JEE Advanced · 2006 · 09 Apr 2006 · Rotational Mechanics

For a cylinder rolling without slipping down an incline, the friction is static and provides the torque for rotation. Using $mg\sin \theta -f=ma$ and $fr=I\alpha =\frac{1}{2}m{r}^2\cdot \frac{a}{r}$, we get $f=\frac{1}{3}mg\sin \theta$. This is less than $\mu mg\cos \theta$ (the maximum static friction), so option A is incorrect. Since there is no slipping, friction does no work and is non-dissipative (option B incorrect). As $\theta$ decreases, $\sin \theta$ decreases, so $f=\frac{1}{3}mg\sin \theta$ decreases (option C correct). Friction acts up the incline, opposing the translational motion down the incline, while providing the torque that causes rotation (option D correct).