Question

A square frame is formed by four rods, each of length $l = 60 cm .$ Mass of two rods $A B$ and $B C$ is $m = \frac{25}{18} kg$ each while that of rods $A D$ and $C D$ is $2 kg$ each. The frame is free to rotate about a fixed horizontal axis passing through its geometric centre $O$ shown in figure. A spring is placed on the $A B$ at a distance $a = 15 cm$ from $B .$ The spring is held vertical and a block is placed on upper end of the spring so that rod $A B$ is horizontal.

(i) Calculate mass $M$ of the block,

(ii) If the spring is initially compressed by connecting a thread between its ends and energy stored in it is $76.5$ joule, calculate velocity with which block bounces up when the thread is burnt.

Accepted Answer

Given,
Length of each rod is, $l = 60 m$
Mass of $A B & B C$ is, $m' = \frac{25}{18} kg$
Mass of $A D & D C$ is, $m = 2 kg$
Spring is at $r = 30 - a = 30 - 15 = 15 cm$ from axis passing through $A B & O$ .
Mass on spring is, $M$

(i) As the body is in equilibrium, so to calculate mass $M$ , we have to balance torque about point $O .$

$m' g \times 30 + M g \times 15 = m g \times 30$
$\Rightarrow \frac{25}{18} \times 30 + M \times 15 = 2 \times 30$
$\Rightarrow M = \frac{11}{9} kg$
Torque due to $rod A B$ and $C D$ is zero, as its centre of mass passes through $O$

(ii) When thread is burnt then energy stored in spring is distributed in terms of velocity of block and to rotate the frame.
Since net torque on the system is zero, therefore by angular momentum conservation about $O$ , we have
Initial momentum = Final momentum
$\Rightarrow 0 = M v r - l ω$ , here $v & ω$ represents velocity of mass of block over spring and angular velocity of frame after spring is burnt.
$\Rightarrow l ω = M v r . . . (1)$

Since there is not any external force and frictional force, by mechanical energy conservation, we have
$\frac{1}{2} k x^{2} = \frac{1}{2} M v^{2} + \frac{1}{2} I ω^{2} \dots (2)$

Value of moment of inertia about point $O$ , using parallel axis theorem is,
$I = 2 [m' \times \frac{l^{2}}{12} + m' \times {(\frac{l}{2})}^{2}] + 2 [m \times \frac{l^{2}}{12} + m \times {(\frac{l}{2})}^{2}]$
$\Rightarrow I = 2 [\frac{25}{18} \times \frac{(0.6)^{2}}{12} + \frac{25}{18} \times (0.3)^{2}] + 2 [2 \times \frac{(0.6)^{2}}{12} + 2 \times (0.3)^{2}]$
$\Rightarrow I = \frac{61}{75} kg m^{2}$

So by equation $(1)$ , We have
$\Rightarrow \frac{61}{75} ω = \frac{11}{9} \times v \times 0.15$
$\Rightarrow \frac{61}{75} ω = \frac{11}{9} \times v \times \frac{15}{100}$
$\Rightarrow ω = \frac{5 \times 11}{61 \times 4} v$

Put this value in equation $(2)$
$\Rightarrow 76.5 = \frac{1}{2} \times \frac{61}{75} \times {(\frac{5 \times 11}{61 \times 4})}^{2} v^{2} + \frac{1}{2} \times \frac{11}{9} \times v^{2}$
$\Rightarrow 76.5 = \frac{1}{2} \times \frac{61}{75} \times \frac{25 \times 121}{61 \times 61 \times 16} v^{2} + \frac{1}{2} \times \frac{11}{9} \times v^{2}$
$\Rightarrow 76.5 = v^{2} [\frac{121}{96 \times 61} + \frac{11}{18}]$
$\Rightarrow 76.5 = v^{2} [\frac{2178 + 64416}{105408}]$
$\Rightarrow 76.5 = v^{2} \times \frac{66594}{105408}$
$\Rightarrow v^{2} = 121.08$
$\Rightarrow v \approx 11 m s^{- 1}$

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