Question

A uniform $\text{L}$ shaped lamina with given dimensions is shown in the figure. Origin is taken at $\text{O}$.


If the mass of the lamina is $6\text{kg}$, then the centre of mass of the lamina is located at:

• A. $(\frac{5}{3}\text{m},\frac{5}{6}\text{m})$
• B. $(\frac{5}{4}\text{m},\frac{5}{6}\text{m})$
• C. $(\frac{5}{3}\text{m},\frac{5}{3}\text{m})$
• D. $(\frac{5}{8}\text{m},\frac{5}{3}\text{m})$
  

Answer 1

The correct option is A $(\frac{5}{3}\text{m},\frac{5}{6}\text{m})$

Dividing the given lamina into two rectangular sections of area $A_1$ and $A_2$.

We replace the rectangular section by placing point masses at their respective $\text{COM's}$.


Area of rectangular section $A_1=4{\text{m}}^2$
Area of rectangular section $A_2=2{\text{m}}^2$

Centre of mass of section $1$ $C_1=(x_1,y_1)=(2\text{m},\frac{1}{2}\text{m})$
Centre of mass of section $2$ $C_2=(x_2,y_2)=(1\text{m},\frac{3}{2}\text{m})$

Applying the formula for $x$ and $y$ coordinates of $\text{CM}$ for the combination of rectangular sections:

$x_{\text{CM}}=\frac{A_1{x}_1+A_2{x}_2}{A_1+A_2}=\frac{(4\times 2)+(2\times 1)}{4+2}=\frac{10}{6}=\frac{5}{3}\text{m}$
$\therefore y_{\text{CM}}=\frac{A_1{y}_1+A_2{y}_2}{A_1{A}_2}=\frac{(4\times \frac{1}{2})+(2\times \frac{3}{2})}{4+2}=\frac{5}{6}\text{m}$

The coordinates of the centre of mass for given shape is:
$(x_{\text{CM}},y_{\text{CM}})=(\frac{5}{3}\text{m},\frac{5}{6}\text{m})$