A body of mass ‘m’ connected to a massless and unstretchable string goes in vertical circle of radius ‘R’ under gravity g. The other end of the string is fixed at the center of circle. If velocity at top of circular path is n gR, where, n 1, then ratio of kinetic energy of the body at bottom to that at top of the circle is :
- n2+4n2
- n+4n
- n2n2+4
- nn+4
Answer
(1) n2+4n2
Solution
At the top: vT = n gR KT = 12mn2gR
Energy conservation (bottom is 2R below the top):
KB = KT + mg(2R) = 12mn2gR + 2mgR
KB = 12mgR(n2 + 4)
KBKT = n2+4n2
Answer: (1)