Question

A cord is wound round the circumference of a wheel of radius $r$. The axis of the wheel is horizontal and the moment of inertia about it is $I$. A weight $mg$ is attached to the cord at the end. The weight falls from rest. After falling through a distance $h$, the square of angular velocity of the wheel will be :

$($There is no slipping between the wheel and the cord$)$

• A. $\frac{2gh}{I+m{r}^2}$
• B. $2gh$
• C. $\frac{2mgh}{I+2m{r}^2}$
• D. $\frac{2mgh}{I+m{r}^2}$

Answer 1

The correct option is D $\frac{2mgh}{I+m{r}^2}$

Using energy conservation principle between points $A$ and $B$,

$mgh+0=\frac{1}{2}m(\omega r{)}^2+\frac{1}{2}I{\omega }^2$

$[\because$ $v=\omega r]$

$\Rightarrow 2mgh=(m{r}^2+I){\omega }^2$

$\Rightarrow {\omega }^2=\frac{2mgh}{I+m{r}^2}$