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Sorry for the vague title, but pls read my question below

Imagine a rigid body b with a point-mass tail t attached to its back at joint J. The tail has 1 DoF and can be actuated by a motor with axis of rotation M (parallel to yaw axis). The whole system is in free space (i.e no effect of gravity and air drag) and is in rest (i.e zero initial angular/linear momentum and no external force/torque acting).

Also, suppose that COM of the system is located at joint J and the tail is aligned with the roll axis of the body.

Edit: The COM of the body is assumed to be at b.

The initial configuration is shown below in the image :

intial_configuration

Now, if the tail is rotated, the body will rotate in the opposite direction, since no external torque is being applied and the angular momentum will remain conserved (i.e zero). As there is no external force acting, the position of COM of the system will not change, i.e joint J will be static. This should look like as -

enter image description here

But then, wouldn't the new COM of the system will be at J' (as the COM of the system should be on the line joining the centre of mass of the body and tail). Isn't this incorrect?

My question is, why is this happening? How should the body and tail rotate so as to satisfy both the conservation of angular momentum (no external torque) and centre of mass (no external force).

If possible, answer with relevant equations. Also, pls explain with a diagram, showing the position of body, tail, joint J and the COM of the system assuming the same initial configuration.

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  • $\begingroup$ How did you arrive at your new COM position in the diagram? Why do you think it is correct or incorrect? $\endgroup$
    – Bill N
    May 13 at 17:36
  • $\begingroup$ @BillN I have assumed the COM of the body to be at b [Fig 1]. Then, the COM of the system should be on the line joining the centre of mass of the body and the tail. $\endgroup$
    – Alu
    May 13 at 18:45
  • $\begingroup$ Why did you draw the COM as moving? Why did you keep the lower corner of the box mass at the same $y$ location (if $y$ is vertical in your diagram) ? $\endgroup$
    – Bill N
    May 13 at 18:54
  • $\begingroup$ The COM of the system from the initial condition is at J. So the body will rotate about J. As, the COM of the system shouldn't move (no external forces). But, then if we again calculate the COM of the system, it will be at J' (as explained before). Regarding the lower corner, it has shifted slightly (not visible properly in the figure), J is fixed. $\endgroup$
    – Alu
    May 13 at 19:04
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If you were to fix body b and move the tail, then the COM would be a function of the tail angle. Use a coordinate system where the joint J is at the origin and find the COM as a function of tail angle $\theta$

fig1

$$ \mathbf{COM}(\theta) = \pmatrix{ \frac{ m_b c - m_t \ell \cos \theta}{m_b + m_t} \\ \frac{ - m_t \ell \sin \theta}{m_b + m_t} } $$

such that with $c = \frac{m_t}{m_b} \ell$ and $\theta =0$ the COM is at the origin.

Now when body b is free, kinematically the joint could be at any location. I use the coordinates $\pmatrix{x_b \\ y_b}$ for the location of the joint.

From the kinetics of the problem the center of mass remains fixed at the origin which means the following must be true

$$ \pmatrix{x_b \\ y_b } + \pmatrix{ \frac{ m_b c - m_t \ell \cos \theta}{m_b + m_t} \\ \frac{ - m_t \ell \sin \theta}{m_b + m_t} } = \pmatrix{0 \\ 0} $$

which produces the following solution of the joint position

$$ x_b = -\frac{ m_b c - m_t \ell \cos \theta}{m_b + m_t} $$ $$ y_b = \frac{ m_t \ell \sin \theta}{m_b + m_t} $$

So when the tail displaces with $\theta$ as shown above, the joint moves up and to the left a bit. But the COM remains at (0,0).

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  • $\begingroup$ But since there are no external forces, the COM of the system shouldn't move, right? If it will move, then aren't we violating Newton's second law of motion? $\endgroup$
    – Alu
    May 14 at 10:53
  • $\begingroup$ Also, I am a bit overwhelmed with the equations in ur case, if possible please explain in simple terms. $\endgroup$
    – Alu
    May 14 at 11:02
  • $\begingroup$ @Alu - that is true. The combined COM remains fixed, but he pivot accelerates vertically. Starting from rest, the vertical acceleration vector of the tail is $\frac{M}{m_t \ell}$ while the acceleration of the body COM is $-\frac{M}{m_b \ell}$. Thus the acceleration of the combined COM is $$ m_t (\frac{M}{m_t \ell}) + m_b (- \frac{M}{m_b \ell}) = 0$$ $\endgroup$
    – JAlex
    May 14 at 18:16
  • $\begingroup$ As assumed in the initial condition, the COM is at the pivot J, so theoretically pivot J shouldn't move. Sorry, but I am still unable to get your point. $\endgroup$
    – Alu
    May 14 at 18:45
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    $\begingroup$ For multibody systems the combined COM does not move (or moves with constant velocity), but that does not mean the pivot won't move. The pivot isn't the COM, it just coincides at that instant. Each configuration of pivot angle results in a different COM location relative to the main body (it moves with the tail). $\endgroup$
    – JAlex
    May 17 at 12:48
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The whole object is at rest in an external coordinate system S which is fixed to your piece of drawing paper. That is, the CM of the object (the point J) is at $\vec{x}$ in S and $\vec{x}$ is not changing with time. After a little spring (that is part of the object) rotates the tail with respect to the body, the new CM is at J' like you have drawn it on the body, except you incorrectly moved J' away from J on your drawing paper. J' remains at $\vec{x}$ and you should have redrawn the bent tail+body lower on the paper to make this so.

This is just an elaborate form of Michael Seifert's answer which correctly was "J′ will end up at the original position of J in space". You just should have moved your second drawing so that J' was at the same place as J on your drawing paper.

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If $J$ is fixed in space, then the body is not free, and forces can be exerted on the system allowing the COM to move. Alternately, if $J$ is not fixed, then the COM will remain at rest; and if the "tail" moves relative to the "block", then $J'$ will end up at the original position of $J$ in space.

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  • $\begingroup$ J is not supported by any means i.e it is also free to move as is the whole system which is in free space. But, as the COM of the system lies at J, it won't move as no external force or torque is acting on the system. $\endgroup$
    – Alu
    May 14 at 11:08
  • $\begingroup$ @Alu: My point is that the laws of physics say that the COM remains fixed for an isolated system. As long as $J$ happens to lie at the COM, it will remain fixed. But if the distribution of mass in the body changes, and $J$ no longer lies at the COM, then $J$ must move away from its original location. $\endgroup$ May 14 at 12:06
  • $\begingroup$ May you please, explain with a diagram, showing the position of body, tail, joint J and the COM of the system. Assume the same initial configuration. $\endgroup$
    – Alu
    May 14 at 12:15
  • $\begingroup$ @Alu: I probably won't have time to draw up a diagram over the next few days, but I'll try to remember to return to this when I have more time. Honestly, though, your diagram above would serve pretty well for this if you shifted the new configuration so that $J'$ lies on the old location of $J$. $\endgroup$ May 14 at 12:21

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