Question

A cord is wound around the circumference of wheel of radius ${}^{\prime }{r}^{\prime }$. The axis of the wheel is horizontal and moment of inertia about it is ${}^{\prime }{l}^{\prime }$. The weight ${}^{\prime }m{g}^{\prime }$ is attached to the end of the cord and falls from rest. After falling through a distance ${}^{\prime }{h}^{\prime }$, the angular velocity of the wheel will be

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

Answer 1

Answer: (C)

${\left[\frac{2mgh}{l+m{r}^2}\right]}^{1/2}$

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 $0+0=-mgh+\frac{1}{2}m{v}^2+\frac{1}{2}l{\omega }^2$

$\Rightarrow mgh=\frac{1}{2}m(\omega r{)}^2+\frac{1}{2}l{\omega }^2$
$=\frac{1}{2}m{\omega }^2{r}^2+\frac{1}{2}l{\omega }^2$

$\Rightarrow 2mgh={\omega }^2[m{r}^2+l]$

$\Rightarrow {\omega }^2=\frac{2mgh}{l+m{r}^2}$

$\Rightarrow \omega ={\left[\frac{2mgh}{l+m{r}^2}\right]}^{1/2}$