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

Given
Mass of bead is, $a$
Force constant of string is, $k = \frac{2 m g}{a}$
Length $A R = R O = a$

Case is shown in below figure,

From above figure, we can conclude that
Final length of wire is, $l_{f} = a + \frac{a}{\sin θ}$
Initial length of wire is, $B$
Increment length is, $O C = a$
$\Rightarrow ∆ l = a + \frac{a}{\sin θ} - a$

At position $B$ in equilibrium,
$m g = T \cos θ = k Δ l \cos θ$
$\Rightarrow m g = \frac{2 m g}{a} [a + \frac{a}{\sin θ} - a] \cos θ$
$\Rightarrow 1 = \frac{2}{a} [\frac{a}{\sin θ}] \cos θ$
$\Rightarrow \cot θ = \frac{1}{2}$

(a) $O B = a \cot θ = \frac{a}{2}$

(b) Since there is no external force and frictional force, therefore from conservation of mechanical energy, we have
$K_{C} + U_{C} = K_{O} + U_{O}$
$\Rightarrow 0 + m g a + \frac{1}{2} \times (\frac{2 m g}{a}) {(\sqrt{2} a)}^{2} = \frac{1}{2} m {v_{c}}^{2} + \frac{1}{2} k a^{2}$ , let potential energy at $O$ is zero.
$\Rightarrow 2 m g a = \frac{1}{2} m {v_{c}}^{2}$
$\Rightarrow v_{c} = 2 \sqrt{g a}$

Similarly, from conservation of mechanical energy at $C$ the point $P$ , where velocity of ring again comes to zero.
$K_{C} + U_{C} = K_{P} + U_{p}$
$\Rightarrow m g a + \frac{1}{2} (\frac{2 m g}{a}) {(\sqrt{2} a)}^{2} = - m g x + \frac{1}{2} (\frac{2 m g}{a}) (a^{2} + x^{2})$ , here $x$ is the distance below point $O$ , upto which ring can go.
$\Rightarrow 3 m g a = - m g x + m g a + \frac{m g x^{2}}{a}$
$\Rightarrow 2 m g a = - m g x + \frac{m g x^{2}}{a}$
$\Rightarrow 2 a = - x + \frac{x^{2}}{a}$
$\Rightarrow 2 a^{2} = - a x + x^{2}$
$\Rightarrow x^{2} - a x - 2 a^{2} = 0$
$\Rightarrow (x - 2 a) (x + a) = 0$
$\Rightarrow x = 2 a, - a$
Negative value is not possible, so $x = 2 a$

Embibe Experts Solutions for Chapter: Work, Energy and Power, Exercise 4: Exercise-4

Embibe Experts Physics Solutions for Exercise - Embibe Experts Solutions for Chapter: Work, Energy and Power, Exercise 4: Exercise-4

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