Question

A circular disk of moment of inertia $I_{t}$ is rotating in a horizontal plane, about its symmetry axis, with a constant angular speed $\omega_{i}$. Another disk of moment of inertia $I_{h}$ 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 disc to friction is

• A. $\frac{1}{2}\frac{I^2_b}{(I_t+I_b)}\omega^2_i$
• B. $\frac{1}{2}\frac{I^2_t}{2(I_t+I_b)}\omega^2_i$
• C. $\frac{I_b-I_t}{(I_t+I_b)}\omega^2_i$
• D. $\frac{1}{2}\frac{I_bI_t}{(I_t+I_b)}\omega^2_i$

Answer 1

Answer: (D)

As no external torque is applied to the system, the angular momentum of the system remains conserved.
$\therefore L_{i}=L_{f}$
According to given problem,
$I_{t} \omega_{i}=\left(I_{t}+I_{b}\right) \omega_{f}$
or $\omega_{f}=\frac{I_{t} \omega_{i}}{\left(I_{t}+I_{b}\right)} \ldots(i)$
Initial energy, $E_{i}=\frac{1}{2} I_{t} \omega_{i}^{2} \ldots(i i)$
Final energy, $E_{f}=\frac{1}{2}\left(I_{t}+I_{b}\right) \omega_{f}^{2} \ldots(i i i)$
Substituting the value of $\omega_{f}$ from equation (i) in equation (iii), we get Final energy, $E_{f}=\frac{1}{2}\left(I_{t}+I_{b}\right)\left(\frac{I_{i} \omega_{i}}{I_{t}+I_{b}}\right)^{2}$ $=\frac{1}{2} \frac{I_{t}^{2} \omega_{i}^{2}}{2\left(I_{t}+I_{b}\right)} \ldots(i v)$
Loss of energy, $\Delta E=E_{i}-E_{f}$
$\frac{1}{2} I_{t} \omega_{i}^{2}-\frac{1}{2} \frac{I_{t}^{2} \omega_{i}^{2}}{I_{t}+I_{b}}$ (Using (ii) and (iv))
$\frac{\omega_{i}^{2}}{2}\left(I_{t}-\frac{I_{t}^{2}}{\left(I_{t}+I_{b}\right)}\right) $
$=\frac{\omega_{i}^{2}}{2}\left(\frac{I_{t}^{2}+I_{b} I_{t}-I_{t}^{2}}{\left(I_{t}+I_{b}\right)}\right) $
$=\frac{1}{2} \frac{I_{b} I_{t}}{\left(I_{t}+I_{b}\right)} \omega_{i}^{2}$