Comprehensive Study Guide for Centre of Mass of an Asymmetric Uniform Bar

In the previous calculation, the position vector of the centre of mass was determined as: $\vec{r}_{cm} = \frac{L}{6}\hat{i} + \frac{2L}{3}\hat{j}$. However, it seems there was a misunderstanding or error leading to the conclusion of $\frac{13}{8} L \hat{i} + \frac{5}{8} L \hat{j}$.
To clarify the correct approach to finding the position vector:

• The x-coordinate was calculated as $x_{cm} = \frac{L}{6}$, which comes from the average weighted position of the horizontal segment.
• The y-coordinate was calculated as $y_{cm} = \frac{2L}{3}$, based on the vertical segment's average position.
  If you have a different approach or if additional details substantiate your result, please share them for further discussion.