Question

A square of side $2\text{m}$ has one of its corners fixed at the origin as shown in the figure. If the surface density of the square varies with distance $x$ from one side to another as $\sigma =5x{\text{kg/m}}^2$, find the $x$ coordinate of the position of the centre of mass of the square.

• A. $\frac{4}{3}\text{m}$
• B. $\frac{1}{3}\text{m}$
• C. $\frac{2}{3}\text{m}$
• D. $1\text{m}$

Answer 1

The correct option is A $\frac{4}{3}\text{m}$


Let us consider a very small mass element $dm$ at a distance $x$ from origin. Length of mass element is $dx$

Area of the small element $dA=2dx$;
and mass per unit area $\left(\sigma \right)=\frac{dm}{dA}$
But $\sigma =5x$ (given)
$\Rightarrow 5x=\frac{dm}{2dx}$
$\therefore dm=10xdx$

Let $x_{\text{CM}}$ be the position of $COM$ along $x$ direction. Thus, applying the formula:
$x_{\text{CM}}=\frac{\int xdm}{\int dm}$
The limits for integration are $x=0$ to $x=2\text{m}$
$x_{\text{CM}}=\frac{{\int }_0^{2}x\left(10xdx\right)}{{\int }_0^{2}10xdx}=\frac{10{\int }_0^2{x}^{2}dx}{10{\int }_0^{2}xdx}$
$=\frac{{\left[\frac{x^3}{3}\right]}_0^2}{{\left[\frac{x^2}{2}\right]}_0^2}=\frac{\left(\frac{8}{3}\right)}{\left(\frac{4}{2}\right)}=\frac{4}{3}\text{m}$