A body of mass m is rotating in a vertical circle of radius 'r' with critical speed. The difference in its K.E. at the top and at the bottom is

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A body of mass m is rotating in a vertical circle of radius 'r' with critical speed. The difference in its K.E. at the top and at the bottom is

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2mgr

4mgr

6mgr

3mgr

answer is A.

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Detailed Solution

$∆ K . E = \frac{1}{2} m ({v^{2}}_{bottom} - {v^{2}}_{top})$ =mgh=mg2r

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