With what angular velocity the earth should spin in order that a body lying at `37^@` latitude may become weightless.

Options

A. `(5)/(4) sqrt((g)/(R))`
B. `(25)/(16) sqrt((g)/(R))`
C. `(5)/(3)sqrt((g)/(R))`
D. `(25)/(g) sqrt((g)/(R)`

Explanation

To find the angular velocity at which a body lying at 37 degrees latitude becomes weightless, we can follow these steps:

### Step 1: Understand the Forces Acting on the Body
At latitude \( \theta \), a body experiences two forces: gravitational force \( Mg \) acting downwards and the centripetal force required for circular motion due to the Earth's rotation. The effective weight \( Mg' \) of the body can be expressed as:
\[ Mg' = Mg - M \omega^2 R \cos^2 \theta \]
where:
- \( M \) is the mass of the body,
- \( g \) is the acceleration due to gravity,
- \( \omega \) is the angular velocity of the Earth,
- \( R \) is the radius of the Earth,
- \( \theta \) is the latitude.

### Step 2: Set the Effective Weight to Zero for Weightlessness
For the body to be weightless, the effective weight \( Mg' \) must equal zero:
\[ Mg' = 0 \implies Mg - M \omega^2 R \cos^2 \theta = 0 \]
This simplifies to:
\[ Mg = M \omega^2 R \cos^2 \theta \]

### Step 3: Cancel Out the Mass
Since \( M \) is present on both sides of the equation, we can cancel it out (assuming \( M \neq 0 \)):
\[ g = \omega^2 R \cos^2 \theta \]

### Step 4: Solve for Angular Velocity \( \omega \)
Rearranging the equation gives:
\[ \omega^2 = \frac{g}{R \cos^2 \theta} \]
Taking the square root of both sides, we find:
\[ \omega = \sqrt{\frac{g}{R \cos^2 \theta}} \]

### Step 5: Substitute the Values
Now, we need to substitute the values for \( g \) and \( R \) and the angle \( \theta = 37^\circ \):
- \( g \approx 9.81 \, \text{m/s}^2 \)
- \( R \approx 6.37 \times 10^6 \, \text{m} \)
- \( \cos(37^\circ) \approx 0.7986 \)

Calculating \( \cos^2(37^\circ) \):
\[ \cos^2(37^\circ) \approx (0.7986)^2 \approx 0.6378 \]

Now substituting these values into the equation for \( \omega \):
\[ \omega = \sqrt{\frac{9.81}{6.37 \times 10^6 \times 0.6378}} \]

### Step 6: Perform the Calculation
Calculating the denominator:
\[ 6.37 \times 10^6 \times 0.6378 \approx 4.060 \times 10^6 \]

Now substituting back:
\[ \omega = \sqrt{\frac{9.81}{4.060 \times 10^6}} \approx \sqrt{2.42 \times 10^{-6}} \approx 0.00155 \, \text{rad/s} \]

### Final Answer
Thus, the angular velocity \( \omega \) at which a body lying at \( 37^\circ \) latitude becomes weightless is approximately:
\[ \omega \approx 0.00155 \, \text{rad/s} \]