General Centre of mass derivation

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SUMMARY

The discussion focuses on the derivation of the center of mass for a triangle with its base on the X-axis and symmetric about it. The key formula for the y-coordinate of the center of gravity is provided as \(\frac{\int y \delta(x,y) dxdy}{\int \delta(x, y) dxdy}\). It is established that if the density function \(\delta\) is constant, the center of gravity coincides with the geometric centroid of the triangle. The coordinate system is set with the origin at the center of the base, facilitating the derivation of the equations for the slant sides of the triangle.

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shaggySS
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Hello,
It's my first question here. So I'll try to give as much as information as I know

Actually I am stuck with a problem of centre of mass derivation of a triangle with its base on the X axis and symmetric about it.
The base is b and height is H

As far as I know I have to imagine it as a couple of 1d bodies and integrate them.

Thanks
SAGNIK
 
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Since its symmetric about the X axis the y coordinate needs to be found only.
 
The y- coordinate of the center of gravity of an object with density function \delta(x,y) is
\frac{\int y \delta(x,y) dxdy}{\int \delta(x, y) dxdy}
In particular, if \delta is a constant, it can be factored out of the two integrals and cancelled- that is the "center of gravity" is just the
geometrical "centroid". Here, we can set up a coordinate system so that the origin is at the center of the base of the isosceles triangle, base
along the x- axis, height along the y-axis. The equations of the two slant sides can easily be determined.
 
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