmatrix}}
\newcommand{\vel}[2]{\vvec_{\textrm{#2}} (\textbf{#1})}
\newcommand{\acc}[2]{\vecbf{a}_{\textrm{#2}} (\textbf{#1})}
\newcommand{\accs}[2]{\vecbf{a}_{\textrm{#2}}^{\textrm{s}} (\textbf{#1})}
\newcommand{\accn}[2]{\vecbf{a}_{\textrm{#2}}^{\textrm{n}} (\textbf{#1})}
\newcommand{\velo}[1]{\vvec_{\textrm{#1}}}
\newcommand{\accso}[1]{\vecbf{a}_{\textrm{#1}}^{\textrm{s}}}
\newcommand{\accno}[1]{\vecbf{a}_{\textrm{#1}}^{\textrm{n}}}
\newcommand{\re}[2]{\Re_{\textrm{#2}}(\textbf{#1})}
\newcommand{\psio}{\dot{\psi}_0}
\newcommand{\Pll}{\textbf{P}_\textrm{lliure}}
\newcommand{\agal}[1]{\vecbf{a}_{\textrm{Gal}} (#1)}
\newcommand{\angal}[1]{\vecbf{a}_{\textrm{NGal}} (#1)}
\newcommand{\vgal}[1]{\vecbf{v}_{\textrm{Gal}} (#1)}
\newcommand{\F}[1]{\vec{\Fs}_{#1}}
\newcommand{\Fcal}[2]{\vec{\cal{F}}^{\text{#1}}_{#2}}
\newcommand{\H}[3]{\vec{\mathbf{H}}_{\text{#3}}^{#2}(\textbf{#1})}
</math>

The '''Vector Theorems''' are a tool for solving the dynamics of mechanical systems, and are obtained from the fundamental law of dynamics ([[D1. Foundational laws of Newtonian dynamics#D1.5 Newton’s second law (fundamental law of dynamics)|'''Newton's second law''']]) and the principle of action and reaction ([[D1. Foundational laws of Newtonian dynamics#D1.6 Newton’s third law (action-reaction principle)|'''Newton's third law''']]).

In this course, only the version of the theorems for the case of constant matter systems is presented. Although this includes systems with fluids, the application examples in this course are essentially multibody systems made up of rigid bodies.

When addressing a problem, it is essential to identify the unknowns contained in the system under study and to analyze whether the vector theorems provide a sufficient number of equations to solve them all (one must know whether the problem is determinate or indeterminate!). The unknowns can be classified into three groups:

* Time evolution of the free [[C2. Movement of a mechanical system#C2.7 Degrees of freedom of a mechanical system|'''degrees of freedom (DoF)''']] (not controlled by actuators) of the system. The equations that govern these DoF are called '''equations of motion'''. If the free DoF are described by time derivatives of coordinates (<math>\qs_i</math>, with i=1,2,3...), their time evolution is the second time derivatives of these coordinates (accelerations). Their general aspect is:

<center><math>\dot{\qs}_i = f(\qs_j, \dot{\qs}_j, \ddot{\qs}_j, \text{ dynamic parameters, geometric parameters})</math></center>

The dependence of the equations of motion on the second time derivatives is always linear, while the dependence on the coordinates and velocities can be of any type. The dynamic parameters are the mass of the elements and those associated with their distribution in space, and the parameters associated with the interactions on the system (spring and damper constants, coefficients of friction, gravitational field constants...); the geometric parameters have to do with the shape of the elements of the system (distances and angles).

* [[D2. Interaction forces between particles#D2.6 Interaction through actuators|'''Actuator''']] '''actions''': these are the forces (in the case of linear actuators) and moments (in the case of rotatory actuators) required to ensure predetermined evolutions of the DoF they control. As discussed in [[D2. Interaction forces between particles#D2.6 Interaction through actuators|'''section D2.6''']], in some cases the actuator actions can be considered as data, and then the associated unknowns are the evolutions of the DoF they control.

* [[D3. Interaccions entre sòlids rígids|'''Constraint forces and movements''']]: the number of unknowns associated with the constraints depends on the description given of them ([[D3. Interactions between rigid bodies#D3.4 Direct constraint interactions|'''direct constraints''']], [[D3. Interactions between rigid bodies#D3.5 Indirect constraint interactions: Constraint Auxiliary Elements (CAE)|'''indirect constraints''']]). When a dynamics problem is indeterminate, the indeterminacy always refers to the constraint torsor components, never to the DoF (free or forced).

---------
---------
==D4.1 Linear Momentum Theorem (LMT) in Galilean reference frames==

Let us consider a system of particles with constant matter ('''Figure D4.1'''). The '''Linear Momentum Theorem''' (LMT) is obtained by applying Newton's second law to each particle '''P''' of the system. If the chosen reference frame is Galilean:

<math>\F{\rightarrow\Ps}=\ms_\Ps\acc{P}{RGal}</math>, where <math>\F{\rightarrow\Ps}</math> is the interaction resultant force on '''P'''.
[[File:D4-1-eng.png|thumb|center|300px|link=]]
<center>'''Figure D4.1''' Forces on a particle of a system with constant matter</center>

The forces acting on each particle can be classified into two groups: internal (which come from the interaction with other particles in the system) and external (associated with interactions with elements external to the system): <math>\F{\is\ns\ts\rightarrow\Ps}+\F{\es\xs\ts\rightarrow\Ps}=\ms_\Ps\acc{P}{RGal}</math>. If these equations are added for all particles, the internal forces between pairs of particles cancel out two by two by the principle of action and reaction:

<center><math>\F{\Ps_j\rightarrow\Ps_i}+\F{\Ps_i\rightarrow\Ps_j}=\vec{0}\Rightarrow\sum_\Ps\F{\es\xs\ts\rightarrow\Ps}=\sum_\Ps \ms_\Ps\:\acc{P}{RGal}</math></center>

The term on the left-hand side is the resultant of external forces on the system, and is often written simply as <math>\sum\F{ext}</math>. The term on the right-hand side can be rewritten as <math>\Ms\left[\sum_\Ps\frac{\ms_\Ps}{\Ms}\acc{P}{RGal}\right]</math>, where M is the total mass of the system <math>\left(\Ms\equiv\sum_\Ps \ms_\Ps\right)</math>. The term <math>\left[\sum_\Ps\frac{\ms_\Ps}{\Ms}\acc{P}{RGal}\right]</math> is a weighted acceleration, where the weighting is proportional to the mass of each particle. This acceleration is associated with a point called '''center of mass''' (or '''center of inertia''') of the system, and in this course it is designated by letter G. For any reference frame, then, the kinematics of G are described by the equations ('''Figure D4.2'''):

<center><math>\vec{\Os_\Rs\Gs}=\sum_\Ps\frac{\ms_\Ps}{\Ms}\vec{\Os_\Rs\Ps},\:\:\:\:\vel{G}{R}=\sum_\Ps\frac{\ms_{\Ps}}{\Ms}\vel{P}{R},\:\:\:\:\acc{G}{R}=\sum_\Ps\frac{\ms_\Ps}{\Ms}\acc{P}{R}</math></center>

[[File:D4-2-eng.png|thumb|center|250px|link=]]
<center>'''Figure D4.2''' Centre of mass (r of inertia) of a ystem with constant matter</center>

In continuous sets of particles (such as a set of rigid or deformable bodies, or fluids), the summation for particles is actually an integral:

<center><math>\vec{\Os_\Rs\Gs}=\sum_{i=1}^N\left(\frac{1}{\Ms_\is}\int_{\Ss_\is}\ds\ms(\Ps)\vec{\Os_\Rs\Ps}\right),\:\:\:\:\vel{G}{R}=\sum_{i=1}^N\left(\frac{1}{\Ms_\is}\int_{\Ss_\is}\ds\ms(\Ps)\vel{P}{R}\right),\:\:\:\:\acc{G}{R}=\sum_{i=1}^N\left(\frac{1}{\Ms_\is}\int_{\Ss_\is}\ds\ms(\Ps)\acc{P}{R}\right)</math></center>

Finally, the LMT is written as:

<center><math>\sum\F{ext}=\Ms\:\acc{G}{RGal}</math></center>

This equation is very similar to [[D1. Foundational laws of Newtonian dynamics#D1.5 Newton’s second law (fundamental law of dynamics)|'''Newton's second law''']]: the centre of mass G behaves as if it were a particle of mass equal to the total mass of the system, and on which all the forces external to the system acted. Despite the parallelism between the LMT and the particle dynamics equation, there are two fundamental differences:

:*the mass of the system is not localized at '''G''' ('''G''' might be even located in a massless region of the system, as in the case of a homogeneous ring);

:*the external forces are not applied to '''G''' in general.

The LMT is so named because it allows us to know the evolution, based on the knowledge of the initial [[C1. Configuration of a mechanical system|'''mechanical state''']] and the external interactions, of the linear momentum of the system (which, in any reference frame R, is defined as <math>\sum_\Ps\ms_\Ps\vel{P}{R}=\Ms\vel{G}{R}</math>):

<center><math>\sum\F{ext}=\Ms\acc{G}{RGal}=\dert{\Ms\vel{G}{RGal}}{RGal}</math></center>

The location of the center of mass in the systems under study (whether it is a single rigid body or a multi-body system) is briefly introduced in [[D4. Vector theorems|'''unit D4''']]. For homogeneous bodies with very simple geometry, the position of G can often be deduced from the body symmetries.

In planar problems (with 2D kinematics), only the two components of the LMT contained in the plane of motion are interesting.

------------
------------

==D4.2 LMT: application examples==

<div>
====✏️ EXAMPLE D4.1: calculation of a constraint force ====

------
:
{|
|[[File:D4-Ex1-1-neut.png|thumb|center|180px|link=]]
|The three homogeneous blocks are smooth, and are in contact with each other and with a smooth horizontal ground. We want to investigate the value of the horizontal constraint force between blocks Q and S when a horizontal force F is applied to the block on the left.

All constraints appearing in this system are multiple-point contacts between smooth surfaces. Therefore, each associated torsor, characterized at a contact point, contains a force and a moment component ([[D3. Interactions between rigid bodies#D3.4 Direct constraint interactions|'''example D3.4''']]). However, in the application of the LMT only the forces are involved, so the moments will not be shown in the following figures.

|}

In order for the constraint force <math>\Fs_{\Qs\rightarrow\Ss}</math> to appear in the horizontal component of the LMT, the theorem must be applied to a system where this force is external. For example, to block S: <math>(\rightarrow\Fs_{\Qs\rightarrow\Ss})=(3\ms)[\rightarrow \as_\epsilon(\Gs_\Ss)]</math>.

[[File:D4-Ex1-2-eng.png.png|thumb|center|200px|link=]]

Since the motion of the three blocks caused by force F is the same, the acceleration can be obtained through the LMT applied to the whole system:

<center><math>\Fs = (6\ms)\as_\epsilon(\Gs)\Rightarrow\as_\epsilon(\Gs)=\frac{\Fs}{6\ms}\Rightarrow\Fs_{\Qs\rightarrow\Ss}=3\ms\frac{\Fs}{6\ms}=\frac{\Fs}{2}</math></center>

</div>
<div>

====✏️ EXAMPLE D4.2: initial motion of a system ====
------
:
{|
|[[File:D4-Ex2-1-eng.png|thumb|center|300px|link=]]
|The homogeneous blocks are initially at rest on a rough floor, and connected by a spring compressed with tension <math>\Fs_0=\ms\gs</math> and an inextensible thread. At a certain instant, the thread is cut and the system begins to move.
|}

We want to calculate the acceleration of the center of inertia of the system relative to the ground. This acceleration can be obtained from the horizontal component of the LMT applied to the whole system. The spring is internal, and therefore its force does not appear: <math>\F{\text{ground}\rightarrow 2\ms}+\F{\text{ground}\rightarrow \ms}=(3\ms)\acc{G}{T}</math>.

The forces of the ground on the blocks can be constraint forces (if the blocks do not move, in which case they are unknown), or friction forces (if the blocks move relative to the ground, in which case they are formulable). It may also happen that one block slides and the other does not.

The value of a constraint force adapts itself to ensure a restriction. In the case of blocks, it is necessary to investigate whether these forces can reach the value required to prevent the blocks from sliding on the ground, and are bounded by the value of the coefficient of static friction between blocks and ground:

<math>|\F{\text{ground}\rightarrow 2\ms}\mid\leq\mu_\es\Ns_{\text{ground}\rightarrow 2\ms}?\:\:\:, \mu_\es\Ns_{\text{ground}\rightarrow 2\ms}=0,6\cdot 2\ms\gs=1,2\ms\gs</math>

<math>|\F{\text{ground}\rightarrow \ms}\mid\leq\mu_\es\Ns_{\text{terra}\rightarrow\ms}?\:\:\:, \mu_\es\Ns_{\text{ground}\rightarrow\ms}=0.4\cdot \ms\gs</math>

If we apply the LMT to each block separately, the spring force becomes external. That force (of value mg) can be counteracted by the constraint force in the case of the block of mass 2m (hence, <math>\as_\epsilon(2\ms) = 0</math>), but not in the case of the block of mass m:

[[File:D4-Ex2-2-eng.png|thumb|center|450px|link=]]

Application of the LMT to the block of mass m leads to:

<math>\left.\begin{aligned}
\sum\F{\es\xs\ts\rightarrow2\ms}=\F{\text{ground}\rightarrow2\ms}+\F{\text{spring}\rightarrow2\ms}=2\ms\acc{2m}{T} \\
\sum\F{\es\xs\ts\rightarrow2\ms}=(\rightarrow\ms\gs)+(\leftarrow\ms\gs)=0
\end{aligned}\right\} \Rightarrow\acc{m}{T}=(\rightarrow 0,8\gs)</math>

Finally: <math>\acc{G}{T}=\frac{2\ms\acc{2m}{T}+\ms\acc{m}{T}}{3\ms}=\frac{0,8}{3}\gs</math>.

</div>
<div>
====✏️ EXAMPLE D4.3: limit condition ====
------
:
{|
|[[File:D4-Ex3-1-neut.png|thumb|center|200px|link=]]
|The homogeneous sphere of mass M is in contact with two identical wedges of mass m, which are on the ground. There is no friction between the sphere and the wedges, but friction coefficient between wedges and ground is <math>\mu</math>. The system is initially at rest relative to the ground. We want to determine the maximum value of M, as a function of m, that allows the system to remain at rest.
|}

If nothing moves, the horizontal interaction forces between ground and wedges are constraint forces, and do not exceed the limit value <math>/mu\ns</math> (where N is the normal force that the ground exerts on each wedge). The constraint torsor of the ground on the wedges also contains a resultant moment perpendicular to the figure. If we want to study the possibility of the wedges overturning, this moment is relevant. But in this example, the shape of the wedges guarantees that they do not overturn, so we must take into account the forces.

Applying the LMT to the entire system, to a wedge, and to the sphere leads to the following equations:

[[File:D4-Ex3-3-neut.png|thumb|center|350px|link=]]

<center>
{|class = "wikitable", style = "background-color:#ffffff;border-style:none;"
|- style="vertical-align:top;"
|style="padding: 20px; text-align:center;"|SYSTEM: sphere + wedges

<math>\left(\uparrow 2\Ns\right)+\left[\downarrow(\Ms+2\ms)\gs\right]=\vec{0}</math>

<math>\Rightarrow\Ns=\frac{1}{2}(\Ms+2\ms)\gs</math>
|style="padding: 20px; text-align:center;"|SYSTEM: left wedge

<math>\left(\rightarrow \Ts\right)+\left(\leftarrow\frac{\Ns'}{\sqrt{2}}\right)=\vec{0}</math>

<math>\Rightarrow\Ts=\frac{\Ns'}{\sqrt{2}}</math>
|style="padding: 20px; text-align:center;"|SYSTEM: sphere

<math>\left(\downarrow\Ms\gs\right)+\left(\uparrow 2\frac{\Ns'}{\sqrt{2}}\right)=\vec{0}</math>

<math>\Rightarrow\Ns'=\frac{\Ms\gs}{\sqrt{2}}</math>
|}</center>

Combining the last two equations: <math>\Ts = \Ms\gs/2</math>. The maximum M value for which there is equilibrium corresponds to the situation in which the tangential constraints force T takes the maximum possible value: <math>\Ts = \Ts_{\text{màx}}=\mu\Ns</math>. Taking into account that <math>N = \frac{1}{2}(\Ms_{\text{max}}+2\ms)\gs</math>:

<math>\Ts=\frac{\Ms_{\text{max}}\gs}{2}=\Ts_{\text{max}}=\mu\Ns=\mu\frac{1}{2}(\Ms_{\text{max}}+2\ms)\gs\Rightarrow\Ms_{\text{max}}=2\ms\frac{\mu}{1-\mu}</math>

</div>

------------
------------

==D4.3 Linear Momentum Theorem (LMT) in non Galilean reference frames==

The LMT in a non-Galilean reference NGal is also obtained from [[D1. Foundational laws of Newtonian dynamics#D1.5 Newton’s second law (fundamental law of dynamics)|'''Newton's second law''']] applied to each particle (or each mass differential) of the system in the reference NGal ([[D1. Foundational laws of Newtonian dynamics#D1.7 Particle dynamics in non Galilean reference frames|'''section D1.7''']]). In principle, then, this equation will contain two inertial forces: the transportation force and the Coriolis force ('''Figure D4.3'''):

<math>\F{\rightarrow \Ps} + \Fcal{tr}{NGal\rightarrow\Ps}+\Fcal{Cor}{NGal\rightarrow\Ps} = \ms_\Ps\acc{P}{NGal}</math>,

where <math>\Fcal{tr}{NGal\rightarrow\Ps} = -\ms_\Ps\acc{P}{ar} = -\ms_\Ps\acc{$\Ps\in\text{NGal}$}{RGal}</math> and <math>\Fcal{Cor}{NGal\rightarrow\Ps} = -\ms_\Ps\acc{P}{Cor} = -2\ms_\Ps\velang{NGal}{RGal}\times\vel{P}{NGal}</math>

[[File:D4-3-eng.png|thumb|center|350px|link=]]
<center>'''Figure D4.3:''' Inertia forces in a non Galilean reference frame</center>

Summing the equations for all the particles, we obtain: <math>\sum\F{ext}+\Fcal{tr}{NGal\rightarrow\Gs}+\Fcal{Cor}{NGal\rightarrow\Gs}=\Ms\acc{G}{NGal}</math>, where M is the total mass of the system.

<div>
=====💭 Proof ➕=====
------
:
By the principle of action and reaction, the sum for all particles of the interaction forces leads to:

<math>\sum_{\Ps\in\text{syst}}\F{\rightarrow\Ps} = \sum\F{ext}</math>

As for the inertia forces, as they are proportional to the transportation and Coriolis acceleration of each particle, their sum for all particles corresponds to the weighted kinematics that defines the [[D4. Vector theorems#D4.2 LMT: application examples|'''center of mass''']]:

<math>
\sum_{\Ps\in\text{syst}} \Fcal{tr}{NGal\rightarrow\Ps} = -\sum_{\Ps\in\text{syst}}\ms_\Ps\acc{$\Ps\in$NGal}{RGal} = -\Ms\acc{$\Gs\in$NGal}{RGal} = -\Ms\acc{$\Gs$}{ar} = \Fcal{tr}{NGal\rightarrow\Ps}
</math>

<math>
\sum_{\Ps\in\text{syst}} \Fcal{Cor}{NGal\rightarrow\Ps} = -\sum_{\Ps\in\text{syst}}\ms_\Ps\acc{$\Ps$}{Cor} = -2\sum_{\Ps\in\text{syst}}\ms_\Ps\velang{NGal}{RGal}\times\vel{P}{NGal} = -2\velang{NGal}{RGal}\times\sum_{\Ps\in\text{syst}}\ms_\Ps\vel{P}{NGal} = -2\velang{NGal}{RGal}\times\Ms_\Ps\vel{G}{NGal} = \Fcal{Cor}{NGal\rightarrow\Ps}
</math>

Hence: <math>\sum\F{ext} + \Fcal{tr}{NGal\rightarrow\Gs}+\Fcal{Cor}{NGal\rightarrow\Gs} = \Ms\acc{G}{NGal}</math>

</div>

<div>

====✏️ EXAMPLE D4.4: vibratory displacement====
------
:
{|
|[[File:D4-Ex4-1-eng.png|thumb|center|250px|link=]]
|The block of mass m is initially at rest on a support which oscillates relative to the ground according to the velocity graph shown in the figure. We want to investigate the possibility of sliding between block and support.

The nonsliding condition between block and support (which is a non-Galilean reference since its motion relative to the ground is not uniform) requires:

<math>\sum\F{\text{ext}} + \Fcal{tr}{sup\rightarrow\Gs} = \vec{0} (\Fcal{Cor}{sup\rightarrow\Gs}= \vec{0})</math> perque <math>\velang{sup}{T} = \vec{0}</math>
|}

The transportation force on the block is strictly horizontal. Therefore, the vertical component of that equation leads to <math>N = mg</math>. As for the horizontal component, it will contain the interaction force between the block and the support. If the block does not slide on the support, this force is a constraint force, and its value is bounded: <math>0\leq\mid\Fs_{sup\rightarrow bloc}\mid \leq\mu\Ms\gs = 0,15\Ms\Gs\simeq 1,5(\ms/s^2)\Ms</math>. If it slides, it is a friction force with value <math>\mu\ms\gs</math>, opposite to the sliding velocity.

Between <math>t=0</math> and <math>t = 0,4s</math>, the acceleration of the support relative to the ground is <math>\ddot{y} = \frac{0,4\ms/s}{0,4s}=1\ms/s^2</math>, therefore the drag force is <math>\Fcal{tr}{sup\rightarrow\Gs} = -\ms\acc{$\Gs\in$support}{T}=[\leftarrow 0,1(\ms/s^2)\Ms]</math>. That value is within the range of values allowed for the horizontal constraint force between support and block. Therefore, this force will take the value <math>\F{sup\rightarrow block} = 0,1(\ms/s^2)</math>, will counteract the <math>\Fcal{tr}{sup\rightarrow\Gs}</math> and there will be no motion between the two elements.

Between <math>t=0,4s</math> and <math>t = 0,6s</math>, the acceleration of the support relative to the ground is 2<math>\ddot{y} = -\frac{0,8\ms/s}{0,2s}=-4\frac{\ms}{s^2}</math>. Therefore, <math>\Fcal{tr}{sup\rightarrow\Gs} =[\rightarrow 4(\ms/s^2)\Ms]</math>. As that value is higher than the maximum value of the horizontal constraint force, it cannot be counteracted. The block will begin to slide, and the horizontal interaction force between the support and the block will be a friction force:

<math>\Fcal{tr}{sup\rightarrow\Gs} + \F{sup\rightarrow block}^{friction} = [\rightarrow 4(\ms/s^2)\Ms] + [\leftarrow 1,5(\ms/s^2)\Ms] = [\rightarrow 2,5(\ms/s^2)\Ms]\Rightarrow \acc{G}{sup} = \left(\rightarrow 2,5\frac{\ms}{s^2}\right)</math>

At time <math>t = 0,6s</math>, the sliding velocity of the block on to the support is

<math>\vel{G}{sup} = \left(\rightarrow 2,5\frac{\ms}{s^2}\cdot 0,2s\right) = \left(\rightarrow 0,5\frac{\ms}{s}\right)</math>.

Even if the bock reaches the phase in which <math>\mid\Fcal{tr}{sup\rightarrow G}\mid<\mu\Ms\gs</math>, the horizontal force between support and block is still a friction force <math>(\F{sup\rightarrow block}^{friction} = [\leftarrow 1,5(\ms/s^2)\Ms])</math> until the block stops sliding.

The acceleration of the block relative to the support is:

<math>\acc{G}{sup} = \acc{G}{T} - \acc{G}{tr} = \left(\leftarrow 1,5\frac{\ms}{s^2}\right) - \left( \rightarrow 1\frac{\ms}{s^2}\right) = \left(\leftarrow 2,5\frac{\ms}{s^2}\right)</math>

Since it is a uniformly decelerated motion, it is easy to calculate the time instant <math>t_f</math> when the block stops sliding:

<math>\vel{$\Gs,t_f$}{sup} = 0 = \vel{$\Gs,t_i$}{sup} + \acc{$\Gs,t_f$}{sup}(t_f - t_i)</math>,
<math>t_i = 0,6s \Rightarrow 0 = \left(\rightarrow 0,5\frac{\ms}{s}\right) + \left(\leftarrow 2,5\frac{\ms}{s^2}\right)(t_f-0,6s)\Rightarrow t_i = 0,8s</math>

At that time instant, the force between support and block becomes a constraint force, and we reach a situation analogous to the initial one. Therefore, the slip will not occur again until <math>\ts = 1,4s</math>. The study for the subsequent intervals follows the same steps as that in the interval <math>[0,1,4s]</math>. The figure shows the evolution of the kinematics of the block with respect to the support, and of the forces acting on it.

[[File:D4-Ex4-2-eng.png|thumb|center|450px|link=]]

</div>

-----------
-----------

==D4.4 Angular Momentum Theorem (AMT): general formulation==

The dynamics of a rigid body (and therefore, of a multibody system) is never fully solved with the LMT when the rígid body rotates: the LMT only reports on the movement of a point ('''G'''), and not on the rotation.

The '''Angular Momentum Theorem''' (AMT) is stated in two different ways in the literature. In this course, a formulation parallel to that of the LMT has been chosen: terms on the left-hand side have to do only with external interactions on the system, while that on the right-hand side is the time derivative of a vector that depends only on the mass geometry of the system and its [[C1. Configuration of a mechanical system#|'''mechanical state''']].

As we have already seen when introducing the concept of [[D3. Interactions between rigid bodies#D3.1 Torsor associated with a system of forces|'''torsor of a system of forces''']], the moment of the forces, and not the resultant force, relates to the rotation of a rigid body. Therefore, although the AMT is also derived from Newton's second law applied to each mass differential of the system, the forces appearing in that equation will have to be transformed into the moment (at a point Q) associated with those forces.

These two considerations (parallel formulation to the LMT and the need to choose a point Q to calculate the moment of the forces) are the reason why the starting point to obtain the AMT is Newton's second law formulated in the '''Reference Frame that Translates with Q''' ('''RTQ''') with respect to a Galilean reference frame <math>\velang{RTQ}{RGal} = \vec{0}</math>. If <math>\acc{Q}{RGal}\neq\vec{0}</math>, this reference frame is not Galilean, and therefore the [[D1. Foundational laws of Newtonian dynamics#D1.7 Particle dynamics in non Galilean reference frames|'''inertia transportation forces''']] must be taken into account in principle ('''Figure D4.3''').

[[File:D4-4-eng.png|thumb|center|400px|link=]]
<center>'''Figure D4.4:''' Inertia forces in the Reference Frame that Translates with point Q (RTQ)</center> 

Let us consider a system of particles with constant matter. Newton's second law applied to each particle of the system and in the RTQ states:

<math>\sum \F{\rightarrow \Ps} + \Fcal{tr}{\mathrm{RTQ}\rightarrow \Ps}=\ms_\Ps\acc{P}{RTQ}</math> , 

where <math>\Fcal{tr}{\mathrm{RTQ}\rightarrow \Ps}=-\ms_\Ps \acc{P}{tr}=-\ms_\Ps \overline{\mathbf{a}}_\mathrm{RGal}(\Ps \in \mathrm{RTQ})=-\ms_\Ps \acc{Q}{RGal}</math>, and <math>\Fcal{Cor}{\mathrm{RTQ}\rightarrow \Ps}=\overline{0}</math> since <math>\velang{RTQ}{RGal}=\overline{0}</math>. 

