Center of mass triangle problem

[nicholas1]

After finding the center of mass of a pentagon with five equal sides a but one triangle missing, then iam asked to find the CM of uniform square a with one quadrant missing.
I dont' really grab the idea of CM, please help..

 

[andrevdh]

Wellcome to PF nicholas.

First note that the com will lie along the line of symmetry. This means it will be located along the line through the middle of the pentagon and dividing the missing triangle in half, let's call this line the y-axis. This means that you need to locate the y-coordinate of the combined com of the four triangles along the axis.

To locate the y-coordinate of the combined com of the four triangles you need to take the moments of mass of the com of each of the triangles about a line perpendicular to the y-axis. Let's call it the x-axis. Position it at the base of the pentagon.

For the mass of the coms, and total mass, you just need to consider only the total area of the shape in the moment calculations (that is you assume that the mass can be found by multiplying the area by some constant).

Use the fact that the sum of the mass moments of the coms of the four triangles about the x-axis need to be equal to the moment of the combined com (total area of the four triangles times \bar{y}) about your x-axis- it is in this equation that the above mentioned constant will cancel out.

The circle with one quadrant is actually much easier! Can you proof that the com of one quadrant is located \frac{2}{3}R up from the sharp point?