Question:
The speed with which the earth have to rotate on its axis so that a person on the equator would weigh \((3/5)^{\text{th}\) as much as present. (Radius of earth = 6400 km)}
The speed with which the earth have to rotate on its axis so that a person on the equator would weigh \((3/5)^{\text{th}\) as much as present. (Radius of earth = 6400 km)}
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At equator, centripetal acceleration reduces effective weight. At poles, weight = mg.
Updated On: Apr 23, 2026
- \(4.83 \times 10^{-3}\) rad s\(^{-1}\)
- \(5.41 \times 10^{-3}\) rad s\(^{-1}\)
- \(7.82 \times 10^{-4}\) rad s\(^{-1}\)
- \(8.88 \times 10^{-14}\) rad s\(^{-1}\)
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Concept:
Apparent weight at equator: \(W' = mg - mR\omega^2\). Given \(W' = \frac{3}{5}mg\).
Step 2: Detailed Explanation:
\(mg - mR\omega^2 = \frac{3}{5}mg \Rightarrow mR\omega^2 = mg - \frac{3}{5}mg = \frac{2}{5}mg\).
\(\omega^2 = \frac{2g}{5R} = \frac{2 \times 9.8}{5 \times 6400 \times 10^3} = \frac{19.6}{32 \times 10^6} = 6.125 \times 10^{-7}\).
\(\omega = \sqrt{6.125 \times 10^{-7}} = 7.82 \times 10^{-4}\) rad s\(^{-1}\).
Step 3: Final Answer:
Thus, angular speed = \(7.82 \times 10^{-4}\) rad s\(^{-1}\).
Apparent weight at equator: \(W' = mg - mR\omega^2\). Given \(W' = \frac{3}{5}mg\).
Step 2: Detailed Explanation:
\(mg - mR\omega^2 = \frac{3}{5}mg \Rightarrow mR\omega^2 = mg - \frac{3}{5}mg = \frac{2}{5}mg\).
\(\omega^2 = \frac{2g}{5R} = \frac{2 \times 9.8}{5 \times 6400 \times 10^3} = \frac{19.6}{32 \times 10^6} = 6.125 \times 10^{-7}\).
\(\omega = \sqrt{6.125 \times 10^{-7}} = 7.82 \times 10^{-4}\) rad s\(^{-1}\).
Step 3: Final Answer:
Thus, angular speed = \(7.82 \times 10^{-4}\) rad s\(^{-1}\).
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