A mass m is released from the top of a vertical circular track of radius r with a horizontal speed `v_(0)`. Calculate the angle `theta` with respect o the vertical where it leaves contact with the track.

Explanation

The forces acting on the body are its weight mg and reaction N as shown in Fig.
So for circular motion of the body at any position `theta`
`(mv^(2))/(r)=mgcostheta-NorN=mgcostheta-(mv^(2))/(r)`
The body will have contact where N = 0
i.e., mg `cos theta = (mv^(2))/(r) = 0`
i.e, `cos theta = (v^(2))/(rg) " " .............(2)`
Now applying conservation of mechanical energy between H and P, we get
`(1)/(2)mv^(2)=(1)/(2)mv_(0)^(2)+mgr(1-costheta)`
where .v. is the velocity at .P. `[as y = r(1-cos theta)]`
`or v^(2)=v_(0)^(2)+2gr(1-costheta).........(3)`
substituting the value of `v^(2)` in Equation (2)
`cos theta=(v_(0)^(2))/(rg)+2(1-costheta)`
`i.e., cos theta=[(v_(0)^(2))/(3rg)+(2)/(3)]ortheta=cos^(-1)[(v_(0)^(2))/(3rg)+(2)/(3)]`