Q 11-06-181JEE MainJEE Main 2020 (7 Jan, Shift 1)Medium
As shown in the figure, a bob of mass $m$ is tied to a massless string whose other end portion is wound on a flywheel (disc) of radius $r$ and mass $m$. When released from rest the bob starts falling vertically. When it has covered a distance of $h$, the angular speed of the wheel will be:
Answer: (A) $\dfrac1r\sqrt{\dfrac{4gh}{3}}$
The string does not slip, so $v = \omega r$. Energy conservation:
$$mgh = \frac12mv^2 + \frac12\left(\frac12mr^2\right)\omega^2 = \frac12m\omega^2r^2 + \frac14m\omega^2r^2 = \frac34m\omega^2r^2$$
$$\omega = \frac1r\sqrt{\frac{4gh}{3}}$$
Solution by Sreeraj P, M.Sc Physics