A circular hoop of radius 50 cm and mass 1 | vedclass

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(C) The initial angular momentum of the hoop about the point of contact is $L_i = I_{cm} \omega_0 + m r^2 \omega_0 = m r^2 \omega_0 + m r^2 \omega_0 =

Vedclass · Accepted Answer

$25$

Vedclass · Answer

(C) The initial angular momentum of the hoop about the point of contact is $L_i = I_{cm} \omega_0 + m r^2 \omega_0 = m r^2 \omega_0 + m r^2 \omega_0 = 2 m r^2 \omega_0$. However,considering the angular momentum about the contact point $P$ at the instant it is placed: $L_i = I_P \omega_0 = (I_{cm} + m r^2) \omega_0 = (m r^2 + m r^2) \omega_0 = 2 m r^2 \omega_0$.When the hoop stops slipping,it rolls without slipping,so $v = r \omega$. The final angular momentum about the contact point $P$ is $L_f = I_{cm} \omega + m r v = m r^2 (v/r) + m r v = m r v + m r v = 2 m r v$.Since the frictional force acts through the contact point,the net torque about the contact point is zero. Thus,angular momentum is conserved: $L_i = L_f$.$2 m r^2 \omega_0 = 2 m r v$.Dividing both sides by $2 m r$,we get $v = r \omega_0$.Wait,re-evaluating: $L_i = I_{cm} \omega_0 = m r^2 \omega_0$. The angular momentum about the contact point is $L_i = I_{cm} \omega_0 = m r^2 \omega_0$.$L_f = I_{cm} \omega + m r v = m r^2 (v/r) + m r v = 2 m r v$.Equating $L_i = L_f$: $m r^2 \omega_0 = 2 m r v$.Therefore,$v / \omega_0 = r / 2 = 50 \ cm / 2 = 25 \ cm$.

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