A thin ring of mass 2 kg has a radius of 0.5 | vedclass

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Question

(A) Let the ring have mass $M = 2 \ kg$,radius $R = 0.5 \ m$,and velocity $v = 1 \ m/s$. The ball has mass $m = 0.1 \ kg$,initial velocity $u_x = -20

Vedclass · Accepted Answer

The ring will perform pure rotation about its stationary center of mass.

Vedclass · Answer

(A) Let the ring have mass $M = 2 \ kg$,radius $R = 0.5 \ m$,and velocity $v = 1 \ m/s$. The ball has mass $m = 0.1 \ kg$,initial velocity $u_x = -20 \ m/s$,and final velocity $v_x = 0, v_y = 10 \ m/s$. The collision point is at height $h = 0.75 \ m = 1.5R$. Using impulse-momentum theorem for the ball in the $x$-direction: $J_x = m(v_x - u_x) = 0.1(0 - (-20)) = 2 \ Ns$. The impulse on the ring is $J_x' = -2 \ Ns$. For the ring,the change in linear momentum is $M(v_f - v_i) = J_x'$,so $2(v_f - 1) = -2$,giving $v_f = 0$. The impulse on the ring at height $1.5R$ creates an angular impulse about the center of mass: $	au_{imp} = J_x' 	imes (1.5R - R) = -2 	imes 0.25 = -0.5 \ Nms$. Initial angular momentum $L_i = I\omega = (MR^2)(v/R) = MvR = 2 	imes 1 	imes 0.5 = 1 \ kgm^2/s$. Final angular momentum $L_f = L_i + 	au_{imp} = 1 - 0.5 = 0.5 \ kgm^2/s$. Since $L_f = I\omega_f = (MR^2)\omega_f = (2 	imes 0.5^2)\omega_f = 0.5\omega_f$,we get $0.5 = 0.5\omega_f$,so $\omega_f = 1 \ rad/s$. Since $v_f = 0$ and $\omega_f 
eq 0$,the ring is in pure rotation about its center of mass.

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