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(B) Let $m$ be the mass of the sphere and $k$ be its radius of gyration. The potential energy $PE$ at the top of the incline is converted into kinetic

Vedclass · Accepted Answer

$3R_0/4$

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(B) Let $m$ be the mass of the sphere and $k$ be its radius of gyration. The potential energy $PE$ at the top of the incline is converted into kinetic energy at the bottom.Case $1$: Rolling without slipping.$PE = K.E_{trans} + K.E_{rot} = \frac{1}{2}mv_0^2 + \frac{1}{2}I\omega^2$Since $I = mk^2$ and $\omega = v_0/R_0$,we have:$PE = \frac{1}{2}mv_0^2 + \frac{1}{2}(mk^2)(v_0^2/R_0^2) = \frac{1}{2}mv_0^2(1 + k^2/R_0^2) \dots (i)$Case $2$: Sliding without rolling (frictionless).$PE = K.E_{trans} = \frac{1}{2}m(5v_0/4)^2 = \frac{1}{2}m(25v_0^2/16) \dots (ii)$Equating $(i)$ and $(ii)$ since $PE$ is the same:$\frac{1}{2}mv_0^2(1 + k^2/R_0^2) = \frac{1}{2}m(25v_0^2/16)$$1 + k^2/R_0^2 = 25/16$$k^2/R_0^2 = 25/16 - 1 = 9/16$$k = 3R_0/4$

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