Q. A cord is wound round the circumference of wheel of radius $r$. The axis of the wheel is horizontal and moment of inertia about it is $I$. A weight $mg$ is attached to the end of the cord and falls from the rest. After falling through a distance $h$, the angular velocity of the wheel will be
A) $\sqrt{\frac{2gh}{I + mr}}$
B) $\left\lbrack \frac{2mgh}{I + mr^{2}} \right\rbrack^{1/2}$
C) $\left\lbrack \frac{2mgh}{I + 2mr^{2}} \right\rbrack^{1/2}$
D) $\sqrt{2gh}$
Explanation
(b) We know $v = \sqrt{\frac{2gh}{1 + \frac{k^{2}}{r^{2}}}}\ \ \therefore\omega = \frac{v}{r} = \sqrt{\frac{2gh}{r^{2} + k^{2}}}$
$\Rightarrow \omega = \sqrt{\frac{2mgh}{mr^{2} + mk^{2}}} = \sqrt{\frac{2mgh}{mr^{2} + I}} = \sqrt{\frac{2mgh}{I + mr^{2}}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.