Question

Find the co-ordinates of center of mass of the lamina shown in the figure below

• A. $(0.75m,1.75m)$
• B. $(0.75m,0.75m)$
• C. $(1.25m,1.5m)$
• D. $(1m,1.75m)$

Answer 1

The correct option is A

$(0.75m,1.75m)$

Step 1. given data

Now, by dividing our lamina to get two rectangles $A$ and $B$.

Let $m_A$ and $are{a}_A$ be the mass and area of rectangle $A$, and $m_B$ and $are{a}_B$ be the mass and area of rectangle $B$

Step 2. finding uniform surface density

Assuming that the lamina is of uniform surface density $\sigma$ throughout,

$$\begin{gathered} {\sigma }_A={\sigma }_A \\ \frac{m_A}{are{a}_A}=\frac{m_{\mathbf{B}}}{are{a}_B} \\ \frac{m_A}{2\times 1}=\frac{m_{\mathbf{B}}}{1\times 2} \\ {m}_A=m_B=saym \end{gathered}$$

Step 3. finding centroids of the rectangles

For rectangle $A$, draw the intersecting lines from the both ends, we will get a mid point.

The midpoint values is given as $Co{M}_A=(1,2.5)$

Therefore, The centroid of rectangle $A$ is found to be at $Co{M}_A=(1,2.5)$,

Similarly, For rectangle $B$, draw the intersecting lines from the both ends, we will get a mid point.

The midpoint values is given as $Co{M}_B=(0.5,1)$

Therefore, The centroid of rectangle $B$ is found to be at $Co{M}_B=(0.5,1)$.

Step 4. finding center of mass.

Therefore, the centre of mass of the entire lamina lies somewhere on the line joining these two points.

This is given by $Co{M}_{lamina}=\left(X_{CoM},Y_{CoM}\right)$

Where, $X_{CoM}=\frac{m_A{x}_A+m_B{x}_B}{m_A+m_B}$ and

$Y_{CoM}=\frac{m_A{Y}_A+m_B{Y}_B}{m_A+m_B}$

But we have deduced that $m_A=m_B=m$
$$\begin{gathered} X_{CoM}=\frac{m(x_A+x_B)}{m\left(2\right)} \\ =\frac{x_A+x_B}{2} \\ =\frac{1+0.5}{2} \\ =0.75 \end{gathered}$$

Also,

$$\begin{gathered} Y_{CoM}=\frac{m(Y_A+Y_B)}{m\left(2\right)} \\ =\frac{Y_A+Y_B}{2} \\ =\frac{2.5+1}{2} \\ =1.75 \end{gathered}$$

Therefore, the coordinates of the centre of mass of the lamina will be $\left(0.75,1.75\right)$.

Hence, the correct option is $\left(A\right)$.