System Of Particles And Rotational Motion Ques 36
36. A circular disk of moment of inertia $I_t$ is rotating in a horizontal plane, its symmetry axis, with a constant angular speed $\omega_i$. Another disk of moment of inertia $I_b$ 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 $\omega_f$. The energy lost by the initially rotating disk to friction is:
[2010]
(a) $\frac{1}{2} \frac{I_b^{2}}{(I_t+I_b)} \omega_i^{2}$
(b) $\frac{I_t^{2}}{(I_t+I_b)} \omega_i^{2}$
(c) $\frac{I_b-I_t}{(I_t+I_b)} \omega_i^{2}$
(d) $\frac{1}{2} \frac{I_b I_t}{(I_t+I_b)} \omega_i^{2}$
Show Answer
Answer:
Correct Answer: 36.(d)
Solution:
- (d) By conservation of angular momentum, $I_t \omega_i=(I_t+I_b) \omega_f$
where $\omega_f$ is the final angular velocity of disks
$ \therefore \quad \omega_f=(\frac{I_t}{I_t+I_b}) \omega_i $
Loss in K.E., $\Delta K=$ Initial K.E. - Final K.E.
$ =\frac{1}{2} I_t \omega_i^{2}-\frac{1}{2}(I_t+I_b) \omega_f^{2} $
$=\frac{1}{2} I_t \omega_i^{2}-\frac{1}{2}(I_t+I_b) \frac{I_t^{2}}{(I_t+I_b)^{2}} \omega_i^{2}$
$=\frac{1}{2} \omega_i^{2} \frac{I_t}{I_t+I_b}(I_t+I_b-I_t)=\frac{1}{2} \omega_i^{2} \frac{I_t I_b}{I_t+I_b}$
BETA
