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I want to find the center of mass of thin flat plate with constant density $\delta=3g/cm^{-2}$ as shown in figure. enter image description here

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?

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  • $\begingroup$ It's a triangle: you could use the centroid $\endgroup$ – b00n heT May 27 '19 at 19:07
  • $\begingroup$ @b00nheT Sir I have no information about centroid, How can I use the integral formula for finding the y coordinate of the center of mass written in question. $\endgroup$ – Noor Aslam May 27 '19 at 19:15
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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.

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  • $\begingroup$ Sir you are cent percent write but how can I get the $y-$coordinate of the center of mass through integral approach? $\endgroup$ – Noor Aslam May 27 '19 at 19:26

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