A rope is wound around a hollow cylinder of mass $3 \text{ kg}$ and radius $40 \text{ cm}$. What is the angular acceleration of the cylinder if the rope is pulled with a force of $30 \text{ N}$? (2017-Delhi)
Solution:
For a hollow cylinder, the moment of inertia is $I = MR^2$. The torque provided by the rope is $\tau = F \times R = I\alpha$. Substituting $I$, we get $F \times R = MR^2 \alpha$, which gives $\alpha = \frac{F}{MR} = \frac{30}{3 \times 0.4} = 25 \text{ rad/s}^2$.
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