A thin ring of mass 2 kg and radius 1 m is | vedclass

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(A) Let $M = 2 \ kg$ be the mass of the ring,$R = 1 \ m$ be its radius,and $m = 1 \ kg$ be the mass of the ball.$1$. Conservation of linear momentum a

Vedclass · Accepted Answer

$A$ and $C$

Vedclass · Answer

(A) Let $M = 2 \ kg$ be the mass of the ring,$R = 1 \ m$ be its radius,and $m = 1 \ kg$ be the mass of the ball.$1$. Conservation of linear momentum along the $x$-axis:Initial momentum $P_i = M(-v_{cm}) + m(v_{ball}) = 2(-1) + 1(2) = 0$.Final momentum $P_f = M(v') + m(v'_{ball,x}) = 2(v') + 1(0) = 2v'$.Since no external horizontal force acts during the collision,$P_i = P_f \Rightarrow 0 = 2v' \Rightarrow v' = 0$.Thus,the center of mass of the ring comes to rest.$2$. Conservation of angular momentum about the point of contact $P$:Initial angular momentum $L_i = I_P \omega_i + m v_{ball} r_{\perp, i} = (2MR^2) \omega_i + m v_{ball} (R + h - R) = (2 	imes 2 	imes 1^2) (1) + 1(2)(1.8) = 4 + 3.6 = 7.6 \ kg \cdot m^2/s$.Final angular momentum $L_f = I_P \omega_f + m v'_{ball,y} r_{\perp, f} = (2MR^2) \omega_f + m v'_{ball,y} (R - (1.8 - R)) = (2 	imes 2 	imes 1^2) \omega_f + 1(1)(0.2) = 4\omega_f + 0.2$.Equating $L_i = L_f \Rightarrow 7.6 = 4\omega_f + 0.2 \Rightarrow 4\omega_f = 7.4 \Rightarrow \omega_f = 1.85 \ rad/s$.Since $v_{cm} = 0$ and $\omega_f 
eq 0$,the ring is in pure rotation about its $CM$. The bottommost point has velocity $v = \omega R$ to the right,so friction acts to the left to oppose this motion. Thus,$(A)$ and $(C)$ are correct.

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The torque acting on a body about a point is equal to $\overrightarrow{A} \times \overrightarrow{L}$,where $\overrightarrow{A}$ is a constant vector and $\overrightarrow{L}$ is the angular momentum about that point. This implies that:

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