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. ${\left[\frac{2mgh}{I+2m{r}^2}\right]}^{\frac{1}{2}}$
• C. ${\left[\frac{2mgh}{I+m{r}^2}\right]}^{\frac{1}{2}}$
• D. ${\left[\frac{mgh}{I+m{r}^2}\right]}^{\frac{1}{2}}$

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 $0+0=-mgh+\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2$
$\Rightarrow mgh=\frac{1}{2}m(\omega r{)}^2+\frac{1}{2}I{\omega }^2$
$=\frac{1}{2}m{\omega }^2{r}^2+\frac{1}{2}I{\omega }^2$
$\Rightarrow 2mgh={\omega }^2\left[m{r}^2+I\right]$
$\Rightarrow {\omega }^2=\frac{2mgh}{I+m{r}^2}$
or $\omega ={\left[\frac{2mgh}{I+m{r}^2}\right]}^{1/2}$