Question

The angular speed with which the earth would have to rotate on its axis so that a person on the equator would weight $\left(3/5\right)th$ as much as present will be : (Take the equatorial radius as $6400km$)

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

Answer 1

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

We know that,

True weight at the equator$W=mg$

Observed weight at the equator,

$W^{\prime }=m{g}^{\prime }$

$W^{\prime }=\left(\frac{3}{5}\right)mg$

Now, at the equator

$\varphi =0$

Now,

$m{g}^{\prime }=mg-mR{\omega }^2\cos \varphi$

$m{g}^{\prime }=mg-mR{\omega }^2$

$\left(\frac{3}{5}\right)mg=mg-mR{\omega }^2$

$-\frac{2}{5}g=-R{\omega }^2$

${\omega }^2=\frac{2g}{5R}$

${\omega }^2=\frac{2\times 9.8}{5\times 6400\times {10}^3}$

${\omega }^2=\frac{196}{320}\times {10}^{-6}$

$\omega =7.8\times {10}^{-4}rad/s$

Hence, the angular speed is $7.8\times {10}^{-4}rad/s$