A uniform ring of radius $R$ if given a back spin of angular velocity $V_{0} / 2 R$ and thrown on a horizontal rough surface with velocity of centre the be $V_{0}$ . The velocity of the centre of the centre of the ring when it starts pure rolling will be

\frac{V_{0}}{2}

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\frac{V_{0}}{4}

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\frac{{3 V}_{0}}{4}

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Solution

The correct option is B $\frac{V_{0}}{4}$
Let $V$ and $a$ be the velocity and acceleration respectively of centre of ring after pure rolling. $⟹ V = R w_{f}$
$m a = - F$ ................(1)
$τ = I α$
$F R = m R^{2} α ⟹ F = m R α$ . ........(2)
Now, $V = V_{o} + a t$
$V = V_{o} - \frac{F}{m} t$ (from 1) .........(a)
Also, $w_{f} = - w_{i} + α t$
$w_{f} = - w_{i} + \frac{F}{m R} t$ ..............(b) (from 2)
Eliminating Ft in (a) & (b), $V + R w_{f} = V_{o} - R w_{i}$
$V + V = V_{o} - R \frac{V_{o}}{2 R}$
$⟹ V = \frac{V_{o}}{4}$

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A uniform ring of radius R if given a back spin of angular velocity V0/2R and thrown on a horizontal rough surface with velocity of centre the be V0. The velocity of the centre of the centre of the ring when it starts pure rolling will be

A uniform ring of radius $R$ if given a back spin of angular velocity $V_{0} / 2 R$ and thrown on a horizontal rough surface with velocity of centre the be $V_{0}$ . The velocity of the centre of the centre of the ring when it starts pure rolling will be