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This problem appeared on the 2021 $F=ma$ Test that was held three days ago. However, I'm having trouble understanding other solutions.

Exact wording of problem:

"A uniform solid circular disk of mass $m$ is on a flat, frictionless horizontal table. The center of mass of the disk is at rest and the disk is spinning with angular frequency $\omega_0$. A stone, modeled as a point object also of mass $m,$ is placed on the edge of the disk, with zero initial velocity relative to the table. A rim built into the disk constrains the stone to slide, with friction, along the disk's edge. After the stone stops sliding with respect to the disk, what is the angular frequency of rotation of the disk and the stone together?"

Given answer: $\omega_f=(1/2) \omega_0$.

My solution: In order to apply conservation of angular momentum, we need to have a COR that stays consistent. So that means we should choose the center of mass of the system as our COR throughout the process. The beginning angular momentum with CM=COR would be $(1/2+1/4)mR^2 \omega_0$ by parallel axis theorem. And our final angular momentum with CM=COR should be $\omega_f m(R/2)^2+(1/2+1/4)mR^2 \omega_f$. This gives us $\omega_f=3/4 \omega_0$.

However, in some of the solutions other people have posted, they choose the center of the disk as the beginning COR and then the CM as the final COR. I'm pretty sure this is incorrect since the angular momentum depends on COR.

In the other solutions that do keep COR=CM throughout the process, they say that the beginning angular momentum is $I_{center} \omega_0$ instead of $I_{cm} \omega_0$. Shouldn't angular momentum be calculated with the moment of inertia of the COR as the center?

So far, I have no idea what is conceptually wrong with my solution, so any insights are appreciated.

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This problem has a major difference from many similar textbook problems dealing with the conservation of angular momentum.

In many of the textbook problems the axis of rotation of the disk about its center of mass (CM) is fixed by a constraining force/torque and does not move. If that were the case here, the axis of rotation would not move when the stone is placed (and moves) on the disk. The problem would be addressed considering angular momentum of the disk/stone about the fixed axis of rotation.

Here, the disk is on a frictionless table and its axis of rotation can move, but is always perpendicular to the table surface. The problem is best evaluated considering the angular momentum about the center of mass of the disk/stone system. Denote the center of mass of the system consisting of the disk and the stone as CMS (center of mass of system). Due to conservation of linear momentum the CMS does not move. To keep CMS fixed as the stone moves (rotates on the disk), the CM of the disk moves and rotates about the CMS. Also, the disk rotates about its own CM, and the stone rotates about the CMS. Angular momentum is conserved. The overall motion is as quantified in an earlier answer by @ytlu.

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A conceptual error: The parallel axis theorem is used to find the inertial moment about an rotating axis which is not the CM. In this problem, the disk is not rotating about the CM (for rock and disc joint system) but it rotates around the center of disk.

Therefore, it should be think in decomposing the motion into motion of the center of the disk around the CM, and the rotation around the center of disk.

$$ L_{disc-total} = L_{disc-around-CM} + L_{disc-rotation-around-disc-center}. $$

Therefore, in the beginning, the rock and disk are both not moving w.r.t the CM (between rock and disk center:

$$ L_i = 0 + 0 + \frac{1}{2} mR^2 \omega_0. $$

In the final motion, the rock is rotating around the CM, and the disc center is also rotating around the CM, in order to keep the center of mass fixed. Besides, the disc has additional rotation about the disc center. All these rotations are of a same frequence $\omega_f$ for syncronization.

$$ L_f = L_{rock} + L_{disc} = L_{rock} + \left( L_{disc-center} + L_{disk-rotation} \right),\\ = \frac{1}{4}mR^2 \omega_f +\left( \frac{1}{4}mR^2 \omega_f + \frac{1}{2}mR^2 \omega_f\right) \\ = mR^2 \omega_f. $$

Then applying $L_i = L_f$, render

$$ \omega_f = \frac{1}{2} \omega_0. $$

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    $\begingroup$ Excellent explanation! $\endgroup$
    – John Darby
    Feb 21 at 21:26
  • $\begingroup$ Contrary to many textbook problems, the axis of rotation of the disk about its CM is not constrained and it moves. I added a short answer to point this out. $\endgroup$
    – John Darby
    Feb 23 at 16:12
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In order to apply conservation of angular momentum, we need to have a COR that stays consistent.

