First slide

Variation of g

Question

What should be the angular velocity of earth about own axis so that a person's weight at equator will be $\frac{3}{5}$ of his weight at poles?

Easy

Solution

$\begin{array}{l} mg' = mg - {mω}^{2} {Rcos}^{2} λ \Rightarrow \frac{3}{5} mg = mg - {mω}^{2} R \\ \Rightarrow ω = \sqrt{\frac{2 g}{5 R}} \end{array}$

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