What should be the angular speed with which t | vedclass

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(A) 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

Vedclass · Accepted Answer

$\sqrt{\frac{2 g}{5 R}}$

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(A) 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 speed,and $R$ is the radius of the Earth.The weight of a person at the equator is $W' = m g'$.Given that the new weight is $\frac{3}{5}$ of the present weight,we have $m g' = \frac{3}{5} m g$.Substituting the expression for $g'$:$m(g - \omega^2 R) = \frac{3}{5} m g$Divide both sides by $m$:$g - \omega^2 R = \frac{3}{5} g$Rearranging the terms to solve for $\omega^2 R$:$\omega^2 R = g - \frac{3}{5} g = \frac{2}{5} g$Solving for $\omega$:$\omega^2 = \frac{2 g}{5 R}$$\omega = \sqrt{\frac{2 g}{5 R}}$

What should be the angular speed with which the Earth must rotate on its axis so that a person on the equator would weigh $\frac{3}{5}$ th as much as their present weight?

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