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  4. A body of mass 1 kg is rotating in a vertical circle of radius 1 m. What will be the difference in its kinetic energy at the top and bottom of the circle? (take g=10 ms-2 )

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Q. A body of mass $1 \,kg$ is rotating in a vertical circle of radius $1\,m$. What will be the difference in its kinetic energy at the top and bottom of the circle? (take $ g=10\,m{{s}^{-2}} $ )

ManipalManipal 2008Work, Energy and Power

Solution:

Velocity at the top is $ \sqrt{gr} $ and that at the bottom is $ \sqrt{5gr} $ .
Hence, required difference in kinetic energy
$ =\frac{1}{2}m[5gr-gr]=2gmr $
$ =2\times 10\times 1\times 1=20\,J $