D4. Vector theorems

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The Vector Theorems are a tool for solving the dynamics of mechanical systems, and are obtained from the fundamental law of dynamics (Newton's second law) and the principle of action and reaction (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 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]\displaystyle{ \qs_i }[/math], with i=1,2,3...), their time evolution is the second time derivatives of these coordinates (accelerations). Their general aspect is:

[math]\displaystyle{ \dot{\qs}_i = f(\qs_j, \dot{\qs}_j, \ddot{\qs}_j, \text{ dynamic parameters, geometric parameters}) }[/math]

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).

• 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 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.

• Constraint forces and movements: the number of unknowns associated with the constraints depends on the description given of them (direct constraints, 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]\displaystyle{ \F{\rightarrow\Ps}=\ms_\Ps\acc{P}{RGal} }[/math], where [math]\displaystyle{ \F{\rightarrow\Ps} }[/math] is the interaction resultant force on P.

Figure D4.1 Forces on a particle of a system with constant matter

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]\displaystyle{ \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:

[math]\displaystyle{ \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]

The term on the left-hand side is the resultant of external forces on the system, and is often written simply as [math]\displaystyle{ \sum\F{ext} }[/math]. The term on the right-hand side can be rewritten as [math]\displaystyle{ \Ms\left[\sum_\Ps\frac{\ms_\Ps}{\Ms}\acc{P}{RGal}\right] }[/math], where M is the total mass of the system [math]\displaystyle{ \left(\Ms\equiv\sum_\Ps \ms_\Ps\right) }[/math]. The term [math]\displaystyle{ \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):

[math]\displaystyle{ \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]

Figure D4.2 Centre of mass (r of inertia) of a ystem with constant matter

In continuous sets of particles (such as a set of rigid or deformable bodies, or fluids), the summation for particles is actually an integral:

[math]\displaystyle{ \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]

Finally, the LMT is written as:

[math]\displaystyle{ \sum\F{ext}=\Ms\:\acc{G}{RGal} }[/math]

This equation is very similar to 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 mechanical state and the external interactions, of the linear momentum of the system (which, in any reference frame R, is defined as [math]\displaystyle{ \sum_\Ps\ms_\Ps\vel{P}{R}=\Ms\vel{G}{R} }[/math]):

[math]\displaystyle{ \sum\F{ext}=\Ms\acc{G}{RGal}=\dert{\Ms\vel{G}{RGal}}{RGal} }[/math]

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 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.

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D4.2 LMT: application examples

✏️ EXAMPLE D4.1: calculation of a constraint force

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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 (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]\displaystyle{ \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]\displaystyle{ (\rightarrow\Fs_{\Qs\rightarrow\Ss})=(3\ms)[\rightarrow\Es_\Ts(\Gs_\Ss)] }[/math].

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:

[math]\displaystyle{ \Fs = (6\ms)\as_\Ts(\Gs)\Rightarrow\as_\Ts(\Gs)=\frac{\Fs}{6\ms}\Rightarrow\Fs_{\Qs\rightarrow\Ss}=3\ms\frac{\Fs}{6\ms}=\frac{\Fs}{2} }[/math]

✏️ EXAMPLE D4.2: initial motion of a system

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The homogeneous blocks are initially at rest on a rough floor, and connected by a spring compressed with tension [math]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ |\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]\displaystyle{ |\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]\displaystyle{ \as_\Ts(2\ms) = 0 }[/math]), but not in the case of the block of mass m:

Application of the LMT to the block of mass m leads to:

[math]\displaystyle{ \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]\displaystyle{ \acc{G}{T}=\frac{2\ms\acc{2m}{T}+\ms\acc{m}{T}}{3\ms}=\frac{0,8}{3}\gs }[/math].

