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A bob of mass m attached to one end of a...

A bob of mass `m` attached to one end of a light rod hangs vertically. The rod is tuned throudh `90^(@)` so that it becomed horizontal and then relseased . Calculate the angle between the rod and vertical positior at which the tension in the rod is equal to weight of the bob.

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As the rod moves from horizontal position `OA` to vertical position `OB` suppose `T=mg` at position `P`, where `/_BOP=theta` figure
If `upsilon` velocity of bob at `P` then `(1)/(2)m upsilon^(2)=mg(Lcostheta)`
`upsilon^(2)=2gLcostheta` …(i)
Net force acting on the bob along `PO=T-mg cos theta `
This must provide the necessary centripetal force
`T-mg cos theta=(m upsilon^(2))/(L)=(m)/(L)(2g L costheta)` ... using (i)
`=2mg cos theta`
`T=3mg cos theta`
When `T=mg, mg=3mgcostheta`
`costheta=(1)/(3)=0.3333:. theta=cos^(-1)(0.3333)=70^(@)30'`
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