Question

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

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

Answer 1

Answer: (C)

Applying energy conservation, we have
$U_{i}+K_{i}=U_{f}+K_{f} \,\,\,\,\,\, ...(i)$
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 corresponds to position after falling through height $h$.
Eq. (i) gives $0+0=-m g h+\frac{1}{2} m v^{2}+\frac{1}{2} I \omega^{2}$
$ \Rightarrow \,\,\, m g h =\frac{1}{2} m(\omega r)^{2}+\frac{1}{2} I \omega^{2} $
$=\frac{1}{2} m \omega^{2} r^{2}+\frac{1}{2} I \omega^{2} $
$\Rightarrow 2 m g h=\omega^{2}\left[m r^{2}+I\right] $
$\Rightarrow \,\,\,\omega^{2} =\frac{2 m g h}{I+m r^{2}} $
or $\,\,\, \omega=\left[\frac{2 m g h}{I+m r^{2}}\right]^{1 / 2} $