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Question

(A) For a body sliding down a smooth inclined plane, there is no friction, so all potential energy is converted into translational kinetic energy: $mg

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) For a body sliding down a smooth inclined plane, there is no friction, so all potential energy is converted into translational kinetic energy: $mgh = \frac{1}{2}mV^2$, which gives $V = \sqrt{2gh}$ or $V^2 = 2gh$.For a ring rolling down the same inclined plane, the potential energy is converted into both translational and rotational kinetic energy: $mgh = \frac{1}{2}mv'^2 + \frac{1}{2}I\omega^2$.Since $I = mr^2$ and $\omega = v'/r$ for a ring, we have $mgh = \frac{1}{2}mv'^2 + \frac{1}{2}(mr^2)(v'/r)^2 = \frac{1}{2}mv'^2 + \frac{1}{2}mv'^2 = mv'^2$.Thus, $v'^2 = gh$.Substituting $gh = V^2/2$ into the equation, we get $v'^2 = V^2/2$, which implies $v' = \frac{V}{\sqrt{2}}$.

$A$ 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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