An object slides down a smooth incline and re | vedclass

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Question

(b)

$v=\sqrt{2 g h}$

$m g h=\frac{1}{2} m v^2+\frac{1}{2} m r^2 \frac{v^2}{r^2}$

$v^{\prime}=\sqrt{g h}$

$v^{\prime}=\frac{v}{\sqrt{2}}$

Vedclass · Accepted Answer

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

Vedclass · Answer

(b)

$v=\sqrt{2 g h}$

$m g h=\frac{1}{2} m v^2+\frac{1}{2} m r^2 \frac{v^2}{r^2}$

$v^{\prime}=\sqrt{g h}$

$v^{\prime}=\frac{v}{\sqrt{2}}$

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