A bob of mass m is suspeded 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 semicircular trajectroy in the vertical plane with the string becoming slack only on reaching the topmost point C . Thisis shown in figure .Obtain a expression for (i) `v_(0)` (ii) the ratio of the kinetic energies `(K_(B)/(K_(C )))` at B and C . Comment on the nature of the trajectroy of the bob after it reaches the point C .

Explanation

Two forces acting on bob. (i)Gravity and (ii) the tension T in the string .the work is not done by tension in the string .because dispalcement of bob is always normal to the string .
The potential energy of the bob is thus associated with the gravitational force only and mechanical energy of the system is conserved .
(i) Let potential energy of the system is conserved .
L=0
` :. mgL =0`
Total mechanical energy point A,
`E =1/2 mv_(0)^(2) [ :. "Newton.s Second law"] `
` E =1/2 mv_(0^(2) " "....(1)`
The velocity of bob at point A , is `v_(0)`
Bob moving circular .
` :. ` Resultant force is string = centripetal force `T_(A)-mg =(mv_(0)^(2))/L [ :. "Newton.s Second law"] `
Where `T_(A)` is the tension in the string at A. At the highest point C, the string = centripetal force
`T_(A) -mg =(mv_(0)^(2))/L [ :. "Newton.s Second Law"]`
Where `T_(A)` is the tension in the tring at A. At the highest [point C, the string slackens , as the tension in the string `T_(C)` become zero.
Thus , at C
`E=1/2 mv_(C )^(2) +2mgL " "....(2)`
Where `T_(A)` is the tension in the string t A. At the highest point C, the string slackness, as the tension in the string `T_(C)`becomes zero. Thus at C
` E=1/2 mv_(C)^(2)+2mgL " "...(2)`
`(mv_(C )^(2))/L = mg" " {:. "Newton.s second law "]`
Where `v_(c)` is the speed at C
` :. v_(C)^(2) =mgL " "....(3)`
And `v_(C)^(2) -gL " "...(4)`
` :. ` From equation (2). ,
` E = 1/2 m cc"gL"+2mL`
` :. E=(5mgL)/2 " "...(5)`
`1/2 mv_(0)^(2) =(5mgL)/2 " "` [ From equation (1)]
` :. v_(0)=sqrt(5gL) " "...(6)`
(ii) From equation (4),
`v_(C )=sqrt(gL) " "...(7)`
and at the Btotal mechanical energy .
`E =1/2 mv_(B)^(2) +mgL " " [ :. h = L]`
From equation (6) and (1),
`1/2 mv_(0)^(2) =1/2 mv_(B)^(2) +mgL`
`1/2 m cc 5gL =1/2 mv_(B)^(2)+mgL`
` [ :. "From equation (7)"]`
` :. 1/2mv_(B)^(2)=5/2 mgL-mgL`
`1/2 mv_(B)^(2)=3/2 mgL`
` :. v_(B) =sqrt(3gL)`
(iii)The ratio of the kinetic energies at B and C is :
The ratio of the kinetic energies at B and C is :
`(K_(B))/(K_(C))=(1/2mv_(B)^(2))/(1/2mv_(C)^(2))=(v_(b)^(2))/(v_(C )^(2))`
` :. (K_(B))/(K_(C ))=(3gL)/("gL")=3:1`
At point C , the string becomes slack and the to the left . If the string is cut at this instant , the bob will execute a projectile to motion with horizontal and to the left . If the string is cit at this instant , the bob will execute a projectile to motion with horizontal projection akin to motion with horizontal projection akin to a rock kicked horizontally from the edge of a sliff. Otherwise the bob will continue on its circular pathand complete the revolution .