Three particles of equal mass M, interacting via gravity, lie on the vertices of an equilateral triangle of side d, as shown in the figure below. The whole system is rotating with an angular velocity \(\omega\) about an axis perpendicular to the plane of the system and passing through the center of mass. The value of \(\omega\), for which the distance between the masses remains d, is (G is the universal gravitational constant)
Show Hint
For symmetric systems in circular motion, always calculate the net force on one of the particles. This net force, directed towards the center of rotation, must be equal to the required centripetal force. Correctly identifying the radius of rotation is a crucial first step.
Step 1: Understanding the Concept:
For the system to rotate with a constant distance \(d\) between the particles, the net gravitational force on each particle must provide the necessary centripetal force to maintain its circular motion. We need to calculate the net force on one particle due to the other two and equate it to the centripetal force \(M R \omega^2\). Step 2: Key Formula or Approach:
1. Newton's Law of Gravitation: \(F_g = \frac{Gm_1m_2}{r^2}\).
2. Centripetal Force: \(F_c = M R \omega^2\), where R is the radius of the circular path.
3. Geometry of an equilateral triangle: The center of mass is at the centroid, which is at a distance \(R = \frac{d}{\sqrt{3}}\) from each vertex. Step 3: Detailed Explanation:
1. Find the radius of rotation (R):
The axis of rotation passes through the center of mass, which is the centroid of the equilateral triangle. The distance from the centroid to any vertex is the radius R of the circular path. For an equilateral triangle of side d, this distance is \(R = \frac{d}{\sqrt{3}}\).
2. Calculate the net gravitational force on one particle:
Let's focus on the top particle. It is attracted by the other two particles at the base of the triangle.
The force exerted by the left particle is \(F_1\) with magnitude \(\frac{GM^2}{d^2}\) along the triangle side.
The force exerted by the right particle is \(F_2\) with magnitude \(\frac{GM^2}{d^2}\) along the other side.
The angle between these two force vectors is \(60^\circ\). The net force \(\mathbf{F}_{\text{net}} = \mathbf{F}_1 + \mathbf{F}_2\) will be directed towards the centroid.
We can find the magnitude of the net force using the law of cosines for vector addition or by resolving components. Let's use components. The horizontal components cancel out. The vertical components add up. The angle each force makes with the vertical direction is \(30^\circ\).
\[ F_{\text{net}} = F_1 \cos(30^\circ) + F_2 \cos(30^\circ) = 2 \left(\frac{GM^2}{d^2}\right) \cos(30^\circ) \]
\[ F_{\text{net}} = 2 \left(\frac{GM^2}{d^2}\right) \left(\frac{\sqrt{3}}{2}\right) = \frac{\sqrt{3}GM^2}{d^2} \]
3. Equate net force to centripetal force:
The net gravitational force provides the centripetal force required for rotation.
\[ F_c = F_{\text{net}} \]
\[ M R \omega^2 = \frac{\sqrt{3}GM^2}{d^2} \]
Substitute \(R = \frac{d}{\sqrt{3}}\):
\[ M \left(\frac{d}{\sqrt{3}}\right) \omega^2 = \frac{\sqrt{3}GM^2}{d^2} \]
4. Solve for \(\omega\):
\[ \omega^2 = \frac{\sqrt{3}GM^2}{d^2} \cdot \frac{\sqrt{3}}{Md} \]
\[ \omega^2 = \frac{3GM}{d^3} \]
\[ \omega = \sqrt{\frac{3GM}{d^3}} \]
Step 4: Final Answer:
The required angular velocity is \(\omega = \sqrt{\frac{3GM}{d^3}}\). This matches option (B).
