Question

A wire of uniform cross-section is bent in the shape shown in the figure. The coordinates of the centre of mass of each side are shown in the figure. Origin is taken at $O$. Find the coordinates of the center of mass of the given system.

• A. $(\frac{15l}{14},\frac{6l}{7})$
• B. $(\frac{15l}{14},l)$
• C. $(l,\frac{l}{2})$
• D. $(l,l)$

Answer 1

The correct option is A $(\frac{15l}{14},\frac{6l}{7})$

Let mass of each side of length $2l$ be $m$. Then mass of $3^{rd}$ side with length $l$ will be $\frac{m}{2}$.

The coordinates of ${\text{COM}}^{\prime }s$ of all sides are given below:

$C_1\Rightarrow (l,0)$
$C_2\Rightarrow (2l,l)$
$C_3\Rightarrow (\frac{3l}{2},2l)$
$C_4\Rightarrow (0,l)$

Replacing all the sides with point masses placed at their respective centre of mass & applying the formula for position of $\text{COM}$ of system:
$y_{\text{CM}}=\frac{(m\times 0)+(m\times l)+(\frac{m}{2}\times 2l)+(m\times l)}{m+m+\frac{m}{2}+m}$
$\therefore y_{\text{CM}}=\frac{6l}{7}$

Simillarly;
$x_{\text{CM}}=\frac{(m\times l)+(m\times 2l)+(\frac{m}{2}\times \frac{3l}{2})+(m\times 0)}{m+m+\frac{m}{2}+m}$
$\therefore x_{\text{CM}}=\frac{\frac{15ml}{4}}{\frac{7m}{2}}=\frac{15l}{14}$

$\Rightarrow (x_{\text{CM}},y_{\text{CM}})=(\frac{15l}{14},\frac{6l}{7})$