Question

The frictional force on the ring (as shown in figure) is

• A. $\frac{F}{3}$ towards left
• B. $\frac{2F}{3}$ towards left
• C. $\frac{F}{3}$ towards right
• D. Zero

Answer 1

The correct option is D Zero

Let frictional force $\left(f\right)$, be acting opposite to applied force $F$.


Applying $\sum {\overrightarrow{F}}_{\text{net}}=m{\overrightarrow{a}}_{\text{COM}}$
$\Rightarrow F-f=m{a}_{\text{COM}}...\left(1\right)$
From equation of torque about centre of mass of the ring,
$\sum {\overrightarrow{\tau }}_{\text{net, COM}}=I_{\text{COM}}\alpha$
$\Rightarrow FR+fR=m{R}^2\alpha ...\left(2\right)$
[Both torques tend to produce rotation in clockwise sense, hence will add up.]

$\because \alpha =a_{\text{COM}}/R$ (in case of pure rolling)
$\Rightarrow (F+f)R=m{R}^2\times \frac{a_{\text{COM}}}{R}$
Or, $F+f=m{a}_{\text{COM}}...\left(3\right)$
From Eq $\left(1\right)$ and $\left(3\right)$, we get
$f=0$
Hence no frictional force will be acting on the ring.