A person stands on a spring balance at the equator. a.By what 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 of 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): By what fraction is the balance reading less than his true weight?

1. **Understanding Forces at the Equator**:
- At the equator, a person experiences two forces: gravitational force (weight) acting downwards, \( mg \), and the centrifugal force due to Earth's rotation acting outward.
- The spring balance measures the normal force \( N \) exerted by the balance, which is less than the true weight due to the centrifugal force.

2. **Centrifugal Force**:
- The centrifugal force \( F_c \) can be expressed as:
\[ F_c = m \omega^2 R \]
where:
- \( m \) = mass of the person,
- \( \omega \) = angular velocity of the Earth,
- \( R \) = radius of the Earth (approximately \( 6.4 \times 10^6 \) m).

3. **Angular Velocity**:
- The angular velocity \( \omega \) can be calculated using:
\[ \omega = \frac{2\pi}{T} \]
where \( T \) is the time period for one complete rotation of the Earth (24 hours or 86400 seconds).

4. **Calculating the Normal Force**:
- The normal force \( N \) is given by:
\[ N = mg - F_c = mg - m\omega^2 R \]
- Hence, the reading on the spring balance is:
\[ N = mg(1 - \frac{\omega^2 R}{g}) \]

5. **Fraction of Weight Reduction**:
- The fraction by which the balance reading is less than the true weight is:
\[ \text{Fraction} = \frac{mg - N}{mg} = \frac{m\omega^2 R}{mg} = \frac{\omega^2 R}{g} \]

6. **Substituting Values**:
- Substitute \( \omega = \frac{2\pi}{86400} \) into the fraction:
\[ \text{Fraction} = \frac{\left(\frac{2\pi}{86400}\right)^2 R}{g} \]
- Using \( R \approx 6.4 \times 10^6 \) m and \( g \approx 9.8 \, \text{m/s}^2 \):
\[ \text{Fraction} = \frac{\left(\frac{2\pi}{86400}\right)^2 (6.4 \times 10^6)}{9.8} \]

7. **Calculating the Value**:
- After calculating, we find:
\[ \text{Fraction} \approx 3.5 \times 10^{-3} \]

### Part (b): Length of the day if the balance reading is half the true weight

1. **Setting up the Equation**:
- If the balance reading is half the true weight, we set:
\[ N = \frac{1}{2} mg \]
- From the previous equation for normal force, we have:
\[ mg(1 - \frac{\omega^2 R}{g}) = \frac{1}{2} mg \]

2. **Simplifying the Equation**:
- Cancel \( mg \) from both sides:
\[ 1 - \frac{\omega^2 R}{g} = \frac{1}{2} \]
- Rearranging gives:
\[ \frac{\omega^2 R}{g} = \frac{1}{2} \]

3. **Substituting for Angular Velocity**:
- Substitute \( \omega = \frac{2\pi}{T} \):
\[ \frac{\left(\frac{2\pi}{T}\right)^2 R}{g} = \frac{1}{2} \]

4. **Solving for T**:
- Rearranging gives:
\[ T^2 = \frac{4\pi^2 R}{g} \]
- Substitute \( R \) and \( g \):
\[ T^2 = \frac{4\pi^2 (6.4 \times 10^6)}{9.8} \]

5. **Calculating T**:
- After calculating, we find:
\[ T \approx 7200 \, \text{s} \quad \text{(which is 2 hours)} \]

### Final Answers:
- a) The fraction by which the balance reading is less than his true weight is approximately \( 3.5 \times 10^{-3} \).
- b) If the speed of Earth's rotation is increased such that the balance reading is half the true weight, the length of the day would be approximately 2 hours.