A solid sphere of radius R starts rotating on rough horizontal surface with translational velocity v₀ and initial angular velocity $ω_{0} = 2 v_{0} / 3 R$ . The sphere starts pure
rolling after some time t. Find the angle by which sphere rotates upto the instant at which pure rolling starts, if v is the translational velocity at pure rolling. Assume uniformly accelerated motion up to start of pure rolling.

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$\frac{33}{38} \frac{v t}{R}$

$\frac{37}{38} \frac{v t}{R}$

$\frac{31}{38} \frac{v t}{R}$

$\frac{29}{37} \frac{v t}{R}$

answer is A.

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Detailed Solution

$ω_{0} = \frac{2 v_{0}}{3 R} or, v_{0} = \frac{3}{2} R ω_{0}$

Angular momentum is conserved about A.

$L_{i} = L_{f}$

$\begin{array}{l} \Rightarrow \frac{2}{5} m R^{2} \times \frac{2 v_{0}}{3 R} + m v_{0} R = \frac{2}{5} m R^{2} \frac{v}{R} + m v R \\ \Rightarrow v_{0} = \frac{21}{19} v . . . . . . . . . . . (i) \\ N o w, ω = ω_{0} + α t \\ \Rightarrow v / R = 2 v_{0} / 3 R + α t \\ \Rightarrow α = \frac{5 v}{19 R t} \end{array}$

where t is the time in which pure rolling starts.
Pure rolling starts when sphere rotates by an angle,

$θ = ω_{0} t + \frac{1}{2} α t^{2} = \frac{2 v_{0}}{3 R} t + \frac{1}{2} \cdot \frac{5 v}{19 R t} \cdot t^{2} = \frac{33}{38} \frac{v t}{R}$

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