A stationary homogenous sphere of radius R and mass M is gently placed on the conveyor belt moving with a constant velocity v₀ towards the right. After a time, t from the instant the sphere was placed on the belt, it starts pure rolling, and its CM moves with a velocity v relative to the ground. Coefficient of kinetic friction is $μ$ _k.

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$t = \frac{2 v_{0}}{5 μ_{k} g}$

$v = \frac{2 v_{0}}{7}$

$t = \frac{2 v_{0}}{7 μ_{k} g}$

$v = \frac{2 v_{0}}{5}$

answer is B, D.

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Detailed Solution

$\begin{array}{l} V_{cm} = A_{cm} \times t \\ ω = αt \\ V_{bottom} = V_{cm} + ωR \end{array}$
When pure rolling begin
$\begin{array}{l} V_{bottom} = V_{0} = A_{cm} t + R (αt) \\ V_{0} = t (A_{cm} + αR) \end{array}$
Using dynamics …..
$\begin{array}{l} \begin{aligned} μ_{k} mg = {mA}_{cm} \Rightarrow A_{cm} = μ_{k} g \\ μ_{k} mg \times R = \frac{2}{5} {mR}^{2} α \Rightarrow α = \frac{5}{2} \frac{μ_{k} g}{R} \\ V_{0} = t (μ_{k} g + \frac{5}{2} μ_{k} g) \Rightarrow t = \frac{2 V_{0}}{7 μ_{k} g} \end{aligned} \\ V_{cm} = μ_{k} g \times t = \frac{2 V_{0}}{7} \end{array}$