A bob of mass 'm' is suspended by a light string of length 'L'. It is imparted a minimum horizontal velocity at the lowest point A such that it just completes half circle reaching the top most position B. The ratio of kinetic energies (K.E.)B(K.E.)A is :
(A)3:2
(B)5:1✓ Correct
(C)2:5
(D)1:5
Answer: (B)
Solution
To just complete the vertical circle, the minimum speed at the top B must make gravity alone supply the centripetal force:
mg=LmvB2⟹vB2=gL
Applying energy conservation between the lowest point A and the top B, a height gain of 2L:
21mvA2=21mvB2+mg(2L)
vA2=gL+4gL=5gL
(K.E.)B(K.E.)A=21m(gL)21m(5gL)=15
JEE Main · 2024 · 29 Jan 2024 · Shift 2 · Single correct option