D8. Conservation of dynamic magnitudes

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The vector theorems relate the variation of two dynamic magnitudes ([math]\displaystyle{ \overline{\Ds\Ms} }[/math]) that depend on the mass geometry and the motion of the system (the linear momentum and the angular momentum) with the resultant of the external actions on the system [math]\displaystyle{ (\sum \overline{\As\mathrm{C}_\mathrm{ext}}) }[/math] ( [math]\displaystyle{ \sum \overline{\As\mathrm{C}_\mathrm{ext}} }[/math] includes the external interactions and, if working in a non-Galilean reference frame,the associated inertial actions). In a generic way, these theorems can be written in the following form:

[math]\displaystyle{ \sum \overline{\As\mathrm{C}_\mathrm{ext}}=\dert{\overline{\Ds\Ms}}{R}. }[/math]

When in a direction fixed to the reference frame R(dfR) direction fixed to the reference frame R is conserved:[math]\displaystyle{ \left. \sum \overline{\As\mathrm{C}_\mathrm{ext}}\right]_\mathrm{dfR}=0 \Rightarrow \left. \overline{\Ds\Ms}\right]_\mathrm{dfR}=\text{constant} }[/math].

A conservation is an interesting property: it allows one to ignore the evolution of the system for a finite time and maintain a partial knowledge (if not all the components of [math]\displaystyle{ \overline{\Ds\Ms} }[/math]) are conserved) or a total knowledge (if the conservation occurs in the three directions of the space of R) of the system’s state.

Two important things have to be kept in mind when it comes to invoking conservations:

• We have to be sure that the component of the external actions that is zero corresponds to a direction fixed in the reference frame (a zero value in a direction variable with respect to R means that the [math]\displaystyle{ \overline{\Ds\Ms} }[/math] component in that direction has a constant value, but not a constant direction!).
  

• We have to remember that conservation refers to a dynamic magnitude and not a kinematic one (in principle). When the [math]\displaystyle{ \overline{\Ds\Ms} }[/math] is the linear momentum [math]\displaystyle{ (\overline{\Ds\Ms}=\Ms\vel{G}{R}) }[/math], as it is proportional to the velocity of the center of mass, the corresponding component of [math]\displaystyle{ \vel{G}{R} }[/math] is conserved. When it is the angular momentum of a single rigid body about a point that belongs to that rigid body [math]\displaystyle{ \overline{\Ds\Ms} }[/math] since in general it is not proportional to the angular velocity [math]\displaystyle{ (\overline{\Ds\Ms}=\overline{\mathrm{H}}_\mathrm{RTQ}(\Qs),\Qs \in \mathrm{S}) }[/math] the conservation of [math]\displaystyle{ \overline{\Ds\Ms} }[/math] does not imply that of [math]\displaystyle{ \velang{S}{R} }[/math].
  

Conservations are often the consequence of simplifications in the formulation of problems, such as neglecting friction. In real life, in general nothing is conserved.

💭 Proof ➕

Let’s project the vector thorem in a vector basis with a dierction fixed to the reference frame (for example, direction3):

[math]\displaystyle{ \sum \overline{\As\mathrm{C}_\mathrm{ext}}=\dert{\overline{\Ds\Ms}}{R} \quad \Rightarrow \quad \vector{\sum\As\mathrm{C}_1}{\sum\As\mathrm{C}_2}{\sum\As\mathrm{C}_3}=\left\{\dert{\overline{\Ds\Ms}}{R}\right\}=\vector{\dot{\Ds\Ms_1}}{\dot{\Ds\Ms_2}}{\dot{\Ds\Ms_3}}+\left\{\velang{B}{R}\right\}\times \vector{\Ds\Ms_1}{\Ds\Ms_2}{\Ds\Ms_3} }[/math]

Since direction 3 is fixed to R, the angular velocity of the vector basis relative to R [math]\displaystyle{ \velang{B}{R} }[/math] must have a component in that direction. Therefore:

[math]\displaystyle{ \vector{\sum\As\mathrm{C}_1}{\sum\As\mathrm{C}_2}{\sum\As\mathrm{C}_3}=\left\{\dert{\overline{\Ds\Ms}}{R}\right\}=\vector{\dot{\Ds\Ms_1}}{\dot{\Ds\Ms_2}}{\dot{\Ds\Ms_3}}=\vector{0}{0}{\Omega_3}\times \vector{\Ds\Ms_1}{\Ds\Ms_2}{\Ds\Ms_3}= \vector{\dot{\Ds\Ms_1}-\Omega_3\cdot\Ds\Ms_2}{\dot{\Ds\Ms_2}-\Omega_3\cdot\Ds\Ms_1}{\dot{\Ds\Ms_3}} }[/math]

Si [math]\displaystyle{ \sum \As\mathrm{C}_3=0 \quad \Rightarrow \quad \dot{\Ds\Ms_3}=0 \quad \Rightarrow \quad \Ds\Ms_3=\text{CONSTANT!}. }[/math]

However, if [math]\displaystyle{ \sum \As\mathrm{C}_1 }[/math] or [math]\displaystyle{ \sum \As\mathrm{C}_2 }[/math] are zero, the corresponding components in [math]\displaystyle{ \overline{\Ds\Ms} }[/math] ( [math]\displaystyle{ \Ds\Ms_1 }[/math] or [math]\displaystyle{ \Ds\Ms_2 }[/math]) are not constant in principle:

[math]\displaystyle{ \sum \As\mathrm{C}_1=0 \quad \Rightarrow \quad \dot{\Ds\Ms_1}-\Omega_3\cdot\Ds\Ms_2 =0 \quad \Rightarrow \quad \dot{\Ds\Ms_1}=\Omega_3\cdot\Ds\Ms_2 \quad \Rightarrow \quad \Ds\Ms_1 \neq \text{constant}, }[/math]

[math]\displaystyle{ \sum \As\mathrm{C}_2=0 \quad \Rightarrow \quad \dot{\Ds\Ms_2}+\Omega_3\cdot\Ds\Ms_1 =0 \quad \Rightarrow \quad \dot{\Ds\Ms_2}=-\Omega_3\cdot\Ds\Ms_1 \quad \Rightarrow \quad \Ds\Ms_2 \neq \text{constant} . }[/math]

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D8.1 Examples

✏️ EXAMPLE D8.1: person jumping on a platform

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A person of mass M, moving at speed [math]\displaystyle{ \mathrm{v}_0 }[/math] on a smooth ground [math]\displaystyle{ (\mu=0) }[/math], jumps onto a platform of mass m which is initially at rest with respect to the floor, and comes to rest relative to it. We want to investigate the evolution of the motion of these two elements (person and platform).

