Question

As shown in the figure, a bob of mass $$m$$ is tied by a massless string whose other end portion is wound on a fly wheel (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 1

Correct option is B. $$\dfrac{1}{r}\sqrt{\dfrac{4gh}{3}}$$
Let the bob of mass $$m$$ fall through a height $$h$$.

According to work-energy theorem, the charge in potential energy of the bob is converted to the kinetic energies of the bob and the flywheel.

Rotational kinetic energy of a flywheel is given by

$$\dfrac{1}{2}Iw^2=\dfrac{1}{2}\dfrac{mr^2}{2}w^2=\dfrac{mr^2w^2}{4}$$

Since the string is tied around the flywheel, $$v$$ of the bob is related to

$$w$$ of the flywheel by the relation $$v=wr$$

$$mgh=\dfrac{1}{2}mv^2+\dfrac{1}{2}Iw^2=\dfrac{1}{2}mw^2r^2+\dfrac{mr^2w^2}{4}$$

$$=\dfrac{3mw^2r^2}{4}$$

$$gh=\dfrac{3w^2r^2}{4},w^2=\dfrac{4gh}{3r^2}$$

$$w=\dfrac{1}{r}\sqrt{\dfrac{4gh}{3}}$$