JEE Main · 2019 · 10 Jan 2019 · Shift 1 · Rotational Mechanics

The friction does not all act at one radius, so I build the torque up from rings. A ring of radius $x$ and width $dx$ carries a share of the load in proportion to its area, $dN=\frac{F}{\pi R^2}2\pi xdx$. Friction on it is $\mu dN$ acting at lever arm $x$, so the elementary torque is $d\tau =\mu xdN=\frac{2\mu F}{R^2}{x}^{2}dx$. Integrating from centre to rim, $\tau =\frac{2\mu F}{R^2}{\int }_0^R{x}^{2}dx=\frac{2\mu F}{R^2}\cdot \frac{R^3}{3}=\frac{2}{3}\mu FR$. The factor $\frac{2}{3}$ rather than $\frac{1}{2}$ is the signature of a load spread over the area, since the outer rings carry more of the rubbing.