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

✏️ 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_\epsilon(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{tr}{NGal\rightarrow\Ps}+\Fcal{Cor}{NGal\rightarrow\Ps} = \ms_\Ps\acc{P}{NGal} }[/math],

where [math]\displaystyle{ \Fcal{tr}{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{tr}{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{tr}{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{tr}{sup\rightarrow\Gs} = -\ms\acc{$\Gs\in$support}{T}=[\leftarrow 0,1(\ms/s^2)\Ms] }[/math]. That value is within the range of values allowed for the horizontal constraint force between support and block. Therefore, this force will take the value [math]\displaystyle{ \F{sup\rightarrow block} = 0,1(\ms/s^2) }[/math], will counteract the [math]\displaystyle{ \Fcal{tr}{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.

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]

[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]

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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A vehicle without suspensions moves on a straight road. The mass of the wheels is negligible compared to that of the rest of the elements, and their contact with the ground is considered to be a single-point one. We want to analyse the normal constraint forces ground and wheels. This is a 2D dynamics problem. The external forces on the vehicle are the weight and and the ground constraint forces. If the acceleration of the chassis relative to the ground is a given, the application of the AMT leads to:

[math]\displaystyle{ \sum\F{\text{ext}} = \ms\acc{G}{\epsilon} \Rightarrow \left\{\begin{aligned} \uparrow(\Ns_{\ds\vs}+ \Ns_{\ds\rs})+ (\downarrow\ms\gs) = 0\\ [\rightarrow(\Ts_{\ds\vs}+ \Ts_{\ds\rs})] = (\rightarrow\ms\as_\epsilon) \end{aligned}\right. }[/math] where [math]\displaystyle{ (\Ns_{\ds\vs}, \Ts_{\ds\vs}) }[/math] and [math]\displaystyle{ (\Ns_{\ds\rs}, \Ts_{\ds\rs}) }[/math] are the resultant normal and tangential forces exerted by the ground on the two front wheels and the two rear wheels, respectively. The AMT at G yields: [math]\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] (the angular momentum [math]\displaystyle{ \vec{H}_{RTG}(\Gs) }[/math] is permanently zero beacuse the wheels are massless, and the chassis motion relative to the RTG is zero).

Solving the system of equations: [math]\displaystyle{ \Ns_{\ds\vs} = \frac{\ms\gs\Ls_{\ds\rs}}{\Ls_{\ds\vs} + \Ls_{\ds\rs}} - \frac{\ms\as_\epsilon\hs}{\Ls_{\ds\vs} + \Ls_{\ds\rs}}, \:\:\:\:\Ns_{\ds\rs} = \frac{\ms\gs\Ls_{\ds\vs}}{\Ls_{\ds\vs} + \Ls_{\ds\rs}} + \frac{\ms\as_\epsilon\hs}{\Ls_{\ds\vs} + \Ls_{\ds\rs}}, }[/math]

This result is valid for any value of [math]\displaystyle{ \as_\epsilon }[/math]:

• Static situation ([math]\displaystyle{ \as_\epsilon = 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]; the normal force is higher on the wheels whose axle is closer to G.

Accelerated motion ([math]\displaystyle{ \as_\epsilon \gt 0 }[/math]): [math]\displaystyle{ \Ns_{\ds\vs} = \Ns_{\ds\vs}^{\es\ss\ts} - \frac{\ms|\as_\epsilon|\hs}{\Ls_{\ds\vs}+ \Ls_{\ds\rs}} }[/math], [math]\displaystyle{ \Ns_{\ds\rs} = \Ns_{\ds\rs}^{\es\ss\ts} + \frac{\ms|\as_\epsilon|\hs}{\Ls_{\ds\vs}+ \Ls_{\ds\rs}} }[/math]; there is a force transfer from front to rear (the rear wheels are loaded and the front wheels are unloaded). If the acceleration is sufficiently high, the front normal force becomes negative, which indicates that contact has been lost and the vehicle rolls over counter-clockwise. Then, [math]\displaystyle{ \Ns_{\ds\vs} = 0 }[/math] and the chassis initially has angular acceleration. As a consequence, the centre of mass G acquires vertical acceleration.

• Braking ([math]\displaystyle{ \as_\epsilon \lt 0 }[/math]): [math]\displaystyle{ \Ns_{\ds\vs} = \Ns_{\ds\vs}^{\es\ss\ts} - \frac{\ms|\as_\epsilon|\hs}{\Ls_{\ds\vs}+ \Ls_{\ds\rs}} }[/math], [math]\displaystyle{ \Ns_{\ds\rs} = \Ns_{\ds\rs}^{\es\ss\ts} - \frac{\ms|\as_\epsilon|\hs}{\Ls_{\ds\vs}+ \Ls_{\ds\rs}} }[/math]; there is a force transfer from the rear to the front (the rear wheels are unloaded and the front wheels are loaded). As in the previous case, there is a critical value that causes the vehicle to tip over, this time clockwise.

