Question

A uniform steel rod of $1m$ in length is bent in a ${90}^{\circ }$ angle at its mid point. Determine the position of its center of mass from the corner at bent.

Answer 1

To find:

Location of corner from origin

$x_{com}=\frac{m_1{x}_1+m_2{x}_2}{m_1+m_2}$, $y_{com}=\frac{m_1{y}_1+m_2{y}_2}{m_1+m_2}$

$x_{com}=\frac{\frac{m}{2}(-\frac{L}{4})+\left(\frac{M}{2}\right)\left(0\right)}{M},$ $y_{com}=\frac{\left(\frac{M}{2}\right)\left(0\right)+\left(\frac{M}{2}\right)\left(\frac{L}{4}\right)}{M}$

$x_{com}=-\frac{ML/8}{M}$, $y_{com}=\frac{ML/8}{M}$

$x_{com},y_{com}=(-\frac{L}{8},\frac{L}{8})$

Distance between origin & com $=\sqrt{{\left(\frac{L}{8}\right)}^2+{\left(\frac{L}{8}\right)}^2}$

$=\frac{L}{8}\sqrt{2}$

$=\frac{L}{4\sqrt{2}}$.
Given $L=1$

$\therefore$ Distance$=\frac{1}{4\sqrt{2}}$