A ring has inner radius R_1 and outer radius | vedclass

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Question

(D) When a body rolls without slipping with a constant angular velocity $\omega$,the center of mass moves with a constant linear velocity $v = R_{cm}\

Vedclass · Accepted Answer

$\frac{R_1}{R_2}$

Vedclass · Answer

(D) When a body rolls without slipping with a constant angular velocity $\omega$,the center of mass moves with a constant linear velocity $v = R_{cm}\omega$. For any particle at a distance $r$ from the center of the ring,the centripetal force experienced by the particle is given by $F = m\omega^2r$. Since the ring is rolling with a constant angular velocity $\omega$,the centripetal force acting on a particle of mass $m$ at distance $r$ from the center is $F = m\omega^2r$. For the particle on the inner surface,the radius is $R_1$,so $F_1 = m\omega^2R_1$. For the particle on the outer surface,the radius is $R_2$,so $F_2 = m\omega^2R_2$. The ratio of the forces is $\frac{F_1}{F_2} = \frac{m\omega^2R_1}{m\omega^2R_2} = \frac{R_1}{R_2}$. Therefore,the correct option is $D$.

$A$ ring has inner radius $R_1$ and outer radius $R_2$. It rolls without slipping with a constant angular velocity $\omega$. What is the ratio of the forces $F_1/F_2$ experienced by two particles located on the inner and outer surfaces of the ring?

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