A Cone Balancing a Cylinder

**Swlabr** · April 20th 2010, 06:20 AM

I am wanting to balance a cone with a cylinder, in the way that Archimedes did in `The Method' with other objects. The plan is to find the volume of the cone using the volume of the cylinder. However, I seem to be having trouble finding a relationship between the sections of the cone and the sections of the cylinder I am balancing it with.

I know that the centre of gravity of the cone is 1/3 of the way along it, and the centre of gravity of the cylinder is half way along it. Having thought about this problem for a while I was convinced that if we take some proportion $\alpha \in [0, 1]$ then the sections to the left and to the right of the respective centres of gravity by this proportion would balance (so the sections at $\frac{\alpha}{2}h$ and $(1-\frac{\alpha}{2})h$ of the cylinder would balance with the sections $\frac{\alpha}{3}h$ and $(1-\frac{2\alpha}{3})h$ of the cone).

I could not get this to work, and in retrospect my guess was clearly wrong (just looking at the extremities gives a contradiction!)

Does anyone have any ideas?

Note that as I know what the volume of a cone is, I was attempting to use a cylinder of radius $\frac{\sqrt{2}}{3}$ . This should, I believe, work.

**Grandad** · April 20th 2010, 07:34 AM

Hello Swlabr

The centroid of a cone is $\tfrac14$ of the way along its axis from the base.

Grandad

**Swlabr** · April 20th 2010, 09:21 AM

Originally Posted by Grandad

Hello Swlabr

The centroid of a cone is $\tfrac14$ of the way along its axis from the base.

Grandad

Hmmm...for some reason, I took 3:1 to be 1/3...thanks, I'll give it another crack!

**Swlabr** · April 21st 2010, 01:42 AM

Originally Posted by Grandad

Hello Swlabr

The centroid of a cone is $\tfrac14$ of the way along its axis from the base.

Grandad

Although a problem, it is not as fundamental as it might at first appear. My problem is not that my numbers don't match up, it is that the radius of my cylinder should be a function of x (aka the cylinder does not have straight edges).

So, any more help would be appreciated...

**Grandad** · April 21st 2010, 02:03 AM

Hello Swlabr

I'm not entirely sure what you're hoping to do here. I think you're talking about a solid formed by attaching a cone to the top of a cylinder (with the same radius and density as the cone), and you want the ratio of the heights of the cylinder and the cone, so that the resulting solid's centre of gravity lies in the plane where the cylinder and the cone meet.

If this is so, let the heights of the cylinder and the cone be $h_y$ and $h_o$ respectively. Then, with common radius $r$ , their respective volumes are

$\pi r^2h_y$ and $\tfrac13\pi r^2 h_o$

and the distances of their respective CG's from the plane interface are

$\tfrac12h_y$ and $\tfrac14h_o$

If the solid's CG lies in this plane, the moments of the two separate parts about a line in this plane are equal. So:

$\pi r^2h_y\cdot\tfrac12h_y = \tfrac13\pi r^2 h_o\cdot\tfrac14h_o$

$\Rightarrow \tfrac12h_y^2=\tfrac{1}{12}h_o^2$

$\Rightarrow h_y:h_o = 1:\sqrt 6$

Is that what you were looking for?

Grandad

**Swlabr** · April 21st 2010, 02:16 AM

Originally Posted by Grandad

Hello Swlabr

I'm not entirely sure what you're hoping to do here. I think you're talking about a solid formed by attaching a cone to the top of a cylinder (with the same radius and density as the cone), and you want the ratio of the heights of the cylinder and the cone, so that the resulting solid's centre of gravity lies in the plane where the cylinder and the cone meet.

If this is so, let the heights of the cylinder and the cone be $h_y$ and $h_o$ respectively. Then, with common radius $r$ , their respective volumes are

$\pi r^2h_y$ and $\tfrac13\pi r^2 h_o$

and the distances of their respective CG's from the plane interface are

$\tfrac12h_y$ and $\tfrac14h_o$

If the solid's CG lies in this plane, the moments of the two separate parts about a line in this plane are equal. So:

$\pi r^2h_y\cdot\tfrac12h_y = \tfrac13\pi r^2 h_o\cdot\tfrac14h_o$

$\Rightarrow \tfrac12h_y^2=\tfrac{1}{12}h_o^2$

$\Rightarrow h_y:h_o = 1:\sqrt 6$

Is that what you were looking for?

Grandad

No, not at all!

The plan is to use Archimedes `Method' (see the book `God Created the Integers', or Wikipedia) to find the volume of a cone.

The technique is roughly as follows:

Take a number of objects, one of which you do not know the volume of (in this case, a cone) and balance them on a lever. You can then equate moments, and solve to find the unknown volume. (p124-125 of GCtI have a good explanation of the technique).

The hard bit is finding a way of balancing the objects, and that is what I am struggling to do here. I am wanting to balance a cone with...some objects which I know the volume of. Most probably, I will want to use things with a round base (as we need to get the $\pi$ from somewhere) but the fact that a cone is sloping seems to be hindering me quite a bit.

I hope that all clarifies what I'm trying to do.

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