Question

A particle of mass $m = 5 kg$ is free to slide on a smooth ring of radius $r = 20 cm$ fixed in a vertical plane. The particle 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 $∠ O C A = 60 °,$ $C$ being the centre of the ring. The natural length of the spring is also equal to $r = 20 cm$ . After the particle is released and slides down the ring the contact force between the particle and the ring becomes zero when it reaches the lowest position $B$ (diametrically opposite to $O$ ). Determine the force constant of the spring.

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Accepted Answer

At position $A$ , the free body diagram of ring can be depicted as

Let spring constant be $k$ and velocity at lowest point be $v$ .

When particle is released from rest, it moves downward due to gravity, as result there will be loss in gravitational potential energy and so gain in the kinetic energy and spring will also get start to stretched.

From work energy theorem,

Work done by gravity $+$ work done by spring $=$ change in kinetic energy of the ring

Here, initially spring is in natural length and the 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 $= h = P C + C B = 20 \cos (60) + 20 = 30 cm$

Now, work done by gravity $= m g h = 5 \times 10 \times 0.30 = 15 J$ ,

Work done by spring $= \frac{1}{2} k {x_{i}}^{2} - \frac{1}{2} k {x_{f}}^{2}$ $= 0 - \frac{1}{2} k \times {(0.20)}^{2} = - 0.02 k$
Therefore, work done by gravity $+$ work done by spring $=$ $\frac{1}{2} m v^{2} - 0$

$\Rightarrow 15 - 0.02 k = 2.5 v^{2} \dots (i)$

At bottom position, forces acting on particle is normal reaction, spring force, its weight and centrifugal force so,

$\Rightarrow N + k x - m g = \frac{m v^{2}}{R}$

But as normal reaction at lowest position is zero, $∴ N = 0$

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

$\Rightarrow 0.008 k - 2 = v^{2} \dots (ii)$

Substituting $(ii)$ in $(i)$

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

$\Rightarrow 15 - 0.02 k = 0.02 k - 5$

$\Rightarrow k = 500 N m^{- 1}$

Therefore, the spring constant is $500 N m^{- 1}$ .

BM Sharma Solutions for Chapter: Work, Power, and Energy, Exercise 6: DPP

BM Sharma Physics Solutions for Exercise - BM Sharma Solutions for Chapter: Work, Power, and Energy, Exercise 6: DPP

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