Question

A solid cylinder of mass $3kg$ is placed on a rough inclined plane of inclination ${30}^o$. If $g=10m{s}^{-2}$, then the minimum frictional force required for it to roll without slipping down the plane is

• A. $2N$
• B. $5N$
• C. $15N$
• D. $18N$

Answer 1

Answer: (B)

$5N$

For translational motion: $F=Ma$

$mgsin\theta -f=ma$ ..............................(1)

For rotational motion: Net torque about O, $\tau =I\alpha$

$fR=I\alpha$

Also body rolls without slipping, thus $a=r\alpha$

$fR=I\frac{a}{R}$

Put it in (1), $mgsin\theta -f=m\frac{f{R}^2}{I}$

$⟹f=\frac{mgsin\theta }{1+\frac{m{R}^2}{I}}$

Given: $m=3kg;\theta ={30}^o$

For solid cylinder, $I=\frac{1}{2}m{R}^2$

Thus $f=\frac{3\left(10\right)\left(0.5\right)}{1+\frac{m{R}^2}{0.5m{R}^2}}$

$⟹f=5N$