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D4.7 Dynamics of Constraint Auxiliary Elements

The application of the vector theorems to Constraint Auxiliary Elements (CAE) is simple since the terms on the right (change in linear momentum and change in angular momentum) are zero because the CAE mass is zero:

[math]\displaystyle{ \ms_\mathrm{CAE} \approx 0 \Rightarrow \sum \mathbf{\overline{F}}_\mathrm{ext} \approx \overline{0} \quad , \quad \sum \mathbf{\overline{M}}_\mathrm{ext} (\mathrm{qualsevol} \hspace{0.2cm} \mathrm{punt}) \approx \overline{0}. }[/math]

For the case of a CAE (rígid body S) connecting two rigid bodies S1 and S2 (Figure D4.5), these equations allow to prove that the torsor exerted on S1 can be obtained from an analytical characterization equation where the kinematics of the characterization point P (which has to be a point fixed to S1) is assessed from the S2 reference frame:

[math]\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]

(When dealing with velocities, it is important to specify which rigid body does P belong to. However, this is irrelevant when dealing with moments: [math]\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]).

Figure D4.5:Constraint Auxiliary Element between to rigid bodies with nonzero mass

💭 Proof ➕

The vector theorems applied to the CAE imply that the constraint torsors exerted by S1 and S2 at the same point P must sum up to zero:

[math]\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]

On the other hand, those torsores fulfill the equation of analytical characterization:

[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]

If all the above equations are combined, we obtain:

[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]

A composition of movements allows us to rewrite the above equation in a more compact way:

[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]

✏️ EXAMPLE D4.8: longitudinal dynamics of a vehicle

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Let’s assume that the vehicle of example D4.7 has rear traction. This means that the vehicle's engine acts between the chassis and the rear wheels, while the front wheels are only subject to constraint interactions (articulation with the chassis and contact with the ground). If the front wheels are treated as CAE, the kinematic analysis of the chassis relative to the ground for the characterization of the constraint through the front wheels leads to a torsor at the wheel centre (which is fixed to the chassis) with only one force component:

The existence of a horizontal component of force between the rear wheels (necessary to accelerate or brake the vehicle!) is associated with the engine torque. If the vector theorems are applied to the rear wheels, the external interactions to be taken into account are the connection with the ground, the articulation with the chassis and the engine torque:

[math]\displaystyle{ \ms_\mathrm{wheels} \approx 0 \Rightarrow \sum \overline{\mathbf{F}}_\mathrm{ext} \approx \overline{0} , \sum \overline{\mathbf{M}}_\mathrm{ext} (\Cs) \approx \overline{0} \Rightarrow \left\{\begin{array}{l} \mathrm{LMT }: \hspace{1.4cm} \Ns_\mathrm{dr}=\Ns'_\mathrm{dr} , \Ts_\mathrm{dr}=\Ts'_\mathrm{dr} \\ \mathrm{AMT } \hspace{0.2cm} \mathrm{a} \hspace{0.2cm} \Cs: \quad \Ts_\mathrm{dr}\rs=\Gamma \Rightarrow \Ts_\mathrm{dr}=\Gamma / \rs \end{array}\right. }[/math]

The tangential force on the driven wheels is directly proportional to the engine torque applied to them. However, it should be remembered that, as a constraint tangential force, its value is bounded by [math]\displaystyle{ \mu_\mathrm{s}\Ns_\mathrm{dr} }[/math]. This allows to calculate the maximum acceleration that the vehicle can acquire (as long as the front wheel does not lose contact with the ground): [math]\displaystyle{ \left.\as_\epsilon (\mathrm{vehicle})\right]_\mathrm{max}= \Ts_\mathrm{dr,max}/\ms \quad , \quad \Ts_\mathrm{dr}=\Gamma / \rs \quad , \quad \Ts_\mathrm{dr} \leq \mu_\mathrm{e}\Ns_\mathrm{dr} }[/math].

On low friction (low [math]\displaystyle{ \mu_\ss }[/math] value) terrains, an engine capable of delivering a high maximum torque is useless: what limits acceleration is the value of the friction coefficient [math]\displaystyle{ \mu_\es }[/math]: [math]\displaystyle{ \left.\as_\epsilon (\mathrm{vehicle})\right]_\mathrm{màx}= \mu_\mathrm{e}\Ns_\mathrm{dr} /\ms }[/math].

