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When we rotate a right angled triangle about its perpendicular we get a solid cone. For a right-angled triangle the centre of mass is at its centroid i.e., at $(\frac h3, \frac b3)$. If we consider a solid cone to be made up of many right-angled triangles, the centre of mass should be at a height of $\frac h3$ from the base whereas it is at $\frac h4$ from the base for a solid cone. I have obtained the proper answer through integration but I am unable to figure out where I am going wrong in this logic. Kindly point out where I am going wrong.


marked as duplicate by Aaron Stevens, Jon Custer, PM 2Ring, Thomas Fritsch, ZeroTheHero Aug 19 at 23:49

This question was marked as an exact duplicate of an existing question.


Note that this question has already been asked and answered on Mathematics SE. I think it is a really good answer,so I will just summarize it here for future reference as well as add some of my own thoughts.

Essentially areas and volumes cannot always be treated on equal footing. Areas and volumes scale differently. For the cone there is proportionally more volume located near the base of the solid compared to how much area is proportionally near the base of the triangle. This is why the center of mass of the cone is located closer to the base than it is for the triangle.

To relate this more to physics, and to add some new information, I will say that this issue comes up in other places as well. For example, you could imagine a thin disk as just a bunch of thin rods aligned so that they are all concentric. Then one could wonder why the moment of inertia of a thin rod of mass $M$ and length $L$ about its center is $\frac13ML^2=\frac43M(L/2)^2$ but the moment of inertia of a thin disk of mass $m$ and radius $r$ is $\frac12mr^2$. You would expect both of the constant factors out front to be the same, but this is not what happens.$^*$

I suppose this goes to show that there are differences between countably and uncountably infinite sets of objects. In order to move from the triangle to the cone or from the rod to the cylinder you have to give the smaller elements some thickness in the new dimension you are considering. This moves you from a countable set of objects to an uncountable set of objects. Hence we get different results.

$^*$ However, you are free to think of the disk as a bunch of really small wedges. But in this case you are not moving between the "number of dimensions" since the wedges and the disk are both areas. A similar fix for the cone would be to imagine using a bunch of "triangle wedges" to construct your cone. No issues would then arise.

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    $\begingroup$ On a similar note, there's the staircase paradox. $\endgroup$ – PM 2Ring Aug 19 at 20:47
  • $\begingroup$ @PM2Ring Yeah I like that one too. I'll on occasion randomly start thinking about it. $\endgroup$ – Aaron Stevens Aug 19 at 21:16

Rotating is not a uniform operation. The inside travels a distance of $0$, while the outside moves through $2\pi b$. These thing matter, esp. when finding moments (the center of mass is just the zeroth moment of the mass distribution).

It's better to work in the orthogonal direction. For the 2D triangle all the does is swap $(h, b)$, but now you can sum a bunch of disks, but before doing that, note the 2D --> $h/3$ and 3D --> $h/4$, maybe we can just guess that $n$D --> $h/(n+1)$, and then work it out (scaled so $b=h$):

$$ z_{cm} = \frac{\int_0^h{a(h-z)^{(n-1)}zdz}}{\int^h_0{a(h-z)^{(n-1)}dz}}$$

where $a$ is the area/volume/hyper-volume of a disk, sphere, hypersphere and so on. We don't need to know it, it cancel. After subbing $t = h-z$ up top:

$$z_{cm} = \frac{ (h\frac{t^n}n - \frac{t^{n+1}}{n+1})|_{t=0}^{t=h} } { \frac{z^n}n|^{z=h}_{z=0}} $$

$$z_{cm} = \frac{\frac{h^{n+1}}n -\frac{h^{n+1}}{n+1}}{\frac{h^n}n} $$

$$z_{cm} =h\big [\frac{\frac 1 n - \frac 1 {n+1}}{\frac 1 n }\big] $$

$$z_{cm} = h[1-\frac n {n+1}] = h/(n+1) $$


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