Question

A massless cord is wound round the circumference of the wheel of radius $r$. The axis of the wheel is horizontal and the moment of inertia about the centre is $I$. A weight $mg$ is attached to the cord at the end. The weight falls from rest and performs pure rolling. After falling through distance $h$, angular velocity of wheel is

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

Answer 1

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

Loss of potential energy= gain in kinetic energy
Loss of potential energy=mgh
Gain of kinetic energy = translational kinetic energy + rotational kinetic energy
Gain in Kinetic energy = $\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2$
$\therefore$ $mgh=\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2$
on solving above equation, we get
${\omega }^2\left[\frac{m{r}^2+I}{2}\right]=mgh$
${\left[\frac{2mgh}{m{r}^2+I}\right]}^{1/2}=\omega$