✏️ EXAMPLE D4.3: limit condition

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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]\displaystyle{ \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]\displaystyle{ /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:

SYSTEM: sphere + wedges [math]\displaystyle{ \left(\uparrow 2\Ns\right)+\left[\downarrow(\Ms+2\ms)\gs\right]=\vec{0} }[/math] [math]\displaystyle{ \Rightarrow\Ns=\frac{1}{2}(\Ms+2\ms)\gs }[/math]

SYSTEM: left wedge [math]\displaystyle{ \left(\rightarrow \Ts\right)+\left(\leftarrow\frac{\Ns'}{\sqrt{2}}\right)=\vec{0} }[/math] [math]\displaystyle{ \Rightarrow\Ts=\frac{\Ns'}{\sqrt{2}} }[/math]

SYSTEM: sphere [math]\displaystyle{ \left(\downarrow\Ms\gs\right)+\left(\uparrow 2\frac{\Ns'}{\sqrt{2}}\right)=\vec{0} }[/math] [math]\displaystyle{ \Rightarrow\Ns'=\frac{\Ms\gs}{\sqrt{2}} }[/math]

Combining the last two equations: [math]\displaystyle{ \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]\displaystyle{ \Ts = \Ts_{\text{màx}}=\mu\Ns }[/math]. Taking into account that [math]\displaystyle{ N = \frac{1}{2}(\Ms_{\text{max}}+2\ms)\gs }[/math]:

[math]\displaystyle{ \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]

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D4.3 Linear Momentum Theorem (LMT) in non Galilean reference frames

The LMT in a non-Galilean reference NGal is also obtained from Newton's second law applied to each particle (or each mass differential) of the system in the reference NGal (section D1.7). In principle, then, this equation will contain two inertial forces: the transportation force and the Coriolis force (Figure D4.3):

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

where [math]\displaystyle{ \Fcal{ar}{NGal\rightarrow\Ps} = -\ms_\Ps\acc{P}{ar} = -\ms_\Ps\acc{$\Ps\in\text{NGal}$}{RGal} }[/math] and [math]\displaystyle{ \Fcal{Cor}{NGal\rightarrow\Ps} = -\ms_\Ps\acc{P}{Cor} = -2\ms_\Ps\velang{NGal}{RGal}\times\vel{P}{NGal} }[/math]

Figure D4.3: Inertia forces in a non Galilean reference frame

Summing the equations for all the particles, we obtain: [math]\displaystyle{ \sum\F{ext}+\Fcal{ar}{NGal\rightarrow\Gs}+\Fcal{Cor}{NGal\rightarrow\Gs}=\Ms\acc{G}{NGal} }[/math], where M is the total mass of the system.

💭 Proof ➕

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By the principle of action and reaction, the sum for all particles of the interaction forces leads to:

[math]\displaystyle{ \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 center of mass:

[math]\displaystyle{ \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{ar}{NGal\rightarrow\Ps} }[/math]

[math]\displaystyle{ \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]\displaystyle{ \sum\F{ext} + \Fcal{tr}{NGal\rightarrow\Gs}+\Fcal{Cor}{NGal\rightarrow\Gs} = \Ms\acc{G}{NGal} }[/math]

✏️ EXAMPLE D4.4: vibratory displacement

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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]\displaystyle{ \sum\F{\text{ext}} + \Fcal{tr}{sup\rightarrow\Gs} = \vec{0} (\Fcal{Cor}{sup\rightarrow\Gs}= \vec{0}) }[/math] perque [math]\displaystyle{ \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]\displaystyle{ 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]\displaystyle{ 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]\displaystyle{ \mu\ms\gs }[/math], opposite to the sliding velocity.

Between [math]\displaystyle{ t=0 }[/math] and [math]\displaystyle{ t = 0,4s }[/math], the acceleration of the support relative to the ground is [math]\displaystyle{ \ddot{y} = \frac{0,4\ms/s}{0,4s}=1\ms/s^2 }[/math], therefore the drag force is [math]\displaystyle{ \Fcal{ar}{sup\rightarrow\Gs} = -\ms\acc{$\Gs\in$suport}{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]\displaystyle{ \F{sup\rightarrow bloc} = 0,1(\ms/s^2) }[/math], will counteract the [math]\displaystyle{ \Fcal{ar}{sup\rightarrow\Gs} }[/math] and there will be no motion between the two elements.

