please solve the given quest (im providing the ans0?

Question

a disk of mass m is rotating with "w'(constant)another disk of mass N of same radius is kept on the first disk if the contact surface are rough then fractional decrease in K.E will be ANS-N/m

Answer 1

The moment of inertia of a disk is : M*R^2 / 2

I can demostrate it, but let's go on with the problem :

The Kinetic energy for the first disk : 1/2*I*w^2

This will be : 1/4*M*R^2*w^2

For the other disk, let's consider that the angular momentum is constant :

1/2*M*R^2*w =1/2*N*R^2*w'

w' = (M / N)*w

then, the kinetic energy for the second disk :

1/4*M*R^2*(M / N)^2*w^2

if we rest the kinetics energy, we will have :

1/4*M*R^2*w^2( 1 - M^2 / N^2 )

If we divide them :

we will have = N / M