Three point masses are at the corners of an equilateral traingle of side `r`. Their separations do not change when the system rotates about the centre of the triangle. For this, the time period of rotation must be proportional to

Options

A. `r^(3//2)`
B. `r`
C. `m`
D. `m^(-1//2)`

Answer

To solve the problem, we need to find the relationship between the time period of rotation of three point masses located at the corners of an equilateral triangle and the parameters involved, namely the side length \( r \) of the triangle and the mass \( m \) of the point masses.

### Step-by-Step Solution:

1. **Understand the System**:
We have three point masses \( m \) located at the corners of an equilateral triangle with side length \( r \). The system rotates about the centroid of the triangle.

2. **Calculate the Gravitational Force**:
The gravitational force between any two masses \( m \) is given by:
\[ F_0 = \frac{G m^2}{r^2} \]
where \( G \) is the gravitational constant.

3. **Determine the Resultant Force**:
Each mass experiences gravitational forces from the other two masses. By symmetry, the resultant force acting on any mass can be calculated. The angle between the forces acting on one mass due to the other two is \( 60^\circ \). The net force \( F_{\text{net}} \) can be calculated as:
\[ F_{\text{net}} = 2 F_0 \cos\left(\frac{60^\circ}{2}\right) = 2 F_0 \cos(30^\circ) \]
Substituting \( \cos(30^\circ) = \frac{\sqrt{3}}{2} \):
\[ F_{\text{net}} = 2 \cdot \frac{G m^2}{r^2} \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3} G m^2}{r^2} \]

4. **Relate to Centripetal Force**:
The net force acts as the centripetal force required for circular motion. If \( A \) is the radius of the circular path for the masses, we can equate the net force to the centripetal force:
\[ F_{\text{net}} = m \omega^2 A \]
where \( \omega \) is the angular velocity.

5. **Express Angular Velocity**:
The angular velocity \( \omega \) is related to the time period \( T \) by:
\[ \omega = \frac{2\pi}{T} \]
Substituting this into the centripetal force equation gives:
\[ F_{\text{net}} = m \left(\frac{2\pi}{T}\right)^2 A \]

6. **Substituting and Rearranging**:
Now substituting \( F_{\text{net}} \):
\[ \frac{\sqrt{3} G m^2}{r^2} = m \left(\frac{2\pi}{T}\right)^2 A \]
Canceling \( m \) from both sides:
\[ \frac{\sqrt{3} G m}{r^2} = \left(\frac{2\pi}{T}\right)^2 A \]

7. **Finding the Radius \( A \)**:
The distance \( A \) from the centroid to a vertex of the triangle can be calculated as:
\[ A = \frac{r}{\sqrt{3}} \]
Substituting this back into the equation:
\[ \frac{\sqrt{3} G m}{r^2} = \left(\frac{2\pi}{T}\right)^2 \cdot \frac{r}{\sqrt{3}} \]

8. **Solving for Time Period \( T \)**:
Rearranging gives:
\[ T^2 = \frac{4\pi^2 r^3}{\sqrt{3} G m} \]
Therefore, we find:
\[ T \propto r^{3/2} m^{-1/2} \]

### Final Result:
The time period \( T \) of rotation is proportional to:
\[ T \propto r^{3/2} m^{-1/2} \]