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.

• A. \(8.7 \times 10^{-7}\,\text{rad/s}\)
• B. \(7.8 \times 10^{-4}\,\text{rad/s}\)
• C. \(6.7 \times 10^{-4}\,\text{rad/s}\)
• D. 7.4 × 10⁻3rad/s

Hint:

At the equator, apparent weight decreases due to centrifugal force. Always use: W = mg - mω² R for rotating frames.

Answer 1

The Correct Option is B

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