A bead of mass m can slide without friction along a vertical ring of radius R . One end of a spring of force constant k = 3 m g R is connected to the bead and the other end is fixed at the centre of t

Question

A bead of mass m can slide without friction along a vertical ring of radius R . One end of a spring of force constant k = 3 m g R is connected to the bead and the other end is fixed at the centre of the ring. Initially, the bead is at the point A and due to a small push it starts sliding down the ring. If the bead momentarily loses contact with the ring at the instant when the spring makes an angle of 60 ° with the vertical, then the natural length of the spring is

TestHub · Accepted Answer

Let's say the bead loses contact with the ring at point $B$ , then $N_{2} = 0$ and

$m g \cos 60 ° + k x = \frac{m v^{2}}{R}$

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

Applying the conservation of mechanical energy principle between $A$ & $B$

$\frac{m v^{2}}{2} = m g R (1 - \cos 60 °) = \frac{m g R}{2}$

$\Rightarrow \frac{m v^{2}}{R} = m g$

$\frac{m g}{2} + k x = m g$

$\Rightarrow x = \frac{m g}{2 k} = \frac{R}{6}$

The natural length of the spring $l = R - x = R - \frac{R}{6} = \frac{5 R}{6}$

Physics - W.E.P. Question with Solution | TestHub

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