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Question

(C) For a body sliding down a smooth inclined plane,the potential energy is converted entirely into translational kinetic energy. Thus,$v = \sqrt{2gh}

Vedclass · Accepted Answer

$\sqrt{\frac{5}{7}} v$

Vedclass · Answer

(C) For a body sliding down a smooth inclined plane,the potential energy is converted entirely into translational kinetic energy. Thus,$v = \sqrt{2gh}$.For a sphere rolling down a rough inclined plane,the potential energy is converted into both translational and rotational kinetic energy. The velocity at the bottom is given by $v_{CM} = \sqrt{\frac{2gh}{1 + \frac{K^2}{R^2}}}$.Substituting $v = \sqrt{2gh}$,we get $v_{CM} = \frac{v}{\sqrt{1 + \frac{K^2}{R^2}}}$.For a solid sphere,the radius of gyration $K$ satisfies $\frac{K^2}{R^2} = \frac{2}{5}$.Substituting this value,$v_{CM} = \frac{v}{\sqrt{1 + \frac{2}{5}}} = \frac{v}{\sqrt{\frac{7}{5}}} = \sqrt{\frac{5}{7}} v$.

$A$ body slides down a smooth inclined plane having angle $\theta$ and reaches the bottom with velocity $v$. If the body is a solid sphere rolling down the same plane,then its linear velocity at the bottom of the plane is

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