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.

Answer

(i) `(3)/(9)` mg N , (ii) `(5 sqrt3)/(8) m//s^(2)`

Explanation

In the figure `angleCPO = 60^(@) = angle OPC , therefore OP = R`
The extension of spring = R = `(3R)/(4) = (R)/(4)`
`therefore " " F = kx = (mg)/(R) xx (R)/(4) = (mg)/(4)`
(i) Resolving the forces along the radius of semicircle , we have
`N + F "cos" 60^(@) = "mg cos" 60^(@) `
or `" " N + (mg)/(4) "cos"60^(@) = "mg cos" 60^(@)`
`therefore " " N = (3)/(8) mg N`
(ii) Tangential force
`F_(t) = " mg cos" 60^(@) + F "sin" 60^(@)`
=`"mg cos" 60^(@) + (mg)/(4) "sin" 60^(@)`
= `( 5 sqrt3 "mg")/(8)`
Thus tangential acceleration `a_(t) = (F_(t))/(m) = (5 sqrt3 g)/(8) m//s^(2)`