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)A(K.E)B is : [figure]
- 5 : 1
- 3 : 2
- 1 : 5
- 2 : 5
Answer
(1) 5 : 1
Solution
Just reaching the top B tension TB=0:
mg= mVH2L VH2=gL
Energy conservation A B: 12mVL2= 12mVH2+mg(2L)
VL2=gL+4gL=5gL
(K.E)A(K.E)B= VL2VH2= 5gLgL=5:1
At the top only gravity supplies the centripetal force when the speed is minimum.
Answer: (1)