E1. Work-Energy Theorem: differential form

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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 unit E2.

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E1.1 Power balance in a system of particles

Figure E1.1 Forces on a particle of a system with constant matter

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]\displaystyle{ \F{\rightarrow \Ps} + \Fcal{}{\text{ar}\rightarrow\Ps} + \Fcal{}{\text{Cor}\rightarrow\Ps} =\ms_\Ps\acc{P}{R} }[/math]

[math]\displaystyle{ \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]\displaystyle{ \F{\rightarrow \Ps}=\F{\rightarrow \Ps}^\text{int}+\F{\rightarrow \Ps}^\text{ext} }[/math]

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

[math]\displaystyle{ \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]\displaystyle{ \Ts_\Rs(\Ps) }[/math] is the kinetic energy of the particle, and is a function of the mechanical state of P. The scalar product of a force and the velocity of its application point is the force power: [math]\displaystyle{ \vel{P}{R} \cdot \F{\rightarrow \Ps} \equiv \dot{\Ws}_\Rs^{\F{\rightarrow \Ps}} }[/math]. Both scalar quantities [math]\displaystyle{ (\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]\displaystyle{ \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]\displaystyle{ \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 unit E2. This unit focuses on the calculation of powers. 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]\displaystyle{ \dot{\Ws}^{\text{int}} }[/math] term of the previous equation).

✏️ EXAMPLE E1-1.1: free particle

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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.

[math]\displaystyle{ \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.

[math]\displaystyle{ \hspace{0.5cm} }[/math] AB:ground, REL:ring [math]\displaystyle{ \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]\displaystyle{ \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.

[math]\displaystyle{ \hspace{0.5cm} }[/math]AB:ground, REL:cabin [math]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ \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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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]\displaystyle{ \Omega_0 }[/math].

Ground reference frame: Since it is a Galilean reference frame, P is only subjected to interaction forces.

[math]\displaystyle{ \dot{\Ws}_\Es^\Ts=(\nearrow\Ls\dot{\theta})\cdot(\nwarrow\Ts)=0 }[/math] [math]\displaystyle{ \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.

AB:ground, REL:platform [math]\displaystyle{ \vel{P}{REL}=\vel{P}{AB}-\vel{P}{ar}=(\nearrow\Ls\dot{\theta})-(\otimes \Ls \Omega_0 \sin \theta) }[/math] [math]\displaystyle{ \dot{\Ws}_\Es^\Ts\left[(\nearrow\Ls\dot{\theta})+(\odot\Ls\Omega_0\sin\theta)\right]\cdot(\nwarrow\Ts)=0 }[/math] [math]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ \dot{\Ts}_\Rs^\text{syst}=\dot{\Ws}_\Rs^\text{ext} + \dot{\Ws}^\text{int} + \dot{\Ws}_\Rs^\text{ar} }[/math] .

💭 PROOF ➕

[math]\displaystyle{ \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 action-reaction principle, [math]\displaystyle{ \overline{\Fs}_{\Ps \rightarrow \Qs}=-\overline{\Fs}_{\Qs \rightarrow \Ps} }[/math]. Therefore: [math]\displaystyle{ \dot{\Ws}^\text{ac-reac}=\overline{\Fs}_{\Ps \rightarrow \Qs}\cdot \left[ \vel{Q}{R}-\vel{P}{R}\right] }[/math] As [math]\displaystyle{ \overline{\Fs}_{\Ps \rightarrow \Qs} }[/math] has the [math]\displaystyle{ \PQvec }[/math] direction, only the components of the velocities in that direction have to be considered: [math]\displaystyle{ \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]\displaystyle{ \biggl( \vel{Q}{R} \Bigr]_{\Ps\Qs}-\ \vel{P}{R} \Bigr]_{\Ps\Qs}\biggl) }[/math] is the separation velocity [math]\displaystyle{ \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]\displaystyle{ \dot{\rho} }[/math] between the two particles). For the previous drawing, [math]\displaystyle{ \dot{\rho}\gt 0 }[/math]. Since this is an attractive interaction, [math]\displaystyle{ \dot{\Ws}^\text{ac-reac} \lt 0 }[/math].

