Rotational Motion — Medium JEE physics MCQ
A disc rotating at $60$ rpm has its radius halved with no external torque. The new angular velocity is:
- A. $240$ rpm
- B. $120$ rpm
- C. $60$ rpm
- D. $30$ rpm
Solution
$L = I\omega = $ constant. For disc, $I \propto R^2$
$I_1\omega_1 = I_2\omega_2 \Rightarrow R^2(60) = (R/2)^2(\omega_2)$
$\omega_2 = 60 \times 4 = 240$ rpm