A solid cylinder is rolling down on an inclined plane of angle $\theta$. The coefficient of static friction between the plane and the cylinder is $\mu_s$. The condition for the cylinder not to slip is

• A

  $\tan \theta \geq 3 \mu_s$
• B

  $\tan \theta\gt3 \mu_s$
• Correct answer
  C

  $\tan \theta \leq 3 \mu_s$
• D

  $\tan \theta \lt 3 \mu_s$

Solution

$\because$ Linear acceleration for rolling, $a=\frac{g \sin \ddot{\theta}}{\sqrt{1+k^2 / R^2}}$
For cylinder, $\frac{k^2}{R^2}=\frac{1}{2}$ $\therefore \quad a_{\text {cylinder }}=\frac{2}{3} g \sin \theta$ For rotation, the torque $F R=I \alpha \cdot\left(M R^2 \alpha\right) / 2$ (where, $F=$ force of friction) But $\quad R \alpha=a \quad \therefore \quad F=\frac{M}{2} a$ $F=\frac{M}{2} \cdot \frac{2}{3} g \sin \theta=\frac{M}{3} g \sin \theta$ $\therefore \quad \mu_s=\frac{F}{N}$, where $N$ is normal reaction, $\mu_s=\frac{\frac{M}{3} g \sin \theta}{M g \cos \theta}=\frac{\tan \theta}{3}$ $\therefore$ For rolling without slipping of a roller down, the inclined plane, $\tan \theta \leq 3 \mu_s$.

Hence, the correct answer is option: 'C'.