Question

A Uniform ring of radius R is given a back spin of angular velocity $\dfrac{{{V_0}}}{{2R}}$ and thrown on a horizontal rough surface with velocity of center to be ${V_0}$. The velocity of the center of the ring when it starts pure rolling will be:
A)$\dfrac{{{V_0}}}{2}$
B) $\dfrac{{{V_0}}}{4}$
C) $\dfrac{{3{V_0}}}{4}$
D) $0$

Answer 1

Angular velocity $\omega$ is analogous to the linear velocity. To get the relationship between angular velocity and linear velocity, consider a rotating CD. This moves an arc length $\Delta s$ in a time $\Delta t$, so it has a linear velocity $v = \dfrac{{\Delta s}}{{\Delta t}}$.
From $\Delta \theta = \dfrac{{\Delta s}}{r}$
$\Rightarrow \Delta s = r\Delta \theta$
We know that $\omega = \dfrac{{\Delta \theta }}{{\Delta t}}$
Then, $v = \dfrac{{r\Delta \theta }}{{\Delta t}} = r\omega$
We get the relation,
$v = r\omega$.
We will be using the formula of relating angular velocity and linear velocity $v = r\omega$.

Complete step by step answer:
It is given that the question stated as, uniform ring of radius R is given a back spin of angular velocity $\dfrac{{{V_0}}}{{2R}}$ and thrown on a horizontal rough surface with velocity of center to be ${V_0}$.
Here we have to find the velocity of the center of the ring when it starts pure rolling
We know about any point P on ground Angular momentum will be conserved. As torque due to friction blows. Let the final state have velocity ’v’ and angular velocity $\omega$ and Let the mass of the ring is m and pure rolling with$\omega$after a certain time.
Hence, we can write it as,
$v = r\omega$
Now we have,${I_{cm}} = M{v^2}$
${v_i} = {v_0}$ , ${v_f} = v$,
$\Rightarrow {\omega _i} = - \dfrac{{{v_0}}}{{2R}}$ , ${\omega _f} = \omega$
$\Rightarrow {\omega _i} = - \dfrac{{{v_0}}}{{2R}}$, The negative means the rotation would be clockwise and the ring will move forward.
Then, ${\overrightarrow L _p} =$Constant
$\Rightarrow$ $m{v_0}r - m{r^2}$ $\times \dfrac{{{v_0}}}{{2r}} = mvr + m{r^2}\omega$
On cancel the equating term and we get,
$\Rightarrow {v_0} - \dfrac{{{v_0}}}{2} = 2v$
Taking LCM,
$\Rightarrow \dfrac{{2{v_0}}}{2} - \dfrac{{{v_0}}}{2} = 2v$
On subtracting the numerator term and we get,
$\Rightarrow$ $\dfrac{{{v_0}}}{2} = 2v$
Let us divide 2 on both sides and we get,
$\Rightarrow v = \dfrac{{{V_0}}}{4}$
Therefore, we get the final velocity of given ring is $v = \dfrac{{{V_0}}}{4}$ in the forward direction.
$\Rightarrow v = \dfrac{{{V_0}}}{4}$
This is the velocity of the center of the ring when it starts pure rolling.

Hence the correct option (B), $\dfrac{{{V_0}}}{4}$.

Note: Angular velocity is also called rotational velocity. It is the rate of velocity at which an object or a particle is rotating around a center or a specific point in a given period of time. The angular velocity is a vector quantity.