Question

A small bead of mass $m$ is free to slide on a fixed smooth vertical wire, as indicated in the diagram. One end of a light elastic string, of unstretched length $a$ and force constant $\frac{2 m g}{a}$ is attached to $B$ . The string passes through a smooth fixed ring $R$ and the other end of the string is attached to the fixed point $A$ , $A R$ being horizontal. The point $O$ on the wire is at same horizontal level as $R$ , and $A R = R O = a$ .

(i) In the equilibrium position, find $O B$ .

(ii) The bead $B$ is raised to a point $C$ of the wire above $O$ , where $O C = a$ , and is released from rest. Find the speed of the bead as it passes $O$ , and find the greatest depth below $O$ of the bead in the subsequent motion.

Accepted Answer

At equilibrium, the length of the elastic string,

$L = A R + R B = a + \frac{a}{\sin (θ)} . . . (1)$

The extension of the string $x = L - a$ and the spring constant $k$ .

The tension in the string

$T = k x = \frac{2 m g}{a} \times (L - a) . . . (2)$

Substituting equation $(1)$ in $(2)$ ,

$T = \frac{2 m g}{a} \times (a + \frac{a}{\sin (θ)} - a) = \frac{2 m g}{\sin (θ)} . . . (3)$

Taking a force balance at equilibrium,

$T \cos (θ) = m g . . . (4)$

From equations $(3)$ and $(4)$ ,

$\frac{2 m g}{\sin (θ)} \times \cos (θ) = m g$

$\Rightarrow \tan (θ) = 2$

(a) From the triangle $O$ ,

$\tan (θ) = \frac{a}{y} = 2$

Hence, distance $O B$ , $y = \frac{a}{2}$

(b) When the bead is raised to point $C$ , the potential energy of the bead and the string with respect to unstretched string and position of mass at $O$ ,

$E = m g a + \frac{1}{2} k x^{2} = m g a + \frac{1}{2} \times \frac{2 m g}{a} \times {(a + \sqrt{2} a - a)}^{2}$

$E = m g a + \frac{1}{2} \times \frac{2 m g}{a} \times {(\sqrt{2} a)}^{2} = m g a + 2 m g a = 3 m g a$

When allowed falling, let it reach a distance $y$ .

The potential energy at that point,

$E_{1} = - m g y + \frac{1}{2} \times \frac{2 m g}{a} \times {(a + \sqrt{a^{2} + y^{2}} - a)}^{2}$

$= - m g y + \frac{m g}{a} \times (a^{2} + y^{2})$

Since kinetic energy is zero at both these positions,

$E = E_{1}$

$\Rightarrow 3 m g a = - m g y + \frac{m g}{a} \times (a^{2} + y^{2})$

$y^{2} - a y - 2 a^{2} = 0$

$\Rightarrow y = \frac{a \pm \sqrt{a^{2} - 4 \times 1 \times (- 2 a^{2})}}{2} = \frac{a \pm 3 a}{2}$

Taking only positive value as solution,

$y = 2 a$

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