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Question

(A) For a body sliding down a smooth inclined plane of height $h$,the potential energy is converted into kinetic energy: $mgh = \frac{1}{2}mV^2$,which

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$\frac{V}{\sqrt{2}}$

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(A) For a body sliding down a smooth inclined plane of height $h$,the potential energy is converted into kinetic energy: $mgh = \frac{1}{2}mV^2$,which gives $V = \sqrt{2gh}$.For a ring rolling down the same inclined plane,the potential energy is converted into both translational and rotational kinetic energy: $mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$.For a ring,the moment of inertia $I = mr^2$ and the rolling condition is $v = r\omega$,so $\omega = v/r$.Substituting these into the energy equation: $mgh = \frac{1}{2}mv^2 + \frac{1}{2}(mr^2)(v/r)^2 = \frac{1}{2}mv^2 + \frac{1}{2}mv^2 = mv^2$.Thus,$v^2 = gh$,which means $v = \sqrt{gh}$.Comparing the two velocities: $v = \sqrt{gh} = \frac{\sqrt{2gh}}{\sqrt{2}} = \frac{V}{\sqrt{2}}$.

$A$ body slides down a smooth inclined plane of inclination $\theta$ and reaches the bottom with velocity $V$. If the same body is a ring which rolls down the same inclined plane,then the linear velocity at the bottom of the plane is:

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