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

$\displaystyle KE_{initial} = \frac{L^2}{2I_{initial}}$

$\displaystyle KE_{final} = \frac{L^2}{2I_{final}}$

Fractional decrease in kinetic energy $\displaystyle = \frac{\Delta KE}{KE_{initial}} $

$\displaystyle = \frac{KE_{initial}-KE_{final}}{KE_{initial}}$

$\displaystyle = 1 – \frac{KE_{final}}{KE_{initial}}$

$\displaystyle = 1 – \frac{I_{initial}}{I_{final}}$

$\displaystyle I_{initial} = \frac{1}{2}m_1 r^2 $

$\displaystyle I_{final} = \frac{1}{2}m_1 r^2 + \frac{1}{2}m_2 r^2$

By putting these values ,

$\displaystyle \frac{\Delta KE}{KE_{initial}} = \frac{m_2}{m_1 + m_2}$