Question

A 2-dimensional bent rod is made out of a material with uniform density.
Its center of mass is located at which position coordinate? Refer to the diagram of the bent rod and ignore the rod's thickness in your calculations.

• A. $(0,0)$
• B. $(\frac{w}{4},\frac{l}{4})$
• C. $(\frac{w}{3},\frac{l}{3})$
• D. $(\frac{w}{2},\frac{l}{2})$
• E. $(\frac{w^2}{2(l+w)},\frac{l^2}{2(l+w)})$

Answer 1

Answer: (E)

$(\frac{w^2}{2(l+w)},\frac{l^2}{2(l+w)})$

Mass of horizontal rod $m_1=\rho w$ where $\rho$ is the density of the rod

Mass of vertical rod $m_2=\rho l$

Centre of mass of an individual rod lies at its geometric centre.

From figure, $x_1=\frac{w}{2}$ $y_2=\frac{l}{2}$ $x_2=y_1=0$

Horizontal direction : $x_{cm}=\frac{m_1{x}_1+m_2{x}_2}{m_1+m_2}$

$\therefore$ $x_{cm}=\frac{\left(\rho w\right)\times \frac{w}{2}+0}{\rho w+\rho l}=\frac{w^2}{2(l+w)}$

Vertical direction : $y_{cm}=\frac{m_1{y}_1+m_2{y}_2}{m_1+m_2}$

$\therefore$ $y_{cm}=\frac{0+\left(\rho l\right)\times \frac{l}{2}}{\rho w+\rho l}=\frac{l^2}{2(l+w)}$

Thus centre of mass coordinates : $(\frac{w^2}{2(l+w)},\frac{l^2}{2(l+w)})$