Between [math]\displaystyle{ t=0,4s }[/math] and [math]\displaystyle{ t = 0,6s }[/math], the acceleration of the support relative to the ground is 2[math]\displaystyle{ \ddot{y} = -\frac{0,8\ms/s}{0,2s}=-4\frac{\ms}{s^2} }[/math]. Therefore, [math]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ t = 0,6s }[/math], the sliding velocity of the block on to the support is

[math]\displaystyle{ \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]\displaystyle{ \mid\Fcal{tr}{sup\rightarrow G}\mid\lt \mu\Ms\gs }[/math], the horizontal force between support and block is still a friction force [math]\displaystyle{ (\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]\displaystyle{ \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]\displaystyle{ t_f }[/math] when the block stops sliding:

[math]\displaystyle{ \vel{$\Gs,t_f$}{sup} = 0 = \vel{$\Gs,t_i$}{sup} + \acc{$\Gs,t_f$}{sup}(t_f - t_i) }[/math], [math]\displaystyle{ 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]\displaystyle{ \ts = 1,4s }[/math]. The study for the subsequent intervals follows the same steps as that in the interval [math]\displaystyle{ [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.

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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 mechanical state.

As we have already seen when introducing the concept of 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]\displaystyle{ \velang{RTQ}{RGal} = \vec{0} }[/math]. If [math]\displaystyle{ \acc{Q}{RGal}\neq\vec{0} }[/math], this reference frame is not Galilean, and therefore the inertia transportation forces must be taken into account in principle (Figure D4.3).

Figure D4.4: Inertia forces in the Reference Frame that Translates with point Q (RTQ)

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]\displaystyle{ \sum \F{\rightarrow \Ps} + \Fcal{tr}{\mathrm{RTQ}\rightarrow \Ps}=\ms_\Ps\acc{P}{RTQ} }[/math] ,

where [math]\displaystyle{ \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]\displaystyle{ \Fcal{Cor}{\mathrm{RTQ}\rightarrow \Ps}=\overline{0} }[/math] since [math]\displaystyle{ \velang{RTQ}{RGal}=\overline{0} }[/math].

If both sides of the equation are multiplied by [math]\displaystyle{ \QPvec }[/math] and summed for all particles (or mass elements) in the system, the AMT at point Q is obtained:

[math]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ \Ss_i }[/math]), the summation for particles is actually an integral:

[math]\displaystyle{ \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].

💭 Proof ➕

Newton's second law for each element of mass multiplied by [math]\displaystyle{ \QPvec }[/math] is:

[math]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ =\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]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ = \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]\displaystyle{ \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]\displaystyle{ \sum \overline{\mathbf{M}}_\mathrm{ext} (\Qs) - \QGvec \times \Ms \acc{Q}{RGal}=\dot{\overline{\mathbf{H}}}_\mathrm{RTQ}(\Qs) }[/math]

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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]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ \QGvec }[/math].

AMT at a fixed point	AMT at G [math]\displaystyle{ \Gs }[/math]	AMT at a moving point [math]\displaystyle{ \Qs }[/math]
[math]\displaystyle{ \quad \Qs \in \mathrm{RGal} \Rightarrow \Qs \equiv \Os , \mathrm{RTQ} = \mathrm{RGal} \quad }[/math] [math]\displaystyle{ \sum \overline{\mathbf{M}}_\mathrm{ext}(\Os) = \dot{\overline{\mathbf{H}}}_\mathrm{RGal} (\Os) }[/math]	[math]\displaystyle{ \quad \Qs=\Gs \Rightarrow \mathrm{RTQ}=\mathrm{RTG} \quad }[/math] [math]\displaystyle{ \sum \overline{\mathbf{M}}_\mathrm{ext}(\Gs) = \dot{\overline{\mathbf{H}}}_\mathrm{RTG} (\Gs) }[/math]	[math]\displaystyle{ \quad \acc{Q}{RGal} \neq \overline{0}\quad }[/math] [math]\displaystyle{ \quad \sum \overline{\mathbf{M}}_\mathrm{ext}(\Qs) - \QGvec \times \Ms\acc{Q}{RGal}= \dot{\overline{\mathbf{H}}}_\mathrm{RTQ} (\Qs) \quad }[/math]

The angular momentum is not easy to calculate in general, and it is presented in unit D5.

El vector moment cinètic [math]\displaystyle{ \overline{\mathbf{H}}_\mathrm{RTQ} (\Qs) }[/math] no és senzill de calcular en general, i es presenta a la unitat D5.

In problems with planar (2D) kinematics, if the angular momentum [math]\displaystyle{ \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.

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).