On high friction terrain, if the maximum torque is low, it is this that limits acceleration: [math]\displaystyle{ \left. \as_\epsilon (\mathrm{vehicle})\right]_\mathrm{max}= \Gamma_\mathrm{max}/\ms \rs }[/math] (if the front wheel does not lose contact with the ground).

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D4.8 Barycentric decomposition of the angular momentum

The general formulation presented for the AMT allows a free choice of the Q point of application. The criterion for choosing it is based on what we want to investigate (a constraint force, an equation of motion...). This criterion is discussed in unit D6.

In some cases, it may be interesting to choose a point Q that does not belong to any element of the system. It is then useful to calculate the angular momentum [math]\displaystyle{ \H{Q}{}{RTQ} }[/math] from the angular momentum at the system center of mass G, [math]\displaystyle{ \H{Q}{}{RTG} }[/math]. This is known as the barycentric decomposition of the angular momentum:

[math]\displaystyle{ \H{Q}{}{RTQ} = \H{G}{}{RTG} + \H{Q}{\oplus}{RTQ} }[/math], where the superscript [math]\displaystyle{ \oplus }[/math] indicates that the angular momentum is to be calculated as if the system had been reduced to a particle concentrated at G with mass equal to the total mass of the system (M):

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

💭 Proof ➕

The definition of the angular momentum of a rigid body is: [math]\displaystyle{ \H{Q}{\Ss}{RTQ} = \int_\Ss\QPvec\times\vel{P}{RTQ}\ds\ms(\Ps) }[/math]. Velocity [math]\displaystyle{ \vel{P}{RTQ} }[/math] can be written as a sum of two terms through a composition of movements:

[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]

Therefore: [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].

The definition of centre of mass leads to rewriting the second term as:

[math]\displaystyle{ \left[\int_\Ss\QPvec\ds\ms(\Ps)\right] = \Ms\QGvec\times\vel{P}{RTQ} = \QGvec\times\Ms\vel{P}{RTQ} }[/math], which coincides with the angular momentum at Q of a particle of mass M located at G: [math]\displaystyle{ \QGvec\times\Ms\vel{G}{RTQ} = \H{Q}{\oplus}{RTQ} }[/math].

In the first term of the [math]\displaystyle{ \H{Q}{\Ss}{RTQ} }[/math] expression, [math]\displaystyle{ \QPvec }[/math] can be decomposed into two terms:

[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]

From the definition of centre of mass: [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].

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

✏️ EXAMPLE D4.9: barycentric decomposition

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The rigid body is made up of two massless bars and four particles (P) with mass m attached to the bars endpoints. The rigid body is attached to a support that can move along a guide fixed to the ground. Its angular momentum at G is: [math]\displaystyle{ \H{G}{}{RTG}\equiv\sum_{\Ps\in\text{syst}}\GPvec\times\ms\vel{P}{RTG} }[/math] The RTG is fixed to the support, and the velocity of the particles relative to that reference frame is proportional to the rotation [math]\displaystyle{ \omega }[/math]: [math]\displaystyle{ |\vel{P}{RTG}|=\Ls\omega }[/math]. Hence: [math]\displaystyle{ \H{G}{}{RTG}\equiv\sum_{\Ps\in\text{syst}}\GPvec\times\ms\vel{P}{RTG} = (\otimes4\ms\Ls^2\omega) }[/math]

The angular momentum at O (fixed to the ground) can be calculated from [math]\displaystyle{ \H{G}{}{RTG} }[/math] through barycentric decomposition:

[math]\displaystyle{ \H{O}{}{RTO} = \H{O}{}{T} = \H{G}{}{RTG} + \H{O}{\oplus}{T} = (\otimes 4\ms\Ls^2\omega) + \OGvec\times 4\ms\vel{G}{T} = }[/math]

[math]\displaystyle{ \hspace{1cm} = (\otimes 4\ms\Ls^2\omega) + [(\rightarrow\xs) + (\downarrow\hs)]\times 4\ms(\rightarrow\dot\xs) = (\otimes4\ms\Ls^2\omega) + (\downarrow\hs)\times 4\ms(\rightarrow\dot\xs) = }[/math] [math]\displaystyle{ \hspace{1cm}= (\otimes4\ms\Ls^2\omega) + (\odot 4\ms\hs\dot\xs) = \otimes 4\ms(\Ls^2\omega - \hs\dot\xs) }[/math]

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