A rope of negligible mass is wound round a hollow cylinder of mass 3 kg and radius 40 cm.

Mass of the hollow cylinder, m = 3 kg

Radius of the hollow cylinder, r = 40 cm = 0.4 m

Applied force, F = 30 N

The moment of inertia of the hollow cylinder about its geometric axis:

$I=m r^{2}$

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

Torque, $\tau=F \times r$

$=30 \times 0.4=12 \mathrm{Nm}$

For angular acceleration $\alpha$, torque is also given by the relation:

$\tau=I \alpha$

$\alpha=\frac{\tau}{I}=\frac{12}{0.48}$

$=25 \mathrm{rad} \mathrm{s}^{-2}$

Linear acceleration $=r \alpha=0.4 \times 25=10 \mathrm{~m} \mathrm{~s}^{-2}$