If both sides of the equation are multiplied by <math>\QPvec</math> and summed for all particles (or mass elements) in the system, the AMT at point '''Q''' is obtained:

<math>\sum \vec{M}_\mathrm{ext} (\Qs) - \QGvec \times \Ms \acc{Q}{RGal} = \dert{\overline{\mathbf{H}}_\mathrm{RTQ}(\Qs)}{RTQ} \equiv \dot{\overline{\mathbf{H}}}_\mathrm{RTQ}(\Qs)</math> 

where M is the total mass of the System, and <math>\overline{\mathbf{H}}_\mathrm{RTQ}(\Qs) \equiv \sum_{\Ps \in \mathrm{syst}} \QPvec \times \ms_\Ps \vel{P}{RTQ}</math> is the '''angular momentum of the System about point Q'''.

If the system contains continuous elements (for example, a system with N rigid bodies <math>\Ss_i</math>), the summation for particles is actually an integral:

<math>\overline{\mathbf{H}}_\mathrm{RTQ}(\Qs)= \sum_{\is =1}^{\Ns} \overline{\mathbf{H}}_\mathrm{RTQ}^\mathrm{Si}(\Qs)= \sum_{\is=1}^{\Ns} \left( \int_{\mathrm{Si}}\QPvec \times \ds\ms (\Ps) \vel{P}{RTQ} \right )</math>.

<div>
=====💭 Proof ➕=====

Newton's second law for each element of mass multiplied by <math>\QPvec</math> is:

<math>\sum\QPvec\times\F{\rightarrow\Ps} + \QPvec\times\Fcal{tr}{RTQ\rightarrow\Ps} = \QPvec\times\ms_\Ps\acc{P}{RTQ}</math>

El terme de la dreta es pot reescriure com a:

<math>\QPvec\times\ms_\Ps\acc{P}{RTQ} = \QPvec\times\ms_\Ps\dert{\vel{P}{RTQ}}{RTQ}</math>

Since '''Q''' is a point of the RTQ, it can be taken as the origin of a position vector of '''P''' in that reference frame. Hence:

<math>\vel{P}{RTQ} = \dert{\QPvec}{RTQ} \Rightarrow\dert{\left[\QPvec\times\ms_\Ps\vel{P}{RTQ}\right]}{RTQ} = \dert{\QPvec}{RTQ} \times\ms_\Ps\vel{P}{RTQ} + \QPvec\times\ms_\Ps\dert{\vel{P}{RTQ}}{RTQ}</math>
<math>=\vel{P}{RTQ}\times\ms_\Ps\vel{P}{RTQ} + \QPvec\times\ms_\Ps\dert{\vel{P}{RTQ}}{RTQ} = \QPvec\times\ms_\Ps\dert{\vel{P}{RTQ}}{RTQ}</math>

Taking into account that the mass is constant:

<math>\QPvec\times\ms_\Ps\acc{P}{RTQ} = \dert{\left[\QPvec\times\ms_\Ps\vel{P}{RTQ}\right]}{RTQ}</math>

Summation for all '''P''' elements leads to:

<math>\sum_{\Ps\in\text{syst}}\QPvec\times\ms_\Ps\dert{\vel{P}{RTQ}}{RTQ} = \sum_{\Ps\in\text{syst}}\dert{\left[\QPvec\times\ms_\Ps\vel{P}{RTQ}\right]}{RTQ} = \dert{\sum_{\Ps\in\text{syst}}\QPvec\times\ms_\Ps\vel{P}{RTQ}}{RTQ} = \dot{\overline{\mathbf{H}}}_\mathrm{RTQ}(\Qs)</math>

If the interaction forces on each particle '''P''' are classified as internal and external, the sum for all the particles in the system on the left-hand side of the first equation becomes:

<math> \sum_{\Ps \in \mathrm{syst}} \QPvec \times \F{\mathrm{int}\rightarrow\Ps} + \sum_{\Ps \in \mathrm{syst}} \QPvec \times \F{\mathrm{ext}\rightarrow\Ps} + \sum_{\Ps \in \mathrm{syst}} \QPvec \times \Fcal{tr}{\mathrm{RTQ}\rightarrow\Ps} = \sum_{\Ps \in \mathrm{syst}} \QPvec \times \F{\mathrm{ext}\rightarrow\Ps} - \sum_{\Ps \in \mathrm{syst}} \QPvec \times \ms_\Ps \acc{Q}{RGal}=</math>

<math>= \sum \overline{\mathbf{M}}_\mathrm{ext} (\Qs) - \left( \sum_{\Ps \in \mathrm{syst}} \ms_\Ps \QPvec \right) \times \acc{Q}{RGal} = \sum \overline{\mathbf{M}}_\mathrm{ext} (\Qs) - \QGvec \times \Ms \acc{Q}{RGal} , </math> 

since the principle of action and reaction guarantees that the total moment of a pair of action and reaction forces is zero:

<math>\QPvec_\is \times \F{\Ps_\js \rightarrow \Ps_\is} + \QPvec _\js \times \F{\Ps_\is \rightarrow \Ps_\js} = \QPvec_\is \times \F{\Ps_\js \rightarrow \Ps_\is} - \QPvec _\js \times \F{\Ps_\js \rightarrow \Ps_\is} = \QPvec_\is \times \F{\Ps_\js \rightarrow \Ps_\is} - \left( \QPvec _\is + \overline{\Ps_\is \Ps_\js}\right)\times \F{\Ps_\js \rightarrow \Ps_\is}= \overline{\Ps_\is \Ps_\js} \times \F{\Ps_\js \rightarrow \Ps_\is}=\overline{0}</math> 

Finally:

<math>\sum \overline{\mathbf{M}}_\mathrm{ext} (\Qs) - \QGvec \times \Ms \acc{Q}{RGal}=\dot{\overline{\mathbf{H}}}_\mathrm{RTQ}(\Qs)</math>

</div>

---------
---------

==D4.5 Angular Momentum Theorem (AMT): particular formulations==

The AMT formulation becomes simpler when the point '''Q''' is fixed to a Galilean reference frame or when it is the center of mass of the system.

* <math>\Qs \in \mathrm{RGal} \Rightarrow \acc{Q}{RGal} = \overline{0} \Rightarrow \sum \overline{\mathbf{M}}_\mathrm{ext} (\Qs) = \dot{\overline{\mathbf{H}}}_\mathrm{RTQ} (\Qs). </math> In short, we call this version “AMT at a fixed point”, where “fixed” has to be understood as “fixed to a Galilean reference frame”. In this case, we will usually choose the letter '''O''' to designate point '''Q'''.

* <math>\Qs=\Gs \Rightarrow \QGvec \times \Ms \acc{Q}{RGal}=\overline{0} \times \Ms \acc{Q}{RGal} = \overline{0} \Rightarrow \sum \overline{\mathbf{M}}_\mathrm{ext} (\Gs) = \dot{\overline{\mathbf{H}}}_\mathrm{RTQ} (\Gs).</math> It should be noted that, although the expression is similar to the fixed point version, the center of mass '''G''' does not have to be fixed to a Galilean reference.

When '''Q''' is neither a fixed point in RGal nor coincides with the center of mass, we usually speak of the “moving point” version of the AMT. Although '''Q''' is in general moving relative to RGal, the term associated with the inertia forces <math>\left( \QGvec \times \Ms \acc{Q}{RGal} \right)</math> can be zero if '''Q''' moves at a constant speed relative to RGal, or if its acceleration relative to RGal is parallel to <math>\QGvec</math>.

<center>
{| class="wikitable"
|+
|-
! '''AMT at a fixed point''' !! '''AMT at G '''<math>\Gs</math> !! '''AMT at a moving point '''<math>\Qs</math>
|-
| <center><math> \quad \Qs \in \mathrm{RGal} \Rightarrow \Qs \equiv \Os , \mathrm{RTQ} = \mathrm{RGal} \quad </math> 

<math>\sum \overline{\mathbf{M}}_\mathrm{ext}(\Os) = \dot{\overline{\mathbf{H}}}_\mathrm{RGal} (\Os)</math> </center>
||
<center>
<math>\quad \Qs=\Gs \Rightarrow \mathrm{RTQ}=\mathrm{RTG} \quad </math> 

<math>\sum \overline{\mathbf{M}}_\mathrm{ext}(\Gs) = \dot{\overline{\mathbf{H}}}_\mathrm{RTG} (\Gs)</math></center>
||
<center>
<math> \quad \acc{Q}{RGal} \neq \overline{0}\quad </math> 

<math>\quad \sum \overline{\mathbf{M}}_\mathrm{ext}(\Qs) - \QGvec \times \Ms\acc{Q}{RGal}= \dot{\overline{\mathbf{H}}}_\mathrm{RTQ} (\Qs) \quad</math></center>

|}
</center>

The angular momentum is not easy to calculate in general, and it is presented in [[D5. Mass distribution|'''unit D5''']].

In problems with planar (2D) kinematics, if the angular momentum <math>\overline{\mathbf{H}}_\mathrm{RTQi}^\mathrm{Si} (\Gs_\is)</math> of each rigid body in the system under study is parallel to the angular velocity of the rigid body (whose direction is orthogonal to the plane of motion), only the component of the AMT perpendicular to that plane is interesting. The problem is then a planar dynamics problem (2D). Planar dynamics can be studied from two LMT components and one AMT component.

<div>
===AMT at a contact point between two rigid bodies===

A link that appears frequently in mechanical systems is the single-point contact between pairs of rígid bodies (S1 and S2, for example). That constraint can introduce between 1 and 3 constraint unknowns (depending on the surface roughness and the contact kinematics– with or without sliding). When we do not want to calculate these forces, it is tempting to apply the AMT at the contact point J (since the moment of those forces at point '''J''' is zero). 

[[File:TMCaJ-0-eng.png|thumb|right|230px|link=]]

The application of the AMT at '''J''' is very tricky. It is necessary to specify which point J has been chosen ([[C5. Cinemàtica plana del sòlid rígid#✏️ Exemple C5-1.8|'''example C5-1.8''']]): is it point '''J''' of rigid body S1 <math>(\Js_{\mathrm{S}1})</math>, of rigid body S2 <math>(\Js_{\mathrm{S}2})</math>, or is it the geometric contact point <math>(\Js_{\mathrm{geom}})</math>. On the one hand, those three points have different kinematics <math> \left(\overline{\as}_\mathrm{Gal}(\Js_{\mathrm{S}1}) \neq \overline{\as}_\mathrm{Gal}(\Js_{\mathrm{S}2}) \neq \overline{\as}_\mathrm{Gal}(\Js_\mathrm{geom}) \right)</math>, and the complementary term associated with the moment of inertia forces <math>\left(\JGvec \times \ms \acc{J}{Gal} \right)</math> is different. On the other hand, if for example the AMT is applied to rigid bdy S1, it is necessary to keep in mind that, although <math>(\Js_{\mathrm{S}1})</math> belongs to S1 and the angular momentum can be calculated from the inertia tensor, <math>(\Js_{\mathrm{S}1})</math> and <math>(\Js_{\mathrm{geom}})</math> not belong to S1, and the barycentric decomposition is compulsory to calculate this vector.

Finally, since the angular momentum has to be differentiated, it has to be calculated in a general configuration (i.e. when '''J''' is not yet the contact point), and only after having done the time derivative can the result be particularized to the configuration in which '''J''' is the contact point. As an illustration of all this, the following diagram shows the application of the AMT to a wheel with planar motion sliding on the ground.

[[File:TMCaJ-1-eng.png|thumb|center|520px|link=]]
<center>
<math>\overline{\Js_\mathrm{wheel}\Gs} \times \ms \overline{\as}_\mathrm{Gal}(\Js_\mathrm{wheel})=\otimes \ms \rs \ddot{\xs} \hspace{0.7cm} \overline{\mathrm{H}}_\mathrm{RTJwheel}(\Js_\mathrm{wheel},\varphi)=\Is\Is (\Js_\mathrm{wheel},\varphi) \overline{\dot{\varphi}} \quad , \quad \left.\dot{\overline{\mathrm{H}}}_\mathrm{RTJwheel}(\Js_\mathrm{wheel},\varphi)\right]_{\varphi=180^\circ}</math>
</center> 

[[File:TMCaJ-2-eng.png|thumb|center|520px|link=]]
<center>
<math>\overline{\Js_\mathrm{ground}\Gs} \times \ms \overline{\as}_\mathrm{Gal}(\Js_\mathrm{ground})= \overline{0} \hspace{0.7cm} \left\{\begin{array}{l}
\overline{\mathrm{H}}_\mathrm{RTJground}(\Js_\mathrm{ground},\xs)=\Is\Is (\Gs) \overline{\dot{\varphi}} + \overline{\Js_\mathrm{ground}\Gs} \times \ms\overline{\vs}_\mathrm{RTJground}(\Gs)=\Is\Is (\Gs)\overline{\dot{\varphi}} + (\otimes \ms\rs\dot{\xs})\\
\left.\dot{\overline{\mathrm{H}}}_\mathrm{RTJground}(\Js_\mathrm{ground},\xs) \right]_{\xs=0}
\end{array}\right.
</math> 

[[File:TMCaJ-3-eng.png.png|thumb|center|520px|link=]]

<center>
<math>\overline{\Js_\mathrm{geom}\Gs} \times \ms \overline{\as}_\mathrm{Gal}(\Js_\mathrm{geom})= (\otimes \ms \rs \ddot{\xs})\hspace{0.7cm} \left\{\begin{array}{l}
\overline{\mathrm{H}}_\mathrm{RTJgeom}(\Js_\mathrm{terra},\xs=0)=\Is\Is (\Gs) \overline{\dot{\varphi}} + \overline{\Js_\mathrm{geom}\Gs} \times \ms\overline{\vs}_\mathrm{RTJgeom}(\Gs)=\Is\Is (\Gs)\overline{\dot{\varphi}}\\
\dot{\overline{\mathrm{H}}}_\mathrm{RTJgeom}(\Js_\mathrm{terra},\xs=0)
\end{array}\right.
</math>
</center></center>
</div>

------
------

==D4.6 AMT: application examples==
<div>
====✏️ EXAMPLE D4.5: static limit condition====
------
:
{|
|[[File:D4-Ex5-1-eng.png|thumb|left|180px|link=]]
|The block hangs from an inclined massless bar through an inextensible thread. The bar ends are in contact with two walls fixed to the ground, one smooth and one rough (with a non-zero friction coefficient <math>\mu_\mathrm{Q}</math>). We want to calculate the minimum <math>\mu_\mathrm{Q}</math> value for equilibrium.

El bloc penja d’una barra inclinada de massa negligible per mitjà d’un fil inextensible. Els extrems de la barra estan en contacte amb dues parets fixes a terra, una llisa i una altra rugosa (amb coeficient de fricció no nul <math>\mu_\mathrm{Q}</math> ). Es tracta de calcular el valor mínim de <math>\mu_\mathrm{Q}</math> que permet l’equilibri. 

It is a 2D dynamics problem. The external forces acting on the system (bar+block+thread) are:
[[File:D4-Ex5-2-neut.png|thumb|center|300px|link=]]
|}

Since the system is at rest relative to the ground, <math>\acc{G}{T}=\overline{0}</math> and the total external force must be zero. Therefore:

<math>\Ns_\Ps=\Ns_\Qs</math> , <math>\ms\gs=\Ts_\Qs</math> .

Since motion is about to occur, the tangential constraint force at '''Q''' has its maximum possible value: <math>\Ts_{\Qs,\mathrm{max}}=\ms\gs=\mu_{\Qs,\mathrm{min}}\Ns_\Qs</math>. Since the above equations do not allow the calculation of <math>\Ns_\Qs</math>, a third equation is needed, which will come from the AMT. Whether applied at '''P''', '''Q''' or '''O''', the angular momentum is zero:

<math>\left.\begin{array}{l}
\mathrm{RTP}=\mathrm{RTQ}=\mathrm{RTO}=\mathrm{ground}(\Ts)\\
\overline{\mathbf{v}}_\Ts(\mathrm{block})=\overline{0}
\end{array}\right\} \Rightarrow \overline{\mathbf{H}}_\mathrm{RTP}(\Ps)=\overline{\mathbf{H}}_\mathrm{RTQ}(\Qs)=\overline{\mathbf{H}}_\mathrm{RTO}(\Os)=\overline{0}</math> 

<math>\sum \overline{\mathbf{M}}_\mathrm{ext}(\Qs)=\overline{0} \quad \Rightarrow \quad (\otimes \Ns_\Ps4\Ls\sin\beta)+(\odot \ms\gs3\Ls\cos\beta)=0 \quad \Rightarrow \quad \Ns_\Ps=\frac{3}{4} \frac{\ms\gs}{\tan\beta}.</math>

Therefore: <math>\Ts_{\Qs,\mathrm{max}}=\ms\gs=\mu_{\Qs,\mathrm{min}}\Ns_\Qs=\mu_{\Qs,\mathrm{min}} \frac{3}{4} \frac{\ms\gs}{\tan\beta} \quad \Rightarrow \mu_{\Qs,\mathrm{min}}=\frac{4}{3} \tan \beta. </math>

</div>
<div>

====✏️ EXAMPLE D4.6: Atwood machine ====
------
:
{|
|[[File:D4-Ex6-1-eng.png|thumb|left|230px|link=]]
|The blocks hang from two inextensible ropes with one end tied to the periphery of two pulleys with radii r and 2r, of negligible mass and articulated to the ceiling. We want to calculate angular acceleration of the pulleys.

It is a 2D dynamics problem. The external forces on the system (blocks + pulleys) and its motion relative to the ground are <math>(\gs \approx 10 \ms/\ss^2)</math>:

[[File:D4-Ex6-1-neut.png|thumb|center|140px|link=]]
|}

In total, there are three unknowns: the two constraint forces associated with the articulation of the pulleys, and the angular acceleration of the pulleys relative to the ground. The two vector theorems applied to the system (blocks + pulleys) provide three equations. Now, since the unknown to be determined is the acceleration, we can just use the AMT at O'''Bold text''' and obtain an equation free of constraint unknowns but including <math>\ddot{\theta}</math>:

<math>\sum \overline{\mathbf{M}}_\mathrm{ext}(\Os) = \dot{\overline{\mathbf{H}}}_\mathrm{RTO}(\Os) \quad \Rightarrow \quad (\odot 100 \Ns \cdot 2\rs) +(\otimes 50\Ns \cdot \rs)=(\odot 150\Ns \cdot \rs)=\dot{\overline{\mathbf{H}}}_\mathrm{RTO}(\Os)</math> 

<math>\overline{\mathbf{H}}_\mathrm{RTO}(\Os)=\overline{\mathbf{H}}_\Ts (\Os)= \int_\mathrm{left \: esp.} \OPvec \times \vel{P}{T} \ds\ms(\Ps)+ \int_\mathrm{right \: block} \OPvec \times \vel{P}{T} \ds\ms(\Ps) =\left(\int_\mathrm{left \: block} \OPvec\ds\ms(\Ps) \right)\times (\downarrow 2\rs\dot{\theta})+ \left(\int_\mathrm{right \: block} \OPvec\ds\ms(\Ps) \right)\times (\uparrow \rs\dot{\theta}) </math> 

By the definition of [[D3. Interactions between rigid bodies#D3.1 Torsor associated with a system of forces|'''center of mass''']]: <math>\int_\mathrm{block} \OPvec\ds\ms(\Ps)=\ms_\mathrm{block}\OGvec_\mathrm{block} </math>. On the other hand, since the blocks have vertical velocity, only the horizontal component of <math>\OGvec_\mathrm{block} \times \overline{\mathbf{v}}_\Ts(\mathrm{block})</math> contributes to the vector product <math>\OGvec_\mathrm{block}</math>. Thus:

<math>\overline{\mathbf{H}}_\mathrm{T}(\Os)=10\ks\gs(\leftarrow 2\rs) \times (\downarrow 2\rs\dot{\theta})+ 5\ks\gs(\rightarrow \rs) \times (\uparrow \rs \dot{\theta})=40\ks\gs(\odot \rs^2\dot{\theta})+5\ks\gs(\odot\rs^2\dot{\theta})=45\ks\gs(\odot \rs^2\dot{\theta})</math> 

<math>\dot{\overline{\mathbf{H}}}_\mathrm{T}(\Os)=45\ks\gs(\odot \rs^2\ddot{\theta})</math> 

<math>\sum \overline{\mathbf{M}}_\mathrm{ext}(\Os) = \dot{\overline{\mathbf{H}}}_\mathrm{RTO}(\Os) \quad \Rightarrow \quad (\odot 150\Ns \cdot \rs)=45\ks\gs(\odot \rs^2\ddot{\theta}) \quad \Rightarrow \quad \ddot{\theta}=\frac{10}{3\rs}</math> 

[[File:D4-Ex6-3-neut.png|thumb|right|150px|link=]]
Alternative solution 

If we apply the AMT at point '''O''' to the system formed by the pulleys and the ropes, the weight of the blocks no longer appears as an external interaction, but instead the tensions of the two ropes appear (which are constraint unknowns).

As the system is massless, its angular momentum is zero:

SYST: pulleys + ropes 

<math>\sum \overline{\mathbf{M}}_\mathrm{ext}(\Os) = \dot{\overline{\mathbf{H}}}_\mathrm{RTO}(\Os)=\overline{0} \quad \Rightarrow \quad (\odot\Ts_1 \cdot\rs) + (\otimes \Ts_2 \cdot 2 \rs)=0</math> 
[[File:D4-Ex6-4-neut.png|thumb|right|220px|link=]]

Applying the AMT to each of the blocks provides two more equations: 

SYST: 10 kg block 

<math>\sum \overline{\mathbf{F}}_\mathrm{ext}(\Os)=(10\ks\gs)\overline{\mathbf{a}}_\Ts(\mathrm{bloc}) \quad \Rightarrow \quad (\downarrow 100\Ns)+(\uparrow \Ts_1)=(10\ks\gs)(\uparrow 2\rs \ddot{\theta})</math> 

SYST: 5 kg block 

<math>\sum \overline{\mathbf{F}}_\mathrm{ext}(\Os)=(5\ks\gs)\overline{\mathbf{a}}_\Ts(\mathrm{bloc}) \quad \Rightarrow \quad (\downarrow 50\Ns)+(\uparrow \Ts_2)=(5\ks\gs)(\uparrow \rs \ddot{\theta})</math> 

Solving the system of equations leads to <math>\ddot{\theta}=\frac{10}{3\rs}</math>. 

The first solution is faster (it just calls for one scalar equation) but involves calculating the angular momentum. This second solution is longer (system of three equations) but does not require calculating any angular momentum.

</div>

<div>

====✏️ EXAMPLE D4.7: longitudinal dynamics of a vehicle====
------
:
{|
|[[File:D4-Ex7-1-eng.png|thumb|center|200px|link=]]
|A vehicle without suspensions moves on a straight road. The mass of the wheels is negligible compared to that of the rest of the elements, and their contact with the ground is considered to be a single-point one. We want to analyse the normal constraint forces ground and wheels.
This is a 2D dynamics problem. The external forces on the vehicle are the weight and and the ground constraint forces. If the acceleration of the chassis relative to the ground is a given, the application of the AMT leads to:
|}

{|
|

<math>\sum\F{\text{ext}} = \ms\acc{G}{\epsilon} \Rightarrow \left\{\begin{aligned}

\uparrow(\Ns_{\ds\vs}+ \Ns_{\ds\rs})+ (\downarrow\ms\gs) = 0\\
[\rightarrow(\Ts_{\ds\vs}+ \Ts_{\ds\rs})] = (\rightarrow\ms\as_\epsilon)
\end{aligned}\right.</math>

where <math>(\Ns_{\ds\vs}, \Ts_{\ds\vs})</math> and <math>(\Ns_{\ds\rs}, \Ts_{\ds\rs})</math> are the resultant normal and tangential forces exerted by the ground on the two front wheels and the two rear wheels, respectively.

The AMT at '''G''' yields: <math>\sum\vec{\Ms}_{ext}(\Gs) = \vec{H}_{RTG}(\Gs)\Rightarrow(\odot\Ls_{\ds\vs}\Ns_{\ds\vs}) + (\otimes\Ls_{\ds\rs}\Ns_{\ds\rs}) + [\odot\hs(\Ts_{\ds\vs} + \Ts_{\ds\rs})] =0</math>
(the angular momentum <math>\vec{H}_{RTG}(\Gs)</math> is permanently zero beacuse the wheels are massless, and the chassis motion relative to the RTG is zero).
|[[File:D4-Ex7-2-eng.png|thumb|center|200px|link=]]
|}

Solving the system of equations: <math>\Ns_{\ds\vs} = \frac{\ms\gs\Ls_{\ds\rs}}{\Ls_{\ds\vs} + \Ls_{\ds\rs}} - \frac{\ms\as_\epsilon\hs}{\Ls_{\ds\vs} + \Ls_{\ds\rs}}, \:\:\:\:\Ns_{\ds\rs} = \frac{\ms\gs\Ls_{\ds\vs}}{\Ls_{\ds\vs} + \Ls_{\ds\rs}} + \frac{\ms\as_\epsilon\hs}{\Ls_{\ds\vs} + \Ls_{\ds\rs}}, </math>

This result is valid for any value of <math>\as_\epsilon</math>:
:* Static situation (<math>\as_\epsilon = 0</math>): <math>\Ns_{\ds\vs}^{\es\ss\ts} = \frac{\ms\gs\Ls_{\ds\rs}}{\Ls_{\ds\vs}+ \Ls_{\ds\rs}}</math>, <math>\Ns_{\ds\rs}^{\es\ss\ts} = \frac{\ms\gs\Ls_{\ds\vs}}{\Ls_{\ds\vs}+ \Ls_{\ds\rs}}</math>; the normal force is higher on the wheels whose axle is closer to '''G'''.
:Accelerated motion (<math>\as_\epsilon > 0</math>): <math>\Ns_{\ds\vs} = \Ns_{\ds\vs}^{\es\ss\ts} - \frac{\ms|\as_\epsilon|\hs}{\Ls_{\ds\vs}+ \Ls_{\ds\rs}}</math>, <math>\Ns_{\ds\rs} = \Ns_{\ds\rs}^{\es\ss\ts} + \frac{\ms|\as_\epsilon|\hs}{\Ls_{\ds\vs}+ \Ls_{\ds\rs}}</math>; there is a force transfer from front to rear (the rear wheels are loaded and the front wheels are unloaded). If the acceleration is sufficiently high, the front normal force becomes negative, which indicates that contact has been lost and the vehicle rolls over counter-clockwise. Then, <math>\Ns_{\ds\vs} = 0</math> and the chassis initially has angular acceleration. As a consequence, the centre of mass '''G''' acquires vertical acceleration.

:*Braking (<math>\as_\epsilon < 0</math>): <math>\Ns_{\ds\vs} = \Ns_{\ds\vs}^{\es\ss\ts} - \frac{\ms|\as_\epsilon|\hs}{\Ls_{\ds\vs}+ \Ls_{\ds\rs}}</math>, <math>\Ns_{\ds\rs} = \Ns_{\ds\rs}^{\es\ss\ts} - \frac{\ms|\as_\epsilon|\hs}{\Ls_{\ds\vs}+ \Ls_{\ds\rs}}</math>; there is a force transfer from the rear to the front (the rear wheels are unloaded and the front wheels are loaded). As in the previous case, there is a critical value that causes the vehicle to tip over, this time clockwise.

