A hoop of radius r and mass m rotating with a | vedclass

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(C) When the hoop is placed on the surface,friction acts on it,creating a torque about any point $O$ on the surface. Since the net external torque abo

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) When the hoop is placed on the surface,friction acts on it,creating a torque about any point $O$ on the surface. Since the net external torque about point $O$ is zero,the angular momentum about $O$ is conserved.Initial angular momentum about $O$ is $L_i = I_{cm}\omega_0 + mvr = (mr^2)\omega_0 + m(0)r = mr^2\omega_0$.When the hoop ceases to slip,it undergoes pure rolling,so $v_{cm} = \omega r$.Final angular momentum about $O$ is $L_f = I_{cm}\omega + mv_{cm}r = (mr^2)\omega + m(v_{cm})r = mr^2(\frac{v_{cm}}{r}) + mv_{cm}r = mv_{cm}r + mv_{cm}r = 2mv_{cm}r$.Equating $L_i$ and $L_f$:$mr^2\omega_0 = 2mv_{cm}r$$v_{cm} = \frac{r\omega_0}{2}$.

$A$ hoop of radius $r$ and mass $m$ rotating with an angular velocity $\omega_0$ is placed on a rough horizontal surface. The initial velocity of the centre of the hoop is zero. What will be the velocity of the centre of the hoop when it ceases to slip?

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