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 :

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

Answer 1

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


Applying energy conservation principle we get,
$\text{Loss in P.E. of bob = gain in total K.E. of wheel}$

$mgh=\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2$

where,
$(v=\omega r)$ is the velocity of disc with which the disc is rotating,
$I=\frac{1}{2}m{r}^2$ is the moment of inertia of disc about its centre.

Substituting the values,
$mgh=\frac{1}{2}m[r\omega {]}^2+\frac{1}{2}\frac{m{r}^2}{2}{\omega }^2$

$\Rightarrow gh=r^2{w}^2[\frac{1}{2}+\frac{1}{4}]$

$\Rightarrow gh=\frac{3}{4}{r}^2{w}^2$

$\Rightarrow {\omega }^2=\frac{4gh}{3r^2}$

Thus, Angular speed of wheel , $\omega =\frac{1}{r}\sqrt{\frac{4gh}{3}}$

Hence, option (d) is the correct answer.