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(C) Initially,the sphere is placed on the surface with angular velocity $\omega$ and linear velocity $v_{CM} = 0$. Friction acts at the point of conta

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$2 \omega / 7$

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(C) Initially,the sphere is placed on the surface with angular velocity $\omega$ and linear velocity $v_{CM} = 0$. Friction acts at the point of contact to oppose the slipping.The force of friction $f = \mu_k mg$ acts in the forward direction,causing linear acceleration $a_{CM} = \frac{f}{m} = \mu_k g$.The torque due to friction about the center of mass is $	au = -fR = -\mu_k mgR$. The angular deceleration is $\alpha = \frac{	au}{I} = \frac{-\mu_k mgR}{\frac{2}{5}mR^2} = -\frac{5\mu_k g}{2R}$.After time $t$,the linear velocity is $v_{CM} = a_{CM}t = \mu_k gt$ and the angular velocity is $\omega_f = \omega + \alpha t = \omega - \frac{5\mu_k g}{2R}t$.Pure rolling begins when $v_{CM} = \omega_f R$. Substituting the expressions:$\mu_k gt = \left(\omega - \frac{5\mu_k g}{2R}tight)R$$\mu_k gt = \omega R - \frac{5}{2}\mu_k gt$$\frac{7}{2}\mu_k gt = \omega R \implies \mu_k gt = \frac{2}{7}\omega R$.Substituting this back into the expression for $\omega_f$:$\omega_f = \frac{v_{CM}}{R} = \frac{\mu_k gt}{R} = \frac{1}{R} \left(\frac{2}{7}\omega Right) = \frac{2}{7}\omega$.

$A$ solid sphere spinning about a horizontal axis with an angular velocity $\omega$ is placed on a horizontal surface. Subsequently,it rolls without slipping with an angular velocity of

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