Question

A man has a weight $105\text{N}$ on a weighing machine kept near the north pole of the earth. What will be the reduction in the weight of the man if he moves to the equator? Take $g=9.8{\text{m/s}}^2$, radius of earth ($R)=6400\text{km}$, angular speed of the earth $\omega =7.27\times {10}^{-5}\text{rad/s}$.

• A. $3.62\text{N}$
• B. $0.36\text{N}$
• C. $10\text{N}$
• D. $0\text{N}$

Answer 1

The correct option is B $0.36\text{N}$

Since earth is rotating about a vertical axis passing through the poles, the effect of rotation at poles is negligible and centrifugal force $F_{\text{centrifugal}}=0$ at the poles.
However at the equator, $F_{\text{centrifugal}}$ is maximum.

Given, the Earth is rotating with $\omega =7.27\times {10}^{-5}\text{rad/s}$
$\therefore F_{\text{centrifugal}}=m{\omega }^{2}R$, in radially outward direction at equator.

At poles:


Weight read $=N=mg$
$\therefore m=\frac{N}{g}=\frac{105}{9.8}=10.71\text{kg}$

Similarly at equator:


$mg=N^{\prime }+m{\omega }^{2}R$
$\Rightarrow N^{\prime }=mg-m{\omega }^{2}R$

Here, $N^{\prime }$ is the weight read by the weighing machine at the equator.

$N^{\prime }=mg-m{\omega }^{2}R=105-(10.71\times (7.27\times {10}^{-5}{)}^2\times (6400\times 1000)$
$\therefore N^{\prime }=104.64\text{N}$

Hence, difference in weight will be :
$\Delta N=N-N^{\prime }=105-104.64=0.36\text{N}$