The angular velocity of the earth with which | vedclass

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(A) The effective acceleration due to gravity $g'$ at a latitude $\lambda$ is given by $g' = g - \omega^2 R \cos^2 \lambda$.For $g'$ to be zero at $\l

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$2.5 	imes 10^{-3} \, rad/s$

Vedclass · Answer

(A) The effective acceleration due to gravity $g'$ at a latitude $\lambda$ is given by $g' = g - \omega^2 R \cos^2 \lambda$.For $g'$ to be zero at $\lambda = 60^o$:$0 = g - \omega^2 R \cos^2(60^o)$$0 = g - \omega^2 R (1/2)^2$$0 = g - \frac{\omega^2 R}{4}$$\omega^2 = \frac{4g}{R}$$\omega = 2 \sqrt{\frac{g}{R}}$Given $g = 10 \, m/s^2$ and $R = 6400 \, km = 6.4 	imes 10^6 \, m$:$\omega = 2 \sqrt{\frac{10}{6.4 	imes 10^6}} = 2 \sqrt{\frac{10}{64 	imes 10^5}} = 2 \sqrt{\frac{1}{6.4 	imes 10^6}} = 2 	imes \frac{1}{800} 	imes \sqrt{1} = \frac{2}{800} = 2.5 	imes 10^{-3} \, rad/s$.

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

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