Question

Find the the $y-$ coordinate of the centre of mass of the system of three rods, each of length $6\text{m}$ and two rods, each of length $3\text{m}$ as arranged in the figure shown below. (Assume all rods to be of uniform density)

• A. $\frac{9\sqrt{3}}{8}\text{m}$
• B. $\frac{9\sqrt{3}}{16}\text{m}$
• C. zero
• D. $8\sqrt{3}\text{m}$

Answer 1

The correct option is B $\frac{9\sqrt{3}}{16}\text{m}$

Let the mass of each rod with length $6\text{m}$ be $2m$, then the mass of rods with length $3\text{m}$ will be $m$.

Since all the triangles are equilateral, the angles will all be ${60}^{\circ }$


Taking origin at $\text{O}$, the $y$ coordinates of the ${\text{COM}}^{\prime }s$ of all the rods are:
$AB\to y_1=3\sin {60}^{\circ }=\frac{3\sqrt{3}}{2}\text{m}$
$BC\to y_2=0$
$CA\to y_3=0$
$FE\to y_4=\frac{3\sin {60}^{\circ }}{2}=\frac{3\sqrt{3}}{4}\text{m}$
$FD\to y_5=\frac{3\sin {60}^{\circ }}{2}=\frac{3\sqrt{3}}{4}\text{m}$

The coordinate of centre of mass of system is:
$y_{\text{CM}}=\frac{m_1{y}_1+m_2{y}_2+m_2{y}_3+m_4{y}_4+m_5{y}_5}{m_1+m_2+m_3+m_4+m_5}$
$=\frac{(2m\times \frac{3\sqrt{3}}{2})+(2m\times 0)+(2m\times 0)+(m\times \frac{3\sqrt{3}}{4})+(m\times \frac{3\sqrt{3}}{4})}{2m+2m+2m+m+m}$

$\therefore y_{\text{CM}}=\frac{9\sqrt{3}}{16}\text{m}$