A uniform rod of length $5L$ is bent at a right angle, keeping one side of length $2L$, and placed as shown in the figure (the $3L$ side along the $y$-axis and the $2L$ side along the $x$-axis, with the bend at the origin). The position of the centre of mass of the system is: (Consider $L = 10\ \text{cm}$; positions in cm)
Answer: (D) $4\hat i + 9\hat j$
The rod is uniform, so mass is proportional to length: take $3m$ for the $3L$ part and $2m$ for the $2L$ part.
- $2L = 20\ \text{cm}$ piece on the $x$-axis: centre at $(10, 0)$.
- $3L = 30\ \text{cm}$ piece on the $y$-axis: centre at $(0, 15)$.
$$x_{cm} = \frac{2m(10) + 3m(0)}{5m} = 4\ \text{cm},\qquad y_{cm} = \frac{2m(0) + 3m(15)}{5m} = 9\ \text{cm}$$
$\vec r_{cm} = 4\hat i + 9\hat j$ (cm).
Solution by Sreeraj P, M.Sc Physics