A rope of negligible mass is wound round a hollow cylinder of mass $3 ext{ kg}$ and radius $40 ext

Question

A rope of negligible mass is wound round a hollow cylinder of mass $3	ext{ kg}$ and radius $40	ext{ cm}$. If the rope is pulled with a steady force of $30	ext{ N}$ and there is no slipping, what is the angular acceleration of the cylinder?

Professor PC Thomas Classes · Answer

The moment of inertia of a hollow cylinder about its axis is $I = MR^2 = 3	ext{ kg} 	imes (0.4	ext{ m})^2 = 0.48	ext{ kg m}^2$. The torque provided by the pull is $	au = F 	imes R = 30	ext{ N} 	imes 0.4	ext{ m} = 12	ext{ Nm}$. Using Newton's second law for rotation ($	au = I\alpha$), the angular acceleration is $\alpha = \frac{	au}{I} = \frac{12}{0.48} = 25	ext{ rad/s}^2$.