Is the linear momentum conserved?.

The person's jump takes place on a smooth ground that does not introduce any horizontal force on the person or on the platform. Therefore, during the jump and for the SYSTEM (person + platform) and the ground reference frame:

[math]\displaystyle{ \left.\sum \overline{\Fs}_\mathrm{ext}\right]_\mathrm{horizontal}=0 \quad \Rightarrow \quad }[/math] CONSTANT horizontal linear momentum relative to the ground!

Before jumping [math]\displaystyle{ (\ts_\mathrm{inicial}) }[/math], the linear momentum (relative to the ground) is only associated with the person: [math]\displaystyle{ (\rightarrow \ms\vs_0) }[/math]. Just after jumping [math]\displaystyle{ (\ts_\mathrm{final}) }[/math], as the person is at rest relative to the platform, both elements move with the same velocity relative to the ground [math]\displaystyle{ \left[\rightarrow (\Ms+\ms)\vs \right]. }[/math].

The conservation of the horizontal linear momentum between these two time instants allows us to calculate the final velocity of the system: [math]\displaystyle{ (\rightarrow \ms\vs_0) = \left[\rightarrow (\Ms+\ms)\vs \right] \quad \Rightarrow \quad \vs=\frac{\ms}{\Ms+\ms}\vs_0. }[/math].

This speed is constant while the system slides on the smooth ground, but as soon as it enters the rough area [math]\displaystyle{ (\mu \neq 0) }[/math], that will change: the friction force of the ground on the platform, horizontal and opposite to [math]\displaystyle{ \overline{\vs}_\Ts }[/math] (platform) will make it decrease. The linear momentum is no longer conserved:

[math]\displaystyle{ \overline{\Fs}_\mathrm{ground \rightarrow syst}=(\leftarrow \Fs_\mathrm{friction})=(\Ms+\ms)\acc{G}{E}. }[/math]

The horizontal linear momentum of the person and the platform (separately) are not conserved during the jump because of the horizontal constraint forces that appear between them when the person-platform contact begins.

ANIMACIONS

✏️ EXAMPLE D8.2: stopping a block on a wagon

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A person stands on a wagon, both initially at rest relative to the ground. The mass of the system (person + wagon) is M, and the wheels of the wagon are massless. The block, with mass m, has an initial velocity [math]\displaystyle{ \vs_0 }[/math] relative to the ground directed towards the person, which stops it relative to the platform. The friction associated with the joints between wheels and wagon is neglected. We want to investigate the evolution of the movement of the system.

Is the linear momentum conserved?.

The wheels are Auxiliary Constraint Elements (ACE) and cannot transmit horizontal forces (see example D3.10). Hence, for the SYSTEM (person + wagon with wheels + block) and the ground reference frame:

[math]\displaystyle{ \left.\sum \overline{\Fs}_\mathrm{ext}\right]_\mathrm{horizontal}=0 \quad \Rightarrow \quad }[/math] CONSTANT horizontal linear momentum (LM) relative to the ground!

Before stopping the block, the LM relative to the ground is associated only with the block: [math]\displaystyle{ (\leftarrow \ms\vs_0) }[/math]. But just after [math]\displaystyle{ (\ts_\mathrm{final}) }[/math], since the person and the block are at rest relative to the wagon, the entire system moves with the same velocity relative to the ground:[math]\displaystyle{ \left[\leftarrow (\Ms+\ms)\vs \right]. }[/math]. The conservation of horizontal LM between these two time instants allows the calculation of the final velocity of the system: [math]\displaystyle{ (\leftarrow \ms\vs_0) = \left[\leftarrow (\Ms+\ms)\vs \right] \quad \Rightarrow \quad \vs=\frac{\ms}{\Ms+\ms}\vs_0. }[/math].

The LM of just the block relative to the ground does not remain constant because of the friction force of the wagon on the block, which tends to stop it. For a time instant between the initial and final ones [math]\displaystyle{ (\ts_\mathrm{initial}\lt \ts\lt \ts_\mathrm{final}) }[/math] when the speed of the block relative to the ground has been reduced to [math]\displaystyle{ \vs'(\lt \vs_0) }[/math], the velocity of the system (person + wagon) can be calculated through the conservation of horizontal LM for the system (person + wagon with wheels + block):

[math]\displaystyle{ (\leftarrow \ms\vs_0) = (\leftarrow \ms\vs') + \left[\leftarrow (\Ms+\ms)\vs'' \right] \quad \Rightarrow \quad \vs''=\frac{\ms}{\Ms+\ms}(\vs_0-\vs'). }[/math]

ANIMACIONS

✏️ EXAMPLE D8.3: skater on ice

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A person is skating on an ice rink. At a certain moment, his arms are symmetrically wide open and he he spins with angular velocity [math]\displaystyle{ \Omega_0 }[/math] In that configuration, the vertical axis through [math]\displaystyle{ \Gs }[/math] is a principal axis of inertia and the corresponding moment of inertia is [math]\displaystyle{ \Is_0 }[/math]. We want to investigate the evolution of the rotation when the configuration of his arms changes assuming that the friction between the ice and the skates is negligible [math]\displaystyle{ \mu=0 }[/math].

Is the angular momentum conserved?.

