Question:
If the Earth has no rotational motion, the weight of a person on the equator is \(W\). Determine the speed with which the earth would have to rotate about its axis so that the person at the equator will weigh \(\frac{3}{4}W\). Radius of the Earth is \(6400\ \mathrm{km}\) and \(g = 10\ \mathrm{m/s^2}\).
If the Earth has no rotational motion, the weight of a person on the equator is \(W\). Determine the speed with which the earth would have to rotate about its axis so that the person at the equator will weigh \(\frac{3}{4}W\). Radius of the Earth is \(6400\ \mathrm{km}\) and \(g = 10\ \mathrm{m/s^2}\).
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At the equator the apparent weight is \(m(g-\omega^2R)\); set it equal to \(\frac34 mg\).
Updated On: Nov 11, 2026
- \(1.1\times10^{-3}\ \mathrm{rad/s}\)
- \(0.83\times10^{-3}\ \mathrm{rad/s}\)
- \(0.63\times10^{-3}\ \mathrm{rad/s}\)
- \(0.28\times10^{-3}\ \mathrm{rad/s}\)
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The Correct Option is C
Solution and Explanation
Apparent weight at the equator: \(W' = m(g-\omega^2 R)\), and \(W = mg\).
\(W'=\frac34 W \Rightarrow g-\omega^2R=\frac34 g \Rightarrow \omega^2R=\frac g4\).
\(\omega^2=\frac{g}{4R}=\frac{10}{4\times6.4\times10^{6}}=3.9\times10^{-7}\ \mathrm{s^{-2}}\).
\(\omega=6.25\times10^{-4}\ \mathrm{rad/s}=0.63\times10^{-3}\ \mathrm{rad/s}\).
Answer: (3) \(0.63\times10^{-3}\ \mathrm{rad/s}\)
\(W'=\frac34 W \Rightarrow g-\omega^2R=\frac34 g \Rightarrow \omega^2R=\frac g4\).
\(\omega^2=\frac{g}{4R}=\frac{10}{4\times6.4\times10^{6}}=3.9\times10^{-7}\ \mathrm{s^{-2}}\).
\(\omega=6.25\times10^{-4}\ \mathrm{rad/s}=0.63\times10^{-3}\ \mathrm{rad/s}\).
Answer: (3) \(0.63\times10^{-3}\ \mathrm{rad/s}\)
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