System of Particles and Rotational Motion - Result Question 35
35. 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: 35. (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} $
$=\frac{1}{2} I_t \omega_i^{2}-\frac{1}{2}(I_t+I_b) \frac{I_t^{2}}{(I_t+I_b)^{2}} \omega_i^{2}$
$=\frac{1}{2} \omega_i^{2} \frac{I_t}{I_t+I_b}(I_t+I_b-I_t)=\frac{1}{2} \omega_i^{2} \frac{I_t I_b}{I_t+I_b}$
(a) In absence of external torque, $L=I \omega$ $=$ constant
$I_1 \omega_1=I_2 \omega_2, I_1=MR^{2}, I_2=MR^{2}+2 mR^{2}$
(Moment of inertia of a thin circular ring about an axis vertical to its plane $=MR^{2}$ )
$\therefore \quad \omega_2=\frac{I_1}{I_2} \omega=\frac{M}{M+2 m} \omega$.
(b) $\vec{\tau}=\overrightarrow{{}r} \times \overrightarrow{{}F} \Rightarrow \overrightarrow{{}r} \cdot \vec{\tau}=0$
$\overrightarrow{{}F} \cdot \vec{\tau}=0$
Since, $\vec{\tau}$ is perpendicular to the plane of $\overrightarrow{{}r}$ and $\overrightarrow{{}F}$, hence the dot product of $\vec{\tau}$ with $\overrightarrow{{}r}$ and $\overrightarrow{{}F}$ is zero.
Radial component of force makes no contribution to the torque. Only transverse component of force contributes to the torque.