Since [math]\displaystyle{ \mu=0 }[/math] between ice and kates, the only external forces on the system (person + skates) are vertical (the weight and the normal forces of the ice on the skates). Those vertical forces cannot generate vertical momento about [math]\displaystyle{ \Gs }[/math]. Hence, for the system (person + skates):

[math]\displaystyle{ \left.\sum\overline{\Ms_\mathrm{ext}}(\Gs)\right]_\mathrm{vertical}=0=\left.\dot{\overline{\Hs}}_\mathrm{RTG}(\Gs)\right]_\mathrm{vertical} \quad \Rightarrow \quad \left.\overline{\Hs}_\mathrm{RTG}(\Gs)\right]_\mathrm{vertical} \text{CONSTANT!} }[/math]

The vertical angular momentum in the initial configuration is [math]\displaystyle{ \overline{\Hs}_\mathrm{RTG}(\Gs,\ts_\mathrm{initial})=(\Uparrow \Is_0 \Omega_0) }[/math]. When approachng or separating the arms from the trunk, the inertia moment of the person about the vertical axis through [math]\displaystyle{ \Gs }[/math] changes. For any value [math]\displaystyle{ \Is }[/math] of this inertia moment, conservation implies: [math]\displaystyle{ (\Uparrow \Is_0 \Omega_0)=(\Uparrow \Is \Omega) }[/math]. When approaching the arms to the trunk, [math]\displaystyle{ \Is\lt \Is_0 }[/math], therefore [math]\displaystyle{ \Omega\gt \Omega_0 }[/math] (the angular velocity increases). For the paricular case [math]\displaystyle{ \Is=\Is_0/2 }[/math], the angular velocity becomes twice the initial value: [math]\displaystyle{ \Omega=2\Omega_0 }[/math].

ANIMACIONS

✏️ EXAMPLE D8.4: collision between two bars

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Two bars, with their mass concentrated at one end, move on a perfectly smooth horizontal ground (the friction coefficient between the ground and the bars is zero, [math]\displaystyle{ \mu=0 }[/math]) towards each other until they collide and become stuck. We want to describe the final motion of the system.

Is the linear momentum conserved?.

If we consider the system formed by the two bars, the external forces on them are strictly vertical (perpendicular to the plane of motion): the weight and the normal forces associated with the ground contact. Therefore, for this system:

[math]\displaystyle{ \left.\sum \overline{\Fs}_\mathrm{ext}\right]_\mathrm{horizontal}=0 \quad \Rightarrow \quad }[/math] CONSTANT horizontal linear momentum (LM) relative to the ground!

Before collision [math]\displaystyle{ (\ts_\mathrm{before}) }[/math]:

[math]\displaystyle{ \left.\overline{\mathrm{QM}}\right]_\mathrm{horizontal}=\Ms \vel{G}{T}=2\ms\overline{\vs}_\Ts(\mathrm{barra P})+\ms\overline{\vs}_\Ts(\mathrm{barra Q})=(\rightarrow \ms\vs_0)+(\leftarrow \ms2\vs_0)=0 }[/math]

After collision [math]\displaystyle{ \ts_\mathrm{after} }[/math], [math]\displaystyle{ \left.\overline{\mathrm{QM}}\right]_\mathrm{horizontal} }[/math] of the system has to be zero, therefore the velocity of the system’s center of inertia is also zero: [math]\displaystyle{ \overline{\vs}_\Ts(\Gs,\ts_\mathrm{after}) }[/math]. Hence, after collision, the rigid body formed by the two bars will have the ICR relative to the ground permanently located at [math]\displaystyle{ \Gs }[/math]:

Position of the inertia center [math]\displaystyle{ \Gs }[/math]: on the line [math]\displaystyle{ \Ps }[/math] and [math]\displaystyle{ \Qs }[/math], at a distance 4L below [math]\displaystyle{ \Qs }[/math].

[math]\displaystyle{ \left.\QGvec\right]_{\uparrow \downarrow}=\frac{\left.\ms \QQvec \right]_{\uparrow \downarrow} +\left. 2\ms \QPvec\right]_{\uparrow \downarrow}}{\ms+2\ms} }[/math]

[math]\displaystyle{ \left.\QGvec\right]_{\uparrow \downarrow}=\frac{2}{3}(\downarrow 6\Ls)=(\downarrow 4\Ls) }[/math]

There is no conservation of the linear momentum for each bar separately: the collision generates very intense forces between them and perpendicular to the bars, that provoke nonzero acceleration of their inertia centres.

Is the angular momentum conserved?

On the other hand, vertical forces (perpendicular to the plane of motion) cannot generate vertical moments about any point. If we apply the AMT at [math]\displaystyle{ \Gs }[/math] to the system formed by the two bars:

[math]\displaystyle{ \left.\sum\overline{\Ms_\mathrm{ext}}(\Gs)\right]_\mathrm{vertical}=0=\left.\dot{\overline{\Hs}}_\mathrm{RTG}(\Gs)\right]_\mathrm{vertical} \quad \Rightarrow \quad \left.\overline{\Hs}_\mathrm{RTG}(\Gs)\right]_\mathrm{vertical} \text{CONSTANT!} }[/math]

[math]\displaystyle{ \left.\overline{\Hs}_\mathrm{RTG}(\Gs,\ts_\mathrm{before})\right]_\mathrm{vertical}=\left.\overline{\Hs}_\mathrm{RTG}(\Gs,\ts_\mathrm{after})\right]_\mathrm{vertical} }[/math]

At time [math]\displaystyle{ \ts_\mathrm{before} }[/math], the two bars have a translational motion, and therefore [math]\displaystyle{ \Gs }[/math] does not belong kinematically to either of them. Its angular momentum has to be calculated through barycentric decomposition . Taking into account that the center of inertia of bar P is [math]\displaystyle{ \Ps }[/math], and that of bar Q is [math]\displaystyle{ \Qs }[/math]:

[math]\displaystyle{ \overline{\Hs}_\mathrm{RTG}(\Gs,\ts_\mathrm{before})=\left.\overline{\Hs}_\mathrm{RTG}(\Gs,\ts_\mathrm{before})\right]_\mathrm{barP}+\left.\overline{\Hs}_\mathrm{RTG}(\Gs,\ts_\mathrm{before})\right]_\mathrm{barQ}= }[/math]

[math]\displaystyle{ \hspace{2.9cm}=\left.\overline{\Hs}_\mathrm{RTP}(\Ps,\ts_\mathrm{before}) \right]_\mathrm{barP}+\GPvec\times 2\ms\vel{P}{RTG}+\left.\overline{\Hs}_\mathrm{RTQ}(\Qs,\ts_\mathrm{before})\right]_\mathrm{barQ}+\GQvec \times 2\ms\vel{Q}{RTG}= }[/math]

