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enter image description hereI am trying to solve for the center of mass of the shape given below. I started by finding the area of the shape, which is

$A = Atriangle + Acircle/4$

$ = 1/2 + pi/4$

$=(2+pi)/4$

then I used the formula $(1/A) \int [(sqrt(x^2-1))^2 - (x-1)^2] dx$ to find the $y-bar$. Now, I am stuck because I do not know how to find the $My$ or $Mx$ using this information. Please help! Thank you!

Niko H
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  • What do you know about finding $M_x$ or $M_y$? Is there any example you can you do it for? How? (The answers to these questions belong in the body of your question above, not in comments below; use the "edit" link to edit the question.) – David K May 30 '19 at 04:42

1 Answers1

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if you do not have to use integration compulsorily you can use the following formula (replace mass m with area).

$y = \dfrac{\Sigma my}{\Sigma m}, x = \dfrac{\Sigma mx}{\Sigma m}$

here $y = \dfrac{m_1y_1+m_2y_2}{m_1+m_2}, x = \dfrac{m_1x_1+m_2x_2}{m_1+m_2}$

Here object 1 is the quarter of the circle and object 2 is the triangle. The coordinates here are the coordinates of centroid of respective areas (you need to know these beforehand).

If you want to use integration from scratch then you should use integration instead of summation.

$y = \dfrac{\int my}{\int m}, x = \dfrac{\int mx}{\int m}$

amitava
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