A cord of negligible mass is wound round the rim of a solid flywheel of mass $20 ext{ kg}$ and radi

Question

A cord of negligible mass is wound round the rim of a solid flywheel of mass $20	ext{ kg}$ and radius $20	ext{ cm}$. A steady pull of $25	ext{ N}$ is applied on the cord. What is the angular acceleration of the wheel?

Professor PC Thomas Classes · Answer

The moment of inertia of the solid flywheel (disc) about its axis is $I = \frac{MR^2}{2} = \frac{20 	imes (0.2)^2}{2} = 0.4	ext{ kg m}^2$. The torque provided by the pull is $	au = F 	imes R = 25 	imes 0.2 = 5.0	ext{ Nm}$. Using $	au = I\alpha$, the angular acceleration $\alpha = \frac{5.0}{0.4} = 12.5	ext{ rad/s}^2$.