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.

Options

A. A-p,B-q,C-s,D-r
B. A-q,B-r,C-q,D-s
C. A-r,B-s,C-q,D-p
D. A-s,B-p,C-r,D-q

Answer

C

Explanation

There are two external forces on the bob: gravity (mg) and tension (T) in the string. The tension does no work as displacement is always perpendicular to the string.Total mechanical energy (E) of the system is conserved.
If we take potential energy of the system to be zero at the lowest point A, then
At A, `E=1/2mv_0^2`...(i)
From Newton's second law
`T_A-mg=(mv_0^2)/L` ...(ii)
where `T_A` is the tension in the string at A.
At the highest point C, to complete the circle tension in string will be minimum (zero).
At C, E =`1/2 mv_C^2`+mg(2L)...(iii)
From Newton's second law
`mg=(mv_C^2)/L` ...(iv)
From (iv), `v_C=sqrt(gL)`,C-q
From (iii) , `E=1/2m(gL)+2mgL=5/2mgL`...(v)
Using (i),`1/2mv_0^2=5/2mgL`,
`v_0=sqrt(5gL), therefore A-r`...(vi)
At B, `E=1/2 mv_B^2+mg(L)`
or `1/2mv_B^2=E-mg(L)=5/2mgL-mgL` (Using (v))
`1/2mv_B^2=3/2mgL`
`therefore v_B=sqrt(3gL), therefore B-s`
The ratio of kinetic energies at B and C is `therefore K_B/K_C=(1/2mv_B^2)/(1/2mv_C^2)=(3gL)/(gL)=3/1, therefore D-p`