Question

A small pulley of radius $30\text{cm}$ and moment of inertia $0.72{\text{kg m}}^2$ is used to hang a $2\text{kg}$ mass with the help of massless string. If the block is released, what will be its acceleration for no slipping condition?

• A. $2{\text{m/s}}^2$
• B. $4{\text{m/s}}^2$
• C. $1{\text{m/s}}^2$
• D. $3{\text{m/s}}^2$

Answer 1

The correct option is A $2{\text{m/s}}^2$

FBD of pulley and block:

For rotational equilibrium of pulley
$TR=I\alpha$
$\Rightarrow TR=I\frac{a}{R}$
or $T=\frac{Ia}{R^2}$
where $a$ is the acceleration of string which is the same as acceleration of block.

For block of mass $M$:
$Mg-T=Ma$
$\Rightarrow Mg=Ma+T=(M+\frac{I}{R^2})a$

$\Rightarrow a=\left(\frac{Mg}{M+\frac{I}{R^2}}\right)$

Here, $M=2\text{kg},I=0.72{\text{kg m}}^2$ & $R=0.3\text{m}$

$\Rightarrow a=\frac{2\times 10}{(2+\frac{0.72}{0.09})}=2{\text{m/s}}^2$