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

The correct option is B $7.8\times {10}^{-4}rad/s$

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