A particle of mass `5kg` is free to slide on a smooth ring of radius `r=20cm` fixed in a vertical plane. The particle is attached to one end of a spring whose other end is fixed to the top point O of the ring. Initially, the particle is at rest at a point A of the ring such that `/_OCA=60^@`, C being the centre of the ring. The natural length of the spring is also equal to `r=20cm`. After the particle is released and slides down the ring, the contact force between the particle and the ring becomes zero when it reaches the lowest position B. Determine the force constant `(i n xx 10^2Nm^-1)` of the spring.

Explanation

At position B, F.B.D. or ring
(since normal contact force is zero) ltbr Equation of motion `kx-mg=(mv_B^(2))/(r)`
`x=OB-` natural length of spring `Z=2r-r`
`x=r`
`Kr-mg=(mv_B^(2))/(r)` .(i)
From conservation of energy.
Total mechanical energy at `A=` total mechanical energy at B
gravitation `PE_A+` spring energy `PE_A^(`)+` Kinetic energy `KE_A`
`=` Gravitation `PE_B+` Spring energy `PE_B^(`)+` Kinetic energy `KE_B`
`mg(BC+CP)+(1)/(2)K(0)^(2)+0`
`=0+(1)/(2)kx^(2)+(1)/(2)mv_B^(2)`
`mg(r+rcos60^@)=(1)/(2)Kr^(2)+(1)/(2)mv_B^(2)`
`implies(3mg)/(2)=(1)/(2)kr^(2)+(1)/(2)mv_B^(2)`
`implies3mg=kr^(2)+mv_B^(2)`
from (i) and (ii)
`Kr^(2)-mgr=3mg-kr^(2)`
`impliesKr^(2)+4mgrimpliesK=(2mg)/(r)`
`impliesK=(2(5)/(10))/((0.2))impliesK=500(N)/(m)`