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

C

Explanation

In `DeltaOCP`, `OC=CP=R`
`:.` `/_COP=/_CPO=60^@implies/_OCP=60^@`
`:.` `DeltaOCP` is an equilateral traingle `impliesOP=R`
`:.` Extension of string `=R-(3R)/(4)=(R)/(4)=x`
The forces acting are shown in the figure (i)
The free body diagram of the ring is shown in fig. (ii)
Force in the tangential direction
`=F cos 30^@ + mg cos 30`
`=[kx+mg]cos 30^@`
`F_t=(5mg)/(8)sqrt3` `:.` `F_t=ma_t implies a_t=(5sqrt3)/(8) g`
Also, when the ring is just reached
`N+F sin 30^@=mg sin 30^@`
`impliesN=(mg-F)sin 30^@=(mg-(mg)/(4))xx1/2=(3mg)/(8)`