If the earth has no rotational motion, the weight of a person on the equation is W. Detrmine the speed with which the earth would have to rotate about its axis so that the person at the equator will weight `(3)/(4)` W. Radius of the earth is 6400 km and g = 10 `m//s^(2)`.

Options

A. `0.63xx10^(-3)` rad/s
B. `0.28 xx 10^(-3)` rad/s
C. `1.1xx10^(-3)` rad/s
D. `0.83xx10^(-3)` rad/s

Answer

To solve the problem step by step, we will analyze the forces acting on a person at the equator of a non-rotating Earth and then determine the speed required for the Earth to rotate such that the person's weight is reduced to \( \frac{3}{4} W \).

### Step 1: Understand the Weight on a Non-Rotating Earth
When the Earth is not rotating, the weight \( W \) of a person is given by:
\[ W = mg \]
where \( m \) is the mass of the person and \( g \) is the acceleration due to gravity.

### Step 2: Introduce the Effect of Earth's Rotation
When the Earth starts rotating, a centrifugal force acts on the person at the equator. The effective weight \( W' \) of the person becomes:
\[ W' = W - F_{\text{centrifugal}} \]
where \( F_{\text{centrifugal}} = m \omega^2 r \) and \( r \) is the radius of the Earth.

### Step 3: Set Up the Equation
According to the problem, the new weight \( W' \) is given as:
\[ W' = \frac{3}{4} W \]
Substituting the expressions for \( W \) and \( W' \):
\[ \frac{3}{4} mg = mg - m \omega^2 r \]

### Step 4: Simplify the Equation
Cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \frac{3}{4} g = g - \omega^2 r \]
Rearranging gives:
\[ \omega^2 r = g - \frac{3}{4} g = \frac{1}{4} g \]

### Step 5: Solve for \( \omega^2 \)
Now, we can express \( \omega^2 \):
\[ \omega^2 = \frac{g}{4r} \]

### Step 6: Substitute Known Values
Given \( g = 10 \, \text{m/s}^2 \) and the radius of the Earth \( r = 6400 \, \text{km} = 6400 \times 10^3 \, \text{m} \):
\[ \omega^2 = \frac{10}{4 \times 6400 \times 10^3} \]
Calculating \( 4 \times 6400 \):
\[ 4 \times 6400 = 25600 \]
Thus,
\[ \omega^2 = \frac{10}{25600 \times 10^3} \]
\[ \omega^2 = \frac{10}{25600000} \]
\[ \omega^2 = \frac{1}{2560000} \]

### Step 7: Find \( \omega \)
Taking the square root:
\[ \omega = \frac{1}{\sqrt{2560000}} = \frac{1}{1600} \, \text{rad/s} \]

### Final Answer
The speed with which the Earth would have to rotate about its axis so that the person at the equator will weigh \( \frac{3}{4} W \) is:
\[ \omega = \frac{1}{1600} \, \text{rad/s} \]