---------
---------
</div>

==D4.7 Dynamics of Constraint Auxiliary Elements==

The application of the vector theorems to [[D3. Interactions between rigid bodies#D3.5 Indirect constraint interactions: Constraint Auxiliary Elements (CAE)|'''Constraint Auxiliary Elements''']] (CAE) is simple since the terms on the right (change in linear momentum and change in angular momentum) are zero because the CAE mass is zero:

<math>\ms_\mathrm{CAE} \approx 0 \Rightarrow \sum \mathbf{\overline{F}}_\mathrm{ext} \approx \overline{0} \quad , \quad \sum \mathbf{\overline{M}}_\mathrm{ext} (\mathrm{qualsevol} \hspace{0.2cm} \mathrm{punt}) \approx \overline{0}.</math> 

For the case of a CAE (rígid body S) connecting two rigid bodies S1 and S2 ('''Figure D4.5'''), these equations allow to prove that the torsor exerted on S1 can be obtained from an analytical characterization equation where the kinematics of the characterization point '''P''' (which has to be a point fixed to S1) is assessed from the S2 reference frame:

<math>\mathbf{\overline{F}}_{\mathrm{S}2 \rightarrow (\mathrm{S}) \rightarrow \mathrm{S}1} \cdot \mathbf{\overline{v}}_{\mathrm{S}2} (\Ps \in \mathrm{S}1) + \mathbf{\overline{M}}_{\mathrm{S}2\rightarrow (\mathrm{S}) \rightarrow \mathrm{S}1} (\Ps) \cdot \velang{S1}{S2} =0.</math> 

(When dealing with velocities, it is important to specify which rigid body does '''P''' belong to. However, this is irrelevant when dealing with moments: <math> \mathbf{\overline{M}} (\Ps \in \mathrm{S}1) = \mathbf{\overline{M}} (\Ps \in \mathrm{S}) + \overline{\Ps_{\mathrm{S}1}\Ps_\mathrm{S}} \times \mathbf{\overline{F}} = \mathbf{\overline{M}} (\Ps \in \mathrm{S}) \equiv \mathbf{\overline{M}} (\Ps)</math>). 

[[File:D4-5-eng.png|center|350px|link=]]
<center>'''Figure D4.5:'''Constraint Auxiliary Element between to rigid bodies with nonzero mass</center> 

<div>
=====💭 Proof ➕=====

The vector theorems applied to the CAE imply that the constraint torsors exerted by S1 and S2 at the same point '''P''' must sum up to zero:

<math>\left.\begin{array}{l}
\sum \overline{\mathbf{F}}_{\mathrm{ext}} \approx \overline{0} \Rightarrow \overline{\mathbf{F}}_{\mathrm{S} 1 \rightarrow \mathrm{S}}+\overline{\mathbf{F}}_{\mathrm{S} 2 \rightarrow \mathrm{S}}=\overline{0} \\
\sum \overline{\mathbf{M}}_{\mathrm{ext}}(\Ps) \approx \overline{0} \Rightarrow \overline{\mathbf{M}}_{\mathrm{S} 1 \rightarrow \mathrm{S}}(\Ps)+\overline{\mathbf{M}}_{\mathrm{S} 2 \rightarrow \mathrm{S}}(\Ps)=\overline{0}
\end{array}\right\} \Rightarrow\left\{\begin{array}{l}
\overline{\mathbf{F}}_{\mathrm{S} 2 \rightarrow \mathrm{S}} =\overline{\mathbf{F}}_{\mathrm{S} \rightarrow \mathrm{S} 1} \equiv \overline{\mathbf{F}}_{\mathrm{S} 2 \rightarrow(\mathrm{S}) \rightarrow \mathrm{S} 1} \\
\overline{\mathbf{M}}_{\mathrm{S} 2 \rightarrow \mathrm{S}}(\Ps)=\overline{\mathbf{M}}_{\mathrm{S} \rightarrow \mathrm{S} 1}(\Ps) \equiv \overline{\mathbf{M}}_{\mathrm{S} 2 \rightarrow(\mathrm{S}) \rightarrow \mathrm{S} 1}(\Ps)
\end{array}\right.</math> 

On the other hand, those torsores fulfill the [[D3. Interactions between rigid bodies#D3.4 Direct constraint interactions|''' equation of analytical characterization''']]:

<math>\mathbf{\overline{F}}_{\mathrm{S}2 \rightarrow \mathrm{S}} \cdot \mathbf{\overline{v}}_{\mathrm{S}2} (\Ps \in \mathrm{S}) + \mathbf{\overline{M}}_{\mathrm{S}2 \rightarrow \mathrm{S}} (\Ps) \cdot \velang{S}{S2} =0 \quad , \quad \mathbf{\overline{F}}_{\mathrm{S} \rightarrow \mathrm{S}1} \cdot \mathbf{\overline{v}}_{\mathrm{S}} (\Ps \in \mathrm{S}1) + \mathbf{\overline{M}}_{\mathrm{S}\rightarrow \mathrm{S}1} (\Ps) \cdot \velang{S1}{S} =0. </math> 

If all the above equations are combined, we obtain: 

<math> \mathbf{\overline{F}}_{\mathrm{S}2 \rightarrow (\mathrm{S}) \rightarrow \mathrm{S}1} \cdot \left[ \mathbf{\overline{v}}_{\mathrm{S}2} (\Ps \in \mathrm{S}) + \mathbf{\overline{v}}_{\mathrm{S}} (\Ps \in \mathrm{S}1) \right] + \mathbf{\overline{M}}_{\mathrm{S}2\rightarrow (\mathrm{S}) \rightarrow \mathrm{S}1} (\Ps) \cdot \left[\velang{S1}{S} + \velang{S}{S2}\right] =0. </math> 

A [[C3. Composition of movements#C3.1 Composition of velocities|'''composition of movements''']] allows us to rewrite the above equation in a more compact way:

<math>\left.\begin{array}{l}
\mathrm{AB}: \mathrm{S}2 \\
\mathrm{REL}: \mathrm{S}
\end{array}\right\} \Rightarrow \velang{S1}{S}=\velang{S1}{S2} - \velang{S}{S2}</math> 

<math>\left.\begin{array}{l}
\mathrm{AB}: \mathrm{S}2 \\
\mathrm{REL}: \mathrm{S}1
\end{array}\right\} \Rightarrow \mathbf{\overline{v}}_{\mathrm{S}2} (\Ps \in \mathrm{S})=\mathbf{\overline{v}}_{\mathrm{S}1} (\Ps \in \mathrm{S}) - \mathbf{\overline{v}}_{\mathrm{S}2} (\Ps \in \mathrm{S}1)</math> 

<math>\left.\begin{array}{l}
\mathrm{AB}: \mathrm{S}1 \\
\mathrm{REL}: \mathrm{S}
\end{array}\right\} \Rightarrow \mathbf{\overline{v}}_{\mathrm{S}} (\Ps \in \mathrm{S}1)=\mathbf{\overline{v}}_{\mathrm{S}1} (\Ps \in \mathrm{S}1) - \mathbf{\overline{v}}_{\mathrm{S}1} (\Ps \in \mathrm{S})= - \mathbf{\overline{v}}_{\mathrm{S}1} (\Ps \in \mathrm{S})</math> 

Finalment: <math>\mathbf{\overline{F}}_{\mathrm{S}2 \rightarrow (\mathrm{S}) \rightarrow \mathrm{S}1} \cdot \mathbf{\overline{v}}_{\mathrm{S}2} (\Ps \in \mathrm{S}1) + \mathbf{\overline{M}}_{\mathrm{S}2\rightarrow (\mathrm{S}) \rightarrow \mathrm{S}1} (\Ps) \cdot \velang{S1}{S2} =0.</math>

</div>
<div>

====✏️ EXAMPLE D4.8: longitudinal dynamics of a vehicle ====
---------
:
{|
|[[File:D4-Ex8-1-eng.png|thumb|left|220px|link=]]
|Let’s assume that the vehicle of example D4.7 has rear traction. This means that the vehicle's engine acts between the chassis and the rear wheels, while the front wheels are only subject to constraint interactions (articulation with the chassis and contact with the ground).

If the front wheels are treated as CAE, the kinematic analysis of the chassis relative to the ground for the characterization of the constraint through the front wheels leads to a torsor at the wheel centre (which is fixed to the chassis) with only one force component:
|}
[[File:D4-Ex8-2-eng.png|thumb|center|370px|link=]]

The existence of a horizontal component of force between the rear wheels (necessary to accelerate or brake the vehicle!) is associated with the engine torque. If the vector theorems are applied to the rear wheels, the external interactions to be taken into account are the connection with the ground, the articulation with the chassis and the engine torque:

[[File:D4-Ex8-3-eng.png|thumb|center|370px|link=]]

<math>\ms_\mathrm{wheels} \approx 0 \Rightarrow \sum \overline{\mathbf{F}}_\mathrm{ext} \approx \overline{0} , \sum
\overline{\mathbf{M}}_\mathrm{ext} (\Cs) \approx \overline{0} \Rightarrow
\left\{\begin{array}{l}
\mathrm{LMT }: \hspace{1.4cm} \Ns_\mathrm{dr}=\Ns'_\mathrm{dr} , \Ts_\mathrm{dr}=\Ts'_\mathrm{dr} \\
\mathrm{AMT } \hspace{0.2cm} \mathrm{a} \hspace{0.2cm} \Cs: \quad \Ts_\mathrm{dr}\rs=\Gamma \Rightarrow \Ts_\mathrm{dr}=\Gamma / \rs
\end{array}\right. </math> 

The tangential force on the driven wheels is directly proportional to the engine torque applied to them. However, it should be remembered that, as a constraint tangential force, its value is bounded by <math>\mu_\mathrm{s}\Ns_\mathrm{dr}</math>. This allows to calculate the maximum acceleration that the vehicle can acquire (as long as the front wheel does not lose contact with the ground): <math>\left.\as_\epsilon (\mathrm{vehicle})\right]_\mathrm{max}= \Ts_\mathrm{dr,max}/\ms \quad , \quad \Ts_\mathrm{dr}=\Gamma / \rs \quad , \quad \Ts_\mathrm{dr} \leq \mu_\mathrm{e}\Ns_\mathrm{dr}</math>. 

On low friction (low <math>\mu_\ss</math> value) terrains, an engine capable of delivering a high maximum torque is useless: what limits acceleration is the value of the friction coefficient <math>\mu_\es</math>: <math> \left.\as_\epsilon (\mathrm{vehicle})\right]_\mathrm{màx}= \mu_\mathrm{e}\Ns_\mathrm{dr} /\ms </math>.

On high friction terrain, if the maximum torque is low, it is this that limits acceleration: <math>\left. \as_\epsilon (\mathrm{vehicle})\right]_\mathrm{max}= \Gamma_\mathrm{max}/\ms \rs</math> (if the front wheel does not lose contact with the ground).


</div>

---------
---------

==D4.8 Barycentric decomposition of the angular momentum==

The general formulation presented for the AMT allows a free choice of the Q point of application. The criterion for choosing it is based on what we want to investigate (a constraint force, an equation of motion...). This criterion is discussed in [[D6. Examples of 2D dynamics|'''unit D6''']].

In some cases, it may be interesting to choose a point '''Q''' that does not belong to any element of the system. It is then useful to calculate the angular momentum <math>\H{Q}{}{RTQ}</math> from the angular momentum at the system center of mass '''G''', <math>\H{Q}{}{RTG}</math>. This is known as the '''barycentric decomposition''' of the angular momentum:

<math>\H{Q}{}{RTQ} = \H{G}{}{RTG} + \H{Q}{\oplus}{RTQ}</math>,
where the superscript <math>\oplus</math> indicates that the angular momentum is to be calculated as if the system had been reduced to a particle concentrated at '''G''' with mass equal to the total mass of the system (M):

<math>\H{Q}{\oplus}{RTQ} = \QGvec\times\Ms\vel{G}{RTQ}</math>.

<div>
=====💭 Proof ➕=====

:The definition of the angular momentum of a rigid body is: <math>\H{Q}{\Ss}{RTQ} = \int_\Ss\QPvec\times\vel{P}{RTQ}\ds\ms(\Ps)</math>. Velocity <math>\vel{P}{RTQ}</math> can be written as a sum of two terms through a composition of movements:

:<math>\left.\begin{array}{l}
\mathrm{AB}:\mathrm{RTQ}\\
\mathrm{REL}:\mathrm{RTG}
\end{array}\right\} \Rightarrow\vel{P}{RTQ} = \vel{P}{RTG} + \vel{P}{ar} = \vel{P}{RTG} + \vel{$\Ps\in\mathbf{RT}\Gs$}{RTQ} = \vel{P}{RTG} + \vel{G}{RTQ}</math> 

:Therefore: <math>\H{Q}{\Ss}{RTQ} = \int_\Ss\QPvec\times\vel{P}{RTQ}\ds\ms(\Ps) + \left[\int_\Ss\QPvec\ds\ms(\Ps)\right]\times\vel{P}{RTQ}</math>.

:The definition of [[D4. Vector theorems#D4.1 Linear Momentum Theorem (LMT) in Galilean reference frames|'''centre of mass''']] leads to rewriting the second term as:

:<math>\left[\int_\Ss\QPvec\ds\ms(\Ps)\right] = \Ms\QGvec\times\vel{P}{RTQ} = \QGvec\times\Ms\vel{P}{RTQ}</math>, which coincides with the angular momentum at '''Q''' of a particle of mass M located at '''G''': <math>\QGvec\times\Ms\vel{G}{RTQ} = \H{Q}{\oplus}{RTQ}</math>.

:In the first term of the <math>\H{Q}{\Ss}{RTQ}</math> expression, <math>\QPvec</math> can be decomposed into two terms:

:<math>\int_\Ss\QPvec\times\vel{P}{RTQ}\ds\ms(\Ps) = \int_\Ss\QGvec\times\vel{P}{RTQ}\ds\ms(\Ps) + \int_\Ss\GPvec\times\vel{P}{RTQ}\ds\ms(\Ps) = \int_\Ss\QGvec\times\vel{P}{RTQ}\ds\ms(\Ps) + \H{G}{\Ss}{RTG}</math>

:From the definition of centre of mass: <math>\int_\Ss\QGvec\times\vel{P}{RTQ}\ds\ms(\Ps) = \QGvec\times\left[\int_\Ss\vel{P}{RTG}\ds\ms(\Ps)\right] = \QGvec\times\Ms\vel{G}{RTG} = \vec{0}</math>.

:Therefore <math>\H{Q}{}{RTQ} = \H{G}{}{RTG} + \H{Q}{\oplus}{RTQ}</math>.

</div>

<div>
===== ✏️ EXAMPLE D4.9: barycentric decomposition =====
----------

{|

|[[File:D4-Ex9-1-eng.png|thumb|center|200px|link=]]
|The rigid body is made up of two massless bars and four particles ('''P''') with mass m attached to the bars endpoints. The rigid body is attached to a support that can move along a guide fixed to the ground.

Its angular momentum at '''G''' is:

<math>\H{G}{}{RTG}\equiv\sum_{\Ps\in\text{syst}}\GPvec\times\ms\vel{P}{RTG}</math>

The RTG is fixed to the support, and the velocity of the particles relative to that reference frame is proportional to the rotation <math>\omega</math>: <math>|\vel{P}{RTG}|=\Ls\omega</math>. Hence:

<math>\H{G}{}{RTG}\equiv\sum_{\Ps\in\text{syst}}\GPvec\times\ms\vel{P}{RTG} = (\otimes4\ms\Ls^2\omega)</math>
|}
:The angular momentum at '''O''' (fixed to the ground) can be calculated from <math>\H{G}{}{RTG}</math> through barycentric decomposition:

:<math>\H{O}{}{RTO} = \H{O}{}{T} = \H{G}{}{RTG} + \H{O}{\oplus}{T} = (\otimes 4\ms\Ls^2\omega) + \OGvec\times 4\ms\vel{G}{T} =</math>
<math>\hspace{1cm} = (\otimes 4\ms\Ls^2\omega) + [(\rightarrow\xs) + (\downarrow\hs)]\times 4\ms(\rightarrow\dot\xs) = (\otimes4\ms\Ls^2\omega) + (\downarrow\hs)\times 4\ms(\rightarrow\dot\xs) =</math>
<math>\hspace{1cm}= (\otimes4\ms\Ls^2\omega) + (\odot 4\ms\hs\dot\xs) = \otimes 4\ms(\Ls^2\omega - \hs\dot\xs)</math>


</div>

© Universitat Politècnica de Catalunya. [[Mecánica:Derechos de autor |Todos los derechos reservados]]
-------------

<center>
[[D3. Interactions between rigid bodies|<<< D3. Interactions between rigid bodies]]

[[D5. Mass distribution|D5. Mass distribution >>>]]
</center>

D4. Vector theorems

2025-05-14T07:14:15Z

Bros: /* D4.8 Barycentric decomposition of the angular momentum */

<div class="noautonum">__TOC__</div>

<math>
\newcommand{\uvec}{\overline{\textbf{u}}}
\newcommand{\vvec}{\overline{\textbf{v}}}
\newcommand{\evec}{\overline{\textbf{e}}}
\newcommand{\Omegavec}{\overline{\mathbf{\Omega}}}
\newcommand{\velang}[2]{\Omegavec^{\textrm{#1}}_{\textrm{#2}}}
\newcommand{\Alfavec}{\overline{\mathbf{\alpha}}}
\newcommand{\accang}[2]{\Alfavec^{\textrm{#1}}_{\textrm{#2}}}
\newcommand{\as}{\textrm{a}}
\newcommand{\ds}{\textrm{d}}
\newcommand{\hs}{\textrm{h}}
\newcommand{\ks}{\textrm{k}}
\newcommand{\Ms}{\textrm{M}}
\newcommand{\Ns}{\textrm{N}}
\newcommand{\Fs}{\textrm{F}}
\newcommand{\ms}{\textrm{m}}
\newcommand{\ns}{\textrm{n}}
\newcommand{\ts}{\textrm{t}}
\newcommand{\Is}{\textrm{I}}
\newcommand{\us}{\textrm{u}}
\newcommand{\qs}{\textrm{q}}
\newcommand{\vs}{\textrm{v}}
\newcommand{\Rs}{\textrm{R}}
\newcommand{\Ts}{\textrm{T}}
\newcommand{\Ls}{\textrm{L}}
\newcommand{\Bs}{\textrm{B}}
\newcommand{\es}{\textrm{e}}
\newcommand{\fs}{\textrm{f}}
\newcommand{\gs}{\textrm{g}}
\newcommand{\is}{\textrm{i}}
\newcommand{\js}{\textrm{j}}
\newcommand{\rs}{\textrm{r}}
\newcommand{\Os}{\textbf{O}}
\newcommand{\Gs}{\textbf{G}}
\newcommand{\Cbf}{\textbf{C}}
\newcommand{\Or}{\Os_\Rs}
\newcommand{\Qs}{\textbf{Q}}
\newcommand{\Js}{\textbf{J}}
\newcommand{\Cs}{\textbf{C}}
\newcommand{\Ps}{\textbf{P}}
\newcommand{\Ss}{\textbf{S}}
\newcommand{\ss}{\textsf{s}}
\newcommand{\P}{\textrm{P}}
\newcommand{\Q}{\textrm{Q}}
\newcommand{\deg}{^\textsf{o}}
\newcommand{\xs}{\textsf{x}}
\newcommand{\ys}{\textsf{y}}
\newcommand{\zs}{\textsf{z}}
\newcommand{\dert}[2]{\left.\frac{\ds{#1}}{\ds\ts}\right]_{\textrm{#2}}}
\newcommand{\ddert}[2]{\left.\frac{\ds^2{#1}}{\ds\ts^2}\right]_{\textrm{#2}}}
\newcommand{\vec}[1]{\overline{#1}}
\newcommand{\vecbf}[1]{\overline{\textbf{#1}}}
\newcommand{\vecdot}[1]{\overline{\dot{#1}}}
\newcommand{\GPvec}{\vec{\Gs\Ps}}
\newcommand{\OQvec}{\vec{\Os\Qs}}
\newcommand{\JGvec}{\vec{\Js\Gs}}
\newcommand{\QGvec}{\vec{\Qs\Gs}}
\newcommand{\QPvec}{\vec{\Qs\Ps}}
\newcommand{\OPvec}{\vec{\Os\textbf{P}}}
\newcommand{\OCvec}{\vec{\Os\Cs}}
\newcommand{\OGvec}{\vec{\Os\Gs}}
\newcommand{\abs}[1]{\left|{#1}\right|}
\newcommand{\braq}[2]{\left\{{#1}\right\}_{\textrm{#2}}}
\newcommand{\vector}[3]{
\begin{Bmatrix}
{#1}\\
{#2}\\
{#3}
\end{Bmatrix}}
\newcommand{\vecdosd}[2]{
\begin{Bmatrix}
{#1}\\
{#2}
\end{Bmatrix}}
\newcommand{\vel}[2]{\vvec_{\textrm{#2}} (\textbf{#1})}
\newcommand{\acc}[2]{\vecbf{a}_{\textrm{#2}} (\textbf{#1})}
\newcommand{\accs}[2]{\vecbf{a}_{\textrm{#2}}^{\textrm{s}} (\textbf{#1})}
\newcommand{\accn}[2]{\vecbf{a}_{\textrm{#2}}^{\textrm{n}} (\textbf{#1})}
\newcommand{\velo}[1]{\vvec_{\textrm{#1}}}
\newcommand{\accso}[1]{\vecbf{a}_{\textrm{#1}}^{\textrm{s}}}
\newcommand{\accno}[1]{\vecbf{a}_{\textrm{#1}}^{\textrm{n}}}
\newcommand{\re}[2]{\Re_{\textrm{#2}}(\textbf{#1})}
\newcommand{\psio}{\dot{\psi}_0}
\newcommand{\Pll}{\textbf{P}_\textrm{lliure}}
\newcommand{\agal}[1]{\vecbf{a}_{\textrm{Gal}} (#1)}
\newcommand{\angal}[1]{\vecbf{a}_{\textrm{NGal}} (#1)}
\newcommand{\vgal}[1]{\vecbf{v}_{\textrm{Gal}} (#1)}
\newcommand{\F}[1]{\vec{\Fs}_{#1}}
\newcommand{\Fcal}[2]{\vec{\cal{F}}^{\text{#1}}_{#2}}
\newcommand{\H}[3]{\vec{\mathbf{H}}_{\text{#3}}^{#2}(\textbf{#1})}
</math>

The '''Vector Theorems''' are a tool for solving the dynamics of mechanical systems, and are obtained from the fundamental law of dynamics ([[D1. Foundational laws of Newtonian dynamics#D1.5 Newton’s second law (fundamental law of dynamics)|'''Newton's second law''']]) and the principle of action and reaction ([[D1. Foundational laws of Newtonian dynamics#D1.6 Newton’s third law (action-reaction principle)|'''Newton's third law''']]).

In this course, only the version of the theorems for the case of constant matter systems is presented. Although this includes systems with fluids, the application examples in this course are essentially multibody systems made up of rigid bodies.

When addressing a problem, it is essential to identify the unknowns contained in the system under study and to analyze whether the vector theorems provide a sufficient number of equations to solve them all (one must know whether the problem is determinate or indeterminate!). The unknowns can be classified into three groups:

* Time evolution of the free [[C2. Movement of a mechanical system#C2.7 Degrees of freedom of a mechanical system|'''degrees of freedom (DoF)''']] (not controlled by actuators) of the system. The equations that govern these DoF are called '''equations of motion'''. If the free DoF are described by time derivatives of coordinates (<math>\qs_i</math>, with i=1,2,3...), their time evolution is the second time derivatives of these coordinates (accelerations). Their general aspect is:

<center><math>\dot{\qs}_i = f(\qs_j, \dot{\qs}_j, \ddot{\qs}_j, \text{ dynamic parameters, geometric parameters})</math></center>

The dependence of the equations of motion on the second time derivatives is always linear, while the dependence on the coordinates and velocities can be of any type. The dynamic parameters are the mass of the elements and those associated with their distribution in space, and the parameters associated with the interactions on the system (spring and damper constants, coefficients of friction, gravitational field constants...); the geometric parameters have to do with the shape of the elements of the system (distances and angles).

* [[D2. Interaction forces between particles#D2.6 Interaction through actuators|'''Actuator''']] '''actions''': these are the forces (in the case of linear actuators) and moments (in the case of rotatory actuators) required to ensure predetermined evolutions of the DoF they control. As discussed in [[D2. Interaction forces between particles#D2.6 Interaction through actuators|'''section D2.6''']], in some cases the actuator actions can be considered as data, and then the associated unknowns are the evolutions of the DoF they control.

* [[D3. Interaccions entre sòlids rígids|'''Constraint forces and movements''']]: the number of unknowns associated with the constraints depends on the description given of them ([[D3. Interactions between rigid bodies#D3.4 Direct constraint interactions|'''direct constraints''']], [[D3. Interactions between rigid bodies#D3.5 Indirect constraint interactions: Constraint Auxiliary Elements (CAE)|'''indirect constraints''']]). When a dynamics problem is indeterminate, the indeterminacy always refers to the constraint torsor components, never to the DoF (free or forced).

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==D4.1 Linear Momentum Theorem (LMT) in Galilean reference frames==

Let us consider a system of particles with constant matter ('''Figure D4.1'''). The '''Linear Momentum Theorem''' (LMT) is obtained by applying Newton's second law to each particle '''P''' of the system. If the chosen reference frame is Galilean:

<math>\F{\rightarrow\Ps}=\ms_\Ps\acc{P}{RGal}</math>, where <math>\F{\rightarrow\Ps}</math> is the interaction resultant force on '''P'''.
[[File:D4-1-eng.png|thumb|center|300px|link=]]
<center>'''Figure D4.1''' Forces on a particle of a system with constant matter</center>

The forces acting on each particle can be classified into two groups: internal (which come from the interaction with other particles in the system) and external (associated with interactions with elements external to the system): <math>\F{\is\ns\ts\rightarrow\Ps}+\F{\es\xs\ts\rightarrow\Ps}=\ms_\Ps\acc{P}{RGal}</math>. If these equations are added for all particles, the internal forces between pairs of particles cancel out two by two by the principle of action and reaction:

<center><math>\F{\Ps_j\rightarrow\Ps_i}+\F{\Ps_i\rightarrow\Ps_j}=\vec{0}\Rightarrow\sum_\Ps\F{\es\xs\ts\rightarrow\Ps}=\sum_\Ps \ms_\Ps\:\acc{P}{RGal}</math></center>

The term on the left-hand side is the resultant of external forces on the system, and is often written simply as <math>\sum\F{ext}</math>. The term on the right-hand side can be rewritten as <math>\Ms\left[\sum_\Ps\frac{\ms_\Ps}{\Ms}\acc{P}{RGal}\right]</math>, where M is the total mass of the system <math>\left(\Ms\equiv\sum_\Ps \ms_\Ps\right)</math>. The term <math>\left[\sum_\Ps\frac{\ms_\Ps}{\Ms}\acc{P}{RGal}\right]</math> is a weighted acceleration, where the weighting is proportional to the mass of each particle. This acceleration is associated with a point called '''center of mass''' (or '''center of inertia''') of the system, and in this course it is designated by letter G. For any reference frame, then, the kinematics of G are described by the equations ('''Figure D4.2'''):

<center><math>\vec{\Os_\Rs\Gs}=\sum_\Ps\frac{\ms_\Ps}{\Ms}\vec{\Os_\Rs\Ps},\:\:\:\:\vel{G}{R}=\sum_\Ps\frac{\ms_{\Ps}}{\Ms}\vel{P}{R},\:\:\:\:\acc{G}{R}=\sum_\Ps\frac{\ms_\Ps}{\Ms}\acc{P}{R}</math></center>

[[File:D4-2-eng.png|thumb|center|250px|link=]]
<center>'''Figure D4.2''' Centre of mass (r of inertia) of a ystem with constant matter</center>

In continuous sets of particles (such as a set of rigid or deformable bodies, or fluids), the summation for particles is actually an integral:

<center><math>\vec{\Os_\Rs\Gs}=\sum_{i=1}^N\left(\frac{1}{\Ms_\is}\int_{\Ss_\is}\ds\ms(\Ps)\vec{\Os_\Rs\Ps}\right),\:\:\:\:\vel{G}{R}=\sum_{i=1}^N\left(\frac{1}{\Ms_\is}\int_{\Ss_\is}\ds\ms(\Ps)\vel{P}{R}\right),\:\:\:\:\acc{G}{R}=\sum_{i=1}^N\left(\frac{1}{\Ms_\is}\int_{\Ss_\is}\ds\ms(\Ps)\acc{P}{R}\right)</math></center>

Finally, the LMT is written as:

<center><math>\sum\F{ext}=\Ms\:\acc{G}{RGal}</math></center>

This equation is very similar to [[D1. Foundational laws of Newtonian dynamics#D1.5 Newton’s second law (fundamental law of dynamics)|'''Newton's second law''']]: the centre of mass G behaves as if it were a particle of mass equal to the total mass of the system, and on which all the forces external to the system acted. Despite the parallelism between the LMT and the particle dynamics equation, there are two fundamental differences:

:*the mass of the system is not localized at '''G''' ('''G''' might be even located in a massless region of the system, as in the case of a homogeneous ring);

:*the external forces are not applied to '''G''' in general.