The application of the AMT at J is very tricky. It is necessary to specify which point J has been chosen (example C5-1.8): is it point J of rigid body S1 [math]\displaystyle{ (\Js_{\mathrm{S}1}) }[/math], of rigid body S2 [math]\displaystyle{ (\Js_{\mathrm{S}2}) }[/math], or is it the geometric contact point [math]\displaystyle{ (\Js_{\mathrm{geom}}) }[/math]. On the one hand, those three points have different kinematics [math]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ (\Js_{\mathrm{S}1}) }[/math] belongs to S1 and the angular momentum can be calculated from the inertia tensor, [math]\displaystyle{ (\Js_{\mathrm{S}1}) }[/math] and [math]\displaystyle{ (\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.

[math]\displaystyle{ \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]

[math]\displaystyle{ \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]

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[math]\displaystyle{ \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]

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D4.6 AMT: application examples

✏️ EXAMPLE D4.5: static limit condition

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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]\displaystyle{ \mu_\mathrm{Q} }[/math]). We want to calculate the minimum [math]\displaystyle{ \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]\displaystyle{ \mu_\mathrm{Q} }[/math] ). Es tracta de calcular el valor mínim de [math]\displaystyle{ \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:

Since the system is at rest relative to the ground, [math]\displaystyle{ \acc{G}{T}=\overline{0} }[/math] and the total external force must be zero. Therefore:

[math]\displaystyle{ \Ns_\Ps=\Ns_\Qs }[/math] , [math]\displaystyle{ \ms\gs=\Ts_\Qs }[/math]
.

Since motion is about to occur, the tangential constraint force at Q has its maximum possible value: [math]\displaystyle{ \Ts_{\Qs,\mathrm{max}}=\ms\gs=\mu_{\Qs,\mathrm{min}}\Ns_\Qs }[/math]. Since the above equations do not allow the calculation of [math]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ \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]

✏️ EXAMPLE D4.6: Atwood machine

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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]\displaystyle{ (\gs \approx 10 \ms/\ss^2) }[/math]:

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 OBold text and obtain an equation free of constraint unknowns but including [math]\displaystyle{ \ddot{\theta} }[/math]:

[math]\displaystyle{ \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]\displaystyle{ \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 center of mass: [math]\displaystyle{ \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]\displaystyle{ \OGvec_\mathrm{block} \times \overline{\mathbf{v}}_\Ts(\mathrm{block}) }[/math] contributes to the vector product [math]\displaystyle{ \OGvec_\mathrm{block} }[/math]. Thus:

[math]\displaystyle{ \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]\displaystyle{ \dot{\overline{\mathbf{H}}}_\mathrm{T}(\Os)=45\ks\gs(\odot \rs^2\ddot{\theta}) }[/math]

[math]\displaystyle{ \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]

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]\displaystyle{ \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]
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Applying the AMT to each of the blocks provides two more equations:

SYST: 10 kg block

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

✏️ EXAMPLE D4.7: longitudinal dynamics of a vehicle

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El vehicle sense suspensions es mou sobre una carretera rectilínia. La massa de les rodes és negligible comparada amb la de la resta dels elements, i es considera que el seu contacte amb el terra es puntual. Es tracta d’analitzar les forces normals d’enllaç entre terra i rodes. És un problema de dinàmica plana. Les forces externes sobre el vehicle es redueixen al pes i a l’enllaç amb el terra. Si l’acceleració del xassís respecte del terra és una dada, l’aplicació del TQM condueix a:

[math]\displaystyle{ \sum\F{\text{ext}} = \ms\acc{G}{T} \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_\Ts) \end{aligned}\right. }[/math] on [math]\displaystyle{ (\Ns_{\ds\vs}, \Ts_{\ds\vs}) }[/math] i [math]\displaystyle{ (\Ns_{\ds\rs}, \Ts_{\ds\rs}) }[/math] són les forces normal i tangencial que en total reben les dues rodes del davant i les dues rodes del darrere, respectivament. El TMC a G condueix a: [math]\displaystyle{ \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] (el moment cinètic [math]\displaystyle{ \vec{H}_{RTG}(\Gs) }[/math] és permanentment nul ja que les rodes no tenen massa, i la resta del vehicle té moviment nul respecte de la RTG).

