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Centre of mass is defined as,

$$ \overline{x} = \int x \rho dA$$

for a semi circle, above the x axis,

$$ \overline{x} = \rho \int_{0}^{R} \int_{-\sqrt{1-y^2} }^{\sqrt{1-y^2} } x dx dy$$

This becomes (my origin is at center of semi circle)

$$ \overline{x} = \rho \int_{0}^{R} (R^2-y^2) dy = \rho [ R^2 y -\frac{y^3}{3} ] = \frac{ \rho}{3} [ 3R^3 -R^3] = \frac{2 \rho}{3} R^3$$

Now, I'm certain something is 'wrong' because the actual answer is suppoed to be '0' for center of mass of semi circle along 'x'.. however it's coming non zero. Where exactly have I made a mistake?

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The image I have shown is the idea behind what I did, first when I did integral along 'x', I got the centre of mass of a thin rod inside the semi circle parallel to horizontal as a function of 'y' , add up the centre of mass of these rods I should get centre of mass of circle but I got something non zero (?)

tryst with freedom
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1 Answers1

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You are integrating an odd function over a symmetric interval around $0$ so it must be $0$. The calculations of the inner integral are:

$$\int_{-\sqrt{1-y^{2}}}^{\sqrt{1-y^{2}}} x\,dx=\frac{1-y^{2}}{2}-\frac{1-y^{2}}{2}=0$$