Question

Two uniform rectangular plates having mass density ($\text{mass/area}$) $2{\text{kg/m}}^2$ and $1{\text{kg/m}}^2$ are joined together to form the $\text{L}$- shaped lamina as shown in the figure. Find the coordinates of centre of mass of the $\text{L}$-shaped lamina.

• A. $(\frac{17}{11},\frac{83}{22})\text{m}$
• B. $(\frac{26}{7},\frac{43}{14})\text{m}$
• C. $(\frac{26}{7},\frac{83}{22})\text{m}$
• D. $(\frac{13}{7},\frac{43}{14})\text{m}$

Answer 1

The correct option is D $(\frac{13}{7},\frac{43}{14})\text{m}$


Area of plate $1$ , $A_1=(6\times 1)=6{\text{m}}^2$
Area of plate $2$ , $A_2=(8\times 2)=16{\text{m}}^2$
Mass of plate $1$ , $m_1=(A_1\times {\sigma }_1)=6\times 2=12\text{kg}$
Mass of plate $2$ , $m_2=(A_2\times {\sigma }_2)=16\times 1=16\text{kg}$

Centre of mass of a uniform rectangular plate is given by: $(\frac{l}{2},\frac{b}{2})$
$\text{COM}$ of plate $1$ $(x_1,y_1)=(3,0.5)\text{m}$
$\text{COM}$ of plate $2$ $(x_2,y_2)=(1,5)\text{m}$

Now, replacing the respective plates by point masses placed at their respective $\text{COM}$ :
$x$- coordinate of com of $\text{L}$- shaped lamina:
$x_{\text{CM}}=\frac{m_1{x}_1+m_2{x}_2}{m_1+m_2}=\frac{12\times 3+16\times 1}{12+16}=\frac{52}{28}\therefore x_{\text{CM}}=\frac{13}{7}\text{m}$
and $y$- coordinate of com of $\text{L}$- shaped lamina:
$y_{\text{CM}}=\frac{m_1{y}_1+m_2{y}_2}{m_1+m_2}=\frac{12\times 0.5+16\times 5}{12+16}=\frac{86}{28}\therefore y_{\text{CM}}=\frac{43}{14}\text{m}$
Hence, $\text{COM}$ of $L$- shaped lamina is given by
$(x_{\text{CM}},y_{\text{CM}})=(\frac{13}{7},\frac{43}{14})\text{m}$