How do I find the center of mass of a cone?

any particular reason you can't research that too?

 

take a cross-section though the axis. how would you find the center of mass of the resulting triangle?

 

So did my question not make sense to you? Finding the center of gravity of a cone is exceedingly trivial and you do not need anything but the Pythagorean theorem. You certainly don't need any calculus, although I'm sure there are lots of hard ways to do it if you WANT to do it a hard way.

 

This is a fairly common question:
https://www.physicsforums.com/showthread.php?t=95510

 

So I take it you disagree w/ my statement that you do not need calculus?

 

No, I never said that. To determine the centroid of a general figure, you certainly need Calculus. But, you can then use a formula for a given type of figure- there is a standard formula for the centroid of a cone. Perhaps the OP could just look that up.

 

agreed

 

phinds,
could you elaborate on your method of using only the Pythagorean theorem? I'm not getting it.

 

The center of gravity of a cone is at the exact same spot as the center of the area of a triangle created if you pass a plane through the axis of the cone. Divide that triangle into an upper triangle (I'm thinking of the cone pointing downward) and a lower trapazoid, with the areas of the two being equal. Pythagorous (and an understanding of similar triangles) will give you the rest.

 

I think you are right if you are considering a hollow cone. I was assuming the OP asked about a solid cone, but of course I might be wrong.

For me, it is not very intuitive that the cross-sectional triangle trick will work if a hollow cone is considered. How did you come to that conclusion?

 

I have no idea why you think the cone should be hollow for this to work; I'm assuming a solid cone. If the upper triangle has the same area as the lower trapezoid then will not the resulting rotated figures have the same volumes?

OH ... oops, maybe they don't

 

LATER: I checked it out and it does work out correctly in this case.

 

The median of an isosceles triangle is H/3 while the CoM of a cone is at H/4. I'm not quite sure how your method works out.

 

I don't know what the median has to do with anything and the CoM of a cone, if I have it right is at H/(1-SQRT(2)) assuming the cone is pointed up

 

But the CoM of the cone is at H/4, assuming the cone is pointed up. There seems to be some miscommunication here. Edit: 1-sqrt(2) isn't even positive.

 

Yeah, sorry about that, got rushed. Meant to say 1-(1/sqrt(2)), or in other words, .707.

I'm running a "stack of disks" computer program just now but seem to have something messed up since instead of .707, or the .75 you say is the right answer, I'm getting .794

Is the center of mass of a cone in a different place than the center of area of a triangle created by intersecting a plane through the axis of the cone? If it is, then I've got it wrong 'cause that's what I'm basing it on.

 

Was your assumption that the CoM of the cone is vertically at the same location of the CoM of the triangle? If so then that's not right. When you rotate the triangle, more weight is prescribed to the larger end, so intuitively the CoM should move towards the larger end.

 

The com of the section area as almost nothing to do with the com of the solid.
For example consider two cubes: the first with volume 1 and the other with volume 8.
The com of the section is at 1/5 the distance from the bigger one; the real com is at 1/9.

If you write the integral you will see that the com of the cone is related to that of the 2d figure which has two parabolas as borders.