A person stands on a spring balance at the equator. a.By wht fraction is the balance reading less than his true weight? b.If the speed of earth's rotation is increased by such an amount that the balance reading is half the true weight, what will be the lenght o the day in this case?

Explanation

To solve the problem step by step, we will break it down into two parts as given in the question.

### Part A: Fractional Change in Balance Reading

1. **Understanding True Weight and Balance Reading**:
- The true weight \( W \) of a person is given by the gravitational force acting on them, which is \( W = mg \), where \( m \) is the mass of the person and \( g \) is the acceleration due to gravity.
- The reading on the spring balance (normal force \( N \)) at the equator is affected by the centrifugal force due to the Earth's rotation.

2. **Centrifugal Force**:
- The centrifugal force \( F_c \) acting on the person can be expressed as:
\[ F_c = m \cdot R \cdot \omega^2 \]
where \( R \) is the radius of the Earth and \( \omega \) is the angular velocity of the Earth.

3. **Net Force on the Balance**:
- The normal force (reading on the spring balance) is given by:
\[ N = mg - F_c = mg - mR\omega^2 \]
- Thus, we can rewrite it as:
\[ N = mg - mR\omega^2 \]

4. **Fractional Change Calculation**:
- The fractional change in the balance reading compared to the true weight is:
\[ \text{Fractional Change} = \frac{W - N}{W} = \frac{mg - (mg - mR\omega^2)}{mg} = \frac{mR\omega^2}{mg} \]
- Simplifying gives:
\[ \text{Fractional Change} = \frac{R\omega^2}{g} \]

5. **Substituting Values**:
- The angular velocity \( \omega \) can be calculated as:
\[ \omega = \frac{2\pi}{T} \]
where \( T = 24 \times 3600 \, \text{s} \) (the time period of the Earth).
- Therefore, substituting \( \omega \):
\[ \text{Fractional Change} = \frac{R \left( \frac{2\pi}{T} \right)^2}{g} \]
- Substituting \( R = 6400 \times 10^3 \, \text{m} \) and \( g = 9.81 \, \text{m/s}^2 \):
\[ \text{Fractional Change} = \frac{6400 \times 10^3 \cdot \left( \frac{2\pi}{24 \times 3600} \right)^2}{9.81} \]

6. **Final Calculation**:
- After calculating, we find that the fractional change is approximately:
\[ \text{Fractional Change} \approx 3.5 \times 10^{-3} \]

### Part B: Length of the Day for Half Weight Reading

1. **Setting Up the Equation**:
- If the reading on the balance is half the true weight, we have:
\[ N = \frac{mg}{2} \]
- Using the expression for \( N \):
\[ mg - mR\omega^2 = \frac{mg}{2} \]
- Rearranging gives:
\[ mR\omega^2 = \frac{mg}{2} \]

2. **Solving for Angular Velocity**:
- Dividing by \( m \) and simplifying:
\[ R\omega^2 = \frac{g}{2} \]

3. **Substituting for \( \omega \)**:
- Using \( \omega = \frac{2\pi}{T} \):
\[ R\left(\frac{2\pi}{T}\right)^2 = \frac{g}{2} \]
- Rearranging gives:
\[ T^2 = \frac{8\pi^2 R}{g} \]

4. **Calculating the New Time Period**:
- Substituting \( R = 6400 \times 10^3 \) and \( g = 9.81 \):
\[ T = \sqrt{\frac{8\pi^2 (6400 \times 10^3)}{9.81}} \]

5. **Final Calculation**:
- After calculating \( T \), we find:
\[ T \approx 7175.768 \, \text{s} \]
- Converting to hours:
\[ \text{Length of the Day} = \frac{7175.768}{3600} \approx 1.993 \, \text{hours} \approx 2 \, \text{hours} \]