Question:
What should be the velocity of rotation of earth due to rotation about its own axis so that the weight of a person becomes (3)/(5) of the present weight at the equator. Equatorial radius of the earth is 6400km.
What should be the velocity of rotation of earth due to rotation about its own axis so that the weight of a person becomes (3)/(5) of the present weight at the equator. Equatorial radius of the earth is 6400km.
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At the equator, apparent weight decreases due to centrifugal force.
Always use: W = mg - mω² R for rotating frames.
Updated On: Oct 17, 2026
- \(8.7 \times 10^{-7}\,\text{rad/s}\)
- \(7.8 \times 10^{-4}\,\text{rad/s}\)
- \(6.7 \times 10^{-4}\,\text{rad/s}\)
- 7.4 × 10⁻3rad/s
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The Correct Option is B
Solution and Explanation
Step 1: Let the angular velocity of earth be ω. The apparent weight at the equator is: W' = mg - mω² R
Step 2: Given that the weight becomes (3)/(5) of the original weight: mg - mω² R = (3)/(5)mg
Step 3: Simplifying: mω² R = mg - (3)/(5)mg = (2)/(5)mg ω² = (2g)/(5R)
Step 4: Substituting values g = 9.8m/s², R = 6.4 × 10⁶m: ω² = (2 × 9.8)/(5 × 6.4 × 10⁶) = 6.125 × 10⁻7 ω = √6.125 × 10⁻7 ≈ 7.8 × 10⁻4rad/s
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