A solid cylinder of mass $M$ and radius $R$ rolls without slipping down an inclined plane of angle $\theta$. The acceleration of the center of mass of the cylinder is:

Solution

1. Equations are $Mg\sin (\theta )-f=Ma$ and $fR=I\alpha$.
2. For a solid cylinder, $I=\frac{1}{2}M{R}^2$.
3. Using $a=\alpha R$, we get $$\begin{array}{rl}fR& =\frac{1}{2}M{R}^2(a/R)\\ & =\frac{1}{2}Ma\end{array}$$.
4. Substituting $f$ into the force equation: $Mg\sin (\theta )-\frac{1}{2}Ma=Ma$.
5. Solving for $a$ gives $a=\frac{2}{3}g\sin (\theta )$.
6. Hence the answer is (B).