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(B) At $P_{1}$,the ring is moving down and its angular velocity $\omega$ is increasing to satisfy the rolling condition $(v_{CM} = R\omega)$. The fric

Vedclass · Accepted Answer

It is opposite to $v_{CM}$ at $P_{1}$ and in the same direction as $v_{CM}$ at $P_{3}$.

Vedclass · Answer

(B) At $P_{1}$,the ring is moving down and its angular velocity $\omega$ is increasing to satisfy the rolling condition $(v_{CM} = R\omega)$. The friction force $f$ acts upwards (opposite to $v_{CM}$) to provide the necessary torque to increase $\omega$.At $P_{2}$,the velocity $v_{CM}$ is horizontal and the acceleration of the center of mass is purely centripetal. There is no tangential acceleration,so the frictional force is zero.At $P_{3}$,the ring is moving up and its angular velocity $\omega$ is decreasing to satisfy the rolling condition. The friction force $f$ acts upwards (in the same direction as $v_{CM}$) to provide the torque required to decrease $\omega$.Therefore,the frictional force is opposite to $v_{CM}$ at $P_{1}$ and in the same direction as $v_{CM}$ at $P_{3}$.

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