A bob of mass ma suspended by a massle3ss inextensible string is rotated in a vertical circle of radius r as shoen in the figure. Obtain an expenssion for speeds and kinetic energies at A, B and C, Comment on the nature of trajectory of the bob after it reaches the point C.

Explanation

Let horizontal velocity at lowest pint A be `V_(A)` such that string becomes slack only on reaching highest point C. Taking the potential energy of the system to be zero at A, the total mechanical energy, E
At A:
`E=1/2mv_(A)^(2)" "...(i)(E=KE+PE=KEasPE=0)`
`T_(A)-mg=(mv_(A)^(2))/(r)" "...(ii)("Using Newron's"II^(nd)Law)`
Where `T_(A)` is the tension in the string at Agt
At the highest point C, the string slackent, so `T_(c)=0`
At C:
`E=1/2mv_(c)^(2)+2mgr" "...(ii) (Wherev_(c)"is the velocity at C.")`
`mg=(mv_(c)^(2))/(r)" "...(iv)("Nowron's"II^(nd)law)`
From equation (iv),`1/2mv_(c)^(2)=1/2mgr`
Put this value in equation (iii)
`E=1/2mg+2mgr`
`E=5/2mgr" "...(v)("This is alos KE at A as"PE_(A)=0)`
From equations(i) & (v)
`1/2mv_(A)^(2)=5/2mgr`
`v_(A)=sqrt(5gr)" "...(vi)`
At B:
`E=1/2mv_(B)^(2)+mgr" "(x)" "("Where"v_(B) "is the velocity at B")`
This should be same as teh energy at A.
So, `1/2mv_(B)^(2)+mgr=1/2mv_(A)^(2)=5/2mgr("Using equation(viii)")`
`1/2mv_(B)^(2)=3/2mgr`
`impliesv_(B)=sqrt(3gr)" "...(xi)`
kinetic energy at `B=3/2mgr" "...(xii).`
At C, the string becomes slack and velocity of bob is horizontal and the left, Bob cntinus on its circular path and completes the revolution.