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Resolent el sistema d’equacions: [math]\displaystyle{ \Ns_{\ds\vs} = \frac{\ms\gs\Ls_{\ds\rs}}{\Ls_{\ds\vs} + \Ls_{\ds\rs}} - \frac{\ms\as_\Ts\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_\Ts\hs}{\Ls_{\ds\vs} + \Ls_{\ds\rs}}, }[/math]

Aquest resultat és vàlid per a qualsevol valor de [math]\displaystyle{ \as_\Ls }[/math]:

• Situació estàtica ([math]\displaystyle{ \as_\Ts = 0 }[/math]): [math]\displaystyle{ \Ns_{\ds\vs}^{\es\ss\ts} = \frac{\ms\gs\Ls_{\ds\rs}}{\Ls_{\ds\vs}+ \Ls_{\ds\rs}} }[/math], [math]\displaystyle{ \Ns_{\ds\rs}^{\es\ss\ts} = \frac{\ms\gs\Ls_{\ds\vs}}{\Ls_{\ds\vs}+ \Ls_{\ds\rs}} }[/math]; la força normal és més gran a les rodes que tenen l’eix més proper a [math]\displaystyle{ \Gs }[/math].

• Accelerant ([math]\displaystyle{ \as_\Ls \gt 0 }[/math]): [math]\displaystyle{ \Ns_{\ds\vs} = \Ns_{\ds\vs}^{\es\ss\ts} - \frac{\ms|\as_\Ts|\hs}{\Ls_{\ds\vs}+ \Ls_{\ds\rs}} }[/math], [math]\displaystyle{ \Ns_{\ds\rs} = \Ns_{\ds\rs}^{\es\ss\ts} + \frac{\ms|\as_\Ts|\hs}{\Ls_{\ds\vs}+ \Ls_{\ds\rs}} }[/math] ; hi ha un transvasament de força del davant cap al darrere (es carreguen les rodes del darrere i es descarreguen les del davant). Si l’acceleració [math]\displaystyle{ |\as_\Ts| }[/math] és prou elevada, la força normal al davant passa a ser negativa, la qual cosa indica que s’ha perdut el contacte i el vehicle bolca en sentit antihorari. Llavors, [math]\displaystyle{ \Ns_{\ds\vs} = 0 }[/math], el xassís passa a tenir, en principi, acceleració angular, i com a conseqüència el centre de masses [math]\displaystyle{ \Gs }[/math] passa a tenir acceleració vertical.

• Frenant ([math]\displaystyle{ \as_\Ls \lt 0 }[/math]): [math]\displaystyle{ \Ns_{\ds\vs} = \Ns_{\ds\vs}^{\es\ss\ts} - \frac{\ms|\as_\Ts|\hs}{\Ls_{\ds\vs}+ \Ls_{\ds\rs}} }[/math], [math]\displaystyle{ \Ns_{\ds\rs} = \Ns_{\ds\rs}^{\es\ss\ts} - \frac{\ms|\as_\Ts|\hs}{\Ls_{\ds\vs}+ \Ls_{\ds\rs}} }[/math]; hi ha un transvasament de força del darrere cap al davant (es descarreguen les rodes del darrere i es carreguen les del davant). Com en el cas precedent, hi ha un valor crític de [math]\displaystyle{ |\as_\Ts| }[/math] que provoca el bolcament, ara en sentit horari.

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D4.7 Dinàmica dels Sòlids Auxiliars d'Enllaç

L’aplicació dels teoremes vectorials als Sòlids Auxiliars d’Enllaç (SAE) és senzilla ja que els termes de la dreta (variació de quantitat de moviment i variació de moment cinètic) són nuls en ser nul·la la seva massa:

[math]\displaystyle{ \ms_\mathrm{SAE} \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]

Per al cas d’un SAE (sòlid S) connectant dos sòlids rígids S1 i S2 (Figura D4.5), aquestes equacions permeten demostrar que el torsor que actua sobre S1 es pot obtenir a partir d’una equació de caracterització analítica on la cinemàtica del punt [math]\displaystyle{ \Ps }[/math] de caracterització (que ha de pertànyer al sòlid S1) s’avalua a la referència solidària a S2:

[math]\displaystyle{ \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]

(Quan es tracta de velocitats, és important especificar a quin sòlid pertany el punt [math]\displaystyle{ \Ps }[/math]. Això, en canvi, és irrellevant quan es tracta de moments: [math]\displaystyle{ \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]).

