A thin rod of length 6 m is lying along the x-axis with its ends at x=0 and x=6m. Its linear density *mass/length ) varies with x as `kx^(4)`. Find the position of centre of mass of rod in meters.

Options

A. 5m
B. 3m
C. 2m
D. 4m

Explanation

To find the position of the center of mass of a thin rod with a varying linear density, we can follow these steps:

### Step-by-Step Solution:

1. **Define the Linear Density**:
The linear density of the rod is given as \( \lambda(x) = kx^4 \), where \( k \) is a constant.

2. **Consider a Small Element**:
Take a small element of the rod at position \( x \) with a small length \( dx \). The mass of this small element \( dm \) can be expressed as:
\[ dm = \lambda(x) \, dx = kx^4 \, dx \]

3. **Set Up the Center of Mass Formula**:
The formula for the center of mass \( x_{cm} \) of the rod is given by:
\[ x_{cm} = \frac{\int_0^L x \, dm}{\int_0^L dm} \]
where \( L \) is the length of the rod (6 m in this case).

4. **Calculate the Numerator**:
Substitute \( dm \) into the numerator:
\[ \int_0^6 x \, dm = \int_0^6 x \, (kx^4 \, dx) = k \int_0^6 x^5 \, dx \]
Now, calculate the integral:
\[ \int_0^6 x^5 \, dx = \left[ \frac{x^6}{6} \right]_0^6 = \frac{6^6}{6} = \frac{46656}{6} = 7776 \]
Thus, the numerator becomes:
\[ k \cdot 7776 \]

5. **Calculate the Denominator**:
Now calculate the total mass \( M \) of the rod:
\[ \int_0^6 dm = \int_0^6 kx^4 \, dx = k \int_0^6 x^4 \, dx \]
Calculate the integral:
\[ \int_0^6 x^4 \, dx = \left[ \frac{x^5}{5} \right]_0^6 = \frac{6^5}{5} = \frac{7776}{5} \]
Thus, the denominator becomes:
\[ k \cdot \frac{7776}{5} \]

6. **Combine the Results**:
Now substitute the results back into the center of mass formula:
\[ x_{cm} = \frac{k \cdot 7776}{k \cdot \frac{7776}{5}} = \frac{7776 \cdot 5}{7776} = 5 \]

7. **Final Answer**:
Therefore, the position of the center of mass of the rod is:
\[ x_{cm} = 5 \, \text{meters} \]