A rod of length L is placed along the x-axis between `x=0` and `x=L`. The linear mass density (mass/length) `rho` of the rod varies with the distance x from the origin as `rho=a+bx`. Here, a and b are constants. Find the position of centre of mass of this rod.

Explanation

To find the position of the center of mass of a rod of length \( L \) with a linear mass density that varies with position, we can follow these steps:

### Step 1: Define the linear mass density
The linear mass density \( \rho \) of the rod is given by:
\[ \rho(x) = a + bx \]
where \( a \) and \( b \) are constants.

### Step 2: Express the mass of a small element
Consider a small element of the rod of length \( dx \) located at position \( x \). The mass \( dm \) of this small element can be expressed as:
\[ dm = \rho(x) \, dx = (a + bx) \, dx \]

### Step 3: Set up the formula for the center of mass
The position of the center of mass \( x_{cm} \) is given by the formula:
\[ x_{cm} = \frac{\int_0^L x \, dm}{\int_0^L dm} \]

### Step 4: Calculate the numerator
Substituting \( dm \) into the numerator:
\[ \int_0^L x \, dm = \int_0^L x (a + bx) \, dx = \int_0^L (ax + bx^2) \, dx \]
Now, we can compute the integrals:
\[ \int_0^L ax \, dx = a \left[ \frac{x^2}{2} \right]_0^L = \frac{aL^2}{2} \]
\[ \int_0^L bx^2 \, dx = b \left[ \frac{x^3}{3} \right]_0^L = \frac{bL^3}{3} \]
Thus, the numerator becomes:
\[ \int_0^L x \, dm = \frac{aL^2}{2} + \frac{bL^3}{3} \]

### Step 5: Calculate the denominator
Now, we calculate the denominator:
\[ \int_0^L dm = \int_0^L (a + bx) \, dx = \int_0^L a \, dx + \int_0^L bx \, dx \]
Calculating these integrals:
\[ \int_0^L a \, dx = aL \]
\[ \int_0^L bx \, dx = b \left[ \frac{x^2}{2} \right]_0^L = \frac{bL^2}{2} \]
Thus, the denominator becomes:
\[ \int_0^L dm = aL + \frac{bL^2}{2} \]

### Step 6: Combine the results
Now we can combine the results to find \( x_{cm} \):
\[ x_{cm} = \frac{\frac{aL^2}{2} + \frac{bL^3}{3}}{aL + \frac{bL^2}{2}} \]

### Step 7: Simplify the expression
To simplify, we can multiply the numerator and denominator by \( 6 \) to eliminate the fractions:
\[ x_{cm} = \frac{3aL^2 + 2bL^3}{6aL + 3bL^2} \]

### Final Result
Thus, the position of the center of mass of the rod is given by:
\[ x_{cm} = \frac{3aL + 2bL^2}{6a + 3bL} \]