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## Homework Statement

A point object of mass m is connected to an inertialess string of length L, the other end of which is connected to a fixed point O. At time t=0, the object is assumed to move horizontally in a vertical plane from the bottom point A (OA=L). in the clockwise direction with an initial speed v

_{0}as seen in the figure.

http://img208.imageshack.us/img208/5223/figure.jpg [Broken]

If [tex]\sqrt{2gL}[/tex]<v

_{0}<[tex]\sqrt{5gL}[/tex], then at a point B (the angle between OB and the horizontal direction is designated [tex]\theta[/tex]) the magnitude of the force acting on the object from the string becomes zero, where OB=L and the velocity is perpendicular to OB. v being the magnitude of the vector velocity. If 0<[tex]\theta[/tex]<[tex]\pi[/tex]/2.

1) the speed v is given by....?

2) the initial speed is ....?

3) From the point B, for a while, the object takes a parabolic orbit till C where OC=L . The maximum elevation (with respect to location B) is expressed as ....?

4) In the case [tex]\theta[/tex]=[tex]\pi[/tex]/3, the angle [tex]\phi[/tex] measured as in fig. specifying the point C becomes. .....?

5) and finally the angle , the angle between the object velocity at the point C and the horizontal direction is ...?

## Homework Equations

for no 1)

mv

^{2}/L-mgsin(90+[tex]\theta[/tex]) = 0

for no 2)

decrease in Kinetic Energy = Gain in Potential Energy

for no 3)

H = v

^{2}sin

^{2}(90-[tex]\theta[/tex])/2g

I am not able to do for no 4 and 5.

## The Attempt at a Solution

I've found out the answers for question 1,2 and 3 which are

[tex]\sqrt{gLsin\theta}[/tex], [tex]\sqrt{(2+3sin\theta)gL}[/tex] and v

^{2}cos

^{2}[tex]\theta[/tex]/2g respectively.

I don't have any ideas to solve the last two ones.

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