If the angular velocity of Earth's spin is increased such that the bodies at the equator start float

Q: Why do we use the centrifugal force here?

Using the centrifugal force allows us to analyze the problem from the rotating frame of reference of the Earth, turning a complex dynamics problem into a simple force-balancing statics problem.

Q: Why does the normal force become zero when floating?

Floating means the object is just about to lose physical contact with the ground. Since the normal force is a contact force, it drops to exactly zero at the moment contact is lost.

Q: Does this floating effect happen at the poles too?

No! At the poles, the radius of rotation is zero, so the centrifugal force ($m\omega^2 R$) is zero. Your weight at the poles is completely unaffected by the Earth's rotation.

Q: Why did we convert the radius to meters?

To maintain dimensional consistency. Since gravity $g$ is given in $\text{m/s}^2$, the radius must be in meters to calculate $\omega$ in standard units ($\text{rad/s}$).

Q: What happens if the Earth spins even faster than this?

If $\omega$ exceeds $\sqrt{g/R}$, the outward centrifugal force would be greater than the inward gravitational pull. Objects at the equator would literally fly off the surface into space!

Question

If the angular velocity of Earth's spin is increased such that the bodies at the equator start floating, the duration of the day would be approximately
(Take $g = 10 	ext{ ms}^{-2}$, the radius of Earth, $R = 6400 	imes 10^3 	ext{ m}$, take $\pi = 3.14$)

Options:

(A) 60 min

(B) does not change

(C) 1200 min

(D) 84 min

Sigma Percentile · Accepted Answer

### The Setup: Earth on Overdrive

Imagine standing at the equator. You feel perfectly grounded, right? That's because gravity is pulling you down much harder than the Earth's rotation is trying to fling you off. But what if the Earth started spinning faster and faster?

Let's analyze the forces acting on you from a rotating frame of reference. Gravity pulls you towards the center with a force $mg$. The ground pushes back with a normal force $N$. And because the Earth is spinning, there's an outward centrifugal force, $m\omega^2 R$.

Balancing these forces, we get the equation:
$$mg = N + m\omega^2 R$$

### The Physics of Floating

Now, the question presents a fascinating scenario: the body starts floating. What does floating mean physically? It means you are just about to lose contact with the ground. At that exact moment, the normal force $N$ becomes exactly zero!

With $N = 0$, our equation simplifies beautifully. The gravitational force is now entirely used up just to provide the required centripetal force to keep you moving in a circle.
$$mg = m\omega^2 R$$

The mass $m$ cancels out, which means this effect happens to everything simultaneously, regardless of how heavy it is. Solving for the angular velocity $\omega$, we get:
$$\omega = \sqrt{\frac{g}{R}}$$

### Crunching the Numbers

Let's plug in the given values. Acceleration due to gravity $g$ is $10 	ext{ m/s}^2$. The radius of the Earth $R$ is $6400 	ext{ km}$, which we must convert to standard units: $6.4 	imes 10^6 	ext{ m}$.

Substituting these into our expression:
$$\omega = \sqrt{\frac{10}{6.4 	imes 10^6}}$$

To make the math easier, we can write $10 / 6.4$ as $100 / 64$.
$$\omega = \sqrt{\frac{100}{64 	imes 10^6}} = \frac{10}{8 	imes 10^3}$$

Simplifying this, we get an angular velocity of:
$$\omega = \frac{1}{800} 	ext{ rad/s}$$

### The 84-Minute Day

We need the duration of the day, which is simply the time period $T$ of Earth's rotation. The formula is:
$$T = \frac{2\pi}{\omega}$$

Substituting our value of $\omega$ and taking $\pi = 3.14$:
$$T = 2 	imes 3.14 	imes 800$$

Multiplying these numbers gives us $5024 	ext{ seconds}$. To convert this into minutes, we divide by $60$.
$$T = \frac{5024}{60} \approx 83.73 	ext{ min}$$

This is closest to **84 minutes**. So, if the Earth spun fast enough to make us float, a day would only last 84 minutes! Think about that—you'd have a sunrise and sunset every hour and a half. And if it spun even a tiny bit faster? We'd all fly off into space!

Animated Solution for Physics - Gravitation: If the angular velocity of Earth's spin is increased such that the bodies at the equator start floating, the duration of the day would be approximately (Take $g = 10 ms^{- 2}$ , the radius of Earth, $R = 6400 \times 1 0^{3} m$ , take $π = 3.14$ )

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Visualized Solution

$Visualizing the Setup$

$Forces at the Equator$

$Condition for Floating$

$Critical Angular Velocity$

$Substituting Values$

$Calculating ω$

$Time Period Formula$

$Final Calculation$

$The Way Forward$

The Sigma Insight: Acceleration due to Gravity and its Variation

The Setup

The Physics of Floating

Crunching the Numbers

The 84-Minute Day

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visualizing the Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Forces at the Equator

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Condition for Floating

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Critical Angular Velocity

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting Values

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating ω

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Time Period Formula

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Calculation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Way Forward