IIT JEE , AIEEE Forum - Physics , Mechanics

lokeshsardana

First part is easy, I think have done it.......

2nd part: bob will move in a circle of radius L/2

let the string becomes slag when it makes an angle theta with vertical.
now, velocity at bottom of this circular motion,v_b={gLcos(theta)}/2..........(1)

now ,when string is slag its tension is zero....

so, mgcos(theta)= (2mv²)/L.............(2)

also by conservation of energy

1/2mv² + mgL(1+ cos(theta)) = 1/2mv_b²........(3)

solve (1) (2)and (3) to get value of cos(theta) which will come out to be 2/3

so, height reached above the bottom most point is
L/2(1+cos(theta)) = 5L/6

3rd part is very similiar to this.............

if you have understood this concept,you can easily solve it......

IS IT CLEAR NOW?

CHEERS!!!

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