A body weights 49N on a spring balance at the north pole. What will be its weight recorded on the same weighing machine, if it is shifted to the equator?
[Use g = GM/R² = 9.8 ms⁻² and radius of earth, R = 6400 km.]

Solution

Physics sir · 4 steps

At the pole the spin does nothing, but at the equator part of gravity keeps the body moving in a circle.

1. Find the mass. At the north pole the body sits on the axis of rotation, so the balance reads the full mg. That gives m = 49/9.8 = 5 kg.
2. Effect of rotation. At the equator the body moves in a circle of radius R. Part of mg provides the centripetal force mω² R, so the spring is pressed less.

   W_(eq) = mg - mω² R

3. Put in numbers. ω = 2π/(24 × 3600) = 7.27 × 10⁻⁵ rad s⁻¹, so ω² R = 5.29 × 10⁻⁹ × 6.4 × 10⁶ = 0.0338 m s⁻². Then mω² R = 5 × 0.0338 = 0.17 N.
4. So. W_(eq) = 49 - 0.17 = 48.83 N, which is option (c).

The idea to keep. A spring balance reads the normal force, not mg. At latitude λ, g' = g - ω² R cos² λ: no change at the poles, the largest drop at the equator.

Answer from the NTA final answer key. Solution by PrepWiser.