Question

The angular velocity of the earth about its polar axis so that the weight of the body at the equator will be zero is:
A) $1.25 \times {10^{ - 3}}\,rad\,{s^{ - 1}}$
B) $2\,rad\,{s^{ - 1}}$
C) $1\,rad\,{s^{ - 1}}$
D) $0$

Answer 1

The question is simply asking the angular velocity of earth $\left( \omega \right)$ when the weight of the body will be zero, i.e. the acceleration due to gravity is zero. So, apply the formula of variation of acceleration due to gravity of earth when the position of the body and angular velocity of the earth change.

Complete step by step solution:
We will approach the solution to the question exactly as explained in the hint section of the solution. The question is ultimately asking us to find out the angular velocity when the acceleration due to gravity at the equator is zero. To find this, we first need to know the formula of variance of the acceleration due to gravity as the position of the body and angular velocity of earth change.
The formula is given as:
$g' = g - R{\omega ^2}\cos \phi$
Where, $g'$ is the resultant acceleration due to gravity,
$g$ is the acceleration due to gravity at the surface of earth, numerically, it is taken as: $g = 9.8\,m{s^{ - 2}}$
$R$ is the radius of earth, which is taken as: $R = 6400\,km = 64 \times {10^5}\,m$
$\omega$ is the angular velocity of earth about its polar axis, which we need to find out,
$\phi$ is the angle that the body makes with the equator of the earth
In the question, the body is at equator itself, hence, we can see that:
$\phi = 0$
Which implies:
$\cos \phi = \cos {0^ \circ } = 1$
So, the equation becomes:
$g' = g - R{\omega ^2}$
It is also said that the weight of the body becomes zero, i.e. the resultant acceleration due to gravity becomes zero, or:
$g' = 0$
Putting this into the equation, we get:
$g - R{\omega ^2} = 0$
Upon transposing, we can easily write that:
$\omega = \sqrt {\dfrac{g}{R}}$
Substituting in the values of $g$ and $R$ , we get:
$\omega = \sqrt {\dfrac{{9.8}}{{64 \times {{10}^5}}}}$
Upon solving, we get:
$\omega = 1.25 \times {10^{ - 3}}\,rad\,{s^{ - 1}}$

Hence, option (A) is the correct answer of the question.

Note: The point was to figure out the fact that at the equator, the value of $\phi$ becomes $0$ . Many students get confused and take the value of $\phi$ as ${90^ \circ }$ which is wrong as this value should be taken only for poles.