Question

A cord is wound around the circumference of a wheel of radius $r$. The axis of the wheel is horizontal and the 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 square of the angular velocity of the wheel will be: (there is no slipping between the wheel and the cord).

• A. $\frac{2gh}{\left(1+m{r}^2\right)}$
• B. $2gh$
• C. $\frac{2mgh}{\left(1+m{r}^2\right)}$
• D. $\frac{2mgh}{\left(1+2m{r}^2\right)}$

Answer 1

The correct option is C

$\frac{2mgh}{\left(1+m{r}^2\right)}$

Step 1. Given data and free body diagram:

$U_i=$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

$r=$ Radius of wheel

$I=$ Moment of inertia of wheel

$h=$ Distance through which weight falls from A to B.

Step 2. Applying energy conservation between point A and B, we get:

$U_i+K_i\hspace{0.17em}=U_f+K_f$

$\begin{array}{rcl}& \Rightarrow & 0+0\hspace{0.17em}=-mgh+\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2\\ & \Rightarrow & mgh=\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2\end{array}$ $\left[Initiallyweightisinrest,U_i=K_i=0\right]$

$\begin{array}{rcl}& \Rightarrow & mgh=\frac{1}{2}m{\left(r\omega \right)}^2+\frac{1}{2}I{\omega }^2\end{array}$ $\left[v=r\omega \right]$

$\begin{array}{rcl}& \Rightarrow & 2mgh=m{r}^2{\omega }^2+I{\omega }^2\\ & \Rightarrow & 2mgh={\omega }^2\left(m{r}^2+I\right)\\ \therefore {\omega }^2& =& \frac{2mgh}{\left(m{r}^2+I\right)}\end{array}$

Thus, the square of the angular velocity of the wheel will be, $\begin{array}{rcl}{\omega }^2& =& \frac{2mgh}{\left(m{r}^2+I\right)}\end{array}$.

Hence, correct option is (C).