Center of mass of a curved surface

In summary, the surface ##\Sigma## given by ##f(x,y)=\frac{1}{2a}(x^2+y^2)## above circle ##K((0,0),a)## can be written as a cone paraboloid with a center of mass at ##(\frac{3}{10}a, \frac{3}{10}a, \frac{3}{10}a)##. The z-coordinate is of interest and can be calculated using integration.
  • #1
brkomir
34
0

Homework Statement


Let ##\Sigma## be surface given as ##f(x,y)=\frac{1}{2a}(x^2+y^2)## above circle ##K((0,0),a)## for ##a>0##. Calculate the center of mass.


Homework Equations



If you are not familiar with ##K((0,0),a)## , that is a circle in point ##(0,0)## with radius a.

The Attempt at a Solution



I would attempt to get a solution if I would be able to imagine ##\Sigma##. If I understand correctly, ##f(x,y)## is also a circle with radius ##\sqrt{2a}## ? or...?

My question here is...: What does the fact, that the ##f(x,y)## is above a circle tell me? o_O
 
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  • #2
I think the surface is meant as 3-dimensional object, so you can write $$z=f(x,y)=\frac{1}{2a}(x^2+y^2)$$ That's a [strike]cone[/strike] paraboloid. The center of mass is trivial for the x- and y-coordinate, and the z-coordinate is interesting.

Edit: fixed, thanks.
 
Last edited:
  • #3
mfb said:
I think the surface is meant as 3-dimensional object, so you can write $$z=f(x,y)=\frac{1}{2a}(x^2+y^2)$$ That's a cone. The center of mass is trivial for the x- and y-coordinate, and the z-coordinate is interesting.

Actually, it's a paraboloid.
 
  • #4
Ok, if the surface is 3-dimensional than i can write ##x=rcos\varphi ## and ##y=rsin\varphi ## for ##r\in \left [ 0,a \right ]## and ##\varphi \in \left [ 0,2\pi \right ]##

Therefore ##f(r,\varphi )=\frac{1}{2r}r^2=\frac{r}{2}##

So ##z=\frac{\int_{0}^{2\pi }\int_{0}^{a}\frac{r^2}{4}drd\varphi }{\int_{0}^{2\pi }\int_{0}^{a}\frac{r}{2}drd\varphi }=\frac{a}{3}##

Or is this not so easy? :D
 
  • #5
Why did you replace a by r in the denominator for f?
And I think there are factors of r missing in the integration (r dr dφ).
 
  • #6
Blah, I don't know. I shouldn't have done that!

##x=rcos\varphi ## and ##y=rsin\varphi ## for ##r\in \left [ 0,a \right ]## and ##\varphi \in \left [ 0,2\pi \right ]##

Therefore ##f(r,\varphi )=\frac{1}{2a}r^2##

So ##z=\frac{\int_{0}^{2\pi }\int_{0}^{a}\frac{r^4}{4a^2}drd\varphi }{\int_{0}^{2\pi }\int_{0}^{a}\frac{r^2}{2a}drd\varphi }=\frac{3}{10}a##
 

1. What is the center of mass of a curved surface?

The center of mass of a curved surface is the point at which the entire mass of the surface can be considered to be concentrated. It is the point where the surface would balance if suspended.

2. How is the center of mass of a curved surface calculated?

The center of mass of a curved surface can be calculated by dividing the surface into small infinitesimal elements and finding the center of mass of each element. The overall center of mass can then be found by integrating the individual centers of mass over the entire surface.

3. Does the shape of the curved surface affect the location of its center of mass?

Yes, the shape of the curved surface does affect the location of its center of mass. The center of mass is typically closer to the heavier side of the surface, and it may also be affected by the distribution of mass within the surface.

4. Can the center of mass of a curved surface be outside of the surface itself?

Yes, it is possible for the center of mass of a curved surface to be outside of the surface itself. This can happen when the surface is not symmetrical or when there is a concentration of mass in one area of the surface.

5. Why is the concept of center of mass important in physics?

The concept of center of mass is important in physics because it allows us to simplify complex systems and analyze them using simple point mass models. It also helps us understand the stability and balance of objects, and it is crucial in studying the motion and dynamics of objects.

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