The LMT is so named because it allows us to know the evolution, based on the knowledge of the initial [[C1. Configuration of a mechanical system|'''mechanical state''']] and the external interactions, of the linear momentum of the system (which, in any reference frame R, is defined as <math>\sum_\Ps\ms_\Ps\vel{P}{R}=\Ms\vel{G}{R}</math>):

<center><math>\sum\F{ext}=\Ms\acc{G}{RGal}=\dert{\Ms\vel{G}{RGal}}{RGal}</math></center>

The location of the center of mass in the systems under study (whether it is a single rigid body or a multi-body system) is briefly introduced in [[D4. Vector theorems|'''unit D4''']]. For homogeneous bodies with very simple geometry, the position of G can often be deduced from the body symmetries.

In planar problems (with 2D kinematics), only the two components of the LMT contained in the plane of motion are interesting.

------------
------------

==D4.2 LMT: application examples==

<div>
====✏️ EXAMPLE D4.1: calculation of a constraint force ====

------
:
{|
|[[File:D4-Ex1-1-neut.png|thumb|center|180px|link=]]
|The three homogeneous blocks are smooth, and are in contact with each other and with a smooth horizontal ground. We want to investigate the value of the horizontal constraint force between blocks Q and S when a horizontal force F is applied to the block on the left.

All constraints appearing in this system are multiple-point contacts between smooth surfaces. Therefore, each associated torsor, characterized at a contact point, contains a force and a moment component ([[D3. Interactions between rigid bodies#D3.4 Direct constraint interactions|'''example D3.4''']]). However, in the application of the LMT only the forces are involved, so the moments will not be shown in the following figures.

|}

In order for the constraint force <math>\Fs_{\Qs\rightarrow\Ss}</math> to appear in the horizontal component of the LMT, the theorem must be applied to a system where this force is external. For example, to block S: <math>(\rightarrow\Fs_{\Qs\rightarrow\Ss})=(3\ms)[\rightarrow \as_\epsilon(\Gs_\Ss)]</math>.

[[File:D4-Ex1-2-eng.png.png|thumb|center|200px|link=]]

Since the motion of the three blocks caused by force F is the same, the acceleration can be obtained through the LMT applied to the whole system:

<center><math>\Fs = (6\ms)\as_\epsilon(\Gs)\Rightarrow\as_\epsilon(\Gs)=\frac{\Fs}{6\ms}\Rightarrow\Fs_{\Qs\rightarrow\Ss}=3\ms\frac{\Fs}{6\ms}=\frac{\Fs}{2}</math></center>

</div>
<div>

====✏️ EXAMPLE D4.2: initial motion of a system ====
------
:
{|
|[[File:D4-Ex2-1-eng.png|thumb|center|300px|link=]]
|The homogeneous blocks are initially at rest on a rough floor, and connected by a spring compressed with tension <math>\Fs_0=\ms\gs</math> and an inextensible thread. At a certain instant, the thread is cut and the system begins to move.
|}

We want to calculate the acceleration of the center of inertia of the system relative to the ground. This acceleration can be obtained from the horizontal component of the LMT applied to the whole system. The spring is internal, and therefore its force does not appear: <math>\F{\text{ground}\rightarrow 2\ms}+\F{\text{ground}\rightarrow \ms}=(3\ms)\acc{G}{T}</math>.

The forces of the ground on the blocks can be constraint forces (if the blocks do not move, in which case they are unknown), or friction forces (if the blocks move relative to the ground, in which case they are formulable). It may also happen that one block slides and the other does not.

The value of a constraint force adapts itself to ensure a restriction. In the case of blocks, it is necessary to investigate whether these forces can reach the value required to prevent the blocks from sliding on the ground, and are bounded by the value of the coefficient of static friction between blocks and ground:

<math>|\F{\text{ground}\rightarrow 2\ms}\mid\leq\mu_\es\Ns_{\text{ground}\rightarrow 2\ms}?\:\:\:, \mu_\es\Ns_{\text{ground}\rightarrow 2\ms}=0,6\cdot 2\ms\gs=1,2\ms\gs</math>

<math>|\F{\text{ground}\rightarrow \ms}\mid\leq\mu_\es\Ns_{\text{terra}\rightarrow\ms}?\:\:\:, \mu_\es\Ns_{\text{ground}\rightarrow\ms}=0.4\cdot \ms\gs</math>

If we apply the LMT to each block separately, the spring force becomes external. That force (of value mg) can be counteracted by the constraint force in the case of the block of mass 2m (hence, <math>\as_\epsilon(2\ms) = 0</math>), but not in the case of the block of mass m:

[[File:D4-Ex2-2-eng.png|thumb|center|450px|link=]]

Application of the LMT to the block of mass m leads to:

<math>\left.\begin{aligned}
\sum\F{\es\xs\ts\rightarrow2\ms}=\F{\text{ground}\rightarrow2\ms}+\F{\text{spring}\rightarrow2\ms}=2\ms\acc{2m}{T} \\
\sum\F{\es\xs\ts\rightarrow2\ms}=(\rightarrow\ms\gs)+(\leftarrow\ms\gs)=0
\end{aligned}\right\} \Rightarrow\acc{m}{T}=(\rightarrow 0,8\gs)</math>

Finally: <math>\acc{G}{T}=\frac{2\ms\acc{2m}{T}+\ms\acc{m}{T}}{3\ms}=\frac{0,8}{3}\gs</math>.

</div>
<div>
====✏️ EXAMPLE D4.3: limit condition ====
------
:
{|
|[[File:D4-Ex3-1-neut.png|thumb|center|200px|link=]]
|The homogeneous sphere of mass M is in contact with two identical wedges of mass m, which are on the ground. There is no friction between the sphere and the wedges, but friction coefficient between wedges and ground is <math>\mu</math>. The system is initially at rest relative to the ground. We want to determine the maximum value of M, as a function of m, that allows the system to remain at rest.
|}

If nothing moves, the horizontal interaction forces between ground and wedges are constraint forces, and do not exceed the limit value <math>/mu\ns</math> (where N is the normal force that the ground exerts on each wedge). The constraint torsor of the ground on the wedges also contains a resultant moment perpendicular to the figure. If we want to study the possibility of the wedges overturning, this moment is relevant. But in this example, the shape of the wedges guarantees that they do not overturn, so we must take into account the forces.

Applying the LMT to the entire system, to a wedge, and to the sphere leads to the following equations:

[[File:D4-Ex3-3-neut.png|thumb|center|350px|link=]]

<center>
{|class = "wikitable", style = "background-color:#ffffff;border-style:none;"
|- style="vertical-align:top;"
|style="padding: 20px; text-align:center;"|SYSTEM: sphere + wedges

<math>\left(\uparrow 2\Ns\right)+\left[\downarrow(\Ms+2\ms)\gs\right]=\vec{0}</math>

<math>\Rightarrow\Ns=\frac{1}{2}(\Ms+2\ms)\gs</math>
|style="padding: 20px; text-align:center;"|SYSTEM: left wedge

<math>\left(\rightarrow \Ts\right)+\left(\leftarrow\frac{\Ns'}{\sqrt{2}}\right)=\vec{0}</math>

<math>\Rightarrow\Ts=\frac{\Ns'}{\sqrt{2}}</math>
|style="padding: 20px; text-align:center;"|SYSTEM: sphere

<math>\left(\downarrow\Ms\gs\right)+\left(\uparrow 2\frac{\Ns'}{\sqrt{2}}\right)=\vec{0}</math>

<math>\Rightarrow\Ns'=\frac{\Ms\gs}{\sqrt{2}}</math>
|}</center>

Combining the last two equations: <math>\Ts = \Ms\gs/2</math>. The maximum M value for which there is equilibrium corresponds to the situation in which the tangential constraints force T takes the maximum possible value: <math>\Ts = \Ts_{\text{màx}}=\mu\Ns</math>. Taking into account that <math>N = \frac{1}{2}(\Ms_{\text{max}}+2\ms)\gs</math>:

<math>\Ts=\frac{\Ms_{\text{max}}\gs}{2}=\Ts_{\text{max}}=\mu\Ns=\mu\frac{1}{2}(\Ms_{\text{max}}+2\ms)\gs\Rightarrow\Ms_{\text{max}}=2\ms\frac{\mu}{1-\mu}</math>

</div>

------------
------------

==D4.3 Linear Momentum Theorem (LMT) in non Galilean reference frames==

The LMT in a non-Galilean reference NGal is also obtained from [[D1. Foundational laws of Newtonian dynamics#D1.5 Newton’s second law (fundamental law of dynamics)|'''Newton's second law''']] applied to each particle (or each mass differential) of the system in the reference NGal ([[D1. Foundational laws of Newtonian dynamics#D1.7 Particle dynamics in non Galilean reference frames|'''section D1.7''']]). In principle, then, this equation will contain two inertial forces: the transportation force and the Coriolis force ('''Figure D4.3'''):

<math>\F{\rightarrow \Ps} + \Fcal{tr}{NGal\rightarrow\Ps}+\Fcal{Cor}{NGal\rightarrow\Ps} = \ms_\Ps\acc{P}{NGal}</math>,

where <math>\Fcal{tr}{NGal\rightarrow\Ps} = -\ms_\Ps\acc{P}{ar} = -\ms_\Ps\acc{$\Ps\in\text{NGal}$}{RGal}</math> and <math>\Fcal{Cor}{NGal\rightarrow\Ps} = -\ms_\Ps\acc{P}{Cor} = -2\ms_\Ps\velang{NGal}{RGal}\times\vel{P}{NGal}</math>

[[File:D4-3-eng.png|thumb|center|350px|link=]]
<center>'''Figure D4.3:''' Inertia forces in a non Galilean reference frame</center>

Summing the equations for all the particles, we obtain: <math>\sum\F{ext}+\Fcal{tr}{NGal\rightarrow\Gs}+\Fcal{Cor}{NGal\rightarrow\Gs}=\Ms\acc{G}{NGal}</math>, where M is the total mass of the system.

<div>
=====💭 Proof ➕=====
------
:
By the principle of action and reaction, the sum for all particles of the interaction forces leads to:

<math>\sum_{\Ps\in\text{syst}}\F{\rightarrow\Ps} = \sum\F{ext}</math>

As for the inertia forces, as they are proportional to the transportation and Coriolis acceleration of each particle, their sum for all particles corresponds to the weighted kinematics that defines the [[D4. Vector theorems#D4.2 LMT: application examples|'''center of mass''']]:

<math>
\sum_{\Ps\in\text{syst}} \Fcal{tr}{NGal\rightarrow\Ps} = -\sum_{\Ps\in\text{syst}}\ms_\Ps\acc{$\Ps\in$NGal}{RGal} = -\Ms\acc{$\Gs\in$NGal}{RGal} = -\Ms\acc{$\Gs$}{ar} = \Fcal{tr}{NGal\rightarrow\Ps}
</math>

<math>
\sum_{\Ps\in\text{syst}} \Fcal{Cor}{NGal\rightarrow\Ps} = -\sum_{\Ps\in\text{syst}}\ms_\Ps\acc{$\Ps$}{Cor} = -2\sum_{\Ps\in\text{syst}}\ms_\Ps\velang{NGal}{RGal}\times\vel{P}{NGal} = -2\velang{NGal}{RGal}\times\sum_{\Ps\in\text{syst}}\ms_\Ps\vel{P}{NGal} = -2\velang{NGal}{RGal}\times\Ms_\Ps\vel{G}{NGal} = \Fcal{Cor}{NGal\rightarrow\Ps}
</math>

Hence: <math>\sum\F{ext} + \Fcal{tr}{NGal\rightarrow\Gs}+\Fcal{Cor}{NGal\rightarrow\Gs} = \Ms\acc{G}{NGal}</math>

</div>

<div>

====✏️ EXAMPLE D4.4: vibratory displacement====
------
:
{|
|[[File:D4-Ex4-1-eng.png|thumb|center|250px|link=]]
|The block of mass m is initially at rest on a support which oscillates relative to the ground according to the velocity graph shown in the figure. We want to investigate the possibility of sliding between block and support.

The nonsliding condition between block and support (which is a non-Galilean reference since its motion relative to the ground is not uniform) requires:

<math>\sum\F{\text{ext}} + \Fcal{tr}{sup\rightarrow\Gs} = \vec{0} (\Fcal{Cor}{sup\rightarrow\Gs}= \vec{0})</math> perque <math>\velang{sup}{T} = \vec{0}</math>
|}

The transportation force on the block is strictly horizontal. Therefore, the vertical component of that equation leads to <math>N = mg</math>. As for the horizontal component, it will contain the interaction force between the block and the support. If the block does not slide on the support, this force is a constraint force, and its value is bounded: <math>0\leq\mid\Fs_{sup\rightarrow bloc}\mid \leq\mu\Ms\gs = 0,15\Ms\Gs\simeq 1,5(\ms/s^2)\Ms</math>. If it slides, it is a friction force with value <math>\mu\ms\gs</math>, opposite to the sliding velocity.

Between <math>t=0</math> and <math>t = 0,4s</math>, the acceleration of the support relative to the ground is <math>\ddot{y} = \frac{0,4\ms/s}{0,4s}=1\ms/s^2</math>, therefore the drag force is <math>\Fcal{tr}{sup\rightarrow\Gs} = -\ms\acc{$\Gs\in$support}{T}=[\leftarrow 0,1(\ms/s^2)\Ms]</math>. That value is within the range of values allowed for the horizontal constraint force between support and block. Therefore, this force will take the value <math>\F{sup\rightarrow block} = 0,1(\ms/s^2)</math>, will counteract the <math>\Fcal{tr}{sup\rightarrow\Gs}</math> and there will be no motion between the two elements.

Between <math>t=0,4s</math> and <math>t = 0,6s</math>, the acceleration of the support relative to the ground is 2<math>\ddot{y} = -\frac{0,8\ms/s}{0,2s}=-4\frac{\ms}{s^2}</math>. Therefore, <math>\Fcal{tr}{sup\rightarrow\Gs} =[\rightarrow 4(\ms/s^2)\Ms]</math>. As that value is higher than the maximum value of the horizontal constraint force, it cannot be counteracted. The block will begin to slide, and the horizontal interaction force between the support and the block will be a friction force:

<math>\Fcal{tr}{sup\rightarrow\Gs} + \F{sup\rightarrow block}^{friction} = [\rightarrow 4(\ms/s^2)\Ms] + [\leftarrow 1,5(\ms/s^2)\Ms] = [\rightarrow 2,5(\ms/s^2)\Ms]\Rightarrow \acc{G}{sup} = \left(\rightarrow 2,5\frac{\ms}{s^2}\right)</math>

At time <math>t = 0,6s</math>, the sliding velocity of the block on to the support is

<math>\vel{G}{sup} = \left(\rightarrow 2,5\frac{\ms}{s^2}\cdot 0,2s\right) = \left(\rightarrow 0,5\frac{\ms}{s}\right)</math>.

Even if the bock reaches the phase in which <math>\mid\Fcal{tr}{sup\rightarrow G}\mid<\mu\Ms\gs</math>, the horizontal force between support and block is still a friction force <math>(\F{sup\rightarrow block}^{friction} = [\leftarrow 1,5(\ms/s^2)\Ms])</math> until the block stops sliding.

The acceleration of the block relative to the support is:

<math>\acc{G}{sup} = \acc{G}{T} - \acc{G}{tr} = \left(\leftarrow 1,5\frac{\ms}{s^2}\right) - \left( \rightarrow 1\frac{\ms}{s^2}\right) = \left(\leftarrow 2,5\frac{\ms}{s^2}\right)</math>

Since it is a uniformly decelerated motion, it is easy to calculate the time instant <math>t_f</math> when the block stops sliding:

<math>\vel{$\Gs,t_f$}{sup} = 0 = \vel{$\Gs,t_i$}{sup} + \acc{$\Gs,t_f$}{sup}(t_f - t_i)</math>,
<math>t_i = 0,6s \Rightarrow 0 = \left(\rightarrow 0,5\frac{\ms}{s}\right) + \left(\leftarrow 2,5\frac{\ms}{s^2}\right)(t_f-0,6s)\Rightarrow t_i = 0,8s</math>

At that time instant, the force between support and block becomes a constraint force, and we reach a situation analogous to the initial one. Therefore, the slip will not occur again until <math>\ts = 1,4s</math>. The study for the subsequent intervals follows the same steps as that in the interval <math>[0,1,4s]</math>. The figure shows the evolution of the kinematics of the block with respect to the support, and of the forces acting on it.

[[File:D4-Ex4-2-eng.png|thumb|center|450px|link=]]

</div>

-----------
-----------

==D4.4 Angular Momentum Theorem (AMT): general formulation==

The dynamics of a rigid body (and therefore, of a multibody system) is never fully solved with the LMT when the rígid body rotates: the LMT only reports on the movement of a point ('''G'''), and not on the rotation.

The '''Angular Momentum Theorem''' (AMT) is stated in two different ways in the literature. In this course, a formulation parallel to that of the LMT has been chosen: terms on the left-hand side have to do only with external interactions on the system, while that on the right-hand side is the time derivative of a vector that depends only on the mass geometry of the system and its [[C1. Configuration of a mechanical system#|'''mechanical state''']].

As we have already seen when introducing the concept of [[D3. Interactions between rigid bodies#D3.1 Torsor associated with a system of forces|'''torsor of a system of forces''']], the moment of the forces, and not the resultant force, relates to the rotation of a rigid body. Therefore, although the AMT is also derived from Newton's second law applied to each mass differential of the system, the forces appearing in that equation will have to be transformed into the moment (at a point Q) associated with those forces.

These two considerations (parallel formulation to the LMT and the need to choose a point Q to calculate the moment of the forces) are the reason why the starting point to obtain the AMT is Newton's second law formulated in the '''Reference Frame that Translates with Q''' ('''RTQ''') with respect to a Galilean reference frame <math>\velang{RTQ}{RGal} = \vec{0}</math>. If <math>\acc{Q}{RGal}\neq\vec{0}</math>, this reference frame is not Galilean, and therefore the [[D1. Foundational laws of Newtonian dynamics#D1.7 Particle dynamics in non Galilean reference frames|'''inertia transportation forces''']] must be taken into account in principle ('''Figure D4.3''').

[[File:D4-4-eng.png|thumb|center|400px|link=]]
<center>'''Figure D4.4:''' Inertia forces in the Reference Frame that Translates with point Q (RTQ)</center> 

Let us consider a system of particles with constant matter. Newton's second law applied to each particle of the system and in the RTQ states:

<math>\sum \F{\rightarrow \Ps} + \Fcal{tr}{\mathrm{RTQ}\rightarrow \Ps}=\ms_\Ps\acc{P}{RTQ}</math> , 

where <math>\Fcal{tr}{\mathrm{RTQ}\rightarrow \Ps}=-\ms_\Ps \acc{P}{tr}=-\ms_\Ps \overline{\mathbf{a}}_\mathrm{RGal}(\Ps \in \mathrm{RTQ})=-\ms_\Ps \acc{Q}{RGal}</math>, and <math>\Fcal{Cor}{\mathrm{RTQ}\rightarrow \Ps}=\overline{0}</math> since <math>\velang{RTQ}{RGal}=\overline{0}</math>. 

If both sides of the equation are multiplied by <math>\QPvec</math> and summed for all particles (or mass elements) in the system, the AMT at point '''Q''' is obtained:

<math>\sum \vec{M}_\mathrm{ext} (\Qs) - \QGvec \times \Ms \acc{Q}{RGal} = \dert{\overline{\mathbf{H}}_\mathrm{RTQ}(\Qs)}{RTQ} \equiv \dot{\overline{\mathbf{H}}}_\mathrm{RTQ}(\Qs)</math> 

where M is the total mass of the System, and <math>\overline{\mathbf{H}}_\mathrm{RTQ}(\Qs) \equiv \sum_{\Ps \in \mathrm{syst}} \QPvec \times \ms_\Ps \vel{P}{RTQ}</math> is the '''angular momentum of the System about point Q'''.

If the system contains continuous elements (for example, a system with N rigid bodies <math>\Ss_i</math>), the summation for particles is actually an integral:

<math>\overline{\mathbf{H}}_\mathrm{RTQ}(\Qs)= \sum_{\is =1}^{\Ns} \overline{\mathbf{H}}_\mathrm{RTQ}^\mathrm{Si}(\Qs)= \sum_{\is=1}^{\Ns} \left( \int_{\mathrm{Si}}\QPvec \times \ds\ms (\Ps) \vel{P}{RTQ} \right )</math>.

<div>
=====💭 Proof ➕=====

Newton's second law for each element of mass multiplied by <math>\QPvec</math> is:

<math>\sum\QPvec\times\F{\rightarrow\Ps} + \QPvec\times\Fcal{tr}{RTQ\rightarrow\Ps} = \QPvec\times\ms_\Ps\acc{P}{RTQ}</math>

El terme de la dreta es pot reescriure com a:

<math>\QPvec\times\ms_\Ps\acc{P}{RTQ} = \QPvec\times\ms_\Ps\dert{\vel{P}{RTQ}}{RTQ}</math>

Since '''Q''' is a point of the RTQ, it can be taken as the origin of a position vector of '''P''' in that reference frame. Hence:

<math>\vel{P}{RTQ} = \dert{\QPvec}{RTQ} \Rightarrow\dert{\left[\QPvec\times\ms_\Ps\vel{P}{RTQ}\right]}{RTQ} = \dert{\QPvec}{RTQ} \times\ms_\Ps\vel{P}{RTQ} + \QPvec\times\ms_\Ps\dert{\vel{P}{RTQ}}{RTQ}</math>
<math>=\vel{P}{RTQ}\times\ms_\Ps\vel{P}{RTQ} + \QPvec\times\ms_\Ps\dert{\vel{P}{RTQ}}{RTQ} = \QPvec\times\ms_\Ps\dert{\vel{P}{RTQ}}{RTQ}</math>

Taking into account that the mass is constant:

<math>\QPvec\times\ms_\Ps\acc{P}{RTQ} = \dert{\left[\QPvec\times\ms_\Ps\vel{P}{RTQ}\right]}{RTQ}</math>

Summation for all '''P''' elements leads to:

<math>\sum_{\Ps\in\text{syst}}\QPvec\times\ms_\Ps\dert{\vel{P}{RTQ}}{RTQ} = \sum_{\Ps\in\text{syst}}\dert{\left[\QPvec\times\ms_\Ps\vel{P}{RTQ}\right]}{RTQ} = \dert{\sum_{\Ps\in\text{syst}}\QPvec\times\ms_\Ps\vel{P}{RTQ}}{RTQ} = \dot{\overline{\mathbf{H}}}_\mathrm{RTQ}(\Qs)</math>

If the interaction forces on each particle '''P''' are classified as internal and external, the sum for all the particles in the system on the left-hand side of the first equation becomes:

<math> \sum_{\Ps \in \mathrm{syst}} \QPvec \times \F{\mathrm{int}\rightarrow\Ps} + \sum_{\Ps \in \mathrm{syst}} \QPvec \times \F{\mathrm{ext}\rightarrow\Ps} + \sum_{\Ps \in \mathrm{syst}} \QPvec \times \Fcal{tr}{\mathrm{RTQ}\rightarrow\Ps} = \sum_{\Ps \in \mathrm{syst}} \QPvec \times \F{\mathrm{ext}\rightarrow\Ps} - \sum_{\Ps \in \mathrm{syst}} \QPvec \times \ms_\Ps \acc{Q}{RGal}=</math>

<math>= \sum \overline{\mathbf{M}}_\mathrm{ext} (\Qs) - \left( \sum_{\Ps \in \mathrm{syst}} \ms_\Ps \QPvec \right) \times \acc{Q}{RGal} = \sum \overline{\mathbf{M}}_\mathrm{ext} (\Qs) - \QGvec \times \Ms \acc{Q}{RGal} , </math> 

since the principle of action and reaction guarantees that the total moment of a pair of action and reaction forces is zero:

<math>\QPvec_\is \times \F{\Ps_\js \rightarrow \Ps_\is} + \QPvec _\js \times \F{\Ps_\is \rightarrow \Ps_\js} = \QPvec_\is \times \F{\Ps_\js \rightarrow \Ps_\is} - \QPvec _\js \times \F{\Ps_\js \rightarrow \Ps_\is} = \QPvec_\is \times \F{\Ps_\js \rightarrow \Ps_\is} - \left( \QPvec _\is + \overline{\Ps_\is \Ps_\js}\right)\times \F{\Ps_\js \rightarrow \Ps_\is}= \overline{\Ps_\is \Ps_\js} \times \F{\Ps_\js \rightarrow \Ps_\is}=\overline{0}</math> 

Finally:

<math>\sum \overline{\mathbf{M}}_\mathrm{ext} (\Qs) - \QGvec \times \Ms \acc{Q}{RGal}=\dot{\overline{\mathbf{H}}}_\mathrm{RTQ}(\Qs)</math>

</div>

---------
---------

==D4.5 Angular Momentum Theorem (AMT): particular formulations==

The AMT formulation becomes simpler when the point '''Q''' is fixed to a Galilean reference frame or when it is the center of mass of the system.

* <math>\Qs \in \mathrm{RGal} \Rightarrow \acc{Q}{RGal} = \overline{0} \Rightarrow \sum \overline{\mathbf{M}}_\mathrm{ext} (\Qs) = \dot{\overline{\mathbf{H}}}_\mathrm{RTQ} (\Qs). </math> In short, we call this version “AMT at a fixed point”, where “fixed” has to be understood as “fixed to a Galilean reference frame”. In this case, we will usually choose the letter '''O''' to designate point '''Q'''.

