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 of length \( l \) with a varying density given by \( \rho = \frac{\rho_0 x^2}{l^2} \), we can follow these steps:

### Step 1: Define the density and mass element
The density of the rod at a distance \( x \) from one end is given by:
\[ \rho(x) = \frac{\rho_0 x^2}{l^2} \]
To find the mass of a small element \( dx \) of the rod, we use the relationship:
\[ dm = \rho(x) \, dx = \frac{\rho_0 x^2}{l^2} \, dx \]

### Step 2: Set up the integral for the center of mass
The center of mass \( x_{cm} \) of the rod can be calculated using the formula:
\[ x_{cm} = \frac{\int_0^l x \, dm}{\int_0^l dm} \]
Substituting \( dm \):
\[ x_{cm} = \frac{\int_0^l x \left(\frac{\rho_0 x^2}{l^2}\right) dx}{\int_0^l \left(\frac{\rho_0 x^2}{l^2}\right) dx} \]

### Step 3: Simplify the integrals
The numerator becomes:
\[ \int_0^l x \left(\frac{\rho_0 x^2}{l^2}\right) dx = \frac{\rho_0}{l^2} \int_0^l x^3 \, dx \]
Calculating the integral:
\[ \int_0^l x^3 \, dx = \left[\frac{x^4}{4}\right]_0^l = \frac{l^4}{4} \]
Thus, the numerator is:
\[ \frac{\rho_0}{l^2} \cdot \frac{l^4}{4} = \frac{\rho_0 l^2}{4} \]

The denominator becomes:
\[ \int_0^l \left(\frac{\rho_0 x^2}{l^2}\right) dx = \frac{\rho_0}{l^2} \int_0^l x^2 \, dx \]
Calculating this integral:
\[ \int_0^l x^2 \, dx = \left[\frac{x^3}{3}\right]_0^l = \frac{l^3}{3} \]
Thus, the denominator is:
\[ \frac{\rho_0}{l^2} \cdot \frac{l^3}{3} = \frac{\rho_0 l}{3} \]

### Step 4: Calculate the center of mass
Now substituting the values back into the center of mass formula:
\[ x_{cm} = \frac{\frac{\rho_0 l^2}{4}}{\frac{\rho_0 l}{3}} = \frac{l^2}{4} \cdot \frac{3}{l} = \frac{3l}{4} \]

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