Question

The angular velocity of the earth with which it has to rotate so that acceleration due to gravity on $60^\circ $ latitude becomes zero is ( Radius of the earth $= \, 6400 \, km$ . At the poles $g=10 \, m \, s^{- 2}$ )

• A. $2.5\times 10^{- 3}\, rad\, s^{- 1}$
• B. $5.0\times 10^{- 1}\, rad\, s^{- 1}$
• C. $100\, rad\, s^{- 1}$
• D. $7.8\times 10^{- 2 }\, rad\, s^{- 1}$

Answer 1

Answer: (A)

$g^{\prime}=g-\omega^2 R \cos ^2 \lambda \,\,\,0=g-\omega^2 R \cos ^2 60^{\circ}$ $0=g-\frac{\omega^2 R }{4} \Rightarrow \omega=2 \sqrt{\frac{\bar{g}}{R}}=\frac{1}{400}\, rad\, s ^{-1}=2.5 \times 10^{-3} rad \,s^{-1}$