A circular disk of moment of inertia $I_t$ is rotating in a horizontal plane, about its symmetry axis, with…

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 $I_b$ 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 $\omega_{\mathrm{f}}$. The energy lost by the initially rotating disc due to friction is
  1. $\frac{1}{2} \frac{\mathrm{I}_{\mathrm{b}}^2}{\left(\mathrm{I}_{\mathrm{t}}+\mathrm{I}_{\mathrm{b}}\right)} \omega_{\mathrm{i}}^2$
  2. $\frac{1}{2} \frac{\mathrm{I}_{\mathrm{t}}^2}{\left(\mathrm{I}_{\mathrm{t}}+\mathrm{I}_{\mathrm{b}}\right)} \omega_{\mathrm{i}}^2$
  3. $\frac{1}{2} \frac{\mathrm{I}_{\mathrm{b}}-\mathrm{I}_{\mathrm{t}}}{\left(\mathrm{I}_{\mathrm{t}}+\mathrm{I}_{\mathrm{b}}\right)} \omega_{\mathrm{i}}^2$
  4. $\frac{1}{2} \frac{\mathrm{I}_{\mathrm{b}} \mathrm{I}_{\mathrm{t}}}{\left(\mathrm{L}_{\mathrm{t}}+\mathrm{I}_{\mathrm{b}}\right)} \omega_{\mathrm{i}}^2$

Solution

Loss of energy, $\Delta \mathrm{E}=\frac{1}{2} \mathrm{I}_{\mathrm{t}} \omega_{\mathrm{i}}^2-\frac{1}{2} \frac{\mathrm{I}_{\mathrm{t}}^2 \omega_{\mathrm{i}}^2}{\left(\mathrm{I}_{\mathrm{t}}+\mathrm{I}_{\mathrm{b}}\right)}$ $=\frac{1}{2} \frac{\mathrm{I}_{\mathrm{b}} \mathrm{I}_{\mathrm{t}} \omega_{\mathrm{i}}^2}{\left(\mathrm{I}_{\mathrm{t}}+\mathrm{I}_{\mathrm{b}}\right)}$ .

Asked in: NEET 2010 (Screening)

Practice more Rotational Motion questions on Aicharya