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Q.

A sphere of mass M and radius r shown in the figure, slips on a rough horizontal plane. At some instant it has translational velocity v₀ and rotational velocity about centre v₀/2r. The percentage change of translational velocity after the sphere start pure rolling:

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a

$21 .42 %$

b

$7 .14 %$

c

None

d

$14 .28 %$

answer is A.

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Detailed Solution

Initial angular momentum

$L = \frac{2}{5} {Mr}^{2} (\frac{v_{0}}{2 r}) + {Mrv}_{0} = \frac{6}{5} {Mrv}_{0}$ …(i)

The angular momentum about A is

$\begin{array}{l} = \frac{2}{5} {Mr}^{2} (\frac{v}{r}) + Mrv = \frac{2}{5} {Mr}^{2} (\frac{v}{r}) + Mrv (ii) \\ v = \frac{6}{7} v_{0} \end{array}$

$∴ %$ change of translational velocity

$= [\frac{v_{0} - \frac{6}{7} v_{0}}{v_{0}}] \times 100 = \frac{1}{7} \times 100 = 14 .28 %$

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