Victoriv.K
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- May 5, 2020
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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.
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Coordinates of center of mass curve semicircle with diameter.
- Thread starter Victoriv.K
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HallsofIvy
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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.
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.