Question

A disc of mass m and radius r is free to rotate about its centre . a string is wrapped over its rim and a block of mass m is attachedto the free end of the string .The system is released from rest . what will be the speed of the block as it descends through a height h?

Answer 1

Here in order to find out the final speed we can use energy conservation

As per energy conservation we will have

Loss in the kinetic energy of block = gain in kinetic energy of block + pulley

As per formula

$mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$

now since string is wound over the pulley so angular speed and linear speed is related to each other

$v = R\omega$

now plug in all values

$mgh = \frac{1}{2}mv^2 + \frac{1}{2}I(\frac{v}{r})^2$

here moment of inertia is given as

$I = \frac{1}{2}mR^2$

now we will have

$mgh = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{1}{2}mR^2)(\frac{v}{R})^2$

$mgh = \frac{3}{4}mv^2$

$v =\sqrt{ \frac{4gh}{3}}$

so above is the speed of the block after it dropped by height h

Answer 2

hope it will help you....