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A uniform sphere of radius $R$ is placed on a rough horizontal surface and given a linear velocity $v_{0}$ , and angular velocity $ω_{0}$ as shown. The sphere comes to rest after moving some distance to the right. It follows that:

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$2 v_{0} = 5 ω_{0} R$

$5 v_{0} = 2 ω_{0} R$

$2 v_{0} = ω_{0} R$

$v_{0} = ω_{0} R$

answer is C.

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Detailed Solution

$a = \frac{f}{m} = μ g$

and $α = \frac{τ}{I} = \frac{f R}{I} = \frac{μ m g \times R}{\frac{2}{5} m R^{2}} = \frac{5 μ g}{2 R}$

Using equation of motion, we have

$0 = v_{0} - a t$ and $0 = ω_{0} - α t$

From above equations, we get

$\frac{v_{0}}{ω_{0}} = \frac{a}{α} = \frac{μ g}{\frac{5 μ g}{2 R}} = \frac{2 R}{5} \Rightarrow 5 v_{0} = 2 ω_{0} R$

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