A rope is wound around a hollow cylinder of mass 3 kg and radius 40 cm. What is the angular acceleration of the cylinder if the rope is pulled with a force of 30 N ?

$m=3 \mathrm{kg},r=40 \mathrm{cm}=40\times {10}^{-2} \mathrm{m},F=30 \mathrm{N}$

Moment of inertia of hollow cylinder about its axis 

$=m{r}^2=3 \mathrm{kg}\times (0.4{)}^2{ \mathrm{m}}^2=0.48 \mathrm{kg}{ \mathrm{m}}^2$

The torque is given by, 

$\tau =I\alpha$

where I =  moment of inertia, 

$\alpha =\text{ angular acceleration }$

$\text{ In the given case, }\tau =rF,\text{ as the force is acting }$$\text{ perpendicularly to the radial vector. }$

$$\begin{gathered} \therefore \alpha =\frac{\tau }{I}=\frac{Fr}{m{r}^2}=\frac{F}{mr}=\frac{30}{3\times 40\times {10}^{-2}}=\frac{30\times 100}{3\times 40} \\ \alpha =25{\mathrm{rads}}^{-2} \end{gathered}$$