Imagine an air hockey table where there is a puck P and a rectangular slab S. Both are free to move as there is zero friction. The slab is at rest in the middle of the table. The puck is moving towards the slab and collides with it in an elastic collision. The puck is not spinning before the collision. As this is a closed system, the total momentum, angular momentum and energy of the system should be preserved. But after the collision the slab will be spinning about its centre of mass, but does that mean the puck will also be spinning in order to conserve the angular momentum of the system? I feel like the elastic (frictionless) collision won't be able to impart any spin on the puck. I understand that the puck and the slab will also change their linear momentum and energy but overall in the system these will be conserved. How would you calculate the velocity vectors and angular velocity for the slab and the puck post-collision? Assume the puck and slab are made of the same uniform-density material and the puck has radius r and the slab side lengths a and b.
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$\begingroup$ The numerical value of the conserved angular momentum will depend upon where you pick the pivot to be. The puck should not be rotating at the end of this collision, but the linear motion that it moves at, will in general contribute to the angular momentum about said pivot. The scheme that you should be using to analyse such a system is taught in textbooks. It is a bit long to be describing here. $\endgroup$– naturallyInconsistentCommented Jul 26, 2023 at 4:42
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$\begingroup$ Agree about the textbooks, but this is a fairly basic scenario that seems to get glossed over. Most articles discuss selecting a pivot point (as you have). But in a free system won't the slab spin around its centre of mass? And if the puck does not spin afterwards, then has the angular momentum not been conserved in this closed system? $\endgroup$– JoeBloggsCommented Jul 26, 2023 at 5:33
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$\begingroup$ I specifically made sure to explain in my earlier comment how angular momentum is conserved despite the puck not rotating. $\endgroup$– naturallyInconsistentCommented Jul 26, 2023 at 5:45
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$\begingroup$ It is not so much glossed over as that it is conceptually confusing with very little actual insight to be gained $\endgroup$– naturallyInconsistentCommented Jul 26, 2023 at 5:47
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$\begingroup$ If before the collision neither the puck nor the slab are spinning (total angular momentum = 0), but after the collision the slab is spinning but the puck is not then in what sense has there been any conservation of angular momentum? $\endgroup$– JoeBloggsCommented Jul 26, 2023 at 6:13
1 Answer
I think that you have missed the idea that a body moving with a linear velocity can have angular momentum.
Consider a mass $v$ with velocity $v$ hitting a stationary slab $M$, with centre of mass at $C$, elastically as shown in the left-hand diagram.
Your statement,
If before the collision neither the puck nor the slab are spinning (total angular momentum = 0)
is only true of the angular momentum is found along any position along the extended line $BB'$,eg $B$.
The initial clockwise angular momentum is $mu \times 0$ and the final clockwise angular momentum is $MV \times c - I_{\rm B}\times\omega $ where $I_{\rm B}$ is the moment of inertia of the slab about $B$.
Thus if there are no external torques acting on the system consisting of the slab and the mass, $0 = MV \times c - I_{\rm B}\times\omega $.
So when position $B$ is chosen the final linear momentum $MV$ of the slab contributes to the angular momentum.
Zero initial angular momentum would not be true if position $C$ is chosen.
The initial clockwise angular momentum is $-mu\times c$ and the final clockwise angular momentum is $mv\times c - I_{\rm C}\times \omega$.
Angular momentum is conserved for any arbitrary position, eg $A$ (or $A'$), $-mu\times a = mv\times v- I_{\rm A} \times \omega$.
