- #1
shrrikesh
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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 v0 as seen in the figure.
http://img208.imageshack.us/img208/5223/figure.jpg
If [tex]\sqrt{2gL}[/tex]<v0<[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)
mv2/L-mgsin(90+[tex]\theta[/tex]) = 0
for no 2)
decrease in Kinetic Energy = Gain in Potential Energy
for no 3)
H = v2sin2(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 v2cos2[tex]\theta[/tex]/2g respectively.
I don't have any ideas to solve the last two ones.
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