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Figura D4.5:Sòlid Auxiliar d’Enllaç entre dos sòlids de massa no nul·la

💭 Demostració ➕

Els teoremes vectorials aplicats sobre el SAE impliquen que els torsors d’enllaç que rep de S1 i S2 en un mateix punt [math]\displaystyle{ \Ps }[/math] han de sumar zero:

[math]\displaystyle{ \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]

Per altra banda, aquests torsors compleixen l’equació de caracterització analítica:

[math]\displaystyle{ \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]

Si es combinen totes les equacions anteriors, s’obté:

[math]\displaystyle{ \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]

Una composició de moviments permet reescriure l’equació anterior de manera més compacta:

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

✏️ Exemple D4.8: dinàmica longitudinal d’un vehicle

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En el vehicle de l’exemple D4.7, es considera que la tracció és posterior. Això vol dir que el motor del vehicle actua entre el xassís i les rodes posteriors, en tant que les del davant només estan sotmeses a interaccions d’enllaç (articulació amb el xassís i contacte amb el terra). Si es tracten les rodes del davant com a SAE, l’anàlisi cinemàtica del xassís respecte del terra per a la caracterització de l’enllaç a través de les rodes del davant condueix a un torsor en el centre de la roda (que és fix al xassís) amb només una component de força:

thumb|center|370px|link= L’existència d’una component horitzontal de força entre les rodes posteriors (necessària per accelerar o frenar el vehicle!) està associada al parell motor. Si s'apliquen els teoremes vectorials a les rodes posteriors, les interaccions externes a tenir en compte són l’enllaç amb el terra, l’articulació amb el xassís i el parell motor [math]\displaystyle{ \Gamma }[/math]  : thumb|center|370px|link=

[math]\displaystyle{ \ms_\mathrm{rodes} \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{TQM }: \hspace{1.4cm} \Ns_\mathrm{dr}=\Ns'_\mathrm{dr} , \Ts_\mathrm{dr}=\Ts'_\mathrm{dr} \\ \mathrm{TMC } \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]

La força tangencial sobre les rodes motrius és directament proporcional al parell motor que se’ls aplica. Cal no oblidar, però, que com a força tangencial d’enllaç, el seu valor està limitat per [math]\displaystyle{ \mu_\mathrm{e}\Ns_\mathrm{dr} }[/math] . Això permet calcular l’acceleració màxima que pot adquirir el vehicle (mentre la roda del davant no perdi contacte amb el terra):

[math]\displaystyle{ \left.\as_\Ts (\mathrm{vehicle})\right]_\mathrm{màx}= \Ts_\mathrm{dr,màx}/\ms \quad , \quad \Ts_\mathrm{dr}=\Gamma / \rs \quad , \quad \Ts_\mathrm{dr} \leq \mu_\mathrm{e}\Ns_\mathrm{dr} }[/math].

En terrenys de baix fregament ( [math]\displaystyle{ \mu_\es }[/math] de valor baix), un motor capaç de subministrar un parell màxim alt és inútil: qui posa límit a l’acceleració és el valor del coeficient de fricció [math]\displaystyle{ \mu_\es }[/math] : [math]\displaystyle{ \left.\as_\Ts (\mathrm{vehicle})\right]_\mathrm{màx}= \mu_\mathrm{e}\Ns_\mathrm{dr} /\ms }[/math] . En terrenys d’alt fregament, si el parell màxim és baix és ell qui posa límit a l’acceleració: [math]\displaystyle{ \left. \as_\Ts (\mathrm{vehicle})\right]_\mathrm{màx}= \Gamma_\mathrm{màx}/\ms \rs }[/math] (si la roda del davant no perd contacte amb el terra).

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D4.8 Descomposició baricèntrica del moment cinètic

La versió del TMC que s’ha presentat permet triar lliurement el punt [math]\displaystyle{ \Qs }[/math] d’aplicació. El criteri per triar-lo es basa en el que es vulgui investigar (una força d’enllaç, una equació del moviment...). A la unitat D6 es discuteix aquest criteri. En alguns casos, pot ser interessant triar un punt [math]\displaystyle{ \Qs }[/math] que no pertanyi a cap element material del sistema. Llavors, és útil referir el càlcul del moment cinètic [math]\displaystyle{ \H{Q}{}{RTQ} }[/math] al moment cinètic al centre de masses [math]\displaystyle{ \Gs }[/math] del sistema, [math]\displaystyle{ \H{G}{}{RTG} }[/math]. Això es coneix amb el nom de descomposició baricèntrica del moment cinètic:

