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

Question

Dr. Abubacker · Accepted Answer

(b) Difference in $KE = \frac{1}{2}m\left\lbrack \left( \sqrt{5gr} ight)^{2} - \sqrt{gr} ightbrack^{2}$
$= 2mgr = 2 	imes 1 	imes 10 	imes 1 = 20\ $J

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

Explanation

Reviewed by Dr. Abubacker

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