[math]\displaystyle{ \hspace{2.9cm}=\Is_\mathrm{P}\velang{barP}{RTG} + \GPvec \times 2\ms\vel{P}{RTG}+\Is_\mathrm{Q}\velang{barQ}{RTG}+\GQvec\times 2\ms\vel{Q}{RTG} }[/math]

[math]\displaystyle{ \velang{barP}{RTG}=\velang{barQ}{RTG}=\vec{0} \quad \Rightarrow \quad \overline{\Hs}_\mathrm{RTG}(\Gs,\ts_\mathrm{before})=(\downarrow 2\Ls)\times 2\ms(\rightarrow \vs_0)+(\uparrow 4\Ls)\times \ms (\leftarrow 2\vs_0)=(\otimes 10\ms\vs_0\Ls) }[/math]

After collision, [math]\displaystyle{ \Gs }[/math] is a point fixed to the rigid body formed by the two bars stuck together. Therefore:

[math]\displaystyle{ \overline{\Hs}_\mathrm{RTG}(\Gs,\ts_\mathrm{after})= \Is_\mathrm{G}\velang{}{RTG}=\left[2\ms(2\Ls)^2+\ms(4\Ls)^2\right](\otimes \Omega_\mathrm{T})=(\otimes 24\ms\Ls^2\Omega_\mathrm{T}) }[/math]

Finally: [math]\displaystyle{ (\otimes 10\ms\Ls\vs_0)=(\otimes\ms\Ls^2\Omega_\mathrm{T})\quad \Rightarrow \quad \Omega_\mathrm{T}=\frac{5}{12}\frac{\vs_0}{\Ls} }[/math] .

Important comment

Although, for the system formed by the two bars, the vertical external moment is zero for any point, the angular momentum is not conserved either at [math]\displaystyle{ \Ps }[/math] or at [math]\displaystyle{ \Qs }[/math] because the two points are accelerated (their velocities change abruptly when the collision occurs):

[math]\displaystyle{ \acc{P}{Gal}=\frac{\Delta \vel{P}{Gal}}{\ts_\mathrm{after}-\ts_\mathrm{before}}=\frac{\left[\rightarrow (\vs_0-2\Ls\Omega_\mathrm{T})\right]}{\ts_\mathrm{after}-\ts_\mathrm{before}} }[/math] , [math]\displaystyle{ \acc{Q}{Gal}=\frac{\Delta \vel{Q}{Gal}}{\ts_\mathrm{after}-\ts_\mathrm{before}}=\frac{\left[\rightarrow (2\vs_0-4\Ls\Omega_\mathrm{T})\right]}{\ts_\mathrm{after}-\ts_\mathrm{before}} }[/math] .

Therefore:

[math]\displaystyle{ \left.\sum\overline{\Ms_\mathrm{ext}}(\Ps)\right]_\mathrm{vertical}+ \left.\PGvec \times \ms \acc{P}{Gal}\right]_\mathrm{vertical}=\left.\dot{\overline{\Hs}}_\mathrm{RTP}(\Ps)\right]_\mathrm{vertical} }[/math]

[math]\displaystyle{ (\uparrow 2\Ls)\times \ms \frac{\left[\rightarrow (\vs_0-2\Ls\Omega_\mathrm{T})\right]}{\ts_\mathrm{after}-\ts_\mathrm{before}} =\left[\otimes \frac{2\ms\Ls(\vs_0-2\Ls\Omega_\mathrm{T})}{\ts_\mathrm{after}-\ts_\mathrm{before}} \right]=\left.\dot{\overline{\Hs}}_\mathrm{RTP}(\Ps)\right]_\mathrm{vertical} \quad \Rightarrow \quad \left.\overline{\Hs}_\mathrm{RTP}(\Ps)\right]_\mathrm{vertical}\neq \text{constant} }[/math]

[math]\displaystyle{ \left.\sum\overline{\Ms_\mathrm{ext}}(\Qs)\right]_\mathrm{vertical}+ \left.\QGvec \times \ms \acc{Q}{Gal}\right]_\mathrm{vertical}=\left.\dot{\overline{\Hs}}_\mathrm{RTQ}(\Qs)\right]_\mathrm{vertical} }[/math]

[math]\displaystyle{ (\downarrow 4\Ls)\times \ms \frac{\left[\leftarrow (2\vs_0-4\Ls\Omega_\mathrm{T})\right]}{\ts_\mathrm{after}-\ts_\mathrm{before}} =\left[\otimes \frac{8\ms\Ls(\vs_0-2\Ls\Omega_\mathrm{T})}{\ts_\mathrm{after}-\ts_\mathrm{before}} \right]=\left.\dot{\overline{\Hs}}_\mathrm{RTQ}(\Qs)\right]_\mathrm{vertical} \quad \Rightarrow \quad \left.\overline{\Hs}_\mathrm{RTQ}(\Qs)\right]_\mathrm{vertical}\neq \text{constant} }[/math]

ANIMACIONS

✏️ EXAMPLE D8.5: collision of a ring and an articulated arm

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The ring, with radius L and mass m, moves on a smooth horizontal ground with angular velocity [math]\displaystyle{ \Omega_0=\ns\vs_0/\Ls }[/math] relative to the ground (where n is an integer), and its centre [math]\displaystyle{ \Ps }[/math] approaches the end [math]\displaystyle{ \Qs }[/math] of the arm with velocity [math]\displaystyle{ \vs_0 }[/math]. The arm has a length 2L and a mass M, is articulated at point [math]\displaystyle{ \Os }[/math] fixed to the ground and is initially at rest. Because of the collision, the ring and arm stick together. The friction associated with the joint at [math]\displaystyle{ \Os }[/math] is neglected. We want to investigate the motion of the system after the collision.

Is the linear momentum conserved?.

The linear momentum of each element separately (ring and arm) is not conserved: the collision generates very intense horizontal forces between them that cause acceleration in their centers of inertia. In addition, the arm undergoes an intense force at the [math]\displaystyle{ \Os }[/math]-joint.

There is no conservation of the linear momentum for the system (ring + arm) either: the [math]\displaystyle{ \Os }[/math]-joint introduces forces into the plane of motion responsible for the acceleration of the system’s center of inertia [math]\displaystyle{ \Gs }[/math].

Is the angular momentum conserved?

