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As often happens for the rotational world, the Newton's law can be restated in terms of angles, moment of inertia and torques replacing displacements, masses and forces (this is one source among millions on the web).

While for the translational case, I have no problem to figure out which are the forces due to the Newton's third principle, with a rotating system composed by many bodies, I have some problem.

To overcome my problem, I try to figure out some "mental experiments" which I think may help me to understand the topic. Here is one of these experiments.

Consider a simple DC motor able to generate a torque. We can assume that this motor is a dimensionless object with mass $M_1$. We fix this motor on a horizontal surface (moment of inertia: $I_2$, mass: $M_2$), which can freely rotate around a point.

We put a bar (moment of inertia: $I$, mass: $M_3$) on the top of the motor, like a gramophone.

The scheme is represented in the following figure:

enter image description here

The bar is fixed, so we cannot move it (high forces may destroy it). The only way to move the bar is to turn on the motor.

If we switch on the motor, then it will create a constant torque $\tau$.

The system I'm proposing is composed by $3$ bodies: $2$ rigid bodies (the surface, which is not necessarily fixed on the ground, and the bar) and $1$ dimensionless (the motor).

The question is:

Which are the torques on the bar, on the motor and on the rotating surface due to the Newton's third principle?

We can also consider the floor as a forth body. In that case, which is the torque on the floor?

The intuition suggests me that when I turn on the motor, then both bar and rotating surface will start to rotate.

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closed as unclear what you're asking by sammy gerbil, user197851, ZeroTheHero, Jon Custer, user191954 Mar 22 at 9:37

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From what I understand you have two connected rigid bodies with a 1 degree of freedom revolute joint between them. This joint is capable of producing constant torque and it constrains the location and other orientations of the bodies accordingly.

Newton's third law allows the splitting of the joint such that equal and opposite pairs of forces and moments apply to the two bodies. Say the vector forces and moments at the joint are $\vec{F}$ and $\vec{M}$. If the direction of the joint is $\vec{z}$ then the joint torque is specified with the equation $$\tau = \vec{z} \cdot \vec{M}$$.

FBD

Now the equations of motion for the two bodies (assuming they are free floating otherwise) are evaluated by summing the forces and moments at each center of mass. The joint reactions $\vec{F}$ and moment $\vec{M}$ apply to body (2) and equal and opposite at the joint location A apply to body (1).

$$ \begin{aligned} -\vec{F} & = m_1 \vec{a}_1 & \vec{F} & = m_1 \vec{a}_2 \\ -\vec{M} - (\vec{r}_A - \vec{r}_1) \times \vec{F} & = \mathrm{I}_1 \vec{\alpha}_1 + \vec{\omega}_1 \times \mathrm{I}_1 \vec{\omega}_1 & \vec{M} + (\vec{r}_A - \vec{r}_2) \times \vec{F} & = \mathrm{I}_2 \vec{\alpha}_2 + \vec{\omega} _2\times \mathrm{I}_2 \vec{\omega}_2 \end{aligned} $$

These are 4 vector equations, each with 3 components (12 relationships), but with 6 vector unknown quantities (joint reaction force $\vec{F}$, joint reaction moment $\vec{M}$, body (1) acceleration $\vec{a}_1$ , body (2) acceleration $\vec{a}_2$, body (1) rot. acceleration $\vec{\alpha}_1$, body (2) rot. acceleration $\vec{\alpha}_2$) for a total of 18 unknown values.

Of course, the acceleration of body (2) depends on the acceleration of body (1), using the kinematic relationship (linear and rotational) provides for 2 more vector equations (18 relationships total so far) and introduces one more variable (the joint angle acceleration $\ddot{\theta}$) for a total of 19 variables.

$$ \begin{aligned} \vec{a}_2 + \vec{\alpha}_2 \times ( \vec{r}_A - \vec{r}_2 ) + \ldots & = \vec{a}_1 + \vec{\alpha}_1 \times ( \vec{r}_A - \vec{r}_1 ) + \ldots \\ \vec{\alpha}_2 & = \vec{\alpha}_1 + \vec{z} \ddot{\theta} + \ldots \end{aligned} $$

But wait we know the torque along the joint axis is motor torque $\tau$, which as stated earlier is specified with $\tau = \vec{z} \cdot \vec{M}$ which is one more scalar equation, for a total of 6×3+1 = 19 equations and 2×6 + 6 + 1 = 19 unknowns.

The problem is solvable, yet extremely complex. In general with $n$ connected rigid bodies, there are $13n-7$ equations to solve. Quite fun!

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  • $\begingroup$ Thanks a lot. Therefore, we can say that $\vec{M}$ is an internal force? This is misleading for me, since it seems that it has been created by a generator of torque, and thus it should be a "external". $\endgroup$ – the_candyman Mar 12 at 18:59
  • $\begingroup$ Moreover, does $\vec{\omega}_1 \times \mathrm{I}_1 \vec{\omega}_1$ equal $0$? $\endgroup$ – the_candyman Mar 12 at 19:12
  • $\begingroup$ $\vec{M}$ as a vector is an internal moment. A single component of $\vec{M}$ along the direction of the joint is specified by the generator torque. There are other reaction moments that are needed to keep the two bodies aligned per the joint. $\endgroup$ – ja72 Mar 12 at 20:08
  • $\begingroup$ I general the (gyroscopic) terms $\vec{\omega} \times \mathrm{I} \vec{\omega}$ are zero only if the instant rotation $\vec{\omega}$ is parallel to one of the principal directions of the MMOI. So in general the answer is no. $\endgroup$ – ja72 Mar 12 at 20:13
  • $\begingroup$ Thanks a lot again $\endgroup$ – the_candyman Mar 12 at 20:14

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