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A disc of mass M and radius R is rolling purely with center's velocity $v_{0}$ on a flat horizontal floor when it hits a step in the floor of height R / 4. The corner of the step is sufficiently rough to prevent any slipping of the disc against itself. What is the velocity of the centre of the disc just after impact?

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(1) $4 v_{0} / 5$

(3) $5 v_{0} / 6$

(4) None of these

(2) $4 v_{0} / 7$

answer is C.

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Detailed Solution

Angular momentum will remain conserved at the point of impact P and just after impact it starts rotating about point P.

$L_{i} = L_{f} ∴ {Mv}_{0} (R - \frac{R}{4}) + \frac{1}{2} {MR}^{2} (\frac{v_{0}}{R}) = \frac{3}{2} {MR}^{2} \cdot ω) ∴ ω = \frac{5 v_{0}}{6 R} ∴ v_{com} = ωR = \frac{5 v_{0}}{6}$

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