The density of a thin rod of length l varies with the distance x from one end as `rho=rho_0(x^2)/(l^2)`. Find the position of centre of mass of rod.

Explanation

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

### Step 1: Define the density function
The density of the rod varies with the distance \( x \) from one end and is given by:
\[ \rho(x) = \frac{\rho_0 x^2}{l^2} \]
where \( \rho_0 \) is a constant and \( l \) is the length of the rod.

### Step 2: Express the mass element
The mass element \( dm \) of a small segment of the rod of length \( dx \) at position \( x \) can be expressed as:
\[ dm = \rho(x) \cdot dx = \frac{\rho_0 x^2}{l^2} \cdot dx \]

### Step 3: Calculate the total mass of the rod
To find the total mass \( M \) of the rod, we integrate \( dm \) from \( 0 \) to \( l \):
\[ M = \int_0^l dm = \int_0^l \frac{\rho_0 x^2}{l^2} \, dx \]
Calculating this integral:
\[ M = \frac{\rho_0}{l^2} \int_0^l x^2 \, dx = \frac{\rho_0}{l^2} \left[ \frac{x^3}{3} \right]_0^l = \frac{\rho_0}{l^2} \cdot \frac{l^3}{3} = \frac{\rho_0 l}{3} \]

### Step 4: Calculate the position of the center of mass
The position of the center of mass \( x_{cm} \) is given by:
\[ x_{cm} = \frac{1}{M} \int_0^l x \, dm \]
Substituting for \( dm \):
\[ x_{cm} = \frac{1}{M} \int_0^l x \cdot \frac{\rho_0 x^2}{l^2} \, dx = \frac{1}{M} \cdot \frac{\rho_0}{l^2} \int_0^l x^3 \, dx \]
Calculating the integral:
\[ \int_0^l x^3 \, dx = \left[ \frac{x^4}{4} \right]_0^l = \frac{l^4}{4} \]
Thus,
\[ x_{cm} = \frac{1}{M} \cdot \frac{\rho_0}{l^2} \cdot \frac{l^4}{4} = \frac{1}{M} \cdot \frac{\rho_0 l^2}{4} \]
Now substituting \( M = \frac{\rho_0 l}{3} \):
\[ x_{cm} = \frac{\frac{\rho_0 l^2}{4}}{\frac{\rho_0 l}{3}} = \frac{l^2}{4} \cdot \frac{3}{\rho_0 l} = \frac{3l}{4} \]

### Final Result
The position of the center of mass of the rod is:
\[ x_{cm} = \frac{3l}{4} \]