NEET 2010
- Exam
- NEET
- Year
- 2010
- Subject
- Physics
- Chapter
- Rotational Motion
NEET 2010 · Physics · Rotational Motion
A circular disk of moment of inertia I_t is rotating in a...
A circular disk of moment of inertia is rotating in a horizontal plane, about its symmetry axis, with a constant angular speed . Another disk of moment of inertia 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 . The energy lost by the initially rotating disc due to friction is
Options
- A\frac{1}{2} \frac{\mathrm{I}_{\mathrm{b}}^2}{\left(\mathrm{I}\mathrm{I}_{\mathrm{b}}\right)} \omega_{\mathrm{i}}^2
- B\frac{1}{2} \frac{\mathrm{I}_{\mathrm{t}}^2}{\left(\mathrm{I}\mathrm{I}_{\mathrm{b}}\right)} \omega_{\mathrm{i}}^2
- C\frac{1}{2} \frac{\mathrm{I}\mathrm{I}_{\mathrm{t}}}{\left(\mathrm{I}\mathrm{I}_{\mathrm{b}}\right)} \omega_{\mathrm{i}}^2
- D\frac{1}{2} \frac{\mathrm{I}\mathrm{I}_{\mathrm{t}}}{\left(\mathrm{L}_{\mathrm{t}}+\mathrm{I}_{\mathrm{b}}\right)} \omega_{\mathrm{i}}^2
Correct answer
D. \frac{1}{2} \frac{\mathrm{I}\mathrm{I}_{\mathrm{t}}}{\left(\mathrm{L}_{\mathrm{t}}+\mathrm{I}_{\mathrm{b}}\right)} \omega_{\mathrm{i}}^2
Step-by-step solution
Loss of energy, \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{I}_{\mathrm{b}}\right)}=\frac{1}{2} \frac{_{\mathrm{b}} _{\mathrm{t}} \omega_{\mathrm{i}}^2}{\left(\mathrm{I}\mathrm{I}_{\mathrm{b}}\right)}