A mass m is released from the top of a vertical circular track of radius r with a horizontal speed v_(theta). Calculate the angle theta with respect to 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_(theta). Calculate the angle theta with respect to 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 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 conservationof 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)]`