A smooth semicircular wire-track of radius R is fixed in a vertical plane. One end of a massless spring of natural length `3R//4` is attached to the lowest point O of the wire-track. A small ring of mass m, which can slide on the track, is attached to the other end of the spring. The ring is held staionary at point P such that the spring makes an angle of `60^@` with the vertical. The spring constant `K=mg//R`. Consider the instant when the ring is released, and (i) draw the free body diagram of the ring, (ii) determine the tangential acceleration of the ring and the normal reaction.

Explanation

1. The free-body diagram of the ring is shown in fig. the force acting on the ring are:

a. The weight mg acting vertically downwards
(b) Normal force N by the wire track.
Normal force on the ring could be either radially outwards or radially inwards depending on whether the ring presses against the inner surface or outer surface of the track. To ascertaion whether normal force is inwards or outwards, assume thatm to begin with, it is inwards. Then from `Sigma vec(F)=mvec(a)`, find the value of normal force. If it is positive, it is inwards and if it id negative, it is outwards.
(c ) Force of the spring kx. In the given physical situation, the spring is extended, it will pull the ring. So the spring force kx is along the spring towards O.
2. Lenght of spring in the position shown=R.(CP=CO=R, `/_COP= /_OPC=60^(@), /_COP` is equilateral)
Change in length of the spring `=R-(3/4)R=(R/4)`
`kx=((mg)/(R ))(R/4)=(mg)/(4)`
Now from `F_(t)=ma_(t)`,
`((mg)/(4))cos 30^(@)+mg cos 30^(@)=ma_(t)implies a_(t)=(5sqrt(3))/(8)g`
Now consider in radial direction
`N+kx cos 60^(@)=mg cos 60^(@)`
`N=mg cos 60^(@)-kx cos 60^(@)`
`=mg(1/2)-((mg)/(4))*(1/2)=3/8 mg`.