A uniform ring of radius R is moving on a hor | vedclass

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(B) Since the ring is rolling without any slippage,the total initial kinetic energy is equal to the total final potential energy at height $h$.Total i

Vedclass · Accepted Answer

$v^{2} / g$

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(B) Since the ring is rolling without any slippage,the total initial kinetic energy is equal to the total final potential energy at height $h$.Total initial kinetic energy = Kinetic energy of translation + Kinetic energy of rotation$\Rightarrow K.E. = \frac{1}{2} m v^{2} + \frac{1}{2} I \omega^{2}$For a ring,the moment of inertia $I = m R^{2}$ and the condition for pure rolling is $v = R \omega$,which implies $\omega = v / R$.Substituting these into the energy equation:$K.E. = \frac{1}{2} m v^{2} + \frac{1}{2} (m R^{2}) (v / R)^{2}$$K.E. = \frac{1}{2} m v^{2} + \frac{1}{2} m v^{2} = m v^{2}$At the maximum height $h$,the final kinetic energy is zero,so the total initial kinetic energy is converted into potential energy:$m v^{2} = m g h$$h = \frac{v^{2}}{g}$

$A$ uniform ring of radius $R$ is moving on a horizontal surface with speed $v$,then climbs up a ramp of inclination $30^{\circ}$ to a height $h$. There is no slipping in the entire motion. Then,$h$ is

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