Question

A rectangular lamina $ABCD$ has same density along side $AD$ but has increasing density along side $AB$. The density along side $AB$ is given as $\rho ={\rho }_0+kx$ where ${\rho }_0$ and $k$ are constants and $x$ is distance from the side $AD$. Find the distance of $COM$ from side $AD$.

• A. $\frac{2({\rho }_0+kl)}{3}$
• B. $\frac{(3{\rho }_{0}l+2k{l}^2)}{6{\rho }_0+3kl}$
• C. $\frac{2k{l}^2}{{\rho }_0+3kl}$
• D. $\frac{2{\rho }_{0}l+k{l}^2}{(6{\rho }_0+2kl)}$

Answer 1

The correct option is B $\frac{(3{\rho }_{0}l+2k{l}^2)}{6{\rho }_0+3kl}$


Let us consider a small element $dx$ at distance $x$ from side $AD$
Area of small element $dx$ is $dA=b.dx$
and mass of element is $dm=({\rho }_0+kx)bdx$
$COM$ of this element will be at a distance $x$ from side $AD$.

$x$ coordinate of $COM$ of system will be $x_{com}$
$x_{com}=\frac{\int xdm}{\int dm}$
$=\frac{{\int }_0^{l}x({\rho }_0+kx)bdx}{{\int }_0^l({\rho }_0+kx)bdx}$
$=\frac{b{[\frac{{\rho }_0{x}^2}{2}+\frac{k{x}^3}{3}]}_0^l}{b{[{\rho }_{0}x+\frac{k{x}^2}{2}]}_0^l}$
$=\frac{\frac{{\rho }_0{l}^2}{2}+\frac{k{l}^3}{3}}{{\rho }_{0}l+\frac{k{l}^2}{2}}$
$\therefore x_{com}=\frac{3{\rho }_{0}l+2k{l}^2}{6{\rho }_0+3kl}$ is the distance of COM of the lamina from $AD$.