Question

As shown in the figure, a bob of mass $m$ is tied to 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:

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

Answer 1

Answer: (A)

Using work-energy theorem
$mgh=\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2$
$mgh=\frac{1}{2}m{\omega }^2{r}^2+\frac{1}{2}\frac{m{r}^2}{2}{\omega }^2$ ( since $I=\frac{m{r}^2}{2}$ and $v=r\omega$ )
$mgh=\frac{3}{4}m{\omega }^2{r}^2$
Hence, $\omega =\sqrt{\frac{4gh}{3r^2}}=\frac{1}{r}\sqrt{\frac{4gh}{3}}$