A bob of mass m is suspended by a light string of length L. It is imparted a horizontal velocity `v_(0)` at the lowest point A such that it completes a semi-circular trajectory in the vertical plane with the string becoming slack on reaching the topmost point C, figure, Obtain an expression for (i) `v_(0)` (ii) the speeds at points B and C, (ii) the ration of kinetic energies `(K_(B)//K_(C))` at B and C.
Comment on the nature of the trajectory of the bob after it reahes the poing C.

Explanation

i) There are two external forces on the bob: gravity and the tension (T) in the string. The latter does no work since the displacement of the bob is always normal to the string. The potential energy of the bob is thus associated with the gravitational force only. The total mechanical energy E of the system is conserved. We take the potential energy of the system to be zero at the lowest point A. Thus, at A:
`E=1/2mv_(0)^(2)`
`T_(A) -mg=(mv_(0)^(2))/L ` [Newton's Second Law]
Where `T_(A)` isthe tension in the string at A. At the highest point C, the string slackens, as the tension in the string `(T_(c))` becomes zero.
Thus, at C
`E=1/2mv_(c)^()+2mgL`
`mg=(mv_(c)^(2))/L` [Newton's Second law]
where `v_(c)` is the speed at C.
`E=5/mgL`
Equatting this to the energy at A
`5/2mgL = m/2v_(0)^(2)`
or , `v_(0) = sqrt(5gL)`
ii) It is clear from Eq. (6.14)
`v_(C) = sqrt(5gL)`
At B, the energy is
`1/2mv_(B)^(2)+mgL`
Equating this to the energy at A and employing the result from (i), namely `v_(0)^(2) = 5gL`
`1/2mv_(B)^(2) + mgL = 1/2mv_(0)^(2)`
`5/2mgL`
`therefore v_(B) = sqrt(3gL)`
iii) The ratio of the kinetic energies at B and C is:
`(K_(B))/K_(C) = (1/2mv_(B)^(2))/(1/2mv_(C)^(2)` = `3/1`
At point C, the string becomes slack and the velocity of the bob is horizontal projection akin of a cliff. Otherwise the bob will continue on its circular path and complete the revolution.