The mass per unit length of a non-uniform rod of length L varies as m=λx where λ is constant. Where will be the centre of mass of the rod?

The problem involves finding the centre of mass of a rod whose mass per unit length varies linearly with position $x$ along the rod. The mass per unit length is given as $m = \lambda x$, where $\lambda$ is a constant and $x$ runs from 0 to $L$.

To find the centre of mass $x_{cm}$, we use the definition:

Since $dm = m \, dx = \lambda x \, dx$, we substitute and get:

The $\lambda$ cancels out. Then we calculate the integrals:

Putting these back:

So, the centre of mass is located at $\frac{2L}{3}$ from the end where $x=0$. This makes sense because the rod is heavier towards the larger $x$ end.

Imagine you have a stick that is heavier at one end and lighter at the other. The mass is not spread evenly. To find the balance point (centre of mass), you need to think about how the weight changes along the stick. Since the mass per length increases as you move along the stick, the balance point will be closer to the heavier end. We find this by using a formula that adds up all the tiny weights multiplied by their positions and then divides by the total weight.

Given:

Mass per unit length $m = \lambda x$ where $\lambda$ is constant, length of rod = $L$.

Step 1: Express an infinitesimal mass element:

Step 2: Write the formula for centre of mass:

Step 3: Simplify the expression:

Step 4: Calculate the integrals:

Step 5: Substitute back:

Final answer:

The centre of mass lies at two-thirds the length from the end where $x=0$.