A circular hoop of mass m and radius R rests | vedclass

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(B) Let the center of the hoop be the origin. The bullet strikes the hoop at a point on its circumference. Let this point be $(R, 0)$.The center of ma

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$v/(3R)$

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(B) Let the center of the hoop be the origin. The bullet strikes the hoop at a point on its circumference. Let this point be $(R, 0)$.The center of mass of the hoop is at $(0, 0)$. The bullet is at $(R, 0)$.The center of mass of the system (hoop + bullet) is at $x_{CM} = \frac{m(0) + m(R)}{m + m} = \frac{R}{2}$.By conservation of linear momentum,the velocity of the center of mass $v_{CM}$ after the collision is given by $mv = (m + m)v_{CM}$,so $v_{CM} = v/2$.We calculate the angular momentum about the center of mass of the system. The distance of the bullet from the center of mass is $R/2$,and the distance of the hoop's center from the center of mass is $R/2$.The angular momentum of the bullet before the collision about the center of mass is $L = m v (R/2)$.The moment of inertia of the hoop about its own center is $I_{hoop, cm} = mR^2$. Using the parallel axis theorem,the moment of inertia of the hoop about the system's center of mass is $I_{hoop} = I_{hoop, cm} + m(R/2)^2 = mR^2 + mR^2/4 = 5mR^2/4$.The moment of inertia of the bullet about the system's center of mass is $I_{bullet} = m(R/2)^2 = mR^2/4$.The total moment of inertia of the system about the center of mass is $I = I_{hoop} + I_{bullet} = 5mR^2/4 + mR^2/4 = 6mR^2/4 = 3mR^2/2$.By conservation of angular momentum,$L = I \omega$,so $mvR/2 = (3mR^2/2) \omega$.Solving for $\omega$,we get $\omega = \frac{mvR/2}{3mR^2/2} = \frac{v}{3R}$.

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