Question

A solid cylinder of mass $3\text{kg}$ is placed on a rough inclined plane of inclination ${30}^{\circ }$. The value of frictional force required, for the cylinder to roll down the inclined plane without slipping is:- (Take $g=10{\text{ms}}^{-2}$)

• A. $2\text{N}$
• B. $5\text{N}$
• C. $15\text{N}$
• D. $18\text{N}$

Answer 1

The correct option is B $5\text{N}$

Applying the equation of torque about centre of cylinder,
${\tau }_{net}=I\alpha ...\left(i\right)$, where $\alpha =$angular acceleration of the cylinder
${\tau }_{mg}=0,{\tau }_N=0$, as they pass through the COM of the cylinder.
Torque of friction,
${\tau }_f=fR...\left(ii\right)$
Condition for pure rolling is $a=\alpha R...\left(iii\right)$

From Eq $\left(i\right),\left(ii\right),\&\left(iii\right)$
${\tau }_f+{\tau }_{mg}+{\tau }_N=I\alpha$
$\Rightarrow fR+0+0=\left(\frac{1}{2}m{R}^2\right)\times \frac{a}{R}$
$\therefore f=\frac{1}{2}ma...\left(iv\right)$


Applying Newton's $2^{nd}$ law along the inclined plane and substituting the value of $f$ in it.
$mg\sin \theta -f=ma$
$\Rightarrow mg\sin \theta -\frac{1}{2}ma=ma$
$\Rightarrow \frac{3}{2}ma=mg\sin \theta$
$\Rightarrow \frac{1}{2}ma=\frac{mg\sin \theta }{3}$
From Eq.$\left(iv\right)$,
$f=\frac{ma}{2}=\frac{mg\sin \theta }{3}$
Substituting values,
$\therefore f=\frac{3\times 10\times 0.5}{3}=5\text{N}$