Six identical particles each of mass `m` are arranged at the corners of a regular hexagon of side length `L`. If the mass of one of the particle is doubled, the shift in the centre of mass is

Options

A. `L`
B. `6L//7`
C. `L//7`
D. `L/(sqrt(3))`

Explanation

To find the shift in the center of mass when one of the particles' mass is doubled, we can follow these steps:

### Step 1: Determine the initial center of mass of the system

The center of mass (COM) of a system of particles is given by the formula:

\[ \vec{R}_{\text{COM}} = \frac{1}{M} \sum_{i} m_i \vec{r}_i \]

where \(M\) is the total mass of the system, \(m_i\) is the mass of each particle, and \(\vec{r}_i\) is the position vector of each particle.

For six identical particles, each of mass \(m\), arranged at the corners of a regular hexagon, the total mass \(M\) is:

\[ M = 6m \]

The position vectors of the particles can be represented as follows (assuming the hexagon is centered at the origin):

- Particle 1: \( \vec{r}_1 = (L, 0) \)
- Particle 2: \( \vec{r}_2 = \left(\frac{L}{2}, \frac{L\sqrt{3}}{2}\right) \)
- Particle 3: \( \vec{r}_3 = \left(-\frac{L}{2}, \frac{L\sqrt{3}}{2}\right) \)
- Particle 4: \( \vec{r}_4 = (-L, 0) \)
- Particle 5: \( \vec{r}_5 = \left(-\frac{L}{2}, -\frac{L\sqrt{3}}{2}\right) \)
- Particle 6: \( \vec{r}_6 = \left(\frac{L}{2}, -\frac{L\sqrt{3}}{2}\right) \)

Now, we can calculate the center of mass:

\[ \vec{R}_{\text{COM}} = \frac{1}{6m} \left( \vec{r}_1 + \vec{r}_2 + \vec{r}_3 + \vec{r}_4 + \vec{r}_5 + \vec{r}_6 \right) \]

Calculating the x-coordinates:

\[ x_{\text{COM}} = \frac{1}{6m} \left( L + \frac{L}{2} - \frac{L}{2} - L - \frac{L}{2} + \frac{L}{2} \right) = \frac{1}{6m} \left( 0 \right) = 0 \]

Calculating the y-coordinates:

\[ y_{\text{COM}} = \frac{1}{6m} \left( 0 + \frac{L\sqrt{3}}{2} + \frac{L\sqrt{3}}{2} + 0 - \frac{L\sqrt{3}}{2} - \frac{L\sqrt{3}}{2} \right) = \frac{1}{6m} \left( 0 \right) = 0 \]

Thus, the initial center of mass is at the origin \((0, 0)\).

### Step 2: Modify the mass of one particle

Now, let’s double the mass of one of the particles. We can choose particle 1 for this purpose. The new mass of particle 1 becomes \(2m\), while the other particles remain \(m\).

### Step 3: Calculate the new center of mass

The new total mass \(M'\) of the system is:

\[ M' = 5m + 2m = 7m \]

Now, we recalculate the center of mass with the new mass distribution:

\[ \vec{R}'_{\text{COM}} = \frac{1}{M'} \left( 2m \vec{r}_1 + m \vec{r}_2 + m \vec{r}_3 + m \vec{r}_4 + m \vec{r}_5 + m \vec{r}_6 \right) \]

Calculating the x-coordinates:

\[ x'_{\text{COM}} = \frac{1}{7m} \left( 2mL + m\left(\frac{L}{2}\right) + m\left(-\frac{L}{2}\right) - mL + m\left(-\frac{L}{2}\right) + m\left(\frac{L}{2}\right) \right) \]

This simplifies to:

\[ x'_{\text{COM}} = \frac{1}{7m} \left( 2mL + 0 - mL \right) = \frac{1}{7m} \left( mL \right) = \frac{L}{7} \]

Calculating the y-coordinates:

\[ y'_{\text{COM}} = \frac{1}{7m} \left( 0 + m\left(\frac{L\sqrt{3}}{2}\right) + m\left(\frac{L\sqrt{3}}{2}\right) + 0 - m\left(\frac{L\sqrt{3}}{2}\right) - m\left(\frac{L\sqrt{3}}{2}\right) \right) = 0 \]

Thus, the new center of mass is at \(\left(\frac{L}{7}, 0\right)\).

### Step 4: Calculate the shift in the center of mass

The shift in the center of mass is given by the difference between the new and the old center of mass:

\[ \text{Shift} = \left(\frac{L}{7} - 0, 0 - 0\right) = \left(\frac{L}{7}, 0\right) \]

### Final Answer

The shift in the center of mass is \(\frac{L}{7}\) in the x-direction.

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