Question

What should be the velocity of rotation of earth due to rotation about its own axis so that the weight of a person becomes $\frac{3}{5}$ of the present weight at the equator. Equatorial radius of the earth is $6400\, km$.

• A. $8.7 \times 10^{-7} rad /s $
• B. $7.8 \times 10^{-4} rad /s $
• C. $6.7 \times 10^{-4} rad /s $
• D. $7.4 \times 10^{-3} rad /s $

Answer 1

Answer: (B)

True weight at equator, $W = mg$
Observed weight at equator,
$W ^{\prime}= mg '=\frac{3}{5} mg$
At equator, latitude $\lambda=0$
Using the formula,
$ mg ' = mg - mR \omega^{2} \cos ^{2} \lambda $
$=\frac{3}{5} mg = mg - mR ^{2} \omega^{2} \cos ^{2} 0= mg - mR \omega^{2} $
$\Rightarrow mR \omega^{2}= mg -\frac{3}{5} mg =\frac{2}{5} mg $
$\therefore \omega=\left(\frac{2}{5} \frac{ g }{ R }\right)^{1 / 2} $
$=\left(\frac{2 \times 9.8}{5 \times 6.4 \times 10^{6}}\right)^{1 / 2}$
$=7.8 \times 10^{-4} rad / s$.