A uniform circular disc of radius R, lying on a frictionless horizontal plane is rotating with an angular velocity omega about is its own axis. Another identical circular disc is gently placed on the top of the first disc coaxially. The loss in rotational kinetic energy due to friction between the two discs, as they acquire common angular velocity is (I is moment of inertia of the disc)

Options

A. `(1)/(8)1omega^(2)`
B. `(1)/(4)1omega^(2)`
C. `(1)/(2)1omega^(2)`
D. `1omega^(2)`

Answer

(b)

Explanation

K.E of rotating circular disc `= (1)/(2)Iomega^(2)`
  K.E of linear circular disc,
  `L = (1)/(2)mv^(2)`
  Therefore, the resultant loss`=(1)/(2)Iomega^(2)+(1)/(2)mv^(2)=(1)/(4)Iomega ^(2)`