A solid cylinder of mass $M$ and radius $R$ is rolling without slipping down an inclined plane of angle $\theta$. The minimum coefficient of static friction required is:

Solution

1. For a cylinder rolling down an incline, the equations of motion are $Mg\sin \theta -f=Ma$ and $fR=I\alpha$, where $I=\frac{1}{2}M{R}^2$.
2. Using the rolling condition $a=R\alpha$, we substitute $I$ to find the friction force $$\begin{array}{rl}f& =\frac{1}{2}Ma\\ & =\frac{1}{2}M(a)\end{array}$$, which simplifies to $f=\frac{1}{3}Mg\sin \theta$.
3. The condition for no slipping is $f\le \mu N$, where $N=Mg\cos \theta$, leading to $\mu Mg\cos \theta \ge \frac{1}{3}Mg\sin \theta$.
4. Solving for the coefficient of friction gives $\mu \ge \frac{1}{3}\tan \theta$.
5. Hence the answer is (B).