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.

A mass m is released from the top of a vertical circular track of radi

1.	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.
Answer» Solution :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 posiotion `theta` `(mv^(2))/(r )= mg cos theta - N` or `N = mg cos theta - (mv^(2))/(r )` The body will leave 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-cos theta)` where .v. is the velocity at .P. [as `y=r (1-cos theta)]` or `v^(2)=v_(0)^(2)+2gr (1-cos theta)` ......(3) substituting the value of `v^(2)` in Equation (2) `cos theta = (v_(0)^(2))/(rg)+2(1-cos theta)` i.e., `cos theta = [(v_(0)^(2))/(3rg)+(2)/(3)] or theta = cos^(-1) [(v_(0)^(2))/(3rg)+(2)/(3)]`