For the system (ring + arm), the forces associated with the joint yield a nonzero vertical external moment (orthogonal to the plane of motion) at any point except at [math]\displaystyle{ \Os }[/math]:

[math]\displaystyle{ \left.\sum\overline{\Ms_\mathrm{ext}}(\Os)\right]_\mathrm{vertical}=0=\left.\dot{\overline{\Hs}}_\mathrm{RTG}(\Os)\right]_\mathrm{vertical} \quad \Rightarrow \quad \left.\overline{\Hs}_\mathrm{RTG}(\Os)\right]_\mathrm{vertical} \text{CONSTANT!} }[/math]

Since [math]\displaystyle{ \Os }[/math] is permanently fixed to the ground, it must be the permanent ICR relative to the ground of the rigid formed by the ring and the arm stuck together after the collision. Before the collision, the ring ICR is located at [math]\displaystyle{ \Ls/2 }[/math] below the center [math]\displaystyle{ \Ps }[/math]:

[math]\displaystyle{ \vel{S}{T}=\vel{P}{T}+\velang{}{0}\times \PSvec = (\rightarrow \vs_0) + \left(\otimes \frac{\ns\vs_0}{\Ls}\right)\times (\downarrow \Ls)=\left[\leftarrow (\ns-1)\vs_0\right]. }[/math]

The distance e between [math]\displaystyle{ \Ps }[/math] and the ring ICR relative to the ground before collision can be found from:

[math]\displaystyle{ \left.\begin{array}{l} \left|\vel{P}{T}\right|=\vs_0=\es \Omega_0 \\ \left|\vel{S}{T}\right|=(\ns-1) \vs_0=(\Ls-\es) \Omega_0 \end{array}\right\} \Rightarrow \es=\frac{\Ls}{\ns} . }[/math]

A negative value for n means that the ICR is located at a distance e above [math]\displaystyle{ \Ps }[/math].

Before the collision [math]\displaystyle{ \ts_\mathrm{before} }[/math], [math]\displaystyle{ \Os }[/math] is not a point fixed to the ring in general (it is not its ICR). Its angular momentum has to be calculated thorugh barycentric decomposition . The initial angular momentum of the arm is zero because it does not move:

[math]\displaystyle{ \overline{\Hs}_\mathrm{RTO=T}(\Os,\ts_\mathrm{before})=\left.\overline{\Hs}_\mathrm{T}(\Os,\ts_\mathrm{before})\right]_\mathrm{anella}= \left.\overline{\Hs}_\mathrm{RTP}(\Ps,\ts_\mathrm{before})\right]_\mathrm{anella} + \OPvec \times 2\ms\vel{P}{T}= }[/math]

[math]\displaystyle{ \hspace{3.4cm}=\Is_\mathrm{P} \left(\otimes \frac{\ns\vs_0}{\Ls} \right) + \left(\downarrow 2\Ls \right)\times \ms\left(\rightarrow \vs_0 \right) = \left(\otimes \ms\Ls^2 \frac{\ns\vs_0}{\Ls}\right) + \left(\odot 2\ms\Ls\vs_0 \right) = \left[ \otimes (\ns-2)\ms\Ls\vs_0 \right] }[/math]

After the collision:

[math]\displaystyle{ \overline{\Hs}_\mathrm{RTO}(\Os,\ts_\mathrm{after})= \Is_\mathrm{O}\overline{\Omega}_\Ts=\left( \Is_\mathrm{O}^\mathrm{anella}+ \Is_\mathrm{O}^\mathrm{braç} \right) \overline{\Omega}_\Ts= \left( \Is_\mathrm{G}^\mathrm{anella}+ \Is_\mathrm{O}^{\mathrm{anella}\otimes}+\Is_\mathrm{O}^\mathrm{braç} \right) \overline{\Omega}_\Ts = \left[ \ms\Ls^2 + \ms(2\Ls)^2+ \frac{4}{3} \ms\Ls^2\right]\overline{\Omega}_\Ts }[/math]

[math]\displaystyle{ \left.\begin{array}{l} \bar{\mathrm{H}}_{\text {RTO }}\left(\Os, \mathrm{t}_{\text {after }}\right)=\left(\otimes \frac{19}{3} \ms\Ls^2 \Omega_{\mathrm{T}}\right) \\ \bar{\mathrm{H}}_{\text {RTO }}\left(\Os, \mathrm{t}_{\text {before }}\right)=\left[\otimes(\mathrm{n}-2) \mathrm{mLv}_0\right] \end{array}\right\} \Rightarrow \bar{\Omega}_{\mathrm{T}}=\left[\otimes(\mathrm{n}-2) \frac{19}{3} \frac{\mathrm{v}_0}{\mathrm{~L}}\right] }[/math]

For [math]\displaystyle{ n\gt 2 }[/math], the system rotation is clockwise. For [math]\displaystyle{ n\lt 2 }[/math], it is counterclockwise. For [math]\displaystyle{ n=2 }[/math], the system is at rest.

ANIMACIONS

✏️ EXAMPLE D8.6: free rigid body in space

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The rigid body consists of two homogeneous plates, with the same mass and height but different widths, glued at point [math]\displaystyle{ \Os }[/math]. We want to investigate whether any dynamic magnitude is conserved when it is launched in the air. Aerodynamic interactions are neglected.

Is the linear momentum conserved?

The Earth's gravitational attraction as the only external force on the rigid body. Therefore, the vertical component of the linear momentum relative to the ground is not conserved, but the horizontal components are.

Since the linear momentum relative to the ground and the velocity of the center of inertia [math]\displaystyle{ \vel{G=O}{E} }[/math] are strictly proportional, the horizontal components of [math]\displaystyle{ \vel{G}{E} }[/math] are also constant.

Is the angular momentum conserved?