* <math>\Qs=\Gs \Rightarrow \QGvec \times \Ms \acc{Q}{RGal}=\overline{0} \times \Ms \acc{Q}{RGal} = \overline{0} \Rightarrow \sum \overline{\mathbf{M}}_\mathrm{ext} (\Gs) = \dot{\overline{\mathbf{H}}}_\mathrm{RTQ} (\Gs).</math> It should be noted that, although the expression is similar to the fixed point version, the center of mass '''G''' does not have to be fixed to a Galilean reference.

When '''Q''' is neither a fixed point in RGal nor coincides with the center of mass, we usually speak of the “moving point” version of the AMT. Although '''Q''' is in general moving relative to RGal, the term associated with the inertia forces <math>\left( \QGvec \times \Ms \acc{Q}{RGal} \right)</math> can be zero if '''Q''' moves at a constant speed relative to RGal, or if its acceleration relative to RGal is parallel to <math>\QGvec</math>.

<center>
{| class="wikitable"
|+
|-
! '''AMT at a fixed point''' !! '''AMT at G '''<math>\Gs</math> !! '''AMT at a moving point '''<math>\Qs</math>
|-
| <center><math> \quad \Qs \in \mathrm{RGal} \Rightarrow \Qs \equiv \Os , \mathrm{RTQ} = \mathrm{RGal} \quad </math> 

<math>\sum \overline{\mathbf{M}}_\mathrm{ext}(\Os) = \dot{\overline{\mathbf{H}}}_\mathrm{RGal} (\Os)</math> </center>
||
<center>
<math>\quad \Qs=\Gs \Rightarrow \mathrm{RTQ}=\mathrm{RTG} \quad </math> 

<math>\sum \overline{\mathbf{M}}_\mathrm{ext}(\Gs) = \dot{\overline{\mathbf{H}}}_\mathrm{RTG} (\Gs)</math></center>
||
<center>
<math> \quad \acc{Q}{RGal} \neq \overline{0}\quad </math> 

<math>\quad \sum \overline{\mathbf{M}}_\mathrm{ext}(\Qs) - \QGvec \times \Ms\acc{Q}{RGal}= \dot{\overline{\mathbf{H}}}_\mathrm{RTQ} (\Qs) \quad</math></center>

|}
</center>

The angular momentum is not easy to calculate in general, and it is presented in [[D5. Mass distribution|'''unit D5''']].

In problems with planar (2D) kinematics, if the angular momentum <math>\overline{\mathbf{H}}_\mathrm{RTQi}^\mathrm{Si} (\Gs_\is)</math> of each rigid body in the system under study is parallel to the angular velocity of the rigid body (whose direction is orthogonal to the plane of motion), only the component of the AMT perpendicular to that plane is interesting. The problem is then a planar dynamics problem (2D). Planar dynamics can be studied from two LMT components and one AMT component.

<div>
===AMT at a contact point between two rigid bodies===

A link that appears frequently in mechanical systems is the single-point contact between pairs of rígid bodies (S1 and S2, for example). That constraint can introduce between 1 and 3 constraint unknowns (depending on the surface roughness and the contact kinematics– with or without sliding). When we do not want to calculate these forces, it is tempting to apply the AMT at the contact point J (since the moment of those forces at point '''J''' is zero). 

[[File:TMCaJ-0-eng.png|thumb|right|230px|link=]]

The application of the AMT at '''J''' is very tricky. It is necessary to specify which point J has been chosen ([[C5. Cinemàtica plana del sòlid rígid#✏️ Exemple C5-1.8|'''example C5-1.8''']]): is it point '''J''' of rigid body S1 <math>(\Js_{\mathrm{S}1})</math>, of rigid body S2 <math>(\Js_{\mathrm{S}2})</math>, or is it the geometric contact point <math>(\Js_{\mathrm{geom}})</math>. On the one hand, those three points have different kinematics <math> \left(\overline{\as}_\mathrm{Gal}(\Js_{\mathrm{S}1}) \neq \overline{\as}_\mathrm{Gal}(\Js_{\mathrm{S}2}) \neq \overline{\as}_\mathrm{Gal}(\Js_\mathrm{geom}) \right)</math>, and the complementary term associated with the moment of inertia forces <math>\left(\JGvec \times \ms \acc{J}{Gal} \right)</math> is different. On the other hand, if for example the AMT is applied to rigid bdy S1, it is necessary to keep in mind that, although <math>(\Js_{\mathrm{S}1})</math> belongs to S1 and the angular momentum can be calculated from the inertia tensor, <math>(\Js_{\mathrm{S}1})</math> and <math>(\Js_{\mathrm{geom}})</math> not belong to S1, and the barycentric decomposition is compulsory to calculate this vector.

Finally, since the angular momentum has to be differentiated, it has to be calculated in a general configuration (i.e. when '''J''' is not yet the contact point), and only after having done the time derivative can the result be particularized to the configuration in which '''J''' is the contact point. As an illustration of all this, the following diagram shows the application of the AMT to a wheel with planar motion sliding on the ground.

[[File:TMCaJ-1-eng.png|thumb|center|520px|link=]]
<center>
<math>\overline{\Js_\mathrm{wheel}\Gs} \times \ms \overline{\as}_\mathrm{Gal}(\Js_\mathrm{wheel})=\otimes \ms \rs \ddot{\xs} \hspace{0.7cm} \overline{\mathrm{H}}_\mathrm{RTJwheel}(\Js_\mathrm{wheel},\varphi)=\Is\Is (\Js_\mathrm{wheel},\varphi) \overline{\dot{\varphi}} \quad , \quad \left.\dot{\overline{\mathrm{H}}}_\mathrm{RTJwheel}(\Js_\mathrm{wheel},\varphi)\right]_{\varphi=180^\circ}</math>
</center> 

[[File:TMCaJ-2-eng.png|thumb|center|520px|link=]]
<center>
<math>\overline{\Js_\mathrm{ground}\Gs} \times \ms \overline{\as}_\mathrm{Gal}(\Js_\mathrm{ground})= \overline{0} \hspace{0.7cm} \left\{\begin{array}{l}
\overline{\mathrm{H}}_\mathrm{RTJground}(\Js_\mathrm{ground},\xs)=\Is\Is (\Gs) \overline{\dot{\varphi}} + \overline{\Js_\mathrm{ground}\Gs} \times \ms\overline{\vs}_\mathrm{RTJground}(\Gs)=\Is\Is (\Gs)\overline{\dot{\varphi}} + (\otimes \ms\rs\dot{\xs})\\
\left.\dot{\overline{\mathrm{H}}}_\mathrm{RTJground}(\Js_\mathrm{ground},\xs) \right]_{\xs=0}
\end{array}\right.
</math> 

[[File:TMCaJ-3-eng.png.png|thumb|center|520px|link=]]

<center>
<math>\overline{\Js_\mathrm{geom}\Gs} \times \ms \overline{\as}_\mathrm{Gal}(\Js_\mathrm{geom})= (\otimes \ms \rs \ddot{\xs})\hspace{0.7cm} \left\{\begin{array}{l}
\overline{\mathrm{H}}_\mathrm{RTJgeom}(\Js_\mathrm{terra},\xs=0)=\Is\Is (\Gs) \overline{\dot{\varphi}} + \overline{\Js_\mathrm{geom}\Gs} \times \ms\overline{\vs}_\mathrm{RTJgeom}(\Gs)=\Is\Is (\Gs)\overline{\dot{\varphi}}\\
\dot{\overline{\mathrm{H}}}_\mathrm{RTJgeom}(\Js_\mathrm{terra},\xs=0)
\end{array}\right.
</math>
</center></center>
</div>

------
------

==D4.6 AMT: application examples==
<div>
====✏️ EXAMPLE D4.5: static limit condition====
------
:
{|
|[[File:D4-Ex5-1-eng.png|thumb|left|180px|link=]]
|The block hangs from an inclined massless bar through an inextensible thread. The bar ends are in contact with two walls fixed to the ground, one smooth and one rough (with a non-zero friction coefficient <math>\mu_\mathrm{Q}</math>). We want to calculate the minimum <math>\mu_\mathrm{Q}</math> value for equilibrium.

El bloc penja d’una barra inclinada de massa negligible per mitjà d’un fil inextensible. Els extrems de la barra estan en contacte amb dues parets fixes a terra, una llisa i una altra rugosa (amb coeficient de fricció no nul <math>\mu_\mathrm{Q}</math> ). Es tracta de calcular el valor mínim de <math>\mu_\mathrm{Q}</math> que permet l’equilibri. 

It is a 2D dynamics problem. The external forces acting on the system (bar+block+thread) are:
[[File:D4-Ex5-2-neut.png|thumb|center|300px|link=]]
|}

Since the system is at rest relative to the ground, <math>\acc{G}{T}=\overline{0}</math> and the total external force must be zero. Therefore:

<math>\Ns_\Ps=\Ns_\Qs</math> , <math>\ms\gs=\Ts_\Qs</math> .

Since motion is about to occur, the tangential constraint force at '''Q''' has its maximum possible value: <math>\Ts_{\Qs,\mathrm{max}}=\ms\gs=\mu_{\Qs,\mathrm{min}}\Ns_\Qs</math>. Since the above equations do not allow the calculation of <math>\Ns_\Qs</math>, a third equation is needed, which will come from the AMT. Whether applied at '''P''', '''Q''' or '''O''', the angular momentum is zero:

<math>\left.\begin{array}{l}
\mathrm{RTP}=\mathrm{RTQ}=\mathrm{RTO}=\mathrm{ground}(\Ts)\\
\overline{\mathbf{v}}_\Ts(\mathrm{block})=\overline{0}
\end{array}\right\} \Rightarrow \overline{\mathbf{H}}_\mathrm{RTP}(\Ps)=\overline{\mathbf{H}}_\mathrm{RTQ}(\Qs)=\overline{\mathbf{H}}_\mathrm{RTO}(\Os)=\overline{0}</math> 

<math>\sum \overline{\mathbf{M}}_\mathrm{ext}(\Qs)=\overline{0} \quad \Rightarrow \quad (\otimes \Ns_\Ps4\Ls\sin\beta)+(\odot \ms\gs3\Ls\cos\beta)=0 \quad \Rightarrow \quad \Ns_\Ps=\frac{3}{4} \frac{\ms\gs}{\tan\beta}.</math>

Therefore: <math>\Ts_{\Qs,\mathrm{max}}=\ms\gs=\mu_{\Qs,\mathrm{min}}\Ns_\Qs=\mu_{\Qs,\mathrm{min}} \frac{3}{4} \frac{\ms\gs}{\tan\beta} \quad \Rightarrow \mu_{\Qs,\mathrm{min}}=\frac{4}{3} \tan \beta. </math>

</div>
<div>

====✏️ EXAMPLE D4.6: Atwood machine ====
------
:
{|
|[[File:D4-Ex6-1-eng.png|thumb|left|230px|link=]]
|The blocks hang from two inextensible ropes with one end tied to the periphery of two pulleys with radii r and 2r, of negligible mass and articulated to the ceiling. We want to calculate angular acceleration of the pulleys.

It is a 2D dynamics problem. The external forces on the system (blocks + pulleys) and its motion relative to the ground are <math>(\gs \approx 10 \ms/\ss^2)</math>:

[[File:D4-Ex6-1-neut.png|thumb|center|140px|link=]]
|}

In total, there are three unknowns: the two constraint forces associated with the articulation of the pulleys, and the angular acceleration of the pulleys relative to the ground. The two vector theorems applied to the system (blocks + pulleys) provide three equations. Now, since the unknown to be determined is the acceleration, we can just use the AMT at O'''Bold text''' and obtain an equation free of constraint unknowns but including <math>\ddot{\theta}</math>:

<math>\sum \overline{\mathbf{M}}_\mathrm{ext}(\Os) = \dot{\overline{\mathbf{H}}}_\mathrm{RTO}(\Os) \quad \Rightarrow \quad (\odot 100 \Ns \cdot 2\rs) +(\otimes 50\Ns \cdot \rs)=(\odot 150\Ns \cdot \rs)=\dot{\overline{\mathbf{H}}}_\mathrm{RTO}(\Os)</math> 

<math>\overline{\mathbf{H}}_\mathrm{RTO}(\Os)=\overline{\mathbf{H}}_\Ts (\Os)= \int_\mathrm{left \: esp.} \OPvec \times \vel{P}{T} \ds\ms(\Ps)+ \int_\mathrm{right \: block} \OPvec \times \vel{P}{T} \ds\ms(\Ps) =\left(\int_\mathrm{left \: block} \OPvec\ds\ms(\Ps) \right)\times (\downarrow 2\rs\dot{\theta})+ \left(\int_\mathrm{right \: block} \OPvec\ds\ms(\Ps) \right)\times (\uparrow \rs\dot{\theta}) </math> 

By the definition of [[D3. Interactions between rigid bodies#D3.1 Torsor associated with a system of forces|'''center of mass''']]: <math>\int_\mathrm{block} \OPvec\ds\ms(\Ps)=\ms_\mathrm{block}\OGvec_\mathrm{block} </math>. On the other hand, since the blocks have vertical velocity, only the horizontal component of <math>\OGvec_\mathrm{block} \times \overline{\mathbf{v}}_\Ts(\mathrm{block})</math> contributes to the vector product <math>\OGvec_\mathrm{block}</math>. Thus:

<math>\overline{\mathbf{H}}_\mathrm{T}(\Os)=10\ks\gs(\leftarrow 2\rs) \times (\downarrow 2\rs\dot{\theta})+ 5\ks\gs(\rightarrow \rs) \times (\uparrow \rs \dot{\theta})=40\ks\gs(\odot \rs^2\dot{\theta})+5\ks\gs(\odot\rs^2\dot{\theta})=45\ks\gs(\odot \rs^2\dot{\theta})</math> 

<math>\dot{\overline{\mathbf{H}}}_\mathrm{T}(\Os)=45\ks\gs(\odot \rs^2\ddot{\theta})</math> 

<math>\sum \overline{\mathbf{M}}_\mathrm{ext}(\Os) = \dot{\overline{\mathbf{H}}}_\mathrm{RTO}(\Os) \quad \Rightarrow \quad (\odot 150\Ns \cdot \rs)=45\ks\gs(\odot \rs^2\ddot{\theta}) \quad \Rightarrow \quad \ddot{\theta}=\frac{10}{3\rs}</math> 

[[File:D4-Ex6-3-neut.png|thumb|right|150px|link=]]
Alternative solution 

If we apply the AMT at point '''O''' to the system formed by the pulleys and the ropes, the weight of the blocks no longer appears as an external interaction, but instead the tensions of the two ropes appear (which are constraint unknowns).

As the system is massless, its angular momentum is zero:

SYST: pulleys + ropes 

<math>\sum \overline{\mathbf{M}}_\mathrm{ext}(\Os) = \dot{\overline{\mathbf{H}}}_\mathrm{RTO}(\Os)=\overline{0} \quad \Rightarrow \quad (\odot\Ts_1 \cdot\rs) + (\otimes \Ts_2 \cdot 2 \rs)=0</math> 
[[File:D4-Ex6-4-neut.png|thumb|right|220px|link=]]

Applying the AMT to each of the blocks provides two more equations: 

SYST: 10 kg block 

<math>\sum \overline{\mathbf{F}}_\mathrm{ext}(\Os)=(10\ks\gs)\overline{\mathbf{a}}_\Ts(\mathrm{bloc}) \quad \Rightarrow \quad (\downarrow 100\Ns)+(\uparrow \Ts_1)=(10\ks\gs)(\uparrow 2\rs \ddot{\theta})</math> 

SYST: 5 kg block 

<math>\sum \overline{\mathbf{F}}_\mathrm{ext}(\Os)=(5\ks\gs)\overline{\mathbf{a}}_\Ts(\mathrm{bloc}) \quad \Rightarrow \quad (\downarrow 50\Ns)+(\uparrow \Ts_2)=(5\ks\gs)(\uparrow \rs \ddot{\theta})</math> 

Solving the system of equations leads to <math>\ddot{\theta}=\frac{10}{3\rs}</math>. 

The first solution is faster (it just calls for one scalar equation) but involves calculating the angular momentum. This second solution is longer (system of three equations) but does not require calculating any angular momentum.

</div>

<div>

====✏️ EXAMPLE D4.7: longitudinal dynamics of a vehicle====
------
:
{|
|[[File:D4-Ex7-1-eng.png|thumb|center|200px|link=]]
|A vehicle without suspensions moves on a straight road. The mass of the wheels is negligible compared to that of the rest of the elements, and their contact with the ground is considered to be a single-point one. We want to analyse the normal constraint forces ground and wheels.
This is a 2D dynamics problem. The external forces on the vehicle are the weight and and the ground constraint forces. If the acceleration of the chassis relative to the ground is a given, the application of the AMT leads to:
|}

{|
|

<math>\sum\F{\text{ext}} = \ms\acc{G}{\epsilon} \Rightarrow \left\{\begin{aligned}

\uparrow(\Ns_{\ds\vs}+ \Ns_{\ds\rs})+ (\downarrow\ms\gs) = 0\\
[\rightarrow(\Ts_{\ds\vs}+ \Ts_{\ds\rs})] = (\rightarrow\ms\as_\epsilon)
\end{aligned}\right.</math>

where <math>(\Ns_{\ds\vs}, \Ts_{\ds\vs})</math> and <math>(\Ns_{\ds\rs}, \Ts_{\ds\rs})</math> are the resultant normal and tangential forces exerted by the ground on the two front wheels and the two rear wheels, respectively.

The AMT at '''G''' yields: <math>\sum\vec{\Ms}_{ext}(\Gs) = \vec{H}_{RTG}(\Gs)\Rightarrow(\odot\Ls_{\ds\vs}\Ns_{\ds\vs}) + (\otimes\Ls_{\ds\rs}\Ns_{\ds\rs}) + [\odot\hs(\Ts_{\ds\vs} + \Ts_{\ds\rs})] =0</math>
(the angular momentum <math>\vec{H}_{RTG}(\Gs)</math> is permanently zero beacuse the wheels are massless, and the chassis motion relative to the RTG is zero).
|[[File:D4-Ex7-2-eng.png|thumb|center|200px|link=]]
|}

Solving the system of equations: <math>\Ns_{\ds\vs} = \frac{\ms\gs\Ls_{\ds\rs}}{\Ls_{\ds\vs} + \Ls_{\ds\rs}} - \frac{\ms\as_\epsilon\hs}{\Ls_{\ds\vs} + \Ls_{\ds\rs}}, \:\:\:\:\Ns_{\ds\rs} = \frac{\ms\gs\Ls_{\ds\vs}}{\Ls_{\ds\vs} + \Ls_{\ds\rs}} + \frac{\ms\as_\epsilon\hs}{\Ls_{\ds\vs} + \Ls_{\ds\rs}}, </math>

This result is valid for any value of <math>\as_\epsilon</math>:
:* Static situation (<math>\as_\epsilon = 0</math>): <math>\Ns_{\ds\vs}^{\es\ss\ts} = \frac{\ms\gs\Ls_{\ds\rs}}{\Ls_{\ds\vs}+ \Ls_{\ds\rs}}</math>, <math>\Ns_{\ds\rs}^{\es\ss\ts} = \frac{\ms\gs\Ls_{\ds\vs}}{\Ls_{\ds\vs}+ \Ls_{\ds\rs}}</math>; the normal force is higher on the wheels whose axle is closer to '''G'''.
:Accelerated motion (<math>\as_\epsilon > 0</math>): <math>\Ns_{\ds\vs} = \Ns_{\ds\vs}^{\es\ss\ts} - \frac{\ms|\as_\epsilon|\hs}{\Ls_{\ds\vs}+ \Ls_{\ds\rs}}</math>, <math>\Ns_{\ds\rs} = \Ns_{\ds\rs}^{\es\ss\ts} + \frac{\ms|\as_\epsilon|\hs}{\Ls_{\ds\vs}+ \Ls_{\ds\rs}}</math>; there is a force transfer from front to rear (the rear wheels are loaded and the front wheels are unloaded). If the acceleration is sufficiently high, the front normal force becomes negative, which indicates that contact has been lost and the vehicle rolls over counter-clockwise. Then, <math>\Ns_{\ds\vs} = 0</math> and the chassis initially has angular acceleration. As a consequence, the centre of mass '''G''' acquires vertical acceleration.

:*Braking (<math>\as_\epsilon < 0</math>): <math>\Ns_{\ds\vs} = \Ns_{\ds\vs}^{\es\ss\ts} - \frac{\ms|\as_\epsilon|\hs}{\Ls_{\ds\vs}+ \Ls_{\ds\rs}}</math>, <math>\Ns_{\ds\rs} = \Ns_{\ds\rs}^{\es\ss\ts} - \frac{\ms|\as_\epsilon|\hs}{\Ls_{\ds\vs}+ \Ls_{\ds\rs}}</math>; there is a force transfer from the rear to the front (the rear wheels are unloaded and the front wheels are loaded). As in the previous case, there is a critical value that causes the vehicle to tip over, this time clockwise.

---------
---------
</div>

==D4.7 Dynamics of Constraint Auxiliary Elements==

The application of the vector theorems to [[D3. Interactions between rigid bodies#D3.5 Indirect constraint interactions: Constraint Auxiliary Elements (CAE)|'''Constraint Auxiliary Elements''']] (CAE) is simple since the terms on the right (change in linear momentum and change in angular momentum) are zero because the CAE mass is zero:

<math>\ms_\mathrm{CAE} \approx 0 \Rightarrow \sum \mathbf{\overline{F}}_\mathrm{ext} \approx \overline{0} \quad , \quad \sum \mathbf{\overline{M}}_\mathrm{ext} (\mathrm{qualsevol} \hspace{0.2cm} \mathrm{punt}) \approx \overline{0}.</math> 

For the case of a CAE (rígid body S) connecting two rigid bodies S1 and S2 ('''Figure D4.5'''), these equations allow to prove that the torsor exerted on S1 can be obtained from an analytical characterization equation where the kinematics of the characterization point '''P''' (which has to be a point fixed to S1) is assessed from the S2 reference frame:

<math>\mathbf{\overline{F}}_{\mathrm{S}2 \rightarrow (\mathrm{S}) \rightarrow \mathrm{S}1} \cdot \mathbf{\overline{v}}_{\mathrm{S}2} (\Ps \in \mathrm{S}1) + \mathbf{\overline{M}}_{\mathrm{S}2\rightarrow (\mathrm{S}) \rightarrow \mathrm{S}1} (\Ps) \cdot \velang{S1}{S2} =0.</math> 

(When dealing with velocities, it is important to specify which rigid body does '''P''' belong to. However, this is irrelevant when dealing with moments: <math> \mathbf{\overline{M}} (\Ps \in \mathrm{S}1) = \mathbf{\overline{M}} (\Ps \in \mathrm{S}) + \overline{\Ps_{\mathrm{S}1}\Ps_\mathrm{S}} \times \mathbf{\overline{F}} = \mathbf{\overline{M}} (\Ps \in \mathrm{S}) \equiv \mathbf{\overline{M}} (\Ps)</math>). 

[[File:D4-5-eng.png|center|350px|link=]]
<center>'''Figure D4.5:'''Constraint Auxiliary Element between to rigid bodies with nonzero mass</center> 

<div>
=====💭 Proof ➕=====

The vector theorems applied to the CAE imply that the constraint torsors exerted by S1 and S2 at the same point '''P''' must sum up to zero:

<math>\left.\begin{array}{l}
\sum \overline{\mathbf{F}}_{\mathrm{ext}} \approx \overline{0} \Rightarrow \overline{\mathbf{F}}_{\mathrm{S} 1 \rightarrow \mathrm{S}}+\overline{\mathbf{F}}_{\mathrm{S} 2 \rightarrow \mathrm{S}}=\overline{0} \\
\sum \overline{\mathbf{M}}_{\mathrm{ext}}(\Ps) \approx \overline{0} \Rightarrow \overline{\mathbf{M}}_{\mathrm{S} 1 \rightarrow \mathrm{S}}(\Ps)+\overline{\mathbf{M}}_{\mathrm{S} 2 \rightarrow \mathrm{S}}(\Ps)=\overline{0}
\end{array}\right\} \Rightarrow\left\{\begin{array}{l}
\overline{\mathbf{F}}_{\mathrm{S} 2 \rightarrow \mathrm{S}} =\overline{\mathbf{F}}_{\mathrm{S} \rightarrow \mathrm{S} 1} \equiv \overline{\mathbf{F}}_{\mathrm{S} 2 \rightarrow(\mathrm{S}) \rightarrow \mathrm{S} 1} \\
\overline{\mathbf{M}}_{\mathrm{S} 2 \rightarrow \mathrm{S}}(\Ps)=\overline{\mathbf{M}}_{\mathrm{S} \rightarrow \mathrm{S} 1}(\Ps) \equiv \overline{\mathbf{M}}_{\mathrm{S} 2 \rightarrow(\mathrm{S}) \rightarrow \mathrm{S} 1}(\Ps)
\end{array}\right.</math> 

On the other hand, those torsores fulfill the [[D3. Interactions between rigid bodies#D3.4 Direct constraint interactions|''' equation of analytical characterization''']]:

<math>\mathbf{\overline{F}}_{\mathrm{S}2 \rightarrow \mathrm{S}} \cdot \mathbf{\overline{v}}_{\mathrm{S}2} (\Ps \in \mathrm{S}) + \mathbf{\overline{M}}_{\mathrm{S}2 \rightarrow \mathrm{S}} (\Ps) \cdot \velang{S}{S2} =0 \quad , \quad \mathbf{\overline{F}}_{\mathrm{S} \rightarrow \mathrm{S}1} \cdot \mathbf{\overline{v}}_{\mathrm{S}} (\Ps \in \mathrm{S}1) + \mathbf{\overline{M}}_{\mathrm{S}\rightarrow \mathrm{S}1} (\Ps) \cdot \velang{S1}{S} =0. </math> 

If all the above equations are combined, we obtain: 

<math> \mathbf{\overline{F}}_{\mathrm{S}2 \rightarrow (\mathrm{S}) \rightarrow \mathrm{S}1} \cdot \left[ \mathbf{\overline{v}}_{\mathrm{S}2} (\Ps \in \mathrm{S}) + \mathbf{\overline{v}}_{\mathrm{S}} (\Ps \in \mathrm{S}1) \right] + \mathbf{\overline{M}}_{\mathrm{S}2\rightarrow (\mathrm{S}) \rightarrow \mathrm{S}1} (\Ps) \cdot \left[\velang{S1}{S} + \velang{S}{S2}\right] =0. </math> 

A [[C3. Composition of movements#C3.1 Composition of velocities|'''composition of movements''']] allows us to rewrite the above equation in a more compact way:

<math>\left.\begin{array}{l}
\mathrm{AB}: \mathrm{S}2 \\
\mathrm{REL}: \mathrm{S}
\end{array}\right\} \Rightarrow \velang{S1}{S}=\velang{S1}{S2} - \velang{S}{S2}</math> 

<math>\left.\begin{array}{l}
\mathrm{AB}: \mathrm{S}2 \\
\mathrm{REL}: \mathrm{S}1
\end{array}\right\} \Rightarrow \mathbf{\overline{v}}_{\mathrm{S}2} (\Ps \in \mathrm{S})=\mathbf{\overline{v}}_{\mathrm{S}1} (\Ps \in \mathrm{S}) - \mathbf{\overline{v}}_{\mathrm{S}2} (\Ps \in \mathrm{S}1)</math> 

<math>\left.\begin{array}{l}
\mathrm{AB}: \mathrm{S}1 \\
\mathrm{REL}: \mathrm{S}
\end{array}\right\} \Rightarrow \mathbf{\overline{v}}_{\mathrm{S}} (\Ps \in \mathrm{S}1)=\mathbf{\overline{v}}_{\mathrm{S}1} (\Ps \in \mathrm{S}1) - \mathbf{\overline{v}}_{\mathrm{S}1} (\Ps \in \mathrm{S})= - \mathbf{\overline{v}}_{\mathrm{S}1} (\Ps \in \mathrm{S})</math> 

Finalment: <math>\mathbf{\overline{F}}_{\mathrm{S}2 \rightarrow (\mathrm{S}) \rightarrow \mathrm{S}1} \cdot \mathbf{\overline{v}}_{\mathrm{S}2} (\Ps \in \mathrm{S}1) + \mathbf{\overline{M}}_{\mathrm{S}2\rightarrow (\mathrm{S}) \rightarrow \mathrm{S}1} (\Ps) \cdot \velang{S1}{S2} =0.</math>

</div>
<div>

====✏️ EXAMPLE D4.8: longitudinal dynamics of a vehicle ====
---------
:
{|
|[[File:D4-Ex8-1-eng.png|thumb|left|220px|link=]]
|Let’s assume that the vehicle of example D4.7 has rear traction. This means that the vehicle's engine acts between the chassis and the rear wheels, while the front wheels are only subject to constraint interactions (articulation with the chassis and contact with the ground).

