Question

A uniform ring of radius $R$ is given a back spin of angular velocity $\frac{V_0}{2R}$ and thrown on a horizontal rough surface with velocity of centre to be $V_0$. The velocity of the centre of the ring when it starts pure rolling will be

• A. $\frac{V_0}{2}$
• B. $\frac{V_0}{4}$
• C. $\frac{3V_0}{4}$
• D. $0$

Answer 1

The correct option is B $\frac{V_0}{4}$

Given:
Initial linear velocity $\left(u\right)=V_0$

Initial angular velocity $\left({\omega }_0\right)=\frac{-V_0}{2R}$

Let,

$f=$ frictional force acting on ring
$m=$ mass of the ring
$I=m{R}^2=$ MOI of the ring
$a=$ acceleration of ring
$\alpha =$ angular acceleration of the ring
For translational motion (using FBD),

$f=-ma\Rightarrow a=-\frac{f}{m}....\left(1\right)$

For rotational motion (using FBD),

$I\alpha =fR\Rightarrow \left(m{R}^2\right)\alpha =fR$

$\Rightarrow \alpha =\frac{f}{mR}....\left(2\right)$

Let after time $t$, it starts pure rolling.

So, $V^{\prime }=R{\omega }^{\prime }$

$\Rightarrow u+at=R({\omega }_0+\alpha t)$

from eqs. (1) and (2)

$V_0-\frac{f}{m}t=R(-\frac{V_0}{2R}+\frac{f}{mR}t)$

$\Rightarrow t=\frac{3m{V}_0}{4f}$

So, velocity of ring at time $t$ is

$V^{\prime }=u+at=V_0-(\frac{f}{m})(\frac{3m{V}_0}{4f})$

$\Rightarrow V^{\prime }=V_0-\frac{3}{4}{V}_0=\frac{V_0}{4}$

Hence, option $\left(b\right)$ is correct.