Rotation: angular momentum and torque

[Avi Nandi]

Homework Statement

Mass m is attached to a cylindrical post of rdius R by a string. Initially it is distance r from the centre of the post and is moving tangentially with speed v_{0}.

In case (a) the string passes through a hole in the centre of the post at the top. the string is shortened slowly by drawing it through the hole.

In case (b) the string wraps around at the outside of the post.

What quantities are conserved in each case? Find the final speed of the mass when it heats the post for each case..

The Attempt at a Solution

I am considering the origin of the reference frame on the axis of the cylinder.

Now in the first case i applied angular momentum conservation and the speed in question comes out.

But in the second case angular momentum is not conserved since the force acts along the string which is not passing through the centre and there is a torque in action. Let consider at a certain instant of time length of rope l, velocity of mass v, and distance from the origin r'

angular momentum = mvl/r' and torque = -Fl = -mv^{2} (F= mv^{2}/l)

 

[tiny-tim]

Hi Avi Nandi!

… But in the second case angular momentum is not conserved since the force acts along the string which is not passing through the centre …

Hint: what else might be conserved?

(and what is the condition for that?)

 

[Avi Nandi]

Force is always acting perpendicular to the velocity. So no work is done. Thats why kinetic energy is conserved. Velocity is constant. :) thank you tiny tim. Your help is not tiny at all.