Question

Find the angular velocity of the earth at which it should rotate so that weight of a body becomes zero at the equator.

• A. $1.25rad{s}^{-1}$
• B. $1.25\times {10}^{-1}rad{s}^{-1}$
• C. $1.25\times {10}^{-2}rad{s}^{-1}$
• D. $1.25\times {10}^{-3}rad{s}^{-1}$

Answer 1

The correct option is D $1.25\times {10}^{-3}rad{s}^{-1}$

At the equator, $mg-m{g}^{\prime }=m{\omega }^{2}R$
$\Rightarrow g^{\prime }=m(g-{\omega }^{2}R)$
Since the weight of the body is zero, $g^{\prime }=0$
$\Rightarrow g={\omega }^{2}R$

$\Rightarrow \omega =\sqrt{\frac{g}{R}}=\sqrt{\frac{9.8}{6400\times {10}^3}}=1.25\times {10}^{-3}rad{s}^{-1}$