With a drawing, it is easy to see that:

• attractive interaction and particle approach [math]\displaystyle{ (\dot{\rho}\lt 0) }[/math] : [math]\displaystyle{ \dot{\Ws}^\text{ac-reac} \gt 0 }[/math] ;
• repulsive interaction and particle separation [math]\displaystyle{ (\dot{\rho}\gt 0) }[/math] : [math]\displaystyle{ \dot{\Ws}^\text{ac-reac} \gt 0 }[/math]
• repulsive interaction and particle approach [math]\displaystyle{ (\dot{\rho}\lt 0) }[/math] : [math]\displaystyle{ \dot{\Ws}^\text{ac-reac} \lt 0 }[/math]

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]\displaystyle{ \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}\lt 0; }[/math]

• Constraint action-reaction pair between P and Q: from the analytical characterization equation of the constraint forces between two points P and Q belonging to two rigid bodies [math]\displaystyle{ \mathrm{S}_\Ps }[/math] and [math]\displaystyle{ \mathrm{S}_\Qs }[/math], it is easy to see that the associated power is zero:

[math]\displaystyle{ \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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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]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ \mathbf{ICR}^\text{wheel}_\Es }[/math] is point [math]\displaystyle{ \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]\displaystyle{ \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]

Calculation of the friction power: [math]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ \dot{\Ws}_\text{belt}^{\text{fric }\rightarrow \text{wheel}}=(\rightarrow \mu\ms\gs)\cdot(\leftarrow \vs_0)=-\mu\ms\gs\vs_0 , }[/math]

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

[math]\displaystyle{ \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

Figure E1.3: Kinematic torsor of a rigid body in a reference frame R, and forces on different points (dm) of that rigid body

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 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]\displaystyle{ (\overline{\Fs},\overline{\Ms}(\Os)) }[/math]. Taking into account rigid body kinematics (Figure E1.3): [math]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ \dot{\Ws}_\Rs^{\Gamma\rightarrow \mathrm{S}} = \overline{\Gamma} \cdot \velang{S}{R} }[/math].

The power balance for a rigid body is: [math]\displaystyle{ \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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A homogeneous roller, with mass m and radius r, slides down a slope. The friction coefficient between the ground and the roller is [math]\displaystyle{ \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:

[math]\displaystyle{ \hspace{0.5cm}\vel{J}{E}=\vel{C}{T} + \velang{}{E} \times \CJvec= }[/math] [math]\displaystyle{ \hspace{1.7cm}=\left( \rightarrow \vs_\Es \right) + \left( \otimes \Omega_\Es\right) \times \left( \downarrow \rs \right) }[/math] [math]\displaystyle{ \hspace{0.5cm}\vel{J}{E} = \left[\rightarrow (\vs_\Es -\rs\Omega_\Es)\right] }[/math] [math]\displaystyle{ \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]\displaystyle{ \Ns =\ms\gs\cos\beta }[/math]. Therefore:

[math]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ \hspace{0.5cm} }[/math] The result is the same.

✏️ EXAMPLE E1-3.2: block sliding on a truck

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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:

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]\displaystyle{ \Ns=\ms\gs }[/math].

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

[math]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ \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:

[math]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ \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

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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:

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:

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]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ \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 example E1-3.2. The friction and braking torques are the same as before since they do not depend on the reference frame.

[math]\displaystyle{ \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:

[math]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ \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 example E1-3.2:

[math]\displaystyle{ \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]\displaystyle{ \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]

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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]\displaystyle{ \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

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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]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ \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]

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

[math]\displaystyle{ \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]\displaystyle{ \dot{\Ws}^{\mathrm{freno}}= -\Gamma \frac{\vs_0}{\rs} = -\ms\gs\vs_0. }[/math]

Indirect calculation:

[math]\displaystyle{ \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]\displaystyle{ \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

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A bunch of bananas of mass [math]\displaystyle{ (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]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ \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.

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

[math]\displaystyle{ \overline{\vs}_\mathrm{REL}(\mathrm{monkey})=\overline{\vs}_\mathrm{ar}(\mathrm{monkey})=-\overline{\vs}_\mathrm{ar}(\mathrm{monkey})=(\downarrow \rs\Omega_0) }[/math]

[math]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ \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]

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