I came across a question:

A disc of mass 2M and radius R lying in a vertical plane which is free to rotate about a fixed horizontal axis passing from its centre. A particle of mass M is tied to a string of length 2R and another end of the string is attached to the disc as shown in the figure. If the particle is released from the shown position, then what will the angular speed of the disc as the string becomes taut will be?


The little green thing is the particle.

Our teacher said that angular momentum is conserved and that initial angular momentum is MR($\sqrt{4gR}$) which is to be equated to final angular momentum.

I understand that it is mass* velocity *radius and that velocity was found by $1/2mv^2 = mgR$, but how is that the initial momentum? The particle starts from rest, so why is there velocity term there in initial angular momentum?

  • $\begingroup$ "As the string becomes taut" is not a specific instant, so you really can't give an unambiguous answer. You could answer if the question was instead "when the tension is maximum" or "after the particle reaches the end of the string and bounces back elastically". $\endgroup$
    – Ben51
    Feb 22, 2021 at 17:50
  • $\begingroup$ @Ben51 While discussing it, he mentioned that it is before the impulse of the particle bouncing back $\endgroup$
    – sam
    Feb 23, 2021 at 3:01

1 Answer 1


The initial angular momentum is not when the particle is at rest. If you let the particle drop, it will gain some speed just before the string is taut. Your teacher defines the initial momentum to be the momentum of particle just before the string is taught, which is $1/mv^2=mgR$. The final position is when the string is taut and both disc and particle will have some momentum.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.