Question

Find the centre of mass of a uniform $L$ shaped lamina (a thin flate plate) with dimensions as shown in figure. The mass of the lamina is $6\text{kg}$.

• A. $(\frac{5}{6}\text{m},\frac{5}{6}\text{m})$
• B. $(\frac{5}{2}\text{m},\frac{5}{2}\text{m})$
• C. $(\frac{5}{3}\text{m},\frac{5}{3}\text{m})$
• D. $(\frac{5}{4}\text{m},\frac{5}{4}\text{m})$

Answer 1

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

Considering the $x$ & $y$ axis , we have co-ordinates of the vertices of the $L$ shaped lamina as given in the figure.


$L$ - shaped lamina has total $3$ squares each of side $2\text{m}$.
The mass of each lamina is $2\text{k}g$ {lamina is uniform}

The centre of mass of the squares in $(x,y)$ co-ordinates are
$C_1(1,1),C_2(3,1)$ and $C_3(1,3)$

Thus, $x$ coordinate of $COM$ of lamina
$x_{com}=\frac{m_1{x}_1+m_2{x}_2+m_3{x}_3}{m_1+m_2+m_3}$
$x_{com}=\frac{2\left(1\right)+2\left(3\right)+2\left(1\right)}{2+2+2}=\frac{2+6+2}{6}=\frac{10}{6}=\frac{5}{3}\text{m}$
$y$ - coordinate of $COM$ of lamina
$y_{com}=\frac{m_1{y}_1+m_2{y}_2+m_3{y}_3}{m_1+m_2+m_3}=\frac{2\left(1\right)+2\left(1\right)+2\left(3\right)}{2+2+2}$
$=\frac{2+2+6}{6}=\frac{10}{6}=\frac{5}{3}\text{m}$
Hence, centre of mass of $L$ shaped lamina is $(\frac{5}{3}\text{m},\frac{5}{3}\text{m})$