Question

Two forces of magnitude $F$ each are acting on a uniform disc kept on a horizontal surface as shown in the figure. The disc performs rolling without slipping. If the frictional force by the horizontal surface is $nF$. The value of $n$ is

Answer 1

Let $f$ be the frictional force acting on the disc.
As we know, friction opposes the relative motion between the disc and surface. Let us assume the frictional force to be in backward direction.


Let the mass of disc be $M$ and radius be $R$.
From FBD we can say that,
$2F-f=Ma.....\left(1\right)$
Finding the torque due to forces about centre
$(F+f)R=I\alpha ......\left(2\right)$ [since $F$ passing through COM produces zero torque]
For pure rolling, $a=R\alpha$
Moment of inertia of uniform disc about its centre $I=\frac{M{R}^2}{2}$
Using this in $\left(2\right)$, we get
$(F+f)R=\frac{1}{2}M{R}^2\alpha$
$\Rightarrow (F+f)R=\frac{1}{2}M{R}^2\left(\frac{a}{R}\right)[\because \alpha =\frac{a}{R}]$
$\Rightarrow (F+f)=\frac{Ma}{2}.....\left(3\right)$
Solving $\left(1\right)$ and $\left(3\right)$, we get $f=0$
From the data given in the question , $f=nF$
$\Rightarrow n=0$