A uniform ring of mass ‘m’ and radius ‘R’ is projected horizontally with velocity $v_{0}$ on a rough horizontal floor, so that it starts off with a purely sliding motion and it acquires a purely rolling motion after moving a distance d. If the coefficient of friction between the ground and ring is $μ$ , then work done by the friction in the process is

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$- \frac{1}{8} {mv}_{0}^{2}$

$- μmgd$

$μmgd$

$- \frac{1}{4} {mv}_{0}^{2}$

answer is B.

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Detailed Solution

$\begin{array}{l} {mv}_{0} R = I ω + mvR \\ = {mR}^{2} (\frac{v}{R}) + mvR \\ \begin{aligned} v & = \frac{v_{0}}{2} \\ K_{i} & = \frac{1}{2} {mv}_{0}^{2} \end{aligned} \\ K_{f} = \frac{1}{2} {mv}^{2} + \frac{1}{2} ∣ ω^{2} = \frac{1}{2} m {(\frac{v_{0}}{2})}^{2} + \frac{1}{2} {mR}^{2} \frac{v_{0}^{2}}{4 R^{2}} \\ = \frac{1}{4} {mv}_{0}^{2} \end{array}$
So, $W_{f} = - \frac{1}{4} {mv}_{0}^{2}$