The gravitational torsor at the gravity centre [math]\displaystyle{ \Gs }[/math] (which is the same point as the inertia centre [math]\displaystyle{ \Os }[/math]) reduces to a resultant force and no moment. Therefore:

[math]\displaystyle{ \sum\overline{\mathrm{M}}_\mathrm{ext}(\Gs)=\overline{0}=\dot{\overline{\mathrm{H}}}_\mathrm{RTG} (\Gs) \quad \Rightarrow \quad \overline{\mathrm{H}}_\mathrm{RTG} (\Gs) \mathrm{CONSTANT!} }[/math]

For the rigid body under study: [math]\displaystyle{ \overline{\mathrm{H}}_\mathrm{RTG} (\Gs) = \Is\Is (\Gs) \velang{}{RTG} = \Is\Is (\Gs) \velang{}{T}. }[/math]

The angular momentum and the angular velocity are not proportional in general (they are only proportional when the direction of angular velocity is a a principal direction of inertia for the centre of inertia [math]\displaystyle{ \Gs }[/math] (section D5.3)), and the conservation of the former does not imply that of the latter.

Qualitative assessment of the inertia tensor

Since the two plates are symmetrical

[math]\displaystyle{ \left[\Is\Is(\Gs)\right]=\left[\Is\Is(\Gs)\right]_\text{lower plate} + \left[\Is\Is(\Gs)\right]_\text{upper plate} = \diag{\Is_{11}}{\Is_{11} + \Is_{33}}{\Is_{33}}+ \diag{2\Is}{\Is}{\Is} , \text{ amb} \Is_{11}\lt \Is_{33}. }[/math]

Quantitative assessment of the inertia tensor

[math]\displaystyle{ \left[\Is\Is(\Gs)\right]=\frac{1}{3} \ms\Ls^2 \diag{1}{1+4}{4}+\frac{1}{3} \ms\Ls^2 \diag{2}{1}{1}=\frac{1}{3} \ms\Ls^2 \diag{3}{6}{5} \equiv \diag{\Is_\mathrm{low}}{\Is_\mathrm{high}}{\Is_\mathrm{medium}} }[/math]

The directions 1, 2 and 3 are the inertia principal directions for point [math]\displaystyle{ \Gs }[/math].

Angular momentum calculation

[math]\displaystyle{ \left\{\overline{\mathrm{H}}_\mathrm{RTG} (\Gs)\right\}=\diag{\Is_\mathrm{low}}{\Is_\mathrm{high}}{\Is_\mathrm{medium}} \vector{\Omega_1}{\Omega_2}{\Omega_3}=\vector{\Is_\mathrm{low}\Omega_1}{\Is_\mathrm{high}\Omega_2}{\Is_\mathrm{medium}\Omega_3} }[/math], is not proportional to [math]\displaystyle{ \velang{}{T} }[/math] in principle.

If the initial angular velocity is exclusively in one of the three directions (that is, if its direction is a principal direction of inertia for [math]\displaystyle{ \Gs }[/math]), then there is proportionality between [math]\displaystyle{ \overline{\mathrm{H}}_\mathrm{RTG} (\Gs) }[/math] and [math]\displaystyle{ \velang{}{T} }[/math],and the conservation of the former implies that of the latter.

ANIMACIONS

✏️ EXAMPLE D8.7: gyroscope

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The system consists of a homogeneous disk with mass m and radius r, articulated to a massless support, and a fork that can rotate freely about the vertical axis. There is a motor between the support and the fork. The inertia moment of the support with respect to the vertical axis through the center of the disk is [math]\displaystyle{ \Is=(\lambda/2)\ms\rs^2 }[/math]. Initially [math]\displaystyle{ (\ts_\mathrm{incial} }[/math] the disk is parallel to the ground, and rotates with vertical angular velocity [math]\displaystyle{ \velang{disc}{T}=\psio }[/math]. The friction associated with all the joints is neglected. We want to investigate how the support moves when the motor changes the disk orientation relative to the ground.

Is the angular momentum conserved?

For the disk SYSTEM, the resultant moment about its centre of inertia [math]\displaystyle{ \Gs }[/math] is zero in the direction of its axis. The motor can change the orientation of this axis relative to the ground (and relative to any reference frame with a translational motion relative to the ground), and therefore it is not a direction fixed to the ground (so not to the RTG either): the angular momentum is not conserved in this direction.

For the SYSTEM (disk + support + fork), the resulting moment relative to the center of the disk [math]\displaystyle{ \Os }[/math] is zero in the vertical direction, which is fixed to the ground. Therefore:

[math]\displaystyle{ \left.\sum \overline{\Ms}_\mathrm{ext}(\Os)\right]_\mathrm{vert} =0 \quad \Rightarrow \quad \left.\overline{\Hs}_\mathrm{RTO=T}(\Os) \right]_\mathrm{vert} \text{ CONSTANT!} }[/math]

As the motor changes the orientation of the disk plane [math]\displaystyle{ (\dot{\theta} \neq 0) }[/math], the support rotates around the vertical axis [math]\displaystyle{ \dot{\psi} }[/math]. The angular momentum at [math]\displaystyle{ \Os }[/math] is:

[math]\displaystyle{ \overline{\Hs}_\mathrm{RTO=E}(\Os,\ts)=\Is\Is^\mathrm{sup}(\Os)\velang{sup}{E}(\ts)+\Is\Is^\mathrm{disk}(\Os)\velang{disk}{E}(\ts)=\Is\Is^\mathrm{sup}(\Os)\dot{\psi}+\Is\Is^\mathrm{disk}(\Os)\left( \overline{\dot{\psi}}+ \overline{\dot{\theta}}+ \overline{\dot{\varphi}}\right) }[/math]

[math]\displaystyle{ \braq{\overline{\Hs}_\Es^\mathrm{sup}(\Os,\ts)}{B}=\diag{\Is_{11}}{\Is_{22}}{(\lambda/2)\ms\rs^2}\vector{0}{0}{\dot{\psi}}=\vector{0}{0}{(\lambda/2)\ms\rs^2 \dot{\psi}}, }[/math]

[math]\displaystyle{ \braq{\overline{\Hs}_\Es^\mathrm{disk}(\Os,\ts)}{B'}=\frac{1}{4}\ms\rs^2\diag{1}{1}{2}\vector{-\dot{\theta}}{-\dot{\psi} \sin\theta}{\dot{\varphi}+\dot{\psi}\cos\theta}=\frac{1}{4}\ms\rs^2\vector{-\dot{\theta}}{-\dot{\psi} \sin\theta}{2(\dot{\varphi}+\dot{\psi}\cos\theta)} }[/math]

The angular momentum in the vertical direction comes from the projection of components 2 and 3:

