What should be the angular velocity of the Ea | vedclass

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(B) The effective acceleration due to gravity at the equator is given by $g' = g - \omega^2 R$,where $g$ is the acceleration due to gravity at the pol

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$7.91 	imes 10^{-4} \ rad/s$

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(B) The effective acceleration due to gravity at the equator is given by $g' = g - \omega^2 R$,where $g$ is the acceleration due to gravity at the poles (or without rotation),$\omega$ is the angular velocity,and $R$ is the radius of the Earth.Given that the weight at the equator becomes $\frac{3}{5}$ of its initial value,we have $g' = \frac{3}{5}g$.Substituting this into the equation: $\frac{3}{5}g = g - \omega^2 R$.Rearranging the terms: $\omega^2 R = g - \frac{3}{5}g = \frac{2}{5}g$.Thus,$\omega = \sqrt{\frac{2g}{5R}}$.Given $g = 10 \ m/s^2$ and $R = 6400 \ km = 6.4 	imes 10^6 \ m$.$\omega = \sqrt{\frac{2 	imes 10}{5 	imes 6.4 	imes 10^6}} = \sqrt{\frac{20}{32 	imes 10^6}} = \sqrt{\frac{1}{1.6 	imes 10^6}} = \sqrt{0.625 	imes 10^{-6}} \approx 0.791 	imes 10^{-3} \ rad/s = 7.91 	imes 10^{-4} \ rad/s$.

What should be the angular velocity of the Earth due to rotation about its own axis so that the weight at the equator becomes $\left(\frac{3}{5}\right)$ of its initial value? (Radius of Earth at the equator $R = 6400 \ km$,$g = 10 \ m/s^2$,$\cos 0^{\circ} = 1$)

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