Q: A disc of mass m_{1} is freely rotating with constant angular speed ω . Another disc of mass m_{2} & same radius is gently kept on the first disc. If the contacting surfaces are rough, the fractional decrease in kinetic energy of the system will be

(A)m_{1}/m_{2}

(B)m_{2}/(m_{1}+m_{2})

(C)m_{2}/m_{1}

(D) m_{1}/(m_{1}+ m_{2})

Solution: The upper disc speeds up and the lower disc slows down by the accelerating & rotating frictional retarding torques. However, the torque acting on the system is zero. Therefore, the angular momentum of the system remains constant.

L_{initial} = L_{final} = L

= 1-I_{initial}/I_{final}

I_{initial} = ½m_{1}r²

I_{final} = ½m_{1}r² + ½m_{2}r²

By putting these values ,

ΔKE/KE_{initial} = m_{2}/(m_{1}+m_{2})