If the front wheels are treated as CAE, the kinematic analysis of the chassis relative to the ground for the characterization of the constraint through the front wheels leads to a torsor at the wheel centre (which is fixed to the chassis) with only one force component:
|}
[[File:D4-Ex8-2-eng.png|thumb|center|370px|link=]]

The existence of a horizontal component of force between the rear wheels (necessary to accelerate or brake the vehicle!) is associated with the engine torque. If the vector theorems are applied to the rear wheels, the external interactions to be taken into account are the connection with the ground, the articulation with the chassis and the engine torque:

[[File:D4-Ex8-3-eng.png|thumb|center|370px|link=]]

<math>\ms_\mathrm{wheels} \approx 0 \Rightarrow \sum \overline{\mathbf{F}}_\mathrm{ext} \approx \overline{0} , \sum
\overline{\mathbf{M}}_\mathrm{ext} (\Cs) \approx \overline{0} \Rightarrow
\left\{\begin{array}{l}
\mathrm{LMT }: \hspace{1.4cm} \Ns_\mathrm{dr}=\Ns'_\mathrm{dr} , \Ts_\mathrm{dr}=\Ts'_\mathrm{dr} \\
\mathrm{AMT } \hspace{0.2cm} \mathrm{a} \hspace{0.2cm} \Cs: \quad \Ts_\mathrm{dr}\rs=\Gamma \Rightarrow \Ts_\mathrm{dr}=\Gamma / \rs
\end{array}\right. </math> 

The tangential force on the driven wheels is directly proportional to the engine torque applied to them. However, it should be remembered that, as a constraint tangential force, its value is bounded by <math>\mu_\mathrm{s}\Ns_\mathrm{dr}</math>. This allows to calculate the maximum acceleration that the vehicle can acquire (as long as the front wheel does not lose contact with the ground): <math>\left.\as_\epsilon (\mathrm{vehicle})\right]_\mathrm{max}= \Ts_\mathrm{dr,max}/\ms \quad , \quad \Ts_\mathrm{dr}=\Gamma / \rs \quad , \quad \Ts_\mathrm{dr} \leq \mu_\mathrm{e}\Ns_\mathrm{dr}</math>. 

On low friction (low <math>\mu_\ss</math> value) terrains, an engine capable of delivering a high maximum torque is useless: what limits acceleration is the value of the friction coefficient <math>\mu_\es</math>: <math> \left.\as_\epsilon (\mathrm{vehicle})\right]_\mathrm{màx}= \mu_\mathrm{e}\Ns_\mathrm{dr} /\ms </math>.

On high friction terrain, if the maximum torque is low, it is this that limits acceleration: <math>\left. \as_\epsilon (\mathrm{vehicle})\right]_\mathrm{max}= \Gamma_\mathrm{max}/\ms \rs</math> (if the front wheel does not lose contact with the ground).


</div>

---------
---------

==D4.8 Barycentric decomposition of the angular momentum==

The general formulation presented for the AMT allows a free choice of the Q point of application. The criterion for choosing it is based on what we want to investigate (a constraint force, an equation of motion...). This criterion is discussed in [[D6. Examples of 2D dynamics|'''unit D6''']].

In some cases, it may be interesting to choose a point '''Q''' that does not belong to any element of the system. It is then useful to calculate the angular momentum <math>\H{Q}{}{RTQ}</math> from the angular momentum at the system center of mass '''G''', <math>\H{Q}{}{RTG}</math>. This is known as the '''barycentric decomposition''' of the angular momentum: <math>\H{Q}{}{RTQ} = \H{G}{}{RTG} + \H{Q}{\oplus}{RTQ}</math>, where the superscript <math>\oplus</math> indicates that the angular momentum is to be calculated as if the system had been reduced to a particle concentrated at '''G''' with mass equal to the total mass of the system (M): <math>\H{Q}{\oplus}{RTQ} = \QGvec\times\Ms\vel{G}{RTQ}</math>.

<div>
=====💭 Proof ➕=====

:The definition of the angular momentum of a rigid body is: <math>\H{Q}{\Ss}{RTQ} = \int_\Ss\QPvec\times\vel{P}{RTQ}\ds\ms(\Ps)</math>. Velocity <math>\vel{P}{RTQ}</math> can be written as a sum of two terms through a composition of movements:

:<math>\left.\begin{array}{l}
\mathrm{AB}:\mathrm{RTQ}\\
\mathrm{REL}:\mathrm{RTG}
\end{array}\right\} \Rightarrow\vel{P}{RTQ} = \vel{P}{RTG} + \vel{P}{ar} = \vel{P}{RTG} + \vel{$\Ps\in\mathbf{RT}\Gs$}{RTQ} = \vel{P}{RTG} + \vel{G}{RTQ}</math> 

:Therefore: <math>\H{Q}{\Ss}{RTQ} = \int_\Ss\QPvec\times\vel{P}{RTQ}\ds\ms(\Ps) + \left[\int_\Ss\QPvec\ds\ms(\Ps)\right]\times\vel{P}{RTQ}</math>.

:The definition of [[D4. Vector theorems#D4.1 Linear Momentum Theorem (LMT) in Galilean reference frames|'''centre of mass''']] leads to rewriting the second term as:

:<math>\left[\int_\Ss\QPvec\ds\ms(\Ps)\right] = \Ms\QGvec\times\vel{P}{RTQ} = \QGvec\times\Ms\vel{P}{RTQ}</math>, which coincides with the angular momentum at '''Q''' of a particle of mass M located at '''G''': <math>\QGvec\times\Ms\vel{G}{RTQ} = \H{Q}{\oplus}{RTQ}</math>.

:In the first term of the <math>\H{Q}{\Ss}{RTQ}</math> expression, <math>\QPvec</math> can be decomposed into two terms:

:<math>\int_\Ss\QPvec\times\vel{P}{RTQ}\ds\ms(\Ps) = \int_\Ss\QGvec\times\vel{P}{RTQ}\ds\ms(\Ps) + \int_\Ss\GPvec\times\vel{P}{RTQ}\ds\ms(\Ps) = \int_\Ss\QGvec\times\vel{P}{RTQ}\ds\ms(\Ps) + \H{G}{\Ss}{RTG}</math>

:From the definition of centre of mass: <math>\int_\Ss\QGvec\times\vel{P}{RTQ}\ds\ms(\Ps) = \QGvec\times\left[\int_\Ss\vel{P}{RTG}\ds\ms(\Ps)\right] = \QGvec\times\Ms\vel{G}{RTG} = \vec{0}</math>.

:Therefore <math>\H{Q}{}{RTQ} = \H{G}{}{RTG} + \H{Q}{\oplus}{RTQ}</math>.

</div>

<div>
===== ✏️ EXAMPLE D4.9: barycentric decomposition =====
----------

{|

|[[File:D4-Ex9-1-eng.png|thumb|center|200px|link=]]
|The rigid body is made up of two massless bars and four particles ('''P''') with mass m attached to the bars endpoints. The rigid body is attached to a support that can move along a guide fixed to the ground.

Its angular momentum at '''G''' is:

<math>\H{G}{}{RTG}\equiv\sum_{\Ps\in\text{syst}}\GPvec\times\ms\vel{P}{RTG}</math>

The RTG is fixed to the support, and the velocity of the particles relative to that reference frame is proportional to the rotation <math>\omega</math>: <math>|\vel{P}{RTG}|=\Ls\omega</math>. Hence:

<math>\H{G}{}{RTG}\equiv\sum_{\Ps\in\text{syst}}\GPvec\times\ms\vel{P}{RTG} = (\otimes4\ms\Ls^2\omega)</math>
|}
:The angular momentum at '''O''' (fixed to the ground) can be calculated from <math>\H{G}{}{RTG}</math> through barycentric decomposition:

:<math>\H{O}{}{RTO} = \H{O}{}{T} = \H{G}{}{RTG} + \H{O}{\oplus}{T} = (\otimes 4\ms\Ls^2\omega) + \OGvec\times 4\ms\vel{G}{T} =</math>
<math>\hspace{1cm} = (\otimes 4\ms\Ls^2\omega) + [(\rightarrow\xs) + (\downarrow\hs)]\times 4\ms(\rightarrow\dot\xs) = (\otimes4\ms\Ls^2\omega) + (\downarrow\hs)\times 4\ms(\rightarrow\dot\xs) =</math>
<math>\hspace{1cm}= (\otimes4\ms\Ls^2\omega) + (\odot 4\ms\hs\dot\xs) = \otimes 4\ms(\Ls^2\omega - \hs\dot\xs)</math>


</div>

© Universitat Politècnica de Catalunya. [[Mecánica:Derechos de autor |Todos los derechos reservados]]
-------------

<center>
[[D3. Interactions between rigid bodies|<<< D3. Interactions between rigid bodies]]

[[D5. Mass distribution|D5. Mass distribution >>>]]
</center>

E1. Work-Energy Theorem: differential form

2025-05-13T07:37:28Z

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</math>

The '''Work-Energy Theorem''' plays a prominent role in physics, as it bridges all branches of science (electromagnetism, thermodynamics, etc.): all of them deal with energy.

There is a certain parallel between the Vector Theorems and the '''Work-Energy Theorem''': they all allow us to understand the evolution of a magnitude that depends on the mechanical state of the system. In the case of the former, we are dealing with the instantaneous evolution of vector magnitudes (linear momentum, angular momentum), while the latter deals with the finite change (in a finite time interval) of a scalar magnitude ('''mechanical energy''').

This unit presents the differential form of this theorem (the '''power balance'''). The integrated form (which is the Work-Energy Theorem itself) is presented in [[E2. Teorema de la energía: versión integrada|'''unit E2''']].
 
------
------
==E1.1 Power balance in a system of particles==

:[[File:E1-1-eng.png|thumb|right|300px|link=]]
<center>'''Figure E1.1''' Forces on a particle of a system with constant matter</center>

Let us consider a system of constant matter, and let’s apply Newton's second law to each particle (or each mass differential) of this system in a general reference frame R (which may be non-Galilean) ('''Figure E1.1'''):

<math>\F{\rightarrow \Ps} + \Fcal{}{\text{ar}\rightarrow\Ps} + \Fcal{}{\text{Cor}\rightarrow\Ps} =\ms_\Ps\acc{P}{R}</math>

<math>\F{\rightarrow \Ps} </math> is the resultant interaction force on '''P''', and it comes from internal interactions (with particles of the same system) and external ones:

<math>\F{\rightarrow \Ps}=\F{\rightarrow \Ps}^\text{int}+\F{\rightarrow \Ps}^\text{ext}</math>

If both sides of the equation are multiplied by <math>\vel{P}{R}</math>, all forces (interaction and inertia forces) orthogonal to <math>\vel{P}{R}</math> are filtered out. Specifically, since <math>\acc{P}{R}=\accs{P}{R} + \accn{P}{R} </math>, where <math>\accn{P}{R} \perp \vel{P}{R}</math>, and since the Coriolis force is always orthogonal to <math>\vel{P}{R} (\Fcal{}{\text{Cor}\rightarrow\Ps} = -2\ms_\Ps \velang{R}{Gal} \times \vel{P}{R})</math>:

<math>\vel{P}{R} \cdot \left( \F{\rightarrow \Ps}^\text{int}+\F{\rightarrow \Ps}^\text{ext} + \Fcal{}{\text{ar}\rightarrow\Ps} \right) = \ms_\Ps\accs{P}{R} \cdot \vel{P}{R} = \frac{\ds}{\ds \ts} \left(\frac{1}{2}\ms_\Ps \vvec_\Rs^2(\Ps) \right) \equiv \frac{\ds\Ts_\Rs(\Ps)}{\ds\ts}=\dot{\Ts}_\Rs(\Ps)</math>

The scalar function <math>\Ts_\Rs(\Ps)</math> is the '''kinetic energy''' of the particle, and is a function of the [[D1. Foundational laws of Newtonian dynamics#D1.3 Newton’s Principle of Determinacy|'''mechanical state''']] of '''P'''. The scalar product of a force and the velocity of its application point is the '''force power''': <math>\vel{P}{R} \cdot \F{\rightarrow \Ps} \equiv \dot{\Ws}_\Rs^{\F{\rightarrow \Ps}} </math>. Both scalar quantities <math>(\Ts_\Rs(\Ps), \dot{\Ws}_\Rs^{\F{\rightarrow \Ps}})</math> depend on the reference frame. For the same set of interaction forces acting on '''P''', the reference frame determines which are filtered out and which are not.

The equation can be rewritten as <math>\dot{\Ws}_\Rs^{\text{ext}\rightarrow \Ps}+\dot{\Ws}_\Rs^{\text{int}\rightarrow \Ps}+\dot{\Ws}_\Rs^{\text{ar}\rightarrow \Ps} =\dot{\Ts}_\Rs(\Ps)</math>, and constitutes the '''power balance''' for particle '''P''' in the R reference frame. Summing for all particles in the system:

<math>\dot{\Ws}_\Rs^{\text{ext}}+\dot{\Ws}^{\text{int}}+\dot{\Ws}_\Rs^{\text{ar}} =\dot{\Ts}_\Rs^\text{syst}</math>

The calculation of kinetic energy is covered in [[E2. Teorema de la energía: versión integrada|'''unit E2''']]. This unit focuses on the calculation of powers. [[E1. Work-Energy Theorem: differential form#E1.2 Potencia de una pareja de acción-reacción|'''section E1-2''']] shows that the power of internal interactions does not depend on the reference frame (which is why the subscript R does not appear in the <math>\dot{\Ws}^{\text{int}}</math> term of the previous equation).

====✏️ EXAMPLE E1-1.1: free particle====
------
:
{|
|
:[[File:ExE1-1-1-1-eng.jpg|thumb|left|180px|link=]]
|A particle '''P''', with mass m, is released from the ring of a Ferris wheel when it is at its highest point. We want to calculate the power, at this instant, of all forces acting on the particle from three different reference frames: the ground, the ring, and the Ferris wheel cabin.
|}

:Ground reference frame: Since this is a Galilean reference frame, all forces on '''P''' are interaction forces.
{|
|
:[[File:ExE1-1-2- eng.png|thumb|right|230px|link=]]

|<math>\hspace{0.5cm}\sum\dot{\Ws}_\Ts=\dot{\Ws}_\Ts^\text{weight}= \left(\leftarrow \Rs\Omega_0\right) \cdot \left(\downarrow \ms\gs\right)=0</math>
|}

:Ring reference frame: It is a non-Galilean reference, and the transportation force on '''P''' must be taken into account.
{|
|
:[[File:ExE1-1-3-eng.png|thumb|right|250px|link=]]
|<math>\hspace{0.5cm}</math> AB:ground, REL:ring 

<math>\hspace{0.5cm}\vel{P}{REL}=\vel{P}{AB}-\vel{P}{ar}= \left(\leftarrow \Rs\Omega_0\right)-\left(\leftarrow \Rs\Omega_0\right)=\overline{0}</math> 

<math>\hspace{0.5cm}</math>

Since the velocity is zero, the total power is zero.
|}

:Cabin reference frame: It is also a non-Galilean reference, and the transportation force on '''P''' must be taken into account.

{|
|
:[[File:ExE1-1-4-eng.png|thumb|right|250px|link=]]
|<math>\hspace{0.5cm}</math>AB:ground, REL:cabin

<math>\hspace{0.5cm}\vel{P}{REL}=\vel{P}{AB}-\vel{P}{ar}= \left(\leftarrow \Rs\Omega_0\right)-\left(\uparrow \Rs\Omega_0\right)</math>

<math>\hspace{0.5cm}\dot{\Ws}_\text{cabin}^\text{weight}=\left[ \left(\leftarrow \Rs\Omega_0\right)+\left(\downarrow \Rs\Omega_0\right) \right] \cdot \left(\downarrow\ms\gs\right) =\ms\gs\Rs\Omega_0</math>

<math>\hspace{0.5cm}\dot{\Ws}_\text{cabin}^\text{ar}=\left[ \left(\leftarrow \Rs\Omega_0\right)+\left(\downarrow \Rs\Omega_0\right) \right] \cdot \left(\rightarrow\ms\Rs^2\Omega_0^2\right) =-\ms\gs\Rs^2 \Omega_0^3</math>

<math>\hspace{0.5cm}\sum\dot{\Ws}_\text{cabin}=\ms\Rs\Omega_0(\gs-\Rs\Omega_0^2)</math>

|}


====✏️ EXAMPLE E1-1.2: pendulum====
------
:
{|
|
:[[File:ExE1-2-1-eng.png|thumb|left|180px|link=]]
|A particle '''P''' of mass m hangs from an inextensible thread of length L attached to the ceiling. The assembly oscillates in a vertical plane fixed to the ground. We want to calculate the power of all the forces acting on the particle in two different reference frames: the ground and a platform rotating relative to the ground with a constant angular velocity <math>\Omega_0</math>. 

Ground reference frame: Since it is a Galilean reference frame, '''P''' is only subjected to interaction forces.
{|
|
:[[File:ExE1-2-2-neut.png|thumb|left|300px|link=]]

|<math>\dot{\Ws}_\Es^\Ts=(\nearrow\Ls\dot{\theta})\cdot(\nwarrow\Ts)=0</math>

<math>\dot{\Ws}_\Es^\text{weight}=(\nearrow\Ls\dot{\theta}\cdot(\downarrow \ms\gs)=-\ms\gs\Ls\dot{\theta}\sin\theta</math>

|}

Platform reference frame: It is a non Galilean reference frame, and the transportation force on '''P''' has to be taken into account.

{|
|
:[[File:ExE1-2-3-eng.png|thumb|left|400px|link=]]

|AB:ground, REL:platform

<math>\vel{P}{REL}=\vel{P}{AB}-\vel{P}{ar}=(\nearrow\Ls\dot{\theta})-(\otimes \Ls \Omega_0 \sin \theta)</math>

<math>\dot{\Ws}_\Es^\Ts\left[(\nearrow\Ls\dot{\theta})+(\odot\Ls\Omega_0\sin\theta)\right]\cdot(\nwarrow\Ts)=0</math>

<math>\dot{\Ws}_\Es^\text{weight}\left[(\nearrow\Ls\dot{\theta})+(\odot\Ls\Omega_0\sin\theta)\right]\cdot(\downarrow \ms\gs)=-\ms\gs\Ls\dot{\theta}\sin\theta</math>

<math>\dot{\Ws}_\Es^\text{ar}\left[(\nearrow\Ls\dot{\theta})+(\odot\Ls\Omega_0\sin\theta)\right]\cdot(\rightarrow \ms\Ls\Omega_0^2\sin\theta)=\ms\Ls^2\Omega_0^2\dot{\theta}\cos\theta</math>
|}


------
------

==E1.2 Power of an action-reaction pair==

The total power associated with internal interactions in a constant-mass system is not zero in principle (since the two forces of each action-reaction pair are applied to particles that may have different velocities relative to the reference frame in which the calculation is made), and it is independent of the reference frame in which it is evaluated. It only depends on the separation velocity of the two particles. The power balance for the system is therefore written as: <math>\dot{\Ts}_\Rs^\text{syst}=\dot{\Ws}_\Rs^\text{ext} + \dot{\Ws}^\text{int} + \dot{\Ws}_\Rs^\text{ar} </math> .

<div>
=====💭 PROOF ➕=====

{|
|
:[[File:E1-2-neut.png|thumb|left|200px|link=]]
|<math>\dot{\Ws}^\text{ac-reac}=\overline{\Fs}_{\Ps \rightarrow \Qs}\cdot \vel{Q}{R} + \overline{\Fs}_{\Qs \rightarrow \Ps}\cdot \vel{P}{R}= \overline{\Fs}_{\Ps \rightarrow \Qs}\cdot \left[ \vel{Q}{R}-\vel{P}{R}\right]</math>

By the [[D1. Foundational laws of Newtonian dynamics#D1.6 Newton’s third law (action-reaction principle)|'''action-reaction principle''']], <math>\overline{\Fs}_{\Ps \rightarrow \Qs}=-\overline{\Fs}_{\Qs \rightarrow \Ps}</math>. Therefore: <math>\dot{\Ws}^\text{ac-reac}=\overline{\Fs}_{\Ps \rightarrow \Qs}\cdot \left[ \vel{Q}{R}-\vel{P}{R}\right]</math>

As <math>\overline{\Fs}_{\Ps \rightarrow \Qs}</math> has the <math>\PQvec</math> direction, only the components of the velocities in that direction have to be considered:

<math>\dot{\Ws}^\text{ac-reac}=\overline{\Fs}_{\Ps \rightarrow \Qs}\cdot \biggl( \vel{Q}{R} \Bigr]_{||\Ps\Qs}-\ \vel{P}{R} \Bigr]_{||\Ps\Qs}\biggl)</math>
|}

The difference <math>\biggl( \vel{Q}{R} \Bigr]_{\Ps\Qs}-\ \vel{P}{R} \Bigr]_{\Ps\Qs}\biggl)</math> is the separation velocity <math>\dot{\rho}</math> between '''P''' and '''Q''', and it does not dpend on the reference frame since it is the derivative of a scalar (the distance <math>\dot{\rho}</math> between the two particles). For the previous drawing, <math>\dot{\rho}> 0</math>. Since this is an attractive interaction, <math>\dot{\Ws}^\text{ac-reac} < 0</math>.

With a drawing, it is easy to see that:
* attractive interaction and particle approach <math>(\dot{\rho}<0)</math> : <math>\dot{\Ws}^\text{ac-reac} > 0</math> ;
* repulsive interaction and particle separation <math>(\dot{\rho}>0)</math> : <math>\dot{\Ws}^\text{ac-reac} > 0</math>
* repulsive interaction and particle approach <math>(\dot{\rho}<0)</math> : <math>\dot{\Ws}^\text{ac-reac} < 0</math>


</div>

There are two particular cases of interaction forces whose power always has the same sign:

* Friction action-reaction pair between '''P''' and '''Q''': since the friction force always opposes the relative velocity between the two points, its power is always negative:

<math>\hspace{1cm} \dot{\Ws}^\text{friction}=\dot{\Ws}^{\text{fric}\rightarrow\Ps}_\text{RTQ}+\dot{\Ws}^{\text{fric}\rightarrow\Qs}_\text{RTQ}=\overline{\Fs}^{\text{fric}\rightarrow\Ps} \cdot \vel{P}{RTQ} = - \abs{\overline{\Fs}^{\text{fric}\rightarrow\Ps}} \frac{\vel{P}{RTQ}}{\abs{\vel{P}{RTQ}}} \cdot \vel{P}{RTQ}<0;</math>

* Constraint action-reaction pair between '''P''' and '''Q''': from the [[D3. Interactions between rigid bodies#D3.4 Direct constraint interactions|'''analytical characterization equation''']] of the constraint forces between two points '''P''' and '''Q''' belonging to two rigid bodies <math>\mathrm{S}_\Ps</math> and <math>\mathrm{S}_\Qs</math>, it is easy to see that the associated power is zero:

<math>\hspace{1cm} \dot{\Ws}^\text{constraint}=\dot{\Ws}^{\text{constraint}\rightarrow\Ps}_{\mathrm{S}_\Ps}+\dot{\Ws}^{\text{constraint}\rightarrow\Qs}_{\mathrm{S}_\Ps}=\overline{\Fs}^{\text{constraint}\rightarrow\Ps} \cdot \overline{\mathbf{v}}_{\mathrm{S}_\Qs}(\Ps) = - 0;</math>

====✏️ EJEMPLO E1-2.1: potencia de una pareja de acción-reacción de fricción====
------
:
{|
|
:[[Archivo:ExE1-2-1-1-esp.png|thumb|left|180px|link=]]
|Una rueda homogénea, de radio r y masa m, baja por un plano inclinado sin deslizar hasta incorporarse a una cinta transportadora que se mueve con velocidad constante <math>\vs_0</math> respecto al suelo. En el momento de entrar en contacto con la cinta, el centro de la rueda tiene velocidad <math>\vs_0</math> respecto al suelo en el mismo sentido que la de la cinta. Se trata de calcular la potencia de la fricción entre cinta y rueda en este instante.
|}

El <math>\mathbf{CIR}^\text{rueda}_\Ts</math> en este instante es el punto <math>\Js_\text{rueda}</math> que está en contacto con la cinta. Este punto patina hacia atrás respecto a la cinta, y por tanto la fuerza de fricción sobre la rueda es hacia adelante:

<math>\left.\begin{array}{l}
\text { AB: suelo } \\
\text { REL: cinta }
\end{array}\right\} \quad \bar{\vs}_{\text {REL }}\left(\Js_{\text {rueda }}\right)=\bar{\vs}_{\text{AB}}\left(\Js_{\text {rueda }}\right)-\bar{\vs}_{\mathrm{ar}}\left(\Js_{\text {rueda }}\right)=-\bar{\vs}_{\mathrm{ar}}\left(\Js_{\text {rueda }}\right)=-\left(\rightarrow \mathrm{v}_0\right)=\left(\leftarrow \mathrm{v}_0\right) .</math>

[[Archivo:ExE1-2-1-2-esp.png|thumb|center|500px|link=]]:

Cálculo de la potencia de la fricción: <math>\dot{\Ws}^\text{fricción}=\dot{\Ws}_\Rs^{\text{fricció}\rightarrow \text{rueda}}+ \dot{\Ws}_\Rs^{\text{fricción}\rightarrow \text{cinta}}</math>, donde R puede ser cualquier referencia. Para calcular de manera ágil esta potencia, es recomendable escoger una referencia R donde uno de los dos puntos sometidos a la pareja de acción-reacción tenga velocidad nula, y así uno de los dos términos es directamente cero:

<math>\dot{\Ws}^\text{fricción}=\dot{\Ws}_\text{cinta}^{\text{fric}\rightarrow \text{rueda}}+ \dot{\Ws}_\text{cinta}^{\text{fric }\rightarrow \text{cinta}}=\overline{\Fs}^{\text{fric}\rightarrow \text{rueda}} \cdot \bar{\vs}_{\text {cinta}}\left(\Js_{\text {roda }}\right)=(\rightarrow \mu\ms\gs)\cdot(\leftarrow \vs_0)=-\mu\ms\gs\vs_0.</math>

Si el cálculo se hace en cualquier otra referencia, el resultado es el mismo:

<math>\dot{\Ws}^\text{fricción}=\dot{\Ws}_\text{T}^{\text{fric}\rightarrow \text{rueda}}+ \dot{\Ws}_\text{T}^{\text{fric} \rightarrow \text{cinta}}=(\rightarrow \mu\ms\gs)\cdot 0 + (\leftarrow \mu\ms\gs)\cdot(\rightarrow \vs_0)=-\mu\ms\gs\vs_0.</math>

<math>\dot{\Ws}^\text{fricción}=\dot{\Ws}_\text{vehículo}^{\text{fric}\rightarrow \text{rueda}}+ \dot{\Ws}_\text{vehículo}^{\text{fric} \rightarrow \text{cinta}}=(\rightarrow \mu\ms\gs)\cdot (\rightarrow\vs_0) + (\leftarrow \mu\ms\gs) \cdot(\rightarrow 3\vs_0)=-\mu\ms\gs\vs_0.</math>

:[[Archivo:ExE1-2-1-3-esp.png|thumb|center|450px|link=]]

En cambio, si sólo se calcula la potencia de una de las dos fuerzas (por ejemplo, la fricción sobre la rueda), el resultado depende de la referencia y puede tener cualquier signo:

<math>\dot{\Ws}_\text{rueda}^{\text{fric }\rightarrow \text{roda}}=(\rightarrow \mu\ms\gs)\cdot(\leftarrow \vs_0)=-\mu\ms\gs\vs_0 ,</math>

<math>\dot{\Ws}_\text{T}^{\text{fric} \rightarrow \text{rueda}}=(\rightarrow \mu\ms\gs)\cdot 0 = 0 ,</math>

<math>\dot{\Ws}_\text{vehicle}^{\text{fric} \rightarrow \text{rueda}}=(\rightarrow \mu\ms\gs)\cdot (\rightarrow\vs_0)=\mu\ms\gs\vs_0.</math>


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==E1.3 Potencia de un sistema de fuerzas sobre un sólido rígido==

{|
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:[[Archivo:E1-3-esp.png|thumb|left|180px|link=]]
<center>'''Figura E1.3''': Torsor cinemático de un sólido rígido en una referenciay forces sobre los distintos puntos (dm) de este sóldio</center>

|Cuando el sistema sobre el que se calcula el balance de potencias es un sólido rígido S, no hay término asociado a interacciones internas porque todas son de enlace (fuerzas de cohesión) y, como se ha visto en la [[E1. Teorema de la energía: versión diferencial#E1.2 Potencia de un par de acción reacción|'''secciónE1.2''']], su potencia total es nula.

