Question

If the Earth has no rotational motion, the weight of a person on the equator is $W$. Determine the speed with which the earth would have to rotate about its axis so that the person at the equator will weigh $\frac{3}{4} W$ . Radius of the Earth is $6400\, km$ and $g = 10 \, m/s^2$.

• A. $1.1 \times 10^{-3} \, rad / s $
• B. $0.83 \times 10^{-3} \, rad / s $
• C. $0.63 \times 10^{-3} \, rad / s $
• D. $0.28 \times 10^{-3} \, rad / s $

Answer 1

Answer: (C)

$g' = g - \omega^{2}R \,cos^{2}\,\theta$
$\frac{3g}{4} = g - \omega^{2}R$
$w^{2}R = \frac{g}{4}$
$w = \sqrt{\frac{g}{4R}}$
$= \sqrt{\frac{10}{4\times64\omega\times10^{3}}}$
$= \frac{1}{2\times8\times100}$
$= \frac{1}{1600} = \frac{1}{16}\times10^{-2} = 0.6 \times 10^{-3}$