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 $6400km$.

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