La potencia total de un sistema de fuerzas sobre S se puede calcular a partir del torsor del sistema en cualquier punto '''O''' del sólido <math>(\overline{\Fs},\overline{\Ms}(\Os))</math>. Teniendo en cuenta la cinemática de sólido rígido ('''Figura E1.3'''):

<math>\dot{\Ws}_\Rs^{\rightarrow \Ps} =\sum_\Ps\overline{\Fs}^{\rightarrow \Ps} \cdot \vel{P}{R}= \sum_\Ps \left[ \overline{\Fs}^{\rightarrow \Ps} \cdot \vel{O}{R} + \velang{S}{R} \times \OPvec \right] </math>

|}
<math>\dot{\Ws}_\Rs^{\rightarrow \Ps} =\left(\sum_\Ps\overline{\Fs}^{\rightarrow \Ps} \right)\cdot \vel{O}{R} + \sum_\Ps\overline{\Fs}^{\rightarrow \Ps} \cdot \left(\velang{S}{R} \times \OPvec \right) = \left( \sum_\Ps\overline{\Fs}^{\rightarrow \Ps} \right) \cdot \vel{O}{R} + \left[ \sum_\Ps \left( \OPvec \times \overline{\Fs}^{\rightarrow \Ps} \right)\right] \cdot \velang{S}{R} = \overline{\Fs}\cdot \vel{O}{R} + \overline{\Ms}(\Os)\cdot \velang{S}{R} </math>

Este cálculo demuestra también que la potencia de un momento (o un par <math>\Gamma</math>) sobre un sólido en una referencia R es el producto del momento por la velocidad angular del sólido: <math>\dot{\Ws}_\Rs^{\Gamma\rightarrow \mathrm{S}} = \overline{\Gamma} \cdot \velang{S}{R}</math>.

El balance de potencias para un sólido rígido es: <math>\dot{\Ws}_\Rs^{ \rightarrow \mathrm{S}} = \dot{\Ws}_\Rs^{ \mathrm{ext}} + \dot{\Ws}_\Rs^{ \mathrm{ar}}.</math>

====✏️ EJEMPLO E1-3.1: rodillo deslizando sobre un plano inclinado====
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:
{|
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:[[Archivo:ExE1-3-1-1-esp.png|thumb|left|180px|link=]]
|Un rodillo homogéneo, de masa m y radi r, baja deslizando por una pendiente. El coeficiente de fricción entre suelo y rodillo es <math>\mu</math>. trata de calcular las potencias de todas las fuerzas que actúan sobre el rodillo en la referencia del suelo.
|}

:Se trata de un problema plano. Las descripciones cinemática y dinámica son:
{|
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[[Archivo:ExE1-3-1-2-cat-esp.png|thumb|right|250px|link=]]

|<math> \hspace{0.5cm}\vel{J}{T}=\vel{C}{T} + \velang{}{T} \times \CJvec= </math>

<math>\hspace{1.7cm}=\left( \rightarrow \vs_\Ts \right) + \left( \otimes \Omega_\Ts\right) \times \left( \downarrow \rs \right)</math>

<math> \hspace{0.5cm}\vel{J}{T} = \left[\rightarrow (\vs_\Ts -\rs\Omega_\Ts)\right]</math>

<math>\hspace{0.5cm} \dot{\Ws}_\Ts^{\mathrm{ext}} = \dot{\Ws}_\Ts^{\mathrm{peso}} + \dot{\Ws}_\Ts^{\mathrm{fricción}} + \dot{\Ws}_\Ts^{\mathrm{enlace}}= \Bigr[\left(\downarrow \ms\gs\cos\beta\right)+\left(\rightarrow \ms\gs\sin\beta\right)\Bigr] \cdot (\rightarrow \vs_\Ts) + \Bigr[(\leftarrow \mu \Ns) + (\uparrow \Ns ) \Bigr] \cdot \Bigr[\rightarrow (\vs_\Ts -\rs\Omega_\Ts)\Bigr]</math>

|}
:Ya que el centro del rodillo no tiene movimiento perpendicular al plano, el TCM conduce a <math>\Ns =\ms\gs\cos\beta</math>. Por tanto:
{|
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[[Archivo:ExE1-3-1-3-eng.png|thumb|right|140px|link=]]

|<math>\hspace{0.5cm} \dot{\Ws}_\Ts^{\mathrm{ext}} = \ms\gs\vs_\Ts \sin \beta - \ms\gs(\vs_\Ts -\rs\Omega_\Ts)\cos\beta=\ms\gs \Bigr[ \vs_\Ts (\sin\beta-\cos\beta)+\rs\Omega_\Ts \cos\beta \Bigr].</math>

<math>\hspace{0.5cm}</math>

Si en lugar de considerar la fricción en '''J''' se considera el torsor equivalente en '''C''', hay que considerar la fuerza aplicada en '''C''' y el momento respecto a '''C''', que no es nulo:

<math>\hspace{0.5cm} \dot{\Ws}_\Ts^{ \mathrm{fricción}} = (\leftarrow \mu\Ns)\cdot(\rightarrow\vs_\Ts)+(\otimes \mu\Ns\rs)\cdot(\otimes \Omega_\Ts) =\mu\ms\gs\cos\beta(\rs\Omega_\Ts-\vs_\Ts)</math>,

<math>\hspace{0.5cm} \dot{\Ws}_\Ts^{\mathrm{ext}} = \dot{\Ws}_\Ts^{ \mathrm{peso}} + \dot{\Ws}_\Ts^{ \mathrm{fricción}}+ \dot{\Ws}_\Ts^{ \mathrm{enlace}}= \ms\gs\vs_\Ts\sin\beta+\mu\ms\gs\cos\beta (\rs\Omega_\Ts-\vs_\Ts)</math>.

<math>\hspace{0.5cm}</math>El resultado es el mismo.
|}


====✏️ EJEMPLO E1-3.2: bloque deslizando sobre un camión====
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:
{|
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:[[Archivo:ExE1-3-2-1-esp.png|thumb|left|220px|link=]]
|Un camión frena con aceleración '''a''' respecto al suelo en el instante en el que su velocidad (también respecto al suelo) es v. En el mismo instante, el bloque homogéneo, de masa m, desliza con velocidad v' relativa al camión. Se trata de hacer un balance de potencias para el bloque en la referencia del suelo y en la del camión.

|}

:Referencia suelo: Se trata de un problema plano. Las descripciones cinemática y dinámica son:
{|
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[[Archivo:ExE1-3-2-2-cat-esp.png|thumb|right|200px|link=]]

| El enlace entre camión y bloque permite sólo un GL de translación entre los dos. Por tanto, en la aproximación 2D, el torsor asociado tiene dos componentes independientes. Si se caracteriza en el punto medio del contacto entre los dos elementos, son una fuerza normal y un momento perpendicular al plano. El TCM aplicado al bloque conduce a <math>\Ns=\ms\gs</math>.

|}
:Las potencias asociadas a todas las interacciones sobre el bloque en la referencia del suelo son:

:<math>\dot{\Ws}_\Ts^{\rightarrow \mathrm{bloque}}=\dot{\Ws}_\Ts^{\mathrm{ext}}=\dot{\Ws}_\Ts^{\mathrm{peso}}+\dot{\Ws}_\Ts^{\mathrm{fricción}}+\dot{\Ws}_\Ts^{\mathrm{enlace}}= \Bigr[(\downarrow \ms\gs)+(\leftarrow \mu\ms\gs)+(\uparrow \Ns) \Bigr] \cdot \Bigr[\rightarrow(\vs+\vs')\Bigr] + (\odot \Ms_\es)\cdot 0 = -\mu\ms\gs(\vs+\vs')</math>

:Ya que el bloque se traslada respecto al suelo, su energía cinética total es inmediata, y se puede plantear fácilmente el balance de potencias:

:<math>\Ts_\Ts^\mathrm{bloque}=\frac{1}{2}\int_\mathrm{bloque}\ds\ms \vs_\Ts^2(\mathrm{bloque})=\frac{1}{2}\ms \vs_\Ts^2(\mathrm{bloc})\quad \Rightarrow \quad \dot{\Ts}_\Ts^\mathrm{bloque}=\ms\vs_\Ts( \mathrm{bloque})\dot{\vs}_\Ts( \mathrm{bloque}) </math>

:Como la única fuerza horizontal sobre el bloque es la fricción:

<math>\dot{\vs}_\Ts(\mathrm{bloque})=-\mu\gs\quad \Rightarrow \quad \dot{\Ts}_\Ts^\mathrm{bloque}=-\mu\ms\gs(\vs+\vs').</math>

:El balance de potencias es consistente:

:<math>\left.\begin{array}{l}
\dot{\mathrm{W}}_\Ts^{\rightarrow \text { bloque }}=\dot{\mathrm{W}}_{\mathrm{T}}^{\text {ext }}=-\mu \mathrm{mg}\left(\mathrm{v}+\mathrm{v}^{\prime}\right) \\
\dot{\mathrm{T}}_{\mathrm{T}}^{\text {bloque }}=-\mu \mathrm{mg}\left(\mathrm{v}+\mathrm{v}^{\prime}\right)
\end{array}\right\} \Rightarrow \dot{\mathrm{W}}_{\mathrm{T}}^{\rightarrow \text { bloque }}=\dot{\mathrm{T}}_{\mathrm{T}}^{\text {bloque }}</math>

:Referencia camión: Al estar el camión acelerado respecto al suelo, hay que tener en cuenta las fuerzas de arrastre sobre todos los diferenciales de masa (dm) del bloque:

{|
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[[Archivo:ExE1-3-2-3-cat-esp.jpg|thumb|right|250px|link=]]
|<math>\cal{\Fs}^{\rightarrow \ds\ms}_{\mathrm{ar}}=-\ds\ms\overline{\as}_\mathrm{ar}(\ds\ms)=-\ds\ms (\leftarrow \as)=(\rightarrow \ds\ms \cdot \as)</math>.

El torsor resultante de este sistema de fuerzas en el centro de inercia '''G''' se reduce a una fuerza horizontal.

<math>\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{bloque}}=\dot{\Ws}_\mathrm{camión}^{\mathrm{ext}} + \dot{\Ws}_\mathrm{camión}^{\mathrm{ar}} =\dot{\Ws}_\mathrm{camión}^{\mathrm{peso}} + \dot{\Ws}_\mathrm{camión}^{\mathrm{fricción}} + \dot{\Ws}_\mathrm{camión}^{\mathrm{enlace}} + \dot{\Ws}_\mathrm{camión}^{\mathrm{ar}} , </math>

<math>\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{bloque}}= \Bigr[(\downarrow \ms\gs)+(\leftarrow \mu\ms\gs)+(\uparrow \Ns) + (\rightarrow \ms\as)\Bigr] \cdot(\rightarrow \vs')+(\odot\Ms_\es) \cdot0=\ms(\as-\mu\gs)\vs'.</math>
|}

:Se puede comprobar que el balance de potencias es consistente:

:<math>\Ts^\mathrm{bloque}_\mathrm{camión}=\frac{1}{2}\ms\vs^2_\mathrm{camión}(\mathrm{bloque}) \quad \Rightarrow \quad \dot{\Ts}^\mathrm{bloque}_\mathrm{camión} = \ms\vs_\mathrm{camión}(\mathrm{bloque})\dot{\vs}_\mathrm{camión}(\mathrm{bloque})</math>

:<math>\left.\begin{array}{l}
\mathrm{AB}:\mathrm{suelo} \\
\mathrm{REL}:\mathrm{camión}
\end{array}\right\} \quad \dot{\vs}_\mathrm{camión}(\mathrm{bloque})= \dot{\vs}_\mathrm{T}(\mathrm{bloque})-\dot{\vs}_\mathrm{ar}(\mathrm{bloque}) = (-\mu\gs)-(-\as)=\as-\mu\gs \quad \Rightarrow \quad \dot{\Ts}^\mathrm{bloque}_\mathrm{camión}=\ms\vs'(\as-\mu\gs)</math>

:<math>\left.\begin{array}{l}
\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{bloque}}=\dot{\Ws}_\mathrm{camión}^{\mathrm{ext}} + \dot{\Ws}_\mathrm{camión}^{\mathrm{ar}}= \ms(\as-\mu\gs)\vs'\\
\dot{\Ts}^\mathrm{bloque}_\mathrm{camión}=\ms\vs'(\as-\mu\gs)
\end{array}\right\} \Rightarrow\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{bloque}}= \dot{\Ts}^\mathrm{bloque}_\mathrm{camión}</math>


====✏️ EJEMPLO E1-3.3: camión frenando con carga====
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:
{|
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:[[Archivo:ExE1-3-3-1-esp.png|thumb|left|220px|link=]]
|Un camión frena con aceleración '''a''' respecto al suelo en el instante en el que su velocidad (también respecto al suelo) es v. Los pares de frenada en el eje delantero y en el trasero son iguales y valen <math>\Gamma</math>, y las ruedas, de radio r y masa despreciable, no deslizan sobre el suelo. En el mismo instante, el bloque homogéneo, de masa m, desliz con velocidad v' relativa al camión. Se trata de hacer un balance de potencias para el sistema (bloque+camión) en la referencia del suelo y en la del camión.

|}

:Ya que la potencia de las interacciones internas no es nula en principio (no se trata de un único sólido rígido), es conveniente hacer un DGI del sistema para no olvidar ninguna:
{|
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[[Archivo:ExE1-3-3-2-esp.png|thumb|right|250px|link=]]

|
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c.p.s.d.: contacto puntual sin deslizamiento

c.m.c.d.: contacto multipuntual con deslizamiento

|}
:El sistema sobre el que se hace el balance de potencias es todo menos el suelo. Teniendo en cuenta que la potencia de los enlaces internos es siempre nula, las únicas interacciones que hay que tener en cuenta son el peso, los enlaces de las ruedas con el suelo, los pares de frenada y la fricción entre bloque y camión:

[[Archivo:ExE1-3-3-3-esp.png|thumb|center|300px|link=]]

:Referencia suelo:
:Las fuerzas de enlace del suelo sobre las ruedas están aplicadas a puntos de velocidad cero, y por tanto no hacen potencia. El peso es una fuerza vertical, y el movimiento del camión y del bloque es horizontal, por tanto su potencia también es cero.

:<math>\dot{\Ws}_\Ts^{\rightarrow \mathrm{sist}}=\dot{\Ws}_\Ts^{\mathrm{ext}}+\dot{\Ws}^{\mathrm{int}}=\dot{\Ws}_\Ts^{\mathrm{peso}}+\dot{\Ws}_\Ts^{\mathrm{enlace}}+\dot{\Ws}^{\mathrm{fricción}}+\dot{\Ws}^{\Gamma}=\dot{\Ws}^{\mathrm{fricción}}+\dot{\Ws}^{\Gamma}.</math>

:Las potencias de interacciones internas se pueden calcular en cualquier referencia. Por tanto:

:<math>\dot{\Ws}^{\mathrm{fricción}}= \dot{\Ws}_\mathrm{camión}^{\mathrm{fric}\rightarrow \mathrm{bloque}} +\dot{\Ws}_\mathrm{camió}^{\mathrm{fric}\rightarrow \mathrm{camión}} = \dot{\Ws}_\mathrm{camión}^{\mathrm{fric}\rightarrow \mathrm{bloque}}= (\leftarrow \mu\ms\gs)\cdot (\rightarrow \vs')=-\mu\ms\gs\vs'</math>

:<math>\dot{\Ws}^{\Gamma}= \dot{\Ws}_\mathrm{camión}^{\Gamma\rightarrow \mathrm{ruedas}} +\dot{\Ws}_\mathrm{camión}^{\Gamma\rightarrow \mathrm{camión}} = \dot{\Ws}_\mathrm{camión}^{\Gamma\rightarrow \mathrm{ruedas}}= 2 \left(\odot \Gamma \right)\cdot \left(\otimes \frac{\vs}{\rs}\right)=-2\Gamma\frac{\vs}{\rs}.</math>

:Por otro lado: <math>\Ts_\Ts^\mathrm{sist}=\frac{1}{2}\ms\vs_\Ts^2(\mathrm{bloque})+\frac{1}{2}\ms\vs_\Ts^2(\mathrm{camión})\quad \Rightarrow \quad \dot{\Ts}_\Ts^\mathrm{sist}=-\ms(\vs+\vs')\mu\gs-\Ms\vs\as. </math>

:Finalmente: <math>\dot{\Ws}_\Ts^{\rightarrow \mathrm{sist}}=\dot{\Ts}_\Ts^\mathrm{sist} \quad \Rightarrow \quad -\mu\ms\gs\vs'-2\Gamma\frac{\vs}{\rs}=-\ms(\vs+\vs')\mu\gs-\Ms\vs\as.</math>

:Esta ecuación permite calcular el par de frenada: <math>\Gamma=\frac{1}{2}(\mu\ms\gs+\Ms\as)\rs.</math>

:Referencia camión:

:En esta referencia hay que tener en cuenta las fuerzas de arrastre. Por otro lado, las fuerzas de enlace del suelo sobre las ruedas están aplicadas a puntos que ya no tienen velocidad cero, y por tanto hacen potencia. Como antes, la potencia del peso es cero.

:<math>\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{sist}}=\dot{\Ws}_\mathrm{camión}^{\mathrm{ext}}+\dot{\Ws}^{\mathrm{int}}+\dot{\Ws}_\mathrm{camión}^{\mathrm{ar}}=\dot{\Ws}_\mathrm{camión}^{\mathrm{enlace}}+\dot{\Ws}^{\mathrm{fricción}}+\dot{\Ws}^{\Gamma}+\dot{\Ws}_\mathrm{camión}^{\mathrm{ar}}.</math>

:Las fuerzas de arrastre sobre el camión no hacen potencia porque el camión está parado en esta referencia. La de las fuerzas de arrastre sobre el bloque se ha calculado en el [[E1. Teorema de la energía: versión diferencial#✏️ ✏️ EJEMPLO E1-3.2: bloque deslizando sobre un camión|'''ejemplo E1-3.2''']]. La potencia de la fricción y de los pares de frenada es la misma de antes ya que no depende de la referencia.

:<math>\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{sist}}=\dot{\Ws}_\mathrm{camión}^{\mathrm{enllaç}}+\dot{\Ws}^{\mathrm{fricción}}+\dot{\Ws}^{\Gamma}+\dot{\Ws}_\mathrm{camión}^{\mathrm{ar} \rightarrow \mathrm{bloque}}= \dot{\Ws}_\mathrm{camión}^{\mathrm{enlace}} -\mu\ms\gs\vs'-2\Gamma\frac{\vs}{\rs} +\ms\as\vs'.</math>

:Las descripciones cinemática y dinámica de la rueda son:

[[Archivo:ExE1-3-3-4-esp.png|thumb|center|430px|link=]]

:<math>\dot{\Ws}_\mathrm{camión}^{\mathrm{enlace}}=2\Bigr[(\leftarrow \Ts)+(\uparrow \Ns)\Bigr]\cdot(\leftarrow\vs)=2\Ts\vs.</math>

La fuerza tangencial de enlace se puede calcular a partir del TMC aplicado al centro '''C''' de la rueda. Como no tiene masa: <math>\sum\overline{\Ms}_\mathrm{ext}(\Cs)=\overline{0} \quad \Rightarrow \quad (\odot \Gamma)+(\otimes \Ts\rs)=0\quad \Rightarrow \quad \Ts=\Gamma/\rs.</math>

:Por tanto: <math>\dot{\Ws}_\mathrm{camió}^{\rightarrow \mathrm{sist}} =2\frac{\Gamma}{\rs}\vs-\mu\ms\gs\vs'-2\Gamma\frac{\vs}{\rs}+\ms\as\vs'=\ms(\as-\mu\gs)\vs'.</math>

:Por otro lado, como se ha visto en el [[E1. Teorema de la energía: versión diferencial#✏️ ✏️ EJEMPLO E1-3.2: bloque deslizando sobre un camión|'''ejemplo E1-3.2''']]:

:<math>\Ts_\mathrm{camión}^\mathrm{sist}= \Ts_\mathrm{camión}^\mathrm{bloque}=\frac{1}{2}\ms\vs_\mathrm{camión}^2(\mathrm{bloque}) \quad \Rightarrow \quad \dot{\Ts}_\mathrm{camión}^\mathrm{sist}= \dot{\Ts}_\mathrm{camión}^\mathrm{bloque}=\ms\vs'(\as-\mu\gs).</math>

:El balance de potencias es consistente:

:<math>\left.\begin{array}{l}
\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{sist}}= \ms(\as-\mu\gs)\vs'\\
\dot{\Ts}^\mathrm{sist}_\mathrm{camión}=\ms(\as-\mu\gs)\vs'
\end{array}\right\} \Rightarrow\dot{\Ws}_\mathrm{camión}^{\rightarrow \mathrm{sist}}= \dot{\Ts}^\mathrm{sist}_\mathrm{camión}</math>

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==E1.4 Balance de potencias para un sistema multisólido: cálculo directo y cálculo indirecto==

El '''cálculo directo''' (a partir de su definición) de la potencia asociada a una interacción (una fuerza sobre un punto '''P''' o un momento sobre un sólido rígido S) exige el conocimiento de la interacción (fuerza o momento):

<math>\dot{\Ws}_\Rs^{\Fs\rightarrow \Ps}=\overline{\Fs}_{\rightarrow\Ps} \cdot \vel{P}{R} \quad , \quad \dot{\Ws}_\Rs^{\Ms\rightarrow \Ss}=\overline{\Ms}_{\rightarrow\Ss} \cdot \velang{S}{R} </math>

Cuando no se trata de una interacción formulable, hay que recurrir a los teoremas vectoriales para determinarla.

El '''cálculo indirecto''' es una alternativa que no necesita recurrir a estos teoremas. Se basa en la evaluación de un balance de potencias para un sistema y una referencia adecuados de manera que la potencia a determinar sea el único término desconocido de la ecuación (el único que no se puede calcular de manera directa). Cuando el balance incluye más de un término desconocido, se pueden combinar varios balances (para distintos sistemas y referencias) hasta tener un número suficiente de ecuaciones que permitan la determinación de todas las potencias desconocidas.

====✏️ EJEMPLO E1-4.1: polea frenada====
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{|
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:[[Archivo:ExE1-4-1-1-esp.png|thumb|left|150px|link=]]
|Un bloque de masa m cuelga de un hilo inextensible atado a un punto de la periferia de una polea de radio r. Entre soporte (fijo al suelo) y polea actúa un freno que garantiza que el bloque baja a velocidad constante respecto al suelo. Se trata de calcular la potencia del freno.

|}

:Càlculo directo: <math>\dot{\Ws}^{\mathrm{freno}}= \overline{\Gamma}^{\rightarrow \mathrm{polea}}\cdot\velang{polea}{T}=\left(\odot\Gamma\right)\cdot\left( \otimes \frac{\vs_0}{\rs}\right)=-\Gamma \frac{\vs_0}{\rs}.</math>

:El par de frenada no es un dato, y hay que recurrir al TMC para calcularlo.

:<math>\left.\begin{array}{l}
\text{SISTEMA: polea + bloque}\\
\text{TMC en } \Os
\end{array}\right\} \sum\overline{\Ms}_\mathrm{ext}(\Os)=\dot{\overline{\Hs}}_\mathrm{RTO}(\Os)</math>

:El momento cinético proviene de las dos piezas. Ya que '''O''' no es fijo al bloque, hay que hacer descomposición baricéntrica para calcular el momento cinético del bloque:

:<math>\overline{\Hs}_\mathrm{RTO=T} (\Os) =\overline{\Hs}_\Ts^\mathrm{politja}(\Os) + \overline{\Hs}_\Ts^\mathrm{bloc}(\Os)= \overline{\Hs}_\Ts^\mathrm{politja}(\Os) + \overline{\Hs}_\Ts^\mathrm{bloc}(\Gs_\mathrm{bloc})+ \overline{\Hs}_\Ts^{\oplus\mathrm{bloc}}(\Os) = </math>

:<math>\hspace{2.3cm}=\Is_\mathrm{O}^\mathrm{polea} \velang{polea}{T}+\Is_{\Gs_\mathrm{bloque}}^\mathrm{bloque}+\OGvec_\mathrm{bloque}\times \ms\overline{\vs}_\Ts(\Gs_\mathrm{bloque})=\left(\otimes \Is_\mathrm{O}\frac{\vs_0}{\rs}\right)+(\otimes \ms\vs_0\rs).</math>

:[[Archivo:ExE1-4-1-2-esp.png|thumb|right|180px|link=]]
:Se trata de un momento cinético constante en dirección y valor. Por tanto:

:<math>\dot{\overline{\Hs}}_\mathrm{RTO}(\Os)=0 \quad \Rightarrow \quad \sum{\overline{\Ms}}_\mathrm{ext}(\Os)=(\odot\Gamma)+(\otimes \ms\gs\rs) =0\quad \Rightarrow \quad \Gamma= \ms\gs\rs</math>.

