Introduction
This question deals with two different rotational motion concepts:
- (a) Rotational dynamics of the Earth to find the critical angular velocity at which objects at the equator would become weightless.
- (b) Coriolis force, which is an apparent force due to Earth’s rotation, affecting moving bodies in a rotating frame.
Part (a): Angular Velocity for Zero Apparent Weight at Equator
At the equator, a person would fly off the Earth if the normal reaction becomes zero, i.e., when centripetal force due to Earth’s rotation balances the person’s weight.
Let:
- Mass, m = 80 kg
- g = 9.8 m/s²
- Radius of Earth, R = 6.4 × 10⁶ m
- Required centripetal force = Weight = mg
- Centripetal force = m × ω² × R
Set them equal:
mω²R = mg → ω² = g / R → ω = √(g/R)
ω = √(9.8 / 6.4 × 10⁶) ≈ √(1.53125 × 10⁻⁶) ≈ 0.001237 rad/s
Answer (a):
Angular velocity required = 0.001237 rad/s
(This is much higher than Earth’s actual angular velocity: 7.27 × 10⁻⁵ rad/s)
Part (b): Coriolis Force on Moving Ball
Given:
- Mass of ball, m = 60 g = 0.06 kg
- Speed, v = 50 m/s (due south)
- Latitude = 30° N
- Earth’s angular velocity (ω) = 7.27 × 10⁻⁵ rad/s
Coriolis Force (Fc) = 2mωv sin(φ)
φ = latitude = 30° → sin(30°) = 0.5
Fc = 2 × 0.06 × 7.27 × 10⁻⁵ × 50 × 0.5
Fc ≈ 2 × 0.06 × 7.27 × 10⁻⁵ × 25 ≈ 0.0002181 N
Direction:
Since the ball is moving south at 30°N, the Coriolis force will act to the right of motion, i.e., towards the west.
Answer (b):
- Magnitude: ≈ 0.000218 N
- Direction: West
Conclusion
Part (a) shows the rotational speed at which Earth’s gravity would be neutralized at the equator. Part (b) demonstrates how Coriolis force affects moving objects on Earth due to rotation, with a very small magnitude but significant over large distances.