A ball is thrown on a lawn in such a way that | vedclass

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(C) Let the mass of the ball be $m$ and its radius be $R$. The moment of inertia of the ball about its center of mass is $I = \frac{2}{5} mR^2$.Initia

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$\frac{5}{7} v_0$

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(C) Let the mass of the ball be $m$ and its radius be $R$. The moment of inertia of the ball about its center of mass is $I = \frac{2}{5} mR^2$.Initially,the ball has only translational velocity $v_0$ and no angular velocity.As the ball slides,friction acts on it,creating a torque about the point of contact $O$ on the ground. Since the torque due to friction about point $O$ is zero,the angular momentum of the ball about point $O$ is conserved.Initial angular momentum about $O$: $L_i = m v_0 R$.When the ball starts rolling without slipping,its velocity is $v$ and its angular velocity is $\omega = \frac{v}{R}$.Final angular momentum about $O$ (using the parallel axis theorem): $L_f = I_{cm} \omega + m v R = (\frac{2}{5} mR^2) \frac{v}{R} + m v R = \frac{2}{5} m v R + m v R = \frac{7}{5} m v R$.Equating initial and final angular momentum: $m v_0 R = \frac{7}{5} m v R$.Solving for $v$: $v = \frac{5}{7} v_0$.

$A$ ball is thrown on a lawn in such a way that it initially slides with a speed $v_0$ without rolling. It gradually picks up rotational motion. The speed of the ball at which there will be rolling without slipping is

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