A metal coin of mass $5\ \text{g}$ and radius $1\ \text{cm}$ is fixed to a thin stick $AB$ of negligible mass. The stick lies in the plane of the coin and touches its rim, so $AB$ is a tangent to the coin in its plane. The system is initially at rest. The constant torque, that will make the system rotate about $AB$ at $25$ rotations per second in $5\ \text{s}$, is close to:
Answer: (B) $2.0\times10^{-5}\ \text{N m}$
About a tangent in its plane, a disc has $I = \dfrac{MR^2}{4} + MR^2 = \dfrac54MR^2$:
$$I = 1.25\times0.005\times(0.01)^2 = 6.25\times10^{-7}\ \text{kg m}^2$$
$\omega = 2\pi\times25 = 157\ \text{rad/s}$ in $5\ \text{s}$, so $\alpha = 31.4\ \text{rad/s}^2$.
$$\tau = I\alpha = 6.25\times10^{-7}\times31.4 \approx 2.0\times10^{-5}\ \text{N m}$$
Solution by Sreeraj P, M.Sc Physics