Question

The angular velocity of earth at present is $ \omega $ With what angular velocity should it rotate $ \omega $ that weight of a body at the equator appears $ \omega $ be zero?

• A. 289 $ \omega $
• B. 17 $ \omega $
• C. 8 $ \omega $
• D. 2 $ \omega $

Answer 1

Answer: (B)

The body will appear weightless only when the centripetal force equals the weight of the body.
$\therefore \frac{m v^{2}}{R}=m g$
Since, $v = R \omega$
$ \frac{m(R \omega)^{2}}{R} =m g $
$ \Rightarrow m R \omega^{2} = m g$
$\Rightarrow \omega =\sqrt{\frac{g}{R}}$
Standard value $(\omega)$ at present is
$\omega =\frac{2 \pi}{T}=\frac{2 \pi}{86400} rad s ^{-1} $
$=7.3 \times 10^{-5} rad\, s ^{-1}$
and $\omega'=\sqrt{\frac{g}{R}} $
$=\sqrt{\frac{9.8}{6.4 \times 10^{6}}}=1.2 \times 10^{-3}$
$\therefore \omega' \approx 17 \omega$.