A disc of mass m1 is freely rotating with constant angular speed ω . Another disc of mass m2 & same radius is gently…..

Q: A disc of mass m₁ is freely rotating with constant angular speed ω . Another disc of mass m₂ & 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)m₁/m₂

(B)m₂/(m₁+m₂)

(C)m₂/m₁

(D) m₁/(m₁+ m₂)

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 ,

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