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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
    Commented Jan 4, 2021 at 21:36

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This looks like a homework problem, so I will provide some information, but not the solution.

In general, the change in angular momentum of a system of particles with respect to a point Q is $\vec N_Q^{ext} - \vec v_Q \times \vec P$ where $N_Q^{ext}$ is the total external torque about Q, $\vec v_Q$ is the velocity of Q, and $\vec P$ is the total linear momentum of the body. Taking Q as the CM, the change in angular momentum of the system of particles with respect to the CM is $\vec N_Q^{ext}$. See the inertial frame part of my answer to Where does pseudo force act at? on this exchange, and the associated Kochmann reference section 2.2.3 for details.

Your question probably wants the change in angular momentum of the rod about its CM. The problem also probably assumes A strikes B directly (not a glancing collision). The rod is stated to be a rigid body (a special case of a system of particles). You have insufficient information to evaluate the force on the rod from the collision with particle A (the time duration of the impulse is not provided). Therefore, consider the system as both the rod and particle A. Linear momentum of the rod/particle A is conserved. For an elastic collision, kinetic energy of the rod/particle A is conserved. For the rod/particle A system angular momentum is conserved; take the angular momentum with respect to the CM of the rod. You can also evaluate the velocity of the CM after the collision, and the angular velocity of the rod about its CM after the collision.

Hope this helps.

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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
    Commented 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
    Commented 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. $\endgroup$
    – Kristoffer Sjöö
    Commented Jan 5, 2021 at 7:41

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