Question

A rectangular lamina in the $x-y$ plane is shown below:


Density of this lamina is constant along $x$ axis and varies along $y$ axis. If the variation in density is given as $\rho \left(y\right)=16{\rho }_0{y}^2$, then

• A. $x_{COM}=\frac{l}{2}$
• B. $y_{COM}=\frac{b}{2}$
• C. $x_{COM}=\frac{l}{4}$
• D. $y_{COM}=\frac{3b}{4}$

Answer 1

Answer: (A), (D)

The correct options are
A $x_{COM}=\frac{l}{2}$
D $y_{COM}=\frac{3b}{4}$


Taking an element parallel to $x$-axis at a distance $y$ from the $x$ axis.
Area of element $=dA=ldy$
and Mass of element $=dm=\rho dA=16{\rho }_0{y}^{2}ldy$

Therefore, $y_{com}=\frac{{\int }_0^{b}ydm}{{\int }_0^{b}dm}$
$=\frac{{\int }_0^{b}y\times 16{\rho }_0{y}^{2}ldy}{{\int }_0^{b}16{\rho }_0{y}^{2}ldy}$
$=\frac{16{\rho }_{0}l{\int }_0^b{y}^{3}dy}{16{\rho }_{0}l{\int }_0^b{y}^{2}dy}$
$=\frac{{\left[\frac{y^4}{4}\right]}_0^b}{{\left[\frac{y^3}{3}\right]}_0^b}=\frac{3}{4}b$

Density is constant along $x$ - axis
$\Rightarrow x_{com}=\frac{l}{2}$
Thus, coordinates of $COM$ $=[\frac{l}{2},\frac{3b}{4}]$