Question

A particle of mass $5$kg is free to slide on a smooth ring of radius $r=20$cm 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 $\angle$OCA$={60}^o$, C being the centre of the ring. The natural length of the spring is also equal to $r=20$cm. After the particle is released, it slides down the ring, the contact force between the particle & the ring becomes zero when it reaches the lowest position B. Determine the force constant of the spring.

Answer 1

Since particle move in vertical plane then-
Work done by gravity + work done by spring = change in kinetic energy
here initially spring is in natural length and particle finally slides down to bottom position
So, final length of the spring $=20+20=40cm$
extension in the spring $=40-20=20cm=0.20m$
vertically downwards displacement $=h=20\cos 60+20=30cm$
now we have-
1.-
work done by gravity $=mgh=5\times 10\times 0.30=15J$
2.-
Work done by spring $=\frac{1}{2}k{x}_i^2-\frac{1}{2}k{x}_f^2$
work done by spring $=-\frac{1}{2}k\times (0.20{)}^2$
now we have
$W_g+W_{spring}=\frac{1}{2}m{v}^2$
$15-0.02k=2.5v^2$
also we know at this bottom position normal force is zero
$kx-mg=\frac{m{v}^2}{R}$
$k\times 0.20–5\times 10=\frac{5\times v^2}{0.20}$
$0.008k-2=v^2$
now by above equation-
$15-0.02k=2.5(0.008k-2)$
$15-0.02k=0.02k-5$
$k=500N/m$
So spring constant must be $500N/m$.