Question

The angular speed of earth, so that the object on equator may appear weight less, is $ (g=10)\,m/s^{2}$ , radius of earth $6400\,km$

• A. $ 1.25\times 10^{-3}rad/s$
• B. $ 1.56\times 10^{-3}rad/ \sec$
• C. $ 1.25\times 10^{-1}rad/s$
• D. $ 1.56\,rad/ \sec $

Answer 1

The Correct Option is A

Using the relation $W =m\left(g-R \omega^{2}\right)$ $0 =m\left(g-R \omega^{2}\right)$ So, $\omega=\sqrt{\frac{g}{R}}=\sqrt{\frac{10}{6.4 \times 10^{6}}}$ $=\frac{1}{800}$ or $\omega=1.25 \times 10^{-3} rad / \sec$