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Question

(C) When an object of mass $m$ slides down an inclined plane without friction,the potential energy is converted entirely into translational kinetic en

Vedclass · Accepted Answer

$\frac{1}{\sqrt{2}}v$

Vedclass · Answer

(C) When an object of mass $m$ slides down an inclined plane without friction,the potential energy is converted entirely into translational kinetic energy: $mgh = \frac{1}{2}mv^2$,which gives $v = \sqrt{2gh}$.When the object is a ring rolling down the inclined plane,the potential energy is converted into both translational and rotational kinetic energy: $mgh = \frac{1}{2}mv_{ring}^2 + \frac{1}{2}I\omega^2$.For a ring,the moment of inertia $I = mR^2$ and $\omega = \frac{v_{ring}}{R}$.Substituting these,we get: $mgh = \frac{1}{2}mv_{ring}^2 + \frac{1}{2}(mR^2)(\frac{v_{ring}}{R})^2 = \frac{1}{2}mv_{ring}^2 + \frac{1}{2}mv_{ring}^2 = mv_{ring}^2$.Thus,$v_{ring} = \sqrt{gh}$.Since $v = \sqrt{2gh}$,we can write $\sqrt{gh} = \frac{v}{\sqrt{2}}$.Therefore,the velocity of the ring is $\frac{v}{\sqrt{2}}$.

An object of mass $m$ slides down an inclined plane and reaches the bottom with a velocity $v$. If the same object were in the form of a ring and rolled down the same inclined plane,its velocity at the bottom would be:

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