Centre of Mass of a Uniform Rod

Result

The CM of a uniform rod of length $L$ is at its midpoint:

$x_{cm} = \frac{L}{2}$

Derivation

Place the rod along the $x$-axis with the left end at the origin. Linear mass density $\lambda = M/L$.

$x_{cm} = \frac{1}{M}\int_0^L x \, dm = \frac{1}{M}\int_0^L x \cdot \frac{M}{L} \, dx = \frac{1}{L}\int_0^L x \, dx = \frac{1}{L} \cdot \frac{L^2}{2} = \frac{L}{2}$

By symmetry

A uniform rod is symmetric about its midpoint. The CM must lie on the axis of symmetry — at $x = L/2$. The integral confirms this.

Non-uniform rod

If the rod has variable linear density $\lambda(x)$:

$x_{cm} = \frac{\int_0^L x\lambda(x) \, dx}{\int_0^L \lambda(x) \, dx}$

The CM shifts toward the denser end.

Example: Rod with $\lambda = \lambda_0 x$ (density increases linearly from left end):

$x_{cm} = \frac{\int_0^L x \cdot \lambda_0 x \, dx}{\int_0^L \lambda_0 x \, dx} = \frac{\lambda_0 L^3/3}{\lambda_0 L^2/2} = \frac{2L}{3}$

CM shifts to the denser (right) end.