Question

A solid cylinder of mass $M$ and radius $R$ rolls (without slipping) down an inclined plane of inclination $\theta$. The minimum coefficient of friction $\mu$ between the cylinder and the plane so that it rolls without slipping must be

• A. $\tan \theta$
• B. $\frac{\tan \theta }{2}$
• C. $\frac{\tan \theta }{3}$
• D. $\frac{\tan \theta }{4}$

Answer 1

The correct option is C $\frac{\tan \theta }{3}$

The acceleration of the cylinder down the incline is
$a=\frac{g\sin \theta }{1+\frac{I}{M{R}^2}}...\left(i\right)$
Equation for torque about centre of mass of cylinder,
$\tau =I\alpha$
$\Rightarrow fR=I\alpha =I\frac{a}{R}...\left(ii\right)$
[$\because a=\alpha R$ for pure rolling]
From Eq $\left(i\right)\&\left(ii\right)$,
$f=\frac{I}{R^2}\times \frac{g\sin \theta }{(1+\frac{I}{M{R}^2})}$
Putting $I=\frac{M{R}^2}{2}$,
$f=\frac{Mg\sin \theta }{3}...\left(iii\right)$

To avoid slipping, required value of $f$ must be lesser than limiting value of friction i.e $f\le \mu N$ where $N=Mg\cos \theta$
$\Rightarrow \frac{Mg\sin \theta }{3}\le \mu Mg\cos \theta$
or $\mu \ge \frac{\tan \theta }{3}$
$\therefore {\mu }_{min}=\frac{\tan \theta }{3}$