# Thread: Center of mass

1. ## Center of mass

Find the center of mass of the thin semicircular region of consant density,sigma=4 bounded by the x-axis and the curve y=(121-x^2)^1/2

2. The center of mass is simply finding the the mass of the object = density * volume and finding the displacement of the mass to the x-axis and y-axis to find the center of mass, which is usually pair (x',y')
However, This link can explain better than I can

Pauls Online Notes : Calculus II - Center of Mass

3. The computation of the coordinates of the 'center of mass' of a thin region $A$ with density $\sigma (x,y)$ is perfomed is succesive steps...

a) first is computed the 'whole mass' as...

$M= \int \int_{A} \sigma (x,y)\cdot dx\cdot dy$ (1)

b) then the coordinates of the 'center of mass' are computed as...

$x_{g} = \frac{1}{M} \cdot \int \int_{A} x\cdot \sigma(x,y)\cdot dx\cdot dy$

$y_{g} = \frac{1}{M} \cdot \int \int_{A} y\cdot \sigma(x,y)\cdot dx\cdot dy$ (2)

In our case $\sigma(x,y)$ is constant so that it can be $\sigma (x,y)=1$ and $A$ is an half circle of radious $r$ and also for it we can suppose $r=1$. Now we can compute first ...

$M= \int_{-1}^{1} \int_{0}^{\sqrt{1-x^{2}}} dx\cdot dy = \int_{-1}^{1} |y|_{0}^{\sqrt{1-x^{2}}} \cdot dx =$

$= \int_{-1}^{1} \sqrt{1-x^{2}}\cdot dx = \frac{1}{2}\cdot |x\cdot \sqrt{1-x^{2}} + \sin^{-1} x|_{-1}^{1} = \frac{\pi}{2}$ (3)

... and then...

$x_{g} = \frac{2}{\pi}\cdot \int_{-1}^{1} \int_{0}^{\sqrt{1-x^{2}}} x\cdot dx\cdot dy = \frac{2}{\pi}\cdot \int_{-1}^{1} x\cdot \sqrt{1-x^{2}}\cdot dx = 0$ (4)

$y_{g} = \frac{2}{\pi}\cdot \int_{-1}^{1} \int_{0}^{\sqrt{1-x^{2}}} y\cdot dx\cdot dy = \frac{2}{\pi}\cdot \int_{-1}^{1} |\frac{y^{2}}{2}|_{0}^{\sqrt{1-x^{2}}} \cdot dx =$

$= \frac{1}{\pi}\cdot \int_{-1}^{1} (1-x^{2})\cdot dx = \frac{1}{\pi}\cdot |x - \frac{x^{3}}{3}|_{-1}^{1} = \frac{4}{3\pi}$ (5)

Kind regards

$\chi$ $\sigma$