A ring of mass m and radius r is placed on a | vedclass

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(D) Initially,the velocity of the center of mass of the ring is $v_0 = 0$ and the initial angular velocity is $\omega_0$.When the ring stops slipping,

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$\frac{r\omega_0}{2}$

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(D) Initially,the velocity of the center of mass of the ring is $v_0 = 0$ and the initial angular velocity is $\omega_0$.When the ring stops slipping,let the velocity of the center of mass be $v$ and the angular velocity be $\omega$.Since the ring is rolling without slipping,the condition $v = r\omega$ must be satisfied.By the principle of conservation of angular momentum about a point $P$ on the surface:$L_i = L_f$$I_{cm}\omega_0 + mvr_0 = I_{cm}\omega + mvr$Since $I_{cm} = mr^2$ for a ring:$mr^2\omega_0 + 0 = mr^2\omega + mvr$$mr^2\omega_0 = mr^2(\frac{v}{r}) + mvr$$mr^2\omega_0 = mrv + mvr$$mr^2\omega_0 = 2mvr$$v = \frac{r\omega_0}{2}$

$A$ ring of mass $m$ and radius $r$ is placed on a rough surface with an initial angular velocity $\omega_0$. The initial velocity of the center of the ring is zero. What will be the velocity of its center when the ring stops slipping?

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