Coordinates of center of mass curve semicircle with diameter.

[Victoriv.K]

Hello, could anybody help me with that because I have a few more examples but I would like to do them by myself. So could anybody show me how to do it? Pretty please. R-radius and d is diameter.

 

[HallsofIvy]

Help you with what? You say "semicircle". Do you mean the points (x, y) satisfying the equation $\displaystyle x^2+ y^2= R^2$ with $\displaystyle y\ge 0$ (and not the half-disk)?
You say "center of mass" but don't give a mass or density function. Are we to assume constant density (so strictly speaking you want the "centroid")?

Presumably, if you are expected to do a problem like this, you know how to find centroid (your "center of mass") of a curve. How is that defined? Do you know how to find the length of such a curve? To find the length of y= f(x) from x= a to x= b you need to calculate the integral $\displaystyle \int_a^b\sqrt{1+ \left(\frac{dy}{dx}\right)^2} dx$. In the case of a semi-circle of radius R that is, of course, $\displaystyle \pi R$.

The x coordinate of the centroid is $\displaystyle \int_a^b x\sqrt{1+\left(\frac{dy}{dx}\right)^2} dx$ divided by the length. The y coordinate of the centroid is $\displaystyle \int_a^b ysqrt{1+\left(\frac{dy}{dx}\right)^2} dx$ divided by the length.

If you can write the x and y coordinates of the curve in parametric equations, x= f(t), y= g(t) (such as x= Rcos(t), y= Rsin(t)!) Then the length is $\displaystyle \int \sqrt{\left(\frac{dx}{dt}\right)^2+ \left(\frac{dy}{dt}\right)^2} dt$. In the case of the semi-circle that is $\displaystyle \int_0^{\pi}\sqrt{\left(-Rsin(t)\right)^2+ \left(Rcos(t)\right)^2} dt= \pi R$. The x-coordinate of the centroid is $\displaystyle \int x\sqrt{\left(\frac{dx}{dt}\right)^2+ \left(\frac{dy}{dt}\right)^2} dt$ divided by the length of the curve and the y-coordinate is $\displaystyle \int \sqrt{y\left(\frac{dx}{dt}\right)^2+ \left(\frac{dy}{dt}\right)^2} dt$ divided by the length of the curve.