Question

What should be the minimum coefficient of static friction between the plane and the cylinder for the cylinder, for the cylinder not to slip on an inclined plane

• A. $\frac{1}{3}tan\theta$
• B. $\frac{1}{3}sin\theta$
• C. $\frac{2}{3}tan\theta$
• D. $\frac{2}{3}sin\theta$

Answer 1

The correct option is A $\frac{1}{3}tan\theta$

$Let$ :- Mass of cylinder be $m$

Radius be $R$

And , Angle of inclination be $\theta$

$ToFind$ :- Minimum coefficient of static friction ( ${\mu }_s$ )

$Solution$ :- we know, for velocity of point $p$

${\overrightarrow{V}}_p$ = ${\overrightarrow{V}}_{p/cm}$ + ${\overrightarrow{V}}_{cm/g}$

Since, velocity of inclined plane is $0$

$⟹$ $0=\omega R+V$

$\therefore$ $V=\omega R$

$\frac{dV}{dt}=R\frac{d\omega }{dt}$ (Differentiate with respect to time )

$⟹$ $a=R\alpha$ $--------\left(1\right)$

Now , Torque about friction we get ,

$fR=I\alpha$

$fR=\frac{m{R}^2}{2}\alpha$ $(\because I=\frac{m{R}^2}{2})$

$⟹$ $\alpha R=\frac{2f}{m}$

$\therefore$ $ma=2f$ $------\left(2\right)$ ( from 1 )

Now , Along inclined acceleration we get ,

$mgsin\theta -f=ma$

$\therefore$ $f=\frac{mgsin\theta }{3}$ ( from 2)

Now we know , $f_{max}=\mu N$

$\frac{mgsin\theta }{3}={\mu }_{s}mgcos\theta$ $(N=mgcos\theta )$

$\therefore$ $\underset{–}{{\mu }_s=\frac{1}{3}tan\theta }$

Hence , $\underset{–}{OptionA\left(\frac{1}{3}tan\theta \right)iscorrect.}$