A straight rod of length L has one of its end at the origin and the other at `X=L`. If the mass per unit length of the rod is given by `Ax` where A is constant, where is its centre of mass?

Explanation

To find the center of mass of a straight rod of length \( L \) with a mass per unit length given by \( Ax \), where \( A \) is a constant, we can follow these steps:

### Step 1: Define the mass element \( dm \)
The mass per unit length of the rod is given by \( \lambda(x) = Ax \). Therefore, the mass element \( dm \) can be expressed as:
\[ dm = \lambda(x) \, dx = Ax \, dx \]

### Step 2: Set up the formula for the center of mass
The center of mass \( x_{cm} \) for a continuous mass distribution is given by the formula:
\[ x_{cm} = \frac{\int_0^L x \, dm}{\int_0^L dm} \]

### Step 3: Calculate the numerator \( \int_0^L x \, dm \)
Substituting \( dm = Ax \, dx \) into the numerator:
\[ \int_0^L x \, dm = \int_0^L x (Ax) \, dx = A \int_0^L x^2 \, dx \]
Now, calculate \( \int_0^L x^2 \, dx \):
\[ \int_0^L x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^L = \frac{L^3}{3} \]
Thus, the numerator becomes:
\[ \int_0^L x \, dm = A \cdot \frac{L^3}{3} \]

### Step 4: Calculate the denominator \( \int_0^L dm \)
Now, calculate the total mass \( M \) of the rod:
\[ \int_0^L dm = \int_0^L Ax \, dx = A \int_0^L x \, dx \]
Calculating \( \int_0^L x \, dx \):
\[ \int_0^L x \, dx = \left[ \frac{x^2}{2} \right]_0^L = \frac{L^2}{2} \]
Thus, the denominator becomes:
\[ \int_0^L dm = A \cdot \frac{L^2}{2} \]

### Step 5: Substitute into the center of mass formula
Now substitute the values from Steps 3 and 4 into the center of mass formula:
\[ x_{cm} = \frac{A \cdot \frac{L^3}{3}}{A \cdot \frac{L^2}{2}} = \frac{\frac{L^3}{3}}{\frac{L^2}{2}} = \frac{L^3}{3} \cdot \frac{2}{L^2} = \frac{2L}{3} \]

### Conclusion
The center of mass of the rod is located at:
\[ x_{cm} = \frac{2L}{3} \]