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) External forces acting on the bob are (a) Tension (T) in the string (b) force gravity (mg).
At the lowest position A , P.E of the system is taken to be zero .
` :. ` Total M.E at A = K.E
` rArr " " E = 1/2mv_(0)^(2)" "` ...(i)
Using Newton.s second of law
`T_(A)- mg = (mv_(0)^(2))/L " " ....(ii)`
At C , the highest point , the striing slackness . Therefore , tension `T_(C)=0`
If v. is the speed at C , then total M.E at C is
`E = K +U `
`E= 1/2 mv^(2) +mg(2L)`
` = 1/2 mv^(2) +2mgL " " ....(iii)`
Also , applying Newton.s law at point C ,
` mg = (mv^(2))/(L )`
or `mgL = mv^(2) " "....(iv)`
Using (iv) in (iii) , we get
`5/2 mgL = m/2 v_(0)^(2)`
`rArr " " v_(0)= sqrt(5gL)`
(ii) From equation (iv)
`v^(2)= gL or v. = sqrt(gL)`(speed at C)
Total energy at point B is
Since total energy at every point is conserved , using (i) and (vi) , we get
`1/2 mv^(2) +mgL = 1/2 mv_(0)^(2) = 1/2 m xx 5gL `
` 1/2 mv^(2) = 5/2 mgL - mgL = 3/2 mgL `
`rArr " " v" = sqrt(3gL)`
(iii) Ratio of kinetic energies at B and C is
`("K.E at B")/("K.E at C") = (K)/(K) = (1/2mv^(2))/(1/2mv^(2))=(3gL)/(gL) = 3:1`