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Sorry for asking an elementary question. I was asked this by my friend but I for some reason was not able to produce any answer even after thinking about it for a while.

My question is, if a particle strikes, let's say, a rigid ring tangentially with some arbitrary velocity, and sticks to it after the collision then will the system rotate about the center of mass of the original ring or the center of mass of the new system ? (which would be somewhere between the point the particle strikes and the center of the ring)

The ring is kept on a perfectly smooth horizontal surface.

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    $\begingroup$ How did you come to the conclusion that the system rotates around the center of mass of the ring alone? $\endgroup$ Commented Jun 13, 2020 at 15:01
  • $\begingroup$ @Álvaro Luque This was what I assumed intuitively and got the correct answer for the particular question I was trying to solve. I looked at the solution and the same assumption had been made there as well but not explained. To be completely clear, what I mean is that the SYSTEM will rotate about the point that is at the geometric center of the ring $\endgroup$
    – Amadeus
    Commented Jun 13, 2020 at 15:15
  • $\begingroup$ Maybe the mass of the ring was a lot bigger than that of the particle? In that situation both centers of mass are very close. $\endgroup$ Commented Jun 13, 2020 at 15:17
  • $\begingroup$ @Álvaro Luque no such thing was mentioned $\endgroup$
    – Amadeus
    Commented Jun 13, 2020 at 15:19
  • $\begingroup$ I can't see why it is then. $\endgroup$ Commented Jun 13, 2020 at 15:19

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For a rigid object, the equation of motion for the center of mass $\mathbf x_\text{cm}$ of the object is described by Newton's second law $$\mathbf F_\text{net}=m\ddot{\mathbf x}_\text{cm}$$

Since your new composite system (ring + particle) has no forces acting on it after the collision, we have $0=m\ddot{\mathbf x}_\text{cm}$, i.e. the center of mass cannot be accelerating. If the system was rotating about a point that was not the new center of mass, then the new center of mass would be rotating about that point, and thus it would be accelerating. Therefore, if a rigid system is going to be rotating without a net force acting on it, it must be that it rotates about its center of mass.

After the collision, the system rotates about its center of mass, not about the center of mass of just the ring. However, if $m_\text{particle}\ll m_\text{ring}$, then we can essentially assume the center of mass of the entire system is located at the center of mass of the ring.

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  • $\begingroup$ Is this, then, your conclusion: the system after collision does not rotate around the center of the ring. $\endgroup$
    – garyp
    Commented Jun 13, 2020 at 15:34
  • $\begingroup$ Why won't the equation you wrote be with respect to the new center of mass of the system? $\endgroup$
    – Amadeus
    Commented Jun 13, 2020 at 15:34
  • $\begingroup$ @l1mbo There are no outside forces, so the new COM is the same as the old COM. $\endgroup$
    – garyp
    Commented Jun 13, 2020 at 15:35
  • $\begingroup$ @l1mbo It is with respect to the new COM. $\endgroup$ Commented Jun 13, 2020 at 15:37
  • $\begingroup$ @garyp Yes to your first comment. The new COM is not located at where the old COM is. Your second comment is incorrect. $\endgroup$ Commented Jun 13, 2020 at 15:38
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After collision they become a binary system, so they will rotate around common barycenter (not around disk com). However you have made a subtle error, system COM will be not a halfway between particle position and disk center. Barycenter location depends on mass ratio between rotating bodies. And it will be shifted towards more massive body. Ratio between rotation radiuses of bodies from COM is : $$\frac {r_{particle} } {r_{disk} } = \frac {m_{disk} } {m_{particle} }$$

So because disk is a lot more massive than particle, particle will rotate at a lot greater distance from common COM compared to that of disk. Which rotation radius will be very small, so it can be assumed that it is almost zero and thus COM will almost overlap with disk COM.

EDIT

In case ring and particle mass are the same - these radiuses will be the same, so system COM will be half-way of distance between individual bodies com.

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  • $\begingroup$ The OP mentions in the comments that they were in fact told the particle and ring have the same mass $\endgroup$ Commented Jun 13, 2020 at 16:10
  • $\begingroup$ See edit, clarified situation for that case $\endgroup$ Commented Jun 13, 2020 at 16:18

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