Explanation

(c) Linear acceleration for rolling $a = \frac{g\sin\theta}{(1 + K^{2}/R^{2})}$
For cylinder $\frac{K^{2}}{R^{2}} = \frac{1}{2}$
$\therefore a_{cylinder} = \frac{2}{3}g\sin\theta$ …(i)
For rotation, the torque $fR = I\alpha = (MR^{2}\alpha)/2$
(where $f =$ force of friction)
But $R\alpha = a\ \therefore f = \frac{M}{2}a$
[image:7_image171.emf]
$\therefore f = \frac{M}{2}.\frac{2}{3}g\sin{\theta = \frac{M}{3}g\sin\theta}$
$\mu_{s} = f/N$ where $N$ is normal reaction, $Mg\cos\theta$
$\mu_{s} = \frac{\frac{M}{3}g\sin\theta}{Mg\cos\theta} = \frac{\tan\theta}{3}$
$\therefore$ For rolling without slipping of a roller down the inclined plane, $\tan{\theta \geq 3\mu_{s}}$