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Two particles each of mass m are attached by a light rod (massless and cannot bend or stretch) of length l. A particle "A" of same mass strikes B. The collision is perfectly elastic. Find angular momentum of the rod system. There are no other external forces.enter image description here

Now I have done this by conserving angular momentum about a fixed axis in-line to the center of mass (just beside it, fixed to the ground) so angular velocity of COM about that axis becomes zero. I imagined the rod system rotating about the COM with COM itself translating forward.

Now suppose I wanted to find the required answer by conserving angular momentum NOT about an axis about which angular velocity of COM is zero, but the angular velocity of particle C is zero.

I have tried by taking a fixed axis beside C. But even from that axis, the angular momentum comes to be :

L = Iw + Mvr

where I is momentum of inertia about COM, w is angular velocity about it, Mv is linear momentum of COM and r is perpendicular distance from axis, here r = l/2

Is there is such an axis (fixed to the ground) about which I can assume angular velocity of C to be zero? If yes, then how can I find angular momentum?

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  • $\begingroup$ If this collision is elastic, the the rebound motion of particle A must be considered when calculating the final angular momentum. Once mass C is moving it will have an instantaneous angular velocity (and corresponding angular momentum) relative to any fixed axis which is not in-line with its velocity vector. $\endgroup$
    – R.W. Bird
    Jan 4, 2021 at 21:36

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C is a particle, so angular velocity is meaningless for it on its own. Only the B-C system has an angular velocity. And that angular velocity is the same around all axes (all axes perpendicular to the plane, that is).

Your formula is also wrong. I assume you mean "moment of inertia", not "momentum" which doesn't make sense. The equation is correct with only the first term.

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  • $\begingroup$ I think there is some misunderstanding. I agree, C cannot rotate about its own axis as its a particle. What I wanted to know was we generally assume such a body to be rotating about its center of mass, even though it may not, for ease of calculation. So is there a way where I can calculate total angular momentum, not about center of mass, but about C? $\endgroup$
    – Adi
    Jan 4, 2021 at 19:13
  • $\begingroup$ In my opinion I believe the formula did not have any error. Isn't the formula for the angular momentum of a rotating + translating body equal to: [moment of inertia about center of mass * angular velocity] + [Linear momentum of center of mass*perpendicular distance from center of mass to the fixed axis] ? $\endgroup$
    – Adi
    Jan 4, 2021 at 19:19
  • $\begingroup$ I apologize for the confusion, I was talking about the "spin angular momentum" throughout, I.e. the angular momentum around the COM. This must be what the exercise is asking for, because otherwise there's no one correct answer. $\endgroup$ Jan 5, 2021 at 7:41

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