[math]\displaystyle{ \H{Q}{}{RTQ} = \H{G}{}{RTG} + \H{G}{\oplus}{RTQ} }[/math],

on el superíndex [math]\displaystyle{ \oplus }[/math] indica que el moment cinètic s’ha de calcular com si el sistema s’hagués reduït a una partícula concentrada a [math]\displaystyle{ \Gs }[/math] amb massa igual a la massa total del sistema (M):

[math]\displaystyle{ \H{G}{\oplus}{RTQ} = \QGvec\times\Ms\vel{G}{RTQ} }[/math]

💭 Demostració ➕

La definició de moment cinètic és: [math]\displaystyle{ \H{Q}{\Ss}{RTQ} = \int_\Ss\QPvec\times\vel{P}{RTQ}\ds\ms(\Ps) }[/math]. La velocitat [math]\displaystyle{ \vel{P}{RTQ} }[/math] es pot escriure com a suma de dos termes si s’aplica una composició de moviments:

[math]\displaystyle{ \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]

Per tant: [math]\displaystyle{ \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].

La definició de centre de masses porta a reescriure el segon terme com a:

[math]\displaystyle{ \left[\int_\Ss\QPvec\ds\ms(\Ps)\right] = \Ms\QGvec\times\vel{P}{RTQ} = \QGvec\times\Ms\vel{P}{RTQ} }[/math], que coincideix amb el moment cinètic respecte de [math]\displaystyle{ \Qs }[/math] d’una partícula de massa M situada a [math]\displaystyle{ \Gs }[/math]: [math]\displaystyle{ \QGvec\times\Ms\vel{P}{RTQ} = \H{G}{\oplus}{RTQ} }[/math].

En el primer terme de l’expressió de [math]\displaystyle{ \H{Q}{\Ss}{RTQ} }[/math], es pot descompondre [math]\displaystyle{ \QPvec }[/math] en suma de dos termes:

[math]\displaystyle{ \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]

Per definició de centre de masses: [math]\displaystyle{ \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].

Per tant: [math]\displaystyle{ \H{Q}{}{RTQ} = \H{G}{}{RTG} + \H{G}{\oplus}{RTQ} }[/math].

✏️ Exemple D4.9: descomposició baricèntrica

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El sòlid està format per dues barres de massa negligible i quatre partícules ([math]\displaystyle{ \Ps }[/math]) amb massa m unides als extrems de les barres. El sòlid està articulat a un suport que es pot moure al llarg d’una guia fixa a terra. El seu moment cinètic a G és: [math]\displaystyle{ \H{G}{}{RTG}\equiv\sum_{\Ps\in\text{sist}}\GPvec\times\ms\vel{P}{RTG} }[/math] La RTG és fixa al suport, i la velocitat de les partícules respecte d’aquesta referència és proporcional a la rotació [math]\displaystyle{ \omega }[/math]: [math]\displaystyle{ |\vel{P}{RTG}|=\Ls\omega }[/math]. Per tant: [math]\displaystyle{ \H{G}{}{RTG}\equiv\sum_{\Ps\in\text{sist}}\GPvec\times\ms\vel{P}{RTG} = (\otimes4\ms\Ls^2\omega) }[/math]

El moment cinètic a [math]\displaystyle{ \Os }[/math] (fix al terra) es pot calcular a partir de [math]\displaystyle{ \H{G}{}{RTG} }[/math] mitjançant descomposició baricèntrica:

[math]\displaystyle{ \H{O}{}{RTO} = \H{O}{}{T} = \H{G}{}{RTG} + \H{G}{\oplus}{T} = (\otimes 4\ms\Ls^2\omega) + \OGvec\times\Ms\vel{G}{T} = (\otimes 4\ms\Ls^2\omega) + [(\rightarrow\xs) + (\downarrow\hs)]\times 4\ms(\rightarrow\dot\xs) = (\otimes4\ms\Ls^2\omega) + (\downarrow\hs)\times\Ms(\rightarrow\dot\xs) = (\otimes4\ms\Ls^2\omega) + (\odot 4\ms\hs\dot\xs) = \otimes 4\ms(\Ls^2\omega - \hs\dot\xs) }[/math]

© Universitat Politècnica de Catalunya. Tots els drets reservats

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