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Question

(B) For an object sliding down a smooth incline,the potential energy is converted entirely into translational kinetic energy:$mgh = \frac{1}{2}mv^2 \i

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$\frac{v}{\sqrt{2}}$

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(B) For an object sliding down a smooth incline,the potential energy is converted entirely into translational kinetic energy:$mgh = \frac{1}{2}mv^2 \implies v = \sqrt{2gh}$.For a ring rolling down an inclined plane,the potential energy is converted into both translational and rotational kinetic energy:$mgh = \frac{1}{2}mv'^2 + \frac{1}{2}I\omega^2$.Since for a ring $I = mr^2$ and $\omega = \frac{v'}{r}$,we have:$mgh = \frac{1}{2}mv'^2 + \frac{1}{2}(mr^2)(\frac{v'^2}{r^2}) = \frac{1}{2}mv'^2 + \frac{1}{2}mv'^2 = mv'^2$.Equating the two expressions for $mgh$:$mv'^2 = \frac{1}{2}mv^2 \implies v'^2 = \frac{v^2}{2} \implies v' = \frac{v}{\sqrt{2}}$.

An object slides down a smooth incline and reaches the bottom with velocity $v$. If the same mass is in the form of a ring and it rolls down an inclined plane of the same height and angle of inclination,then its velocity at the bottom of the inclined plane will be ............

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