A rigid body consists of a `3 kg` mass connected to a `2 kg` mass by a massless rod. The `3 kg` mass is located at `vec(r )_(1) = (2hat(i) + 5hat(j))m` and the `2 kg` mass at `vec(r )_(2) = (4hat(i) + 2hat(j))m`. Find the length of rod and the coordinates of the centre of mass.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the Length of the Rod

We have two masses connected by a massless rod:
- Mass \( m_1 = 3 \, \text{kg} \) at position \( \vec{r}_1 = (2\hat{i} + 5\hat{j}) \, \text{m} \)
- Mass \( m_2 = 2 \, \text{kg} \) at position \( \vec{r}_2 = (4\hat{i} + 2\hat{j}) \, \text{m} \)

The length of the rod (distance between the two masses) can be calculated using the distance formula:

\[ \text{Length} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]

Substituting the coordinates:

- \( x_1 = 2, y_1 = 5 \)
- \( x_2 = 4, y_2 = 2 \)

\[ \text{Length} = \sqrt{(4 - 2)^2 + (2 - 5)^2} \]
\[ = \sqrt{(2)^2 + (-3)^2} \]
\[ = \sqrt{4 + 9} \]
\[ = \sqrt{13} \, \text{m} \]

### Step 2: Calculate the Coordinates of the Centre of Mass

The coordinates of the center of mass \( \vec{R}_{cm} \) can be calculated using the formula:

\[ \vec{R}_{cm} = \frac{1}{M} \sum m_i \vec{r}_i \]

Where \( M \) is the total mass and \( m_i \) and \( \vec{r}_i \) are the masses and their respective position vectors.

First, calculate the total mass:

\[ M = m_1 + m_2 = 3 \, \text{kg} + 2 \, \text{kg} = 5 \, \text{kg} \]

Now, calculate the x-coordinate of the center of mass:

\[ x_{cm} = \frac{1}{M} (m_1 x_1 + m_2 x_2) = \frac{1}{5} (3 \cdot 2 + 2 \cdot 4) \]
\[ = \frac{1}{5} (6 + 8) = \frac{14}{5} \, \text{m} \]

Now, calculate the y-coordinate of the center of mass:

\[ y_{cm} = \frac{1}{M} (m_1 y_1 + m_2 y_2) = \frac{1}{5} (3 \cdot 5 + 2 \cdot 2) \]
\[ = \frac{1}{5} (15 + 4) = \frac{19}{5} \, \text{m} \]

### Final Result

The length of the rod is \( \sqrt{13} \, \text{m} \) and the coordinates of the center of mass are:

\[ \left( \frac{14}{5} \hat{i}, \frac{19}{5} \hat{j} \right) \, \text{m} \]

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