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?

Options

A. `L/3`
B. `L/2`
C. `(2L)/(3)`
D. `(3L)/4`

Explanation

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

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

### Step 2: Set up the center of mass formula
The center of mass \( x_{cm} \) for a continuous mass distribution is given by the formula:
\[ x_{cm} = \frac{\int x \, dm}{M} \]
where \( M \) is the total mass of the rod.

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

### Step 4: Calculate the integral for \( \int x \, dm \)
Now we need to calculate \( \int x \, dm \):
\[ \int x \, dm = \int_0^L x \, (Ax \, dx) = A \int_0^L x^2 \, dx \]
Calculating this integral:
\[ \int_0^L x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^L = \frac{L^3}{3} \]
Thus,
\[ \int x \, dm = A \cdot \frac{L^3}{3} \]

### Step 5: Substitute into the center of mass formula
Now substituting the values we found into the center of mass formula:
\[ x_{cm} = \frac{\int x \, dm}{M} = \frac{A \cdot \frac{L^3}{3}}{\frac{AL^2}{2}} = \frac{\frac{L^3}{3}}{\frac{L^2}{2}} = \frac{L^3}{3} \cdot \frac{2}{L^2} = \frac{2L}{3} \]

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

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