Question

A solid cylinder of mass $M$ rolls without slipping down on an inclined plane making an angle $\theta$ with the horizontal. The acceleration of the cylinder down the inclined plane is

• A. $\frac{2}{3}g\sin \theta$
• B. $\frac{3}{2}g\sin \theta$
• C. $g\sin \theta$
• D. $g$

Answer 1

The correct option is A $\frac{2}{3}g\sin \theta$

The acceleration, $a=\frac{f\sin \theta }{J+\frac{K^2}{R^2}}$
For a cylinder, $I=\frac{M{R}^2}{2}$
$\therefore K^2=\frac{1}{M}=\frac{M{R}^2}{\frac{2}{M}}$
$\Rightarrow \frac{K^2}{R^2}=\frac{1}{2}$
so, $a=\frac{f\sin \theta }{J+\frac{1}{2}}=\frac{2}{3}g\sin \theta$