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 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, it slides down the ring the contact force between the particle & the ring becomes zero when it reaches the lowest position B. If the force constant of the spring is n × 100 N/m, find k.

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 = 40 cm

extension in the spring = 40 - 20 = 20 cm = 0.20 m

vertically downwards displacement = h = 20cos60 + 20 = 30 cm

now we have

1.

work done by gravity = mgh = 5*10*0.30 = 15 J

2.

Work done by spring = $\frac{1}{2}kx_i^2 - \frac{1}{2}kx_f^2$

work done by spring = $- \frac{1}{2}k*(0.20)^2$

now we have

$W_g + W_{spring} = \frac{1}{2} mv^2$

$15 - 0.02k = 2.5v^2$

also we know at this bottom position normal force is zero

$kx - mg = \frac{mv^2}{R}$

$k*0.20 - 5*10 = \frac{5*v^2}{0.20}$

$0.008k - 2 = v^2$

now by above equation

$15 - 0.02k = 2.5(0.008k - 2)$

$15 - 0.02k = 0.02k - 5$

$k = 500 N/m$

so spring constant must be 500 N/m

Answer 2

Answer:

The correct answer is 500n/m