[math]\displaystyle{ \left.\overline{\Hs}_\Es(\Os,\ts)\right]_\mathrm{vert} =\overline{\Hs}_\Es^\mathrm{sup}(\Os,\ts)+ \left.\overline{\Hs}_\Es^ \mathrm{disk} (\Os,\ts) \right]_{3'}\cos\theta - \left.\overline{\Hs}_\Es^\mathrm{disk}(\Os,\ts)\right]_{2'} \sin\theta , }[/math]

[math]\displaystyle{ \left.\overline{\Hs}_\Es(\Os,\ts)\right]_\mathrm{vert} = \left(\Uparrow \frac{\lambda}{2}\ms\rs^2\dot{\psi}\right)+ \left(\Uparrow \frac{1}{4}\ms\rs^2\left[2\dot{\varphi}\cos\theta+\dot{\psi}(1+\cos^2\theta)\right]\right) }[/math]

Taking into account the conservation of the vertical angular momentum:

[math]\displaystyle{ \left.\begin{array}{ll} \left.\overline{\mathrm{H}}_{\mathrm{E}}(\mathbf{O}, \mathrm{t})\right]_{\text {vert }}=\left(\Uparrow \frac{1}{4} \mathrm{mr}^2\left[2 \dot{\varphi} \cos \theta+\dot{\psi}\left(1+2 \lambda+\cos ^2 \theta\right)\right]\right) \\ \left.\begin{array}{l} \dot{\psi}\left(\mathrm{t}_{\text {initial }}\right)=0 \\ \dot{\varphi}\left(\mathrm{t}_{\text {initial }}\right)=\dot{\varphi}_0 \\ \theta\left(\mathrm{t}_{\text {initial }}\right)=0 \end{array}\right\} \left.\Rightarrow \overline{\mathrm{H}}_{\mathrm{T}}\left(\mathbf{O}, \mathrm{t}_{\text {initial }}\right)\right]_{\text {vert }}=\left[\Uparrow\left(\frac{1}{2} \mathrm{mr}^2 \dot{\varphi}_0\right)\right] \\ \end{array}\right\} \Rightarrow 2 \dot{\varphi} \cos \theta+\dot{\psi}\left(1+2 \lambda+\cos ^2 \theta\right)=2 \dot{\varphi}_0 }[/math]

On the other hand, for the SYSTEM disk, the external moment in the direction of the disk axis (direction 3') is zero. Hence, [math]\displaystyle{ \left.\dot{\overline{\Hs}}_\mathrm{disk,T}(\Os)\right]_{3'}=0 }[/math]

[math]\displaystyle{ \braq{\dot{\overline{\Hs}}_\mathrm{disk,E}(\Os,\ts)}{B'}=\frac{1}{4}\ms\rs^2 \vector{-\ddot{\theta}}{-\ddot{\psi} \sin\theta -\dot{\psi}\dot{\theta}\cos\theta}{2(\ddot{\varphi}+\ddot{\psi}\cos\theta-\dot{\psi}\dot{\theta}\sin\theta)} + \vector{-\dot{\theta}}{-\dot{\psi}\sin\theta}{\dot{\psi}\cos\theta}\times\frac{1}{4}\ms\rs^2\vector{-\dot{\theta}}{-\dot{\psi}\sin\theta}{2(\dot{\varphi}+\dot{\psi}\cos\theta)}= }[/math]

[math]\displaystyle{ \hspace{2.5cm}= \frac{1}{4}\ms\rs^2\vector{-\ddot{\theta}-\dot{\psi}(2\dot{\varphi}+\dot{\psi}\cos\theta)\sin\theta}{-\ddot{\psi}\sin\theta + 2\dot{\theta}\dot{\varphi}}{2(\ddot{\varphi}+\ddot{\psi}\cos\theta-\dot{\psi}\dot{\theta}\sin\theta)} \Rightarrow \ddot{\varphi}+\ddot{\psi}\cos\theta-\dot{\psi}\dot{\theta}\sin\theta=0. }[/math]

The integration of that equation yields: [math]\displaystyle{ \dot{\varphi}+\dot{\psi}\cos\theta=\dot{\varphi}_0 }[/math], where [math]\displaystyle{ \dot{\varphi}_0 }[/math] is the integration constant, which may be found from the initial conditions.

Combination of this result with the previous one yields:

[math]\displaystyle{ \left.\begin{array}{l} 2\dot{\varphi}\cos\theta+\dot{\psi}(1+2\lambda+\cos^2\theta )=2\dot{\varphi}_0\\ \dot{\varphi}+\dot{\psi}\cos\theta=\dot{\varphi}_0 \end{array}\right\} \Rightarrow \frac{\dot{\psi}}{\dot{\varphi}_0}=\frac{2(1-\cos\theta)}{2\lambda+\sin^2\theta} }[/math]

✏️ EXAMPLE D8.8: bar in a smooth rotating circular guide

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The homogeneous bar PQ, with mass m, moves with its two endpoins inside a smooth guide [math]\displaystyle{ (\mu=0) }[/math], with radius r and negligible mass, which can rotate freely around the vertical direction. The POQ angle is [math]\displaystyle{ 120^o }[/math]. We want to find the equation of motion for the coordinate [math]\displaystyle{ \psi }[/math]. The rotation of the bar on its axis (spin [math]\displaystyle{ \dot{\varphi} }[/math]) is neglected.

Is the linear momentum conserved?

The external forces on the bar are not zero: in addition to the weight, there are normal forces from the guide on the bar at [math]\displaystyle{ \Ps }[/math] and [math]\displaystyle{ \Qs }[/math] (directed towards [math]\displaystyle{ \Os }[/math]) and a force component perpendicular to the plane of the guide. Therefore, the resultant external force on the bar has components in the three directions of space, and the linear momentum is not conserved.

If we analyze the external forces on the SYSTEM (bar + guide), the conclusion is the same: in addition to the weight of the bar, there is the constraint force associated with the bearing between the ground and the guide, which has three non-zero components in principle.

Is the angular momentum conserved?

