Rotational Motion Question 67
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_t} $ . 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 disc due to friction is
[AIPMT (S) 2010]
Options:
A) $ \frac{1}{2}\frac{I_b^{2}}{(I _{t}+I _{b})}\omega _i^{2} $
B) $ \frac{1}{2}\frac{I_t^{2}}{(I _{t}+I _{b})}\omega _i^{2} $
C) $ \frac{1}{2}\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: D
Solution:
- Loss of energy,
$ \Delta E=\frac{1}{2}I _{t}\omega _i^{2}-\frac{1}{2}\frac{I_t^{2}\omega _i^{2}}{2(I _{t}+I _{b})} $ $ =\frac{1}{2}\frac{I_b^{{}}I _{t}\omega _i^{2}}{(I _{t}+I _{b})} $
