I want to find the center of mass of thin flat plate with constant density $\delta=3g/cm^{-2}$ as shown in figure.
I know that center of mass formula is: $$(x,y)=\bigg(\frac{\int xdm}{\int dm},\frac{\int ydm}{\int dm}\bigg)$$
For $x-$coordinate of the center of mass, firstly I consider a vertical strip and then compute the moment of the vertical strip about $y-axis$ as:
Length of the vertical strip$=2x$,
width of the vertical strip$=dx$,
area of the vertical strip$=dA=2xdx$, and
mass of the vertical strip$=dm=6xdx$
The moment of the strip about $y-axis$ is: $xdm=6x^2dx$
The moment of the plate about $y-axis$ is therefore: $M_{y}=\int_{0}^{1}6x^2 dx=2g.cm$
The mass of the plate is: $M=\int_{0}^{1}6x^2dx=3g$, so the $x-$coordinate of the center of mass is: $x^{'}=\frac{2}{3}cm.$
My question is how can I compute the $y-$coordinate of the center of mass for this thin flat plate using this vertical strip?
I tried something as:
The $y-$component of the vertical strip center of mass$=\frac{y}{2}=x$
The moment of the plate about $x-$axis$M_{x}=\int_{0}^{1}\frac{y}{2}dm=\int_{0}^{1}6x^2dx=2g.cm$
Then $y-$coordinate of the center of mass$=\frac{2}{3}cm$
Am I right?
Best Answer
The center of mass of a triangle is the intersection of the medians. Its coordinates are the averages of the coordinates of the vertices. In this case the centroid is at $(2/3, 2/3)$, just as in your picture.
There is no need to calculate the integrals.