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Q.

A solid sphere of mass m and radius R is gently placed on a conveyer belt moving with constant velocity $v_{0}$ . If coefficient of friction between belt and sphere is 2 / 7, the distance traveled by the centre of the sphere before it starts pure rolling is

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a

(2) $\frac{2 v_{0}^{2}}{49 g}$

b

(4) $\frac{2 v_{0}^{2}}{7 g}$

c

(3) $\frac{2 v_{0}^{2}}{5 g}$

d

(1) $\frac{v_{0}^{2}}{7 g}$

answer is A.

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Detailed Solution

$a = \frac{μmg}{m} = μg = \frac{2}{7} g$

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$α = \frac{μmgR}{\frac{2}{5} {mR}^{2}} = \frac{5 μg}{2 R} = \frac{5}{7} \frac{g}{R} Pure rolling will start when, v_{p} = v + Rω = v_{0} or at + R (αt) = v_{0} ∴ (\frac{2}{7} gt) + \frac{5}{7} gt = v_{0} ∴ t = \frac{v_{0}}{g} S = \frac{1}{2} {at}^{2} = \frac{1}{2} (\frac{2}{7} g) {(\frac{v_{0}}{g})}^{2} = \frac{v_{0}^{2}}{7 g}$

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