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A uniform disc 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 at t=0. After $t_{0}$ seconds, it acquires a purely rolling motion as shown in figure. The velocity of the centre of mass of the disc at $t_{0}$ is

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$\frac{2}{3} v_{0}$

$\frac{3}{2} v_{0}$

$\frac{3}{4} v_{0}$

$\frac{4}{3} v_{0}$

answer is C.

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Detailed Solution

According to impulse momentum theorem.

$f_{k} \times t_{0} = m (v - v_{0})$ ....(1)

$(f_{k} R) \times t_{0} = Iω$ ....(2), here $I = \frac{{mR}^{2}}{2}$

and while rolling $v = Rω$ .......(3)

By solving (1), (2) and (3), we get $v = \frac{2}{3} v_{0}$

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