Question

Four-point masses, each of $1\,kg$ are joined together by string, which form a square of diagonal $2\,m$. The square is placed on a rotating table, which is rotated at a speed of $5\,rps$. The tension in the string is:
(A) $1\,N$
(B) $1410\,N$
(C) $707\,N$
(D) $1000\,N$

Answer 1

The tension of the string is determined by using the balance equation of the string, for the balance equation, the radius of the revolution I required, the radius of the revolution is the half of the diagonal of the square and the angular frequency are used in the balancing equation, then the tension is determined.

Formula Used: The balance equation of the string is given by,
$2T\cos \theta = mr{\omega ^2}$
Where, $T$ is the tension of the string, $\theta$ is the angle of the diagonal, $m$ is the mass, $r$ is the radius of the revolution and $\omega$ is the angular frequency.

Complete step by step answer: Given that,
The mass of the string is, $m = 1\,kg$,
The diagonal distance is, $d = 2\,m$,
The speed of the rotation is, $5\,rps$.
Assume the length of the square is $a$, then the diagonal will be $\sqrt 2 a$.
By equating the diagonal, then
$\Rightarrow\sqrt 2 a = d$
By substituting the diagonal distance in the above equation, then
$\Rightarrow\sqrt 2 a = 2$
By rearranging the terms in the above equation, then
$\Rightarrow a = \dfrac{2}{{\sqrt 2 }}$
By dividing the terms in the above equation, then
$\Rightarrow a = 1.414\,m$
From the speed of the rotation, then the angular frequency is determined by,
$\Rightarrow\omega = 2\pi \times 5$
By multiplying the terms in the above equation, then
$\Rightarrow\omega = 10\pi$
The radius of the revolution is the half of the length of the square, then
$\Rightarrow r = \dfrac{a}{2}$
By substituting the length of the square in the above equation, then
$\Rightarrow r = \dfrac{{1.414}}{2}$
By dividing the terms in the above equation, then
$\Rightarrow r = 0.707\,m$
Now,
The balance equation of the string is given by,
$\Rightarrow 2T\cos \theta = mr{\omega ^2}$
By substituting the angle, mass radius and angular frequency in the above equation, then
$\Rightarrow 2T\cos {45^ \circ } = 1 \times 0.707 \times {\left( {10\pi } \right)^2}$
The angle of the diagonal in the square is ${45^ \circ }$.
The value of the $\cos {45^ \circ } = \dfrac{1}{{\sqrt 2 }}$, then
$\Rightarrow 2T \times \dfrac{1}{{\sqrt 2 }} = 1 \times 0.707 \times {\left( {10\pi } \right)^2}$
By rearranging the terms in the above equation, then
$\Rightarrow T = \dfrac{{1 \times 0.707 \times {{\left( {10\pi } \right)}^2} \times \sqrt 2 }}{2}$
By multiplying the terms in the above equation, then
$\Rightarrow T = \dfrac{{986.8}}{2}$
By dividing the terms in the above equation, then
$\therefore T = 493.4\,N$

Note: The tension of the string is directly proportional to the mass of the string, radius of the revolution and the square of the angular frequency of the rotating table. If anyone of the mass of the string or radius of the revolution or the square of the angular frequency of the rotating table is increased, the tension of the string also increases.