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 step by step, we will first calculate the length of the rod connecting the two masses and then find the coordinates of the center of mass of the system.

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

The length of the rod can be calculated using the distance formula between the two points given by the position vectors of the masses.

The position vectors are:
- For the 3 kg mass: \(\vec{r}_1 = (2\hat{i} + 5\hat{j}) \, m\)
- For the 2 kg mass: \(\vec{r}_2 = (4\hat{i} + 2\hat{j}) \, m\)

Using the distance formula:
\[ L = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
where \((x_1, y_1) = (2, 5)\) and \((x_2, y_2) = (4, 2)\).

Substituting the values:
\[ L = \sqrt{(4 - 2)^2 + (2 - 5)^2} \]
\[ L = \sqrt{(2)^2 + (-3)^2} \]
\[ L = \sqrt{4 + 9} \]
\[ L = \sqrt{13} \, m \]

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

The coordinates of the center of mass (\(CM\)) for a system of particles is given by the formula:
\[ \vec{r}_{CM} = \frac{\sum m_i \vec{r}_i}{\sum m_i} \]

#### Step 2.1: Calculate the x-coordinate of the Center of Mass

The x-coordinate of the center of mass can be calculated as:
\[ x_{CM} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2} \]
Substituting the values:
- \(m_1 = 3 \, kg\), \(x_1 = 2 \, m\)
- \(m_2 = 2 \, kg\), \(x_2 = 4 \, m\)

\[ x_{CM} = \frac{(3 \cdot 2) + (2 \cdot 4)}{3 + 2} \]
\[ x_{CM} = \frac{6 + 8}{5} \]
\[ x_{CM} = \frac{14}{5} \, m \]

#### Step 2.2: Calculate the y-coordinate of the Center of Mass

The y-coordinate of the center of mass can be calculated as:
\[ y_{CM} = \frac{m_1 y_1 + m_2 y_2}{m_1 + m_2} \]
Substituting the values:
- \(m_1 = 3 \, kg\), \(y_1 = 5 \, m\)
- \(m_2 = 2 \, kg\), \(y_2 = 2 \, m\)

\[ y_{CM} = \frac{(3 \cdot 5) + (2 \cdot 2)}{3 + 2} \]
\[ y_{CM} = \frac{15 + 4}{5} \]
\[ y_{CM} = \frac{19}{5} \, m \]

### Final Result

Thus, the coordinates of the center of mass are:
\[ \vec{r}_{CM} = \left(\frac{14}{5} \hat{i} + \frac{19}{5} \hat{j}\right) \, m \]

### Summary of Results
- Length of the rod: \(\sqrt{13} \, m\)
- Coordinates of the center of mass: \(\left(\frac{14}{5}, \frac{19}{5}\right)\)

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