Question

A cord is wound around the circumference of wheel of radius $r$, the axis of wheel is horizontal and 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 angular velocity of wheel will be

• A. $\sqrt{\frac{2gh}{1+mr}}$
• B. $\sqrt {2gh}$
• C. $\left[\frac{2 mgh }{1+ mr ^{2}}\right]^{1 / 2}$
• D. $\sqrt {\frac {2mgh}{1+2mr^2}}$

Answer 1

Answer: (C)

$\left[\frac{2 mgh }{1+ mr ^{2}}\right]^{1 / 2}$

Answer 2

Applying energy conservation, we have $U _{ i }+ K _{ i }= U _{ f }+ K _{ f }$ Where, $U _{ i }=$ initial potential energy of the (block + pulley) system $U _{ f }=$ final potential energy of the (block+pulley) system $K _{ i }=$ initial kinetic energy of the system $K _{ f }=$ final kinetic energy of the system Here, initial situation corresponds to rest position of the system and final situation correspond to position after falling through height $h$. Eg. (i) gives $0+0=- mgh +\frac{1}{2} mv ^{2}+\frac{1}{2} 1 \omega^{2}$ $\begin{aligned} \Rightarrow mgh &=\frac{1}{2} m (\omega r )^{2}+\frac{1}{2} l \omega^{2} \\\ &=\frac{1}{2} m \omega^{2} r ^{2}+\frac{1}{2} l \omega^{2} \end{aligned}$ $\Rightarrow 2 mgh =\omega^{2}\left[ mr ^{2}+1\right]$ $\Rightarrow \omega^{2}=\frac{2 mgh }{1+ mr ^{2}}$ $\Rightarrow \omega=\left[\frac{2 mgh }{1+ mr ^{2}}\right]^{1 / 2}$