For the SYSTEM (bar + guide), the external moment about any point on the axis of rotation of the guide (in particular, for point [math]\displaystyle{ \Os }[/math]) has a zero vertical component (since the constraint moment of the bearing in that direction is zero, and the weight cannot yield a moment in the vertical direction). On the other hand, the angular acceleration of the guide relative to the ground [math]\displaystyle{ \left(\overline{\ddot{\psi}}\right) }[/math] is vertical. Therefore:

[math]\displaystyle{ \boxed{\left.\text{Roadmap: SYSTEM (bar+guide), AMT at }\Os\right]_\mathrm{vert}} }[/math]

[math]\displaystyle{ \left.\sum\overline{\Ms}_\mathrm{ext}(\Os) \right]_\mathrm{vert}=0 \quad \Rightarrow \quad \left.\overline{\Hs}_\mathrm{RTG}(\Os)\right]_\mathrm{vert} \text{ CONSTANT!} }[/math]

The only element with nonzero mass is the bar, and point [math]\displaystyle{ \Os }[/math] is fixed to it:

[math]\displaystyle{ \overline{\Hs}_\mathrm{RTO}(\Os)=\Is\Is(\Os)\velang{bar}{RTO=E}=\left[\Is\Is(\Gs)+\Is\Is^\bigoplus(\Os)\right]\left(\overline{\dot{\psi}}+\overline{\dot{\theta}}\right) }[/math]

[math]\displaystyle{ \braq{\overline{\Hs}_\mathrm{RTO}(\Os)}{}=\left(\ms(\sqrt{3\rs})^2\diag{1}{0}{1}+\ms\left(\frac{\rs}{2}\right)^2\diag{1}{1}{0}\right)\vector{\dot{\theta}}{\dot{\psi}\sin\theta}{\dot{\psi}\cos\theta} }[/math]

[math]\displaystyle{ \braq{\overline{\Hs}_\mathrm{RTO}(\Os)}{}=\frac{1}{4}\ms\rs^2\vector{14\dot{\theta}}{\dot{\psi}\sin\theta}{13\dot{\psi}\cos\theta} }[/math]

[math]\displaystyle{ \left.\overline{\Hs}_\mathrm{RTO}(\Os)\right]_\mathrm{vert} =\left.\overline{\Hs}_\mathrm{RTO}(\Os)\right]_3\cos\theta+ \left.\overline{\Hs}_\mathrm{RTO}(\Os)\right]_2\sin\theta= \frac{1}{4}\ms\rs^2\dot{\psi}(13\cos^2\theta + \sin^2\theta)= \frac{1}{4}\ms\rs^2 \dot{\psi}(1+12\cos^2\theta) }[/math]

[math]\displaystyle{ \left.\overline{\Hs}_\mathrm{RTO}(\Os)\right]_\mathrm{vert} =\text{constant } \Rightarrow \quad \frac{\ds\left.\overline{\Hs}_\mathrm{RTO}(\Os)\right]_\mathrm{vert}}{\ds\ts}=0=\frac{1}{4}\ms\rs^2 \dot{\psi}(1+12\cos^2\theta)-6\ms\rs^2 \dot{\psi}\dot{\theta}\sin\theta\cos\theta }[/math]

[math]\displaystyle{ \boxed{\ddot{\psi}(3+7\cos^2\theta)-14\dot{\psi}\dot{\theta} \sin\theta \cos\theta =0} }[/math]

Relevant comment

The vertical angular momentum at [math]\displaystyle{ \Qs }[/math] and [math]\displaystyle{ \Ps }[/math] is not conserved since those two points are accelerated:

[math]\displaystyle{ \sum \overline{\bar{\Ms}}_{\text {ext }}(\mathbf{Q})-\overline{\mathbf{P G}} \times \ms \overline{\mathrm{a}}_{\text {Gal }}(\mathbf{Q})=\dot{\overline{\mathrm{H}}}_{\text {RTQ }}(\mathbf{Q}), \quad \overline{\mathbf{QG}} \times \ms \overline{\mathrm{a}}_{\text {Gal }}(\mathbf{Q})=\overline{\mathbf{QG}} \times \mathrm{m}\left[\overline{\mathrm{a}}_{\text {RЕL }}(\mathbf{Q})+\overline{\mathrm{a}}_{\mathrm{tr}}(\mathbf{Q})+\overline{\mathrm{a}}_{\text {cor }}(\mathbf{Q})\right] }[/math]

[math]\displaystyle{ \left\{\overline{\mathbf{Q G}} \times \ms \bar{\mathrm{a}}_{\mathrm{Gal}}(\mathbf{Q})\right\}=\left\{\begin{array}{c} 0 \\ \sqrt{3\rs} \cos \theta \\ \sqrt{3\rs} \sin \theta \end{array}\right\} \times \ms\left[\left\{\begin{array}{c} 0 \\ \mathrm{a}_{\mathrm{REL}}^{\mathrm{s}} \sin \left(30^{\circ}-\theta\right)-\mathrm{a}_{\mathrm{REL}}^{\mathrm{n}} \cos \left(30^{\circ}-\theta\right) \\ \mathrm{a}_{\mathrm{REL}}^{\mathrm{s}} \cos \left(30^{\circ}-\theta\right)+\mathrm{a}_{\mathrm{REL}}^{\mathrm{n}} \sin \left(30^{\circ}-\theta\right) \end{array}\right\}+\left\{\begin{array}{c} \mathrm{a}_{\mathrm{tr}}^{\mathrm{s}}-\mathrm{a}_{\mathrm{Cor}} \\ -\mathrm{a}_{\mathrm{tr}}^{\mathrm{n}} \\ 0 \end{array}\right\}\right] }[/math]

[math]\displaystyle{ \left.\left.\left.\overline{\mathbf{Q G}} \times \ms \overline{\mathrm{a}}_{\text {Gal }}(\mathbf{Q})\right]_{\text {vert }}=\overline{\mathbf{QG}} \times \ms \overline{\mathrm{a}}_{\text {Gal }}(\mathbf{Q})\right]_{3^{\prime}}=\sqrt{3} \mathrm{m}\left(\mathrm{a}_{\text {Cor }}-\mathrm{a}_{\mathrm{tr}}^{\mathrm{s}}\right) \sin \theta \neq 0 \quad \Rightarrow \quad \dot{\bar{\Hs}}_{\text {RTQ }}(\mathbf{Q})\right]_{\text {vert }} \neq 0 }[/math]