$A$ rope of negligible mass is wound round 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 $30\; N$? What is the linear acceleration of the rope? Assume that there is no slipping.

(A) 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 is given by $I = m r^2$.
$I = 3 \times (0.4)^2 = 3 \times 0.16 = 0.48\; kg\; m^2$.
Torque,$\tau = F \times r = 30 \times 0.4 = 12\; Nm$.
Using the relation $\tau = I \alpha$,the angular acceleration $\alpha$ is:
$\alpha = \frac{\tau}{I} = \frac{12}{0.48} = 25\; rad\; s^{-2}$.
The linear acceleration $a$ is given by $a = r \alpha$.
$a = 0.4 \times 25 = 10\; m\; s^{-2}$.