Question

An isosceles triangle is cut out of a square sheet with given dimensions as shown in figure. Find the distance of centre of mass of system from the $x$ axis.


Consider the material to be of uniform density.

• A. $7.29\text{cm}$
• B. $3.81\text{cm}$
• C. $5.25\text{cm}$
• D. $9.5\text{cm}$

Answer 1

The correct option is B $3.81\text{cm}$

Let $\sigma \left({\text{kg/cm}}^2\right)$ be the surface mass density.
Taking mass in $\text{kg}$ and length measured in $\text{cm}.$
$\therefore \sigma =\frac{\text{mass}}{\text{area}}$

$\Rightarrow$ Considering the removed triangular part $(base=height=4\text{cm})$ as the negative mass superimposed on the square part$(side=8\text{cm})$ will give the required remaining shape.

$\therefore$ Mass of the square part
$M_1=\sigma \times 8^2$

$\therefore$ Mass of the triangular part
$M_2=\sigma \times \frac{1}{2}\times 4\times 4$
Now, the COM of a triangle is given $\left(\frac{h}{3}\right)$.

$\therefore$Let the distance of COM of square and triangle from $x$ axis is $y_1$ and $y_2$.
Therefore,
$y_1=4\text{cm}$ and
$y_2=4+\frac{4}{3}\text{cm}$

Mass of the new system$=M_1-M_2$
$\Rightarrow$ The required distance of COM of new system from $x$ axis is the $Y_{\text{CM}}$

The $y-$ coordinate of COM of the new shape is given by:

$Y_{\text{CM}}=\frac{M_1{y}_1-M_2{y}_2}{M_1-M_2}=\frac{\sigma \left(\right(8^2\times 4)-\frac{1}{2}\times (4\times 4\left)(4+\frac{4}{3})\right)}{\sigma \left(\right(8^2)-\frac{1}{2}\times 4\times 4)}$

$\Rightarrow Y_{\text{CM}}=\frac{(256-8\times \frac{16}{3})}{64-8}=\frac{213.334}{56}$

$\therefore Y_{\text{CM}}=3.81\text{cm}$

The cut is symmetric about an axis passing through mid point of square and parallel to $Y-$ axis, so COM of system will lie along that line.