A circular disk of moment of inertia It is rotating in a horizontal plane, about the symmetry axis, with a constant angular speed ωi. Another disk of moment of inertia Ib is dropped coaxially onto the rotating disk. Initially the second disk has zero angular speed. Eventually both the disks rotate with a constant angular speed ωf. The energy loss by the initially rotating disc due to friction is

Q: A circular disk of moment of inertia It is rotating in a horizontal plane, about the symmetry axis, with a constant angular speed ωi. Another disk of moment of inertia Ib is dropped coaxially onto the rotating disk. Initially the second disk has zero angular speed. Eventually both the disks rotate with a constant angular speed ωf. The energy loss by the initially rotating disc due to friction is

(a) $\frac{1}{2} \frac{I_b I_t}{(I_b + I_t)} \omega_i^2 $

(b) $\frac{1}{2} \frac{I_b^2 }{(I_b + I_t)} \omega_i^2 $

(c) $\frac{1}{2} \frac{ I_t^2}{(I_b + I_t)} \omega_i^2 $

(d) $\frac{1}{2} \frac{I_b – I_t}{(I_b + I_t)} \omega_i^2 $

Ans: (a)

Sol: Applying conservation of Angular momentum , L = I ω = constant

$I_1 \omega_1 = I_2 \omega_2 $

$I_t \omega_i = ( I_b + I_t ) \omega_f $

$ \omega_f = \frac{I_t }{ ( I_b + I_t ) }\omega_i  $ ….(i)

Loss of Energy $= K_i – K_f $

$\frac{1}{2}I_t \omega_i^2 – \frac{1}{2}( I_b + I_t) \omega_f^2 $

$\frac{1}{2}I_t \omega_i^2 – \frac{1}{2}( I_b + I_t) (\frac{I_t }{ ( I_b + I_t ) }\omega_i)^2 $

$ = \frac{1}{2} \frac{I_b I_t}{(I_b + I_t)} \omega_i^2 $