0
$\begingroup$

Consider a situation like this: a massless ring is kept fixed at rest on a horizontal plane. A massless thin string attached at its one end to a point on the circumference of the ring while its other point is attached to a certain point mass.then my a certain mechanism we impart a certain initial velocity to the point mass (the velocity direction is at all instants perpendicular to the string).The string wraps around the ring until the mass hits the ring.

I am not looking for a solution to this problem as i just made it up to clarify the situation physically.

My question is:-

would it be right to apply conservation of angular momentum in this case(there being no external forces)? Seeing as the Instantaneous centre of rotation changes with time?

More generally, does angular momentum conservation laws hold even if the centre of rotation changes?

$\endgroup$
  • 1
    $\begingroup$ "there being no external forces" How do you keep the ring fixed in place without an external force? $\endgroup$ – BowlOfRed Jun 7 '15 at 5:06
1
$\begingroup$

I'm afraid I cannot quite understand the specific part of your question. How would the motion always be perpendicular to the string? That would only be true if the mass were moving in a circle, not a decreasing spiral like you are describing.

But to your main question, does the conservation of angular momentum hold if the center of rotation changes..

For any given axis, angular momentum about that axis remains constant for a system. This has nothing to do with a "center of rotation". As long as you pick your axis and stick with it (and include everything in the system), then angular momentum about that axis is constant.

It may be harder to calculate the value if the axis of rotation does not coincide with the chosen axis, but that's a different issue.

$\endgroup$
  • $\begingroup$ I may have failed to express my problem properly through that particular situation.What i meant was does conservation of angular momentum hold even if the axis of rotation changes with time ? $\endgroup$ – Daipayan Mukherjee Jun 7 '15 at 5:54
  • $\begingroup$ That seems to be the exact question that I tried to answer. Where it rotates doesn't matter. Where you choose the axis to calculate angular momentum does. That axis needs to be constant. So you may be calculating angular momentum about an axis that is different than the axis of rotation. $\endgroup$ – BowlOfRed Jun 7 '15 at 5:58
  • $\begingroup$ i know,and i get it,thanks i was just trying to apologise for the incorrect situation i described $\endgroup$ – Daipayan Mukherjee Jun 7 '15 at 6:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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