:Finalmente: <math>\dot{\Ws}^{\mathrm{freno}}= -\Gamma \frac{\vs_0}{\rs} = -\ms\gs\vs_0.</math>

:Cálculo indirecto:

:<math>\left.\begin{array}{l}
\text{SISTEMA: polea + soporte + bloque}\\
\text{REF: suelo(Gal)}
\end{array}\right\} \quad \dot{\Ts}^\mathrm{sist}_\Ts=\dot{\Ws}^\mathrm{ext}_\Ts+ \dot{\Ws}^\mathrm{int}</math>

:<math>\left.\begin{array}{l}
\dot{\Ts}^\mathrm{sist}_\Ts=0\\
\dot{\Ws}^\mathrm{ext}_\Ts=\left(\downarrow \ms\gs\right)\cdot\left(\downarrow \vs_0\right)= \ms\gs\vs_0\\
\dot{\Ws}^\mathrm{int}=\dot{\Ws}^{\mathrm{freno}}
\end{array}\right\} \quad \Rightarrow \quad \dot{\Ws}^{\mathrm{freno}} =-\ms\gs\vs_0</math>



====✏️ EJEMPLO E1-4.2: mono y plátanos====
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:
{|
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:[[Archivo:ExE1-4-2-1-esp.png|thumb|left|180px|link=]]
|Unos plátanos de masa <math>(2/3)\ms</math> cuelgan de una cuerda inextensible que se apoya sin deslizar sobre una polea de radio r y masa despreciable, articulada respecto al suelo. En el otro extremo de la cuerda se agarra un mono de masa m. La polea gira con velocidad angular constante <math>\Omega_0</math> respecto al suelo bajo la acción de un motor. El mono se mueve respecto a la cuerda sin que sus manos se froten contra la cuerda, y de manera que la distancia de su centro de masas al suelo se mantiene constante. Se trata de calcular la potencia del motor.
|}

:Cálculo directo: <math>\dot{\Ws}^{\mathrm{freno}}= \overline{\Gamma}^{\rightarrow \mathrm{polea}}\cdot\velang{polea}{T}=\left(\odot\Gamma\right)\cdot\left( \odot \Omega_0\right)=\Gamma \Omega_0.</math>

:El par motor no es un dato, y hay que recurrir al TMC para calcularlo.

:<math>\left.\begin{array}{l}
\text{SISTEMA: polea + mono + plátanos}\\
\text{TMC en } \Os
\end{array}\right\} \sum\overline{\Ms}_\mathrm{ext}(\Os)=\dot{\overline{\Hs}}_\mathrm{RTO}(\Os)</math>

:El momento cinético proviene del mono y de los plátanos. En primera aproximación, se puede considerar la masa del mono concentrada en su tronco y su cabeza, de manera que no gira y su velocidad respecto al suelo es nula. Teniendo en cuenta que el punto '''O''' no se mueve respecto al mono pero sí respecto a los plátanos:

:<math>\overline{\Hs}_\mathrm{RTO=T} (\Os) =\overline{\Hs}_\Ts^\mathrm{mono}(\Os) + \overline{\Hs}_\Ts^\mathrm{plátanos}(\Os)= \overline{\Hs}_\Ts^\mathrm{mono}(\Os) + \overline{\Hs}_\Ts^\mathrm{plátanos}(\Gs_\mathrm{plát})+ \OGvec_\mathrm{plát} \times \frac{2}{3} \ms \overline{\vs}_\Ts(\Gs_\mathrm{plát})= </math>

:<math>\hspace{2.3cm}=\OGvec_\mathrm{plát}\times \frac{2}{3} \ms \overline{\vs}_\Ts(\Gs_\mathrm{plát}) = \Bigr[(\rightarrow \rs)+(\downarrow \xs)\Bigr] \times\frac{2}{3} \ms \left(\downarrow \rs\Omega_0\right)=\left(\otimes \frac{2}{3}\ms\rs^2\Omega_0\right).</math>

:Se trata de un momento cinético constante en dirección y valor. Por tanto:

:<math>\dot{\overline{\Hs}}_\mathrm{RTO}(\Os)=0 \quad \Rightarrow \quad \sum{\overline{\Ms}}_\mathrm{ext}(\Os)=(\odot\Gamma)+(\otimes \ms\gs\rs) + \left(\otimes \frac{2}{3}\ms\gs\rs\right)=0\quad \Rightarrow \quad \Gamma= \frac{5}{3}\ms\gs\rs</math>.

:Finalmente: <math>\dot{\Ws}^{\mathrm{motor}}= \Gamma \Omega_0 = \frac{5}{3}\ms\gs\rs\Omega_0.</math>

:Cálculo indirecto

:Hay que tener presente que es un problema con dos tipos de actuadores: el motor y la musculatura del mono. Hay diversos modelos biomecánicos para formular las fuerzas de los músculos, pero quedan fuera del ámbito de este curso. Ya que el mono se mueve, sus músculos desarrollan una potencia no nula.

:<math>\left.\begin{array}{l}
\text{SISTEMA: polea + mono + plátano}\\
\text{REF: suelo(Gal)}
\end{array}\right\} \quad \dot{\Ts}^\mathrm{sist}_\Ts=\dot{\Ws}^\mathrm{ext}_\Ts+ \dot{\Ws}^\mathrm{int}</math>

:<math>\left.\begin{array}{l}
\dot{\Ts}^\mathrm{sist}_\Ts=0\\
\dot{\Ws}^\mathrm{ext}_\Ts=\left(\downarrow \frac{2}{3}\ms\gs\right)\cdot\left(\downarrow \rs\Omega_0\right)= \frac{2}{3}\ms\gs\rs\Omega_0\\
\dot{\Ws}^\mathrm{int}=\dot{\Ws}^{\mathrm{musc}}+\dot{\Ws}^{\mathrm{motor}}
\end{array}\right\} \quad \Rightarrow \quad \dot{\Ws}^{\mathrm{musc}} + \dot{\Ws}^{\mathrm{motor}}=-\frac{2}{3}\ms\gs\rs\Omega_0</math>

:Este balance proporciona una ecuación con dos incógnitas. Se puede plantear un segundo balance para un sistema donde una de las dos no aparezca: un sistema sin el mono o un sistema sin el motor. Si se hace un balance de potencias para el sistema (mono + trozo de cuerda), las fuerzas externas a considerar son el peso y la tensión de la cuerda. Si se sigue trabajando en la referencia del suelo, esta tensión (desconocida) hará potencia no nula ya que está aplicada a un punto de velocidad ascendente.
:[[Archivo:ExE1-4-2-2-cat-esp.png|thumb|right|330px|link=]]

:Para evitar que aparezca un término asociado a la tensión de la cuerda, se puede trabajar en la referencia del trozo de cuerda que se ha considerado, que al tener velocidad ascendente constante es galileana. En esta referencia, el mono baja con velocidad constante:

:<math>\left.\begin{array}{l}
\text{REL: trozo de cuerda}\\
\text{AB: suelo}
\end{array}\right\}</math>

:<math>\overline{\vs}_\mathrm{REL}(\mathrm{mono})=\overline{\vs}_\mathrm{ar}(\mathrm{mono})=-\overline{\vs}_\mathrm{ar}(\mathrm{mono})=(\downarrow \rs\Omega_0)</math>

:<math>\left.\begin{array}{l}
\text{SISTEMA: mono+ trozo de cuerda}\\
\text{REF: trozo de cuerda(Gal)}
\end{array}\right\} \quad \dot{\Ts}^\mathrm{sist}_\mathrm{cuerda} =\dot{\Ws}^\mathrm{ext}_\mathrm{cuerda}+ \dot{\Ws}^\mathrm{int}</math>

:<math>\left.\begin{array}{l}
\dot{\Ts}^\mathrm{sist}_\mathrm{cuerda}=0\\
\dot{\Ws}^\mathrm{ext}_\mathrm{cuerda}=\left(\downarrow \ms\gs\right)\cdot\left(\downarrow \rs\Omega_0\right)= \ms\gs\rs\Omega_0\\
\dot{\Ws}^\mathrm{int}=\dot{\Ws}^{\mathrm{musc}}
\end{array}\right\} \quad \Rightarrow \quad \dot{\Ws}^{\mathrm{musc}} =-\ms\gs\rs\Omega_0</math>

:Combinando este resultado con el del balance anterior:

:<math>\dot{\Ws}^\mathrm{motor}=-\frac{2}{3}\ms\gs\rs\Omega_0-\dot{\Ws}^\mathrm{musc}=-\frac{5}{3}\ms\gs\rs\Omega_0</math>



© Universitat Politècnica de Catalunya. [[Mecánica:Derechos de autor |Todos los derechos reservados]]
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[[D8. Conservaciones|<<< D8. Conservaciones]]

[[E2. Teorema de la energía: versión integrada|E2. Teorema de la energía: versión integrada>>>]]
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Main Page

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Bros: /* ENERGETICS */

[[File:Segwaywiki.png|right|top|200px|link=]]
This website is specially built to complement the learning of the '''Mechanics''' course in the bachelor's degrees of '''[https://etseib.upc.edu/en Barcelona School of Industrial Engineering (ETSEIB)]''' of the '''[https://www.upc.edu/en Polytechnic University of Catalonia (UPC) · BarcelonaTech]'''.

This aims to be an '''accessible''' and '''interactive''' '''open tool'''. It's development started on 2022 and it gathers more than 50 years of teaching experience. It's content is organized in brief units which contain the fundamental concepts and some fully worked-out examples. Some simple mathematical proofs are included, but the longer or complex ones are refered to biblography.

The contentent is focused on '''general space movement of rigid bodies and muli-body systems''', but '''particles''' are also considered. Dynamics formulation is vectorial, due to the relevance of the force vector in mechanical engineering. The last units are an introduction to energetics.


'''About the status of the site''':

This project has been developed with limited resources, both technical and human. Nowadays, the server presents some issues, so in case any error may appear, we kindly invite the users to refresh the page and continue enjoying the content 😊.

The best experience will be through a computer or a tablet 📵.

The site is still under construction and some interactive resources and videos are still to be uploaded. Also, the Dynamics and Energetics blocks are still to be published. Having said that, it already is a good tool to help in the process of learning Mechanics 🎯.


© Universitat Politècnica de Catalunya. [[Mechanics:Copyrights |All rights reserved]]
__NOTOC__
------------
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===[[Introduction]]===
:: [[Introduction#I.1 What is mechanics?|I.1 What is mechanics?]]
:: [[Introduction#I.2 Models for material objects|I.2 Models for material objects]]
:: [[Introduction#I.3 Limitations of Newtonian mechanics|I.3 Limitations of Newtonian mechanics]]
:: [[Introduction#I.4 Reference frame|I.4 Reference frame]]

===[[Vector calculus]]===
::[[Vector calculus#V.1 Geometric representation of a vector|V.1 Geometric representation of a vector]]
::[[Vector calculus#V.2 Operations between vectors with geometric representation|V.2 Operations between vectors with geometric representation]]
::::[[Vector calculus#Instantaneous operations: addition, scalar product, vector product|Instantaneous operations: addition, scalar product, vector product]]
::::[[Vector calculus#Operations along time: time derivative|Operations along time: time derivative]]
::[[Vector calculus#V.3 Analytical representation of a vector|V.3 Analytical representation of a vector]]
::[[Vector calculus#V.4 Operations between vectors with analytical representation|V.4 Operations between vectors with analytical representation]]
::::[[Vector calculus#Instantaneous operations: addition, scalar product, vector product|Instantaneous operations: addition, scalar product, vector product]]
::::[[Vector calculus#Operations along time: time derivative|Operations along time: time derivative]]

==KINEMATICS==
===[[C1. Configuration of a mechanical system]]===
:: [[C1. Configuration of a mechanical system#C1.1 Position of a particle|C1.1 Position of a particle]]
:: [[C1. Configuration of a mechanical system#C1.2 Configuration of a rigid body|C1.2 Configuration of a rigid body]]
:: [[C1. Configuration of a mechanical system#C1.3 Orientation of a rigid body with planar motion|C1.3 Orientation of a rigid body with planar motion]]
:: [[C1. Configuration of a mechanical system#C1.4 Orientation of a rigid body moving in space|C1.4 Orientation of a rigid body moving in space]]
::::[[C1. Configuration of a mechanical system#Rotations about fixed axes|Rotations about fixed axes]]
::::[[C1. Configuration of a mechanical system#Euler rotations|Euler rotations]]
:: [[C1. Configuration of a mechanical system#C1.5 Independent coordinates|C1.5 Independent coordinates]]

===[[C2. Movement of a mechanical system]]===
:: [[C2. Movement of a mechanical system#C2.1 Velocity of a particle|C2.1 Velocity of a particle]]
:: [[C2. Movement of a mechanical system#C2.2 Acceleration of a particle|C2.2 Acceleration of a particle]]
:: [[C2. Movement of a mechanical system#C2.3 Intrinsic components of the acceleration|C2.3 Intrinsic components of the acceleration]]
:: [[C2. Movement of a mechanical system#C2.4 Angular velocity of a rigid body|C2.4 Angular velocity of a rigid body]]
::::[[C2. Movement of a mechanical system#Simple rotation|Simple rotation]]
::::[[C2. Movement of a mechanical system#Rotation in space|Rotation in space (Rotacions d'Euler)]]
:: [[C2. Movement of a mechanical system#C2.5 Angular acceleration of a rigid body|C2.5 Angular acceleration of a rigid body]]
:: [[C2. Movement of a mechanical system#C2.6 Particle kinematics versus rigid body kinematicsrígid|C2.6 Particle kinematics versus rigid body kinematics]]
:: [[C2. Movement of a mechanical system#C2.7 Degrees of freedom of a mechanical system|C2.7 Degrees of freedom of a mechanical system]]
:: [[C2. Movement of a mechanical system#C2.8 Usual constraints in mechanical systems|C2.8 Usual constraints in mechanical systems]]
:: [[C2. Movement of a mechanical system#C2.E General examples|C2.E General examples]]

===[[C3. Composition of movements]]===
:: [[C3. Composition of movements#C3.1 Composition of velocities|C3.1 Composition of velocities]]
:: [[C3. Composition of movements#C3.2 Composition of accelerations|C3.2 Composition of accelerations]]
:: [[C3. Composition of movements#C3.3 Composition versus time derivative|C3.3 Composition versus time derivative]]
:: [[C3. Composition of movements#C3.E General examples|C3.E General examples]]

===[[C4. Rigid body kinematics]]===
:: [[C4. Rigid body kinematics#C4.1 Velocity distribution|C4.1 Velocity distribution]]
:: [[C4. Rigid body kinematics#C4.2 Accelerations distribution|C4.2 Accelerations distribution]]
:: [[C4. Rigid body kinematics#C4.3 Geometry of the velocity distribution: Instantaneous Screw Axis (ISA)|C4.3 Geometry of the velocity distribution: Instantaneous Screw Axis (ISA)]]
::[[C4. Rigid body kinematics#C4.4 Fixed axode and moving axode|C4.4 Fixed axode and moving axode]]
::[[C4. Rigid body kinematics#C4.E General examples|C4.E General examples]]

===[[C5. Rigid body kinematics: planar motion]]===
:: [[C5. Rigid body kinematics: planar motion#C5.1 Instantaneous Center of Rotation (ICR)|C5.1 Instantaneous Center of Rotation (ICR)]]
:: [[C5. Rigid body kinematics: planar motion#C5.2 Examples|C5.2 Examples]]
:: [[C5. Rigid body kinematics: planar motion#C5.3 Introduction to vehicle kinematics|C5.3 Introduction to vehicle kinematics]]
:: [[C5. Rigid body kinematics: planar motion#C5.E General examples|C5.E General examples]]

==DYNAMICS==

===[[D1. Foundational laws of Newtonian dynamics#|D1. Foundational laws of Newtonian dynamics]]===

::[[D1. Foundational laws of Newtonian dynamics#D1.1 Galilean reference frames|D1.1 Galilean reference frames]]
::[[D1. Foundational laws of Newtonian dynamics#D1.2 Galileo’s Principle of Relativity|D1.2 Galileo’s Principle of Relativity]]
::[[D1. Foundational laws of Newtonian dynamics#D1.3 Newton’s Principle of Determinacy|D1.3 Newton’s Principle of Determinacy]]
::[[D1. Foundational laws of Newtonian dynamics#D1.4 Newton’s first law (inertia law)|D1.4 Newton’s first law (inertia law)]]
::[[D1. Foundational laws of Newtonian dynamics#D1.5 Newton’s second law (fundamental law of dynamics)|D1.5 Newton’s second law (fundamental law of dynamics)]]
::[[D1. Foundational laws of Newtonian dynamics#D1.6 Newton’s third law (action-reaction principle)|D1.6 Newton’s third law (action-reaction principle)]]
::[[D1. Foundational laws of Newtonian dynamics#D1.7 Particle dynamics in non Galilean reference frames|D1.7 Particle dynamics in non Galilean reference frames]]

===[[D2. Interaction forces between particles#|D2. Interaction forces between particles]]===
::[[D2. Interaction forces between particles#D2.1 Kinematic dependence of interaction forces|D2.1 Kinematic dependence of interaction forces]]
::[[D2. Interaction forces between particles#D2.2 Classification of interaction forces|D2.2 Classification of interaction forces]]
::[[D2. Interaction forces between particles#D2.3 Gravitational attraction|D2.3 Gravitational attraction]]
::[[D2. Interaction forces between particles#D2.4 Interaction through springs|D2.4 Interaction through springs]]
::[[D2. Interaction forces between particles#D2.5 Interaction through dampers|D2.5 Interaction through dampers]]
::[[D2. Interaction forces between particles#D2.6 Interaction through actuators|D2.6 Interaction through actuators]]
::[[D2. Interaction forces between particles#D2.7 Constraint interactions|D2.7 Constraint interactions]]
::[[D2. Interaction forces between particles#D2.8 Friction|D2.8 Friction]]

===[[D3. Interactions between rigid bodies#|D3. Interactions between rigid bodies]]===
::[[D3. Interactions between rigid bodies#D3.1 Torsor associated with a system of forces|D3.1 Torsor associated with a system of forces]]
::[[D3. Interactions between rigid bodies#D3.2 Gravitational attraction|D3.2 Gravitational attraction]]
::[[D3. Interactions between rigid bodies#D3.3 Interaction through linear and torsion springs and dampers|D3.3 Interaction through linear and torsion springs and dampers]]
::[[D3. Interactions between rigid bodies#D3.4 Direct constraint interactions|D3.4 Direct constraint interactions]]
::[[D3. Interactions between rigid bodies#D3.5 Indirect constraint interactions: Constraint Auxiliary Elements (CAE)|D3.5 Indirect constraint interactions: Constraint Auxiliary Elements (CAE)]]
::[[D3. Interactions between rigid bodies#D3.6 Interaccions per mitjà d’actuadors lineals i rotacionals|D3.6 Interaction through linear and rotatory actuators]]

===[[D4. Vector theorems#|D4. Vector theorems]]===
::[[D4. Vector theorems#D4.1 Linear Momentum Theorem (LMT) in Galilean reference frames|D4.1 Linear Momentum Theorem (LMT) in Galilean reference frames]]
::[[D4. Vector theorems#D4.2 LMT: application examples|D4.2 LMT: application examples]]
::[[D4. Vector theorems#D4.3 Linear Momentum Theorem (LMT) in non Galilean reference frames|D4.3 Linear Momentum Theorem (LMT) in non Galilean reference frames]]
::[[D4. Vector theorems#D4.4 Angular Momentum Theorem (AMT): general formulation|D4.4 Angular Momentum Theorem (AMT): general formulation]]
::[[D4. Vector theorems#D4.5 Angular Momentum Theorem (AMT): particular formulations|D4.5 Angular Momentum Theorem (AMT): particular formulations]]
::[[D4. Vector theorems#D4.6 AMT: application examples|D4.6 AMT: application examples]]
::[[D4. Vector theorems#D4.7 Dynamics of Constraint Auxiliary Elements|D4.7 Dynamics of Constraint Auxiliary Elements]]
::[[D4. Vector theorems#D4.8 Barycentric decomposition of the angular momentum|D4.8 Barycentric decomposition of the angular momentum]]

===[[D5. Mass distribution#|D5. Mass distribution]]===
::[[D5. Mass distribution#D5.1 Centre of masses|D5.1 Centre of masses]]
::[[D5. Mass distribution#D5.2 Inertia tensor|D5.2 Inertia tensor]]
::[[D5. Mass distribution#D5.3 Some relevant properties of the inertia tensor|D5.3 Some relevant properties of the inertia tensor]]
::[[D5. Mass distribution#D5.4 Steiner’s Theorem|D5.4 Steiner’s Theorem]]
::[[D5. Mass distribution#D5.5 Change of vector basis|D5.5 Change of vector basis]]

===[[D6. Examples of 2D dynamics#|D6. Examples of 2D dynamics]]===
::[[D6. Examples of 2D dynamics#D6.1 2D kinematics and 2D dynamics|D6.12D kinematics and 2D dynamics]]
::[[D6. Examples of 2D dynamics#D6.2 Free-body diagram (FBD) and roadmap|D6.2 Free-body diagram (FBD) and roadmap]]
::[[D6. Examples of 2D dynamics#D6.3 Examples with just one rigid body|D6.3 Examples with just one rigid body]]
::[[D6. Examples of 2D dynamics#D6.4 General diagram of interactions (GDI)|D6.4 General diagram of interactions (GDI)]]
::[[D6. Examples of 2D dynamics#D6.5 Examples of multibody systems|D6.5 Examples of multibody systems]]

===[[D7. Examples of 3D dynamics#|D7. Examples of 3D dynamics]]===
::[[D7. Examples of 3D dynamics#D7.1 Analysis of the equations of motion|D7.1 Analysis of the equations of motion]]
::[[D7. Examples of 3D dynamics#D7.2 General examples|D7.2 General examples]]

===[[D8. Conservation of dynamic magnitudes#|D8. Conservation of dynamic magnitudes]]===
::[[D8. Conservation of dynamic magnitudes#D8.1 Examples|D8.1 Examples]]

==ENERGETICS==
===[[E1. Work-Energy Theorem: differential form#|E1. Work-Energy Theorem: differential form]]===
::[[E1. Work-Energy Theorem: differential form#E1.1 Power balance in a system of particles|E1.1 Power balance in a system of particles]]
::[[E1. Work-Energy Theorem: differential form#E1.2 Power of an action-reaction pair|E1.2 Power of an action-reaction pair]]
::[[E1. Work-Energy Theorem: differential form#E1.3 Power of a system of forces on a rigid body|E1.3 Power of a system of forces on a rigid body]]
::[[E1. Work-Energy Theorem: differential form#E1.4 Power balance in a multibody system: direct and indirect calculation|E1.4 Power balance in a multibody system: direct and indirect calculation]]

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==Authors==

<center><gallery>
File:Autors Ana.png|alt=Ana Barjau Condomines|Ana Barjau Condomines|link=https://directori.upc.edu/directori/dadesPersona.jsp?id=1001000
File:Lluis Ros2.png|alt=Lluís Ros Giralt|Lluís Ros Giralt|link=https://directori.upc.edu/dadesPersona/1002527
File:Autors Ernest4.jpg|alt=Ernest Bosch Soldevila|Ernest Bosch Soldevila|link=https://www.linkedin.com/in/ernest-bosch-soldevila/
</gallery></center>

::'''Ilustrations:'''
<center><gallery>
File:Autors_Joaquim.png|alt=Joaquim Agulló i Batlle|Joquim Agulló i Batlle|link=https://www.agullobatlle.cat/
</gallery></center>

::'''Collaborators:'''
<center><gallery>
File:Autors_Daniel.png|alt=Daniel Clos Costa|Daniel Clos Costa|link=https://directori.upc.edu/directori/dadesPersona.jsp?id=1002252
File:Autors_Rosa.png|alt=Rosa Pàmies Vilà|Rosa Pàmies Vilà|link=https://directori.upc.edu/directori/dadesPersona.jsp?id=1066910
File:Autors_Albert.png|alt=Albert Peiret Giménez|Albert Peiret Giménez|link=https://directori.upc.edu/directori/dadesPersona.jsp?id=1115007
File:Autors_Javier.png|alt=Javier Sistiaga Vidal-Ribas|Javier Sistiaga Vidal-Ribas|link=https://directori.upc.edu/directori/dadesPersona.jsp?id=1114855
</gallery></center>

::'''Editing and interactive animations:'''
<center><gallery>
File:Autors_Arnau.png|alt=Arnau Marzábal Gatell|Arnau Marzábal Gatell|link=https://www.linkedin.com/in/arnau-marzabal/
File:Autors_Berta.png|alt=Berta Ros Blanco|Berta Ros Blanco|link=https://www.linkedin.com/in/berta-ros/
File:Anna Pons.png|alt=Ana Pons Ferrer|Ana Pons Ferrer|link=https://www.linkedin.com/in/anaponsferrer/
File:Jana.jpg|alt=Jana Bonet|Jana Bonet Fernández|link=https://www.linkedin.com/in/jana-bonet-fern%C3%A1ndez-980aa51b9
File:Gemma3.jpg|alt=Gemma Izquierdo|Gemma Izquierdo Santamaria|link=https://es.linkedin.com/in/gemma-izquierdo-santamaria-b5a8ba240
</gallery></center>

<center>
[https://www.youtube.com/channel/UCqWvnHTViRPI1wHlUQXqH-Q '''Mechanics Lab'''] ''' - ''' [https://etseib.upc.edu/en '''Barcelona School of Industrial Engineering (ETSEIB)''']

[https://em.upc.edu/en '''Mechanical Engineering Department'''] ''' - ''' [https://www.upc.edu/en '''Polytechic University of Catalonia (UPC) · BarcelonaTech''']
</center>

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==Bibliographic references==
Batlle, J. A., Barjau, A. (2020) “'''Rigid Body Kinematics'''” Cambridge Univerity Press. ISBN: 978-1-108-47907-3

Batlle, J. A., Barjau, A. (2022) “'''Rigid Body Dynamics'''” Cambridge Univerity Press. ISBN: 978-1-108-84213-6

Agulló, J. (2002) “'''Mecànica de la partícula i del sòlid rígid'''" Publicacions OK Punt. ISBN: 84-920850-6-1 (''Disponible en accés obert al [https://www.agullobatlle.cat/activitat-docent '''web de l'autor'''])

Agulló, J. (2000) “'''Mecánica de la partícula i del sólido rígido'''" Publicacions OK Punt. ISBN: 84-920850-5-3 (''Disponible en accés obert al [https://www.agullobatlle.cat/activitat-docent '''web de l'autor'''])

<center>
<gallery>
File:RBK portada.png|alt=Rigid body kinematics|link=https://www.cambridge.org/es/academic/subjects/engineering/engineering-design-kinematics-and-robotics/rigid-body-kinematics?format=HB&isbn=9781108479073
File:RBD portada.png|alt=Rigid body dynamics|link=https://www.cambridge.org/es/academic/subjects/engineering/engineering-design-kinematics-and-robotics/rigid-body-dynamics?format=HB&isbn=9781108842136
File:Llibre verd.png|alt=Mecànica de la partícula i del sòlid rígid|link=https://www.agullobatlle.cat/activitat-docent
File:Llibre vermell.jpg|alt=Mecánica de la partícula i del sólido rígido|link=https://www.agullobatlle.cat/activitat-docent
</gallery>
</center>

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[[File:Logo Lab Mec horitzontal.png|thumb|center|500px|link=https://em.upc.edu/en| ]]