Question

The angular speed of rotation of earth about its axis at which the weight of man standing on the equator becomes half of its weight at the poles is given by:
A. $0.034\,{\text{rad}}\,{{\text{s}}^{ - 1}}$
B. $8.75 \times {10^{ - 4}}\,{\text{rad}}\,{{\text{s}}^{ - 1}}$
C. $1.23 \times {10^{ - 2}}\,{\text{rad}}\,{{\text{s}}^{ - 1}}$
D. $7.65 \times {10^{ - 7}}\,{\text{rad}}\,{{\text{s}}^{ - 1}}$

Answer 1

First of all, we will create a relation between the acceleration due to gravity on the equator and the pole. After that we will find an expression for the effective acceleration due to gravity. We will manipulate the expression accordingly and obtain the result.

Complete step by step solution:
In the given question, we are supplied with the following data:
The earth rotates about an imaginary line, which is known as the axis. The weight of a man standing on the equator becomes half the weight measured at the poles. We are asked to find the angular speed of the earth at which the following situation holds good.

To begin with, we know that weight is the force with which the earth pulls an object towards it. Weight is directly dependent on the acceleration due to gravity. Stronger is the gravity, more is the weight and vice-versa. It is given in the question that the weight of the man at poles is twice the weight measured at the equator.

So, we can write:
${g_{{\text{equator}}}} = \dfrac{{{g_{{\text{pole}}}}}}{2}$ …… (1)
Where,
${g_{{\text{equator}}}}$ indicates the acceleration due to gravity at the equator.
${g_{{\text{pole}}}}$ indicates the acceleration due to gravity at the poles.

We know the expression for the effective acceleration due to gravity is given as:
${g_{{\text{effective}}}} = {g_{{\text{pole}}}} - {\omega ^2}R$ …… (2)
Where,
${g_{{\text{effective}}}}$ indicates the final or effective acceleration due to gravity.
$\omega$ indicates the angular speed.
$R$ indicates the radius of the earth.
We know the approximate radius of the earth is $6400\,{\text{km}}$ .

Now, we will equate equations (1) and (2) and we get:
$\dfrac{{{g_{{\text{pole}}}}}}{2} = {g_{{\text{pole}}}} - {\omega ^2}R \\\ \Rightarrow {\omega ^2}R = {g_{{\text{pole}}}} - \dfrac{{{g_{{\text{pole}}}}}}{2} \\\ \Rightarrow {\omega ^2}R = \dfrac{{{g_{{\text{pole}}}}}}{2} \\\ \Rightarrow \omega = \sqrt {\dfrac{{{g_{{\text{pole}}}}}}{{2 \times R}}}$

We will substitute the required values in the above expression for the angular speed and we get:
$\omega = \sqrt {\dfrac{{{g_{{\text{pole}}}}}}{{2 \times R}}} \\\ \Rightarrow \omega = \sqrt {\dfrac{{10}}{{2 \times 6400 \times {{10}^3}}}} \\\ \therefore \omega = 8.75 \times {10^{ - 4}}\,{\text{rad}}\,{{\text{s}}^{ - 1}}$

Hence, the angular speed of the earth at which the following situation holds good is $8.75 \times {10^{ - 4}}\,{\text{rad}}\,{{\text{s}}^{ - 1}}$. The correct option is B.

Note: While solving this problem, students should remember that the weight of an object purely depends on the acceleration due to gravity as the mass of an object always remains constant. Higher the gravity more is the weight measure and vice-versa. Since, the moon has weaker gravity than the earth, hence we will weigh less in the moon.