Question

The stick applies a force of $2\text{N}$ on the ring$(R=0.5\text{m})$ and the ring rolls without slipping on the ground as shown in figure. The coefficient of friction between the stick and the ring is

• A. $0.4$
• B. $0.3$
• C. $0.2$
• D. $0.5$

Answer 1

The correct option is A $0.4$

Since point of contact always remains at rest in the case of pure rolling, let us take instantaneous axis of rotation passing through the point of contact of the ring and the ground.


On applying equation of torque about instantaneous axis of rotation.
$\sum {\tau }_{\text{ext}}=I\alpha ...\left(i\right)$
Weight of the ring and friction acting at the point of contact will have zero torque about the instantaneous axis of rotation.

From the FBD, we can see that the ring will slip at the contact with the stick, along upward direction (ring is rotating in anti-clockwise direction). Hence, limiting friction $f=\mu N$ will act on the ring in downward direction.

Torque due to normal reaction from the stick,
${\tau }_N=2\times 0.5$
Torque due to friction acting at the contact of the ring with stick,
${\tau }_f=\mu N\times R=\mu \times 2\times 0.5$
MOI about instantaneous axis of rotation
$I=I_0+m{d}^2=m{R}^2+m{R}^2=2m{R}^2$
$\Rightarrow I=2\times 2\times {0.5}^2$
Substituting in Eq.$\left(i\right)$, considering anticlockwise sense as $+ve$, we get
$1-1\mu =(2\times 2\times {0.5}^2)\times \alpha$
$\Rightarrow 1-\mu =1\times \alpha ...\left(ii\right)$
For pure rolling, $a=\alpha R$ where $a=0.3{\text{m/s}}^2,R=0.5\text{m}$
$\therefore \alpha =0.6{\text{rad/s}}^2$
Substituting in Eq.$\left(ii\right)$, we get
$1-\mu =0.6$
$\Rightarrow \mu =0.4$