Question 60 of 280

A cord is wound round the circumference of 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 wheel will be

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

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Answer: (A) $\dfrac{2mgh}{I+m{r}^{2}}$

$mgh=\dfrac{1}{2}I{\omega }^{2}+\dfrac{1}{2}m{v}^{2}$
$v=\omega r$
$mgh=\dfrac{1}{2}I{\omega }^{2}+\dfrac{1}{2}m{\omega }^{2}{r}^{2}$
$\dfrac{2mgh}{\left(I+m{r}^{2}\right)}={\omega }^{2}$