Question

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 the rest, the bob starts falling vertically. When it has covered a distance $h$, the angular speed of the wheel will be:

• A. $r\sqrt{\frac{3}{4gh}}$
• B. $\frac{1}{r}\sqrt{\frac{4gh}{3}}$
• C. $r\sqrt{\frac{3}{2gh}}$
• D. $\frac{1}{r}\sqrt{\frac{2gh}{3}}$

Answer 1

The correct option is B $\frac{1}{r}\sqrt{\frac{4gh}{3}}$

By energy conservation,
$$\begin{array}{cc}& mgh=\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2\\ & \Rightarrow gh=\frac{v^2}{2}+\frac{{\omega }^2{r}^2}{4}& \text{(1)}\end{array}$$
Since the rope is inextensible and also it is not slipping,
$$\begin{array}{cc}& \therefore v=r\omega & & \text{from eq.}\left(1\right)\text{and}\left(2\right)\\ & gh=\frac{{\omega }^2{r}^2}{2}+\frac{{\omega }^2{r}^2}{4}\\ & \Rightarrow gh=\frac{3}{4}{r}^2{\omega }^2\\ & \Rightarrow {\omega }^2=\frac{4gh}{3r^2}\\ & \Rightarrow \omega =\frac{1}{r}\sqrt{\frac{4gh}{3}}\\ \text{(2)}\end{array}$$