I agree that this seems like it should be true, but in reality it's not what the conservation of angular momentum is. The law of conservation of angular momentum states that when no external torque acts on an object, no change of angular momentum will occur.

If you look at the object (rock-disk system), if you choose any non-fixed point you will be introducing linear momentum into the equation, which you cannot do for free (i.e without subtracting from angular momentum)

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  • $\begingroup$ I'm not following your logic in the first paragraph, could you explain it further? Angular momentum is definitely conserved but it still changes with COR. As an analogy, take linear momentum. Linear momentum is conserved in any inertial frame if there aren't external forces, but the linear momentum between two inertial frames aren't equal (ie. a a frame with v=0 m/s and a frame with v=100m/s) I agree with the second paragraph, the COR has to be inertial to apply conservation of linear momentum. That's why I'm choosing the CM to calculate my linear momentum. $\endgroup$
    – Luke Huang
    Feb 21 at 18:57
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In order to apply conservation of angular momentum, we need to have a COR that stays consistent

As senor o has answered this is not true the only condition for angular momentum to be conserved is that there should be no external torque.

This conceptual error made you say this:

The beginning angular momentum with CM=COR would be $(1/2+1/4)mR^2 \omega_0$

What you did here is calculated angular momentum about an axis the disc does not even rotate about, this should seem wrong to you. The $I$ in the angular momentum formula $I\omega$ is the moment of inertia about the axis of rotation. Not the axis of future rotation.

Shouldn't angular momentum be calculated with the moment of inertia of the COR as the center?

I don't really understand this statement, are there any further queries?

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  • $\begingroup$ Yes, I see where I made error calculating the beginning angular momentum. But now, how are we supposed to equate the angular momentum of one axis of rotation to the angular momentum of a different axis of rotation? I think this question is what lead me to make my conceptual error. Equating those two angular momentum seems wrong to me, since the same analogy applied to linear momentum is not true either. $\endgroup$
    – Luke Huang
    Feb 21 at 19:08
  • $\begingroup$ The axis of rotation need not be the same for angular momentum to be conserved, For example, take a spinning disc and send it flying in space the axis is continuously moving. Would you still say angular momentum is not conserved? Physics can be quite non-intuitive sometimes. Your analogy to linear momentum tells me you treat angular momentum as an ideal vector. It is not. More on that. $\endgroup$
    – Linkin
    Feb 22 at 1:45
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Exact wording of problem:

"A uniform solid circular disk of mass $m$ is on a flat, frictionless horizontal table. The center of mass of the disk is at rest and the disk is spinning with angular frequency $\omega_0$. A stone, modeled as a point object also of mass $m,$ is placed on the edge of the disk, with zero initial velocity relative to the table. A rim built into the disk constrains the stone to slide, with friction, along the disk's edge. After the stone stops sliding with respect to the disk, what is the angular frequency of rotation of the disk and the stone together?"

I agree with you, of course, that in the final state the combined object will be rotating around the Common Center of Mass.

I restate the problem as follows: initially the stone is sliding frictionless along the rim. Then it encounters a stop and there is an instantaneous inelastic collision.

That is, I'm explicitly stating that whether or not there is a friction phase does not change the outcome. (I'm guessing you are already proceeding that way, implicitly.)


In the initial state the stone is stationary with respect to the coordinate system, so it has zero angular momentum with respect to any point of that coordinate system.

At the instant that the instantaneous inelastic collision occurs both objects undergo a symmetric shift of their axes of angular velocity.

We can treat the initial angular velocity of the stone as an angular velocity of zero around its own center of mass.

At the instant of collision both axes shift, towards each other, so that in the final state the two axes coincide.


So: my proposal is to capitalize on symmetry. Treat the problem as a problem with initially two axes of rotation. Both axes of rotation shift, towards each other, and come to coincide. During that shift of axes of rotation the two objects extert a torque upon each other. As a matter of principle: around the CCM those two torques are equal and opposite.

Conversely, it would be very tricky to attribute an angular momentum to the disk around the future Common Center of Mass, since in the initial state the disk isn't actually rotating around that axis. Those two axes have an instantaneous velocity with respect to each other that would need to be accounted for.

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