Hi!

Attached picture below.

Problem:Calcuate center of mass of the slender rod.

Solution attempt:

Since the density varies, I did: With $\displaystyle dm = \rho \cdot A \cdot dx $ , because I guess the area is the same throughout the rod.

$\displaystyle \bar{x}=\frac{\int_{0}^{1}x\cdot \rho_{0}(1-\frac{x}{2})A\; dx}{\int_{0}^{1} \rho_{0}(1-\frac{x}{2})A\; dx} = \frac{4}{9}$

Thx for your help!