What is inertia tensor for tapered cylinder (solid and with separate inside and outside radii)?

I need the inertia tensor for tapered cylinders, both solid and hollow, and if possible with independent inner and outer radii on the $x$ and $y$ axes (so the cross-sections of the cylinders can be an ellipse). Unfortunately, computing inertia tensors is well beyond my math abilities.

In case it helps, I have the following related inertia tensors (including a couple duplicates just in case).

The Ixx, Iyy, Izz values are the 0,0 and 1,1 and 2,2 locations in the 3x3 tensor/matrix, with all other tensor/matrix elements being zero. In all the following cases the z-axis is the axis of symmetry and the center of mass is on the z-axis and in the center of the object.

Note that the center of mass of a tapered cylinder (solid or hollow) will be slightly closer to the larger end of the cylinder than the other end. Nonetheless, I need the inertia tensors with the center of mass at the axes origin (just like these other tensors below).

#####  solid cylinder  #####  a = x-radius : b = yradius : h = height (z-axis)
Ixx = ((mass*b*b)/4) + ((mass*h*h)/12)
Iyy = ((mass*a*a)/4) + ((mass*h*h)/12)
Izz = (mass*(a*a)+(b*b))/4

#####  solid cylinder  #####  a = x-radius : b = yradius : h = height (z-axis)
Ixx = (mass*((3*b*b)+(h*h)))/12
Iyy = (mass*((3*a*a)+(h*h)))/12
Izz = (mass*(a*a)+(b*b))/4

#####  hollow cylinder  #####
h = height (z-axis)
Ixx = ((mass*((bi*bi)+(bo*bo)))/4) + ((mass*h*h)/12)
Iyy = ((mass*((ai*ai)+(ao*ao)))/4)
Izz = mass*((ai*bi)+(ao*bo))/2

#####  hollow cylinder  #####
h = height (z-axis)
Ixx = (mass*(3*((bo*bo)+(bi*bi))+(h*h)))/12
Iyy = (mass*(3*((ao*ao)+(ai*ai))+(h*h)))/12
Izz = (mass*((ao*bo)+(ai*bi)))/2

• First and second Izz have parenthesis missing Apr 23 '15 at 8:50
• Do you want the inertia tensor about the center of mass, or about a given geometric position (for example, the center of one of the ends) ? Jan 23 '17 at 6:20
• I need inertia tensors around center of mass (and need equations that tell me where that is, too, for both solid and hollow tapered cylinder). Note also that my z-axis is the axis of symmetry for shapes (so, for example, the z-axis runs up the middle of a tube, or out the tip of a cone). Feb 23 '17 at 10:17

That looks lie a homework style question. So here the necessary hints. The tensors for the base shapes are listed here. It should be clear how to get the tensors if you put two objects together. The same applies if subtracting the inner cone or cylinder for a hollow object. For additional details check the parallel axis theorem

Edit

While I think this is not a calculus help site, I will try to give an almost complete solution.

The dimensions are defined in the image. Note that this is a general solution that covers basically all your cases. For $A=B=0$ it is solid and hollow otherwise. For $B=D$ and $A=C$ it is a (hollow) cylinder. It is simple to verify that for those cases the following results coincide with the special case formulae.

The volume is given by

$$V= 2 \pi \int_0^h \mathrm d z \int_{r_\mathrm i(z)}^{r_\mathrm o(z)} r \mathrm d r$$ where $$r_\mathrm i(z)= A+(C-A)\frac z h \quad r_\mathrm o(z)= B+(D-B)\frac z h$$

This results in

$$V= \frac {\pi}{ 3} \left[(B^2+BC+D^2)-(A^2+AC+C^2)\right]$$

One can see easily that for $A=B=C=0$ and $D=R$ this gives the volume of a simple cone. The centre of mass is at: $$\begin{eqnarray} z_\mathrm{cm}&=&\frac 1 V 2 \pi \int_0^h \mathrm d z \int_{r_\mathrm i(z)}^{r_\mathrm o(z)} r \mathrm d r z\\ &=&\frac h 4 \frac{(B^2+2BD+3D^2)-(A^2+2AC+3C^2)}{(B^2+BD+D^2)-(A^2+AC+C^2)} \end{eqnarray}$$Again we see that in the cone limit this gives the known result of $z_\mathrm{cm}=3 h/4$. Also note that this general formula also allows to put the tip at the top, i.e. only $B\neq 0$ in which case one sees easily that the centre of mass is at $h/4$.

Now for $I_{xx}=I_{yy}$ and $I_{zz}$ (Note, I do this in units of the body's mass, so I divide by the volume, assuming constant density $\rho$):

$$\begin{eqnarray} I_{zz}&=&\frac 1 V 2 \pi \int_0^h \mathrm d z \int_{r_\mathrm i(z)}^{r_\mathrm o(z)} r \mathrm d r (x^2+y^2)\\ &=& \frac 3 {10} \frac{(B^4+B^3D+ B^2D^2+BD^3+D^4)-(A^4+A^3C+A^2C^2+AC^3+C^4)}{(B^2+BD+D^2)-(A^2+AC+C^2)} \end{eqnarray}$$ which is in the above mentioned limit $3/10 R^2$. Finally, we have $$\begin{eqnarray} I_{xx}&=&\frac 1 V \int_0^h \mathrm d z \int_0^{2 \pi} \mathrm d \varphi\int_{r_\mathrm i(z)}^{r_\mathrm o(z)} r \mathrm d r (z^2+y^2)\\ &=& \frac 3 {30} h^2 \frac{(B^2+3BD+6D^2)-(A^2+3AC+6C^2)}{(B^2+BD+D^2)-(A^2+AC+C^2)}+\\ &&+\frac 3 {20} \frac{(B^4+B^3D+ B^2D^2+BD^3+D^4)-(A^4+A^3C+A^2C^2+AC^3+C^4)}{(B^2+BD+D^2)-(A^2+AC+C^2)} \end{eqnarray}$$ which again gives the well known factors $3/5 h^2$ and $3/20 R^2$ in case of $A=B=C=0$ and $D=R$.

Eventually, we need to check the moment in the centre of mass. Now this I will leave open for the OP. Just consider this: This calculation is obviously not in the centre of mass.$I_{zz}$ should not change as we shift along this axes. We also know that having the centre of mass solution, going to the coordinate system we have, is done by using the parallel axis theorem, i.e. $$I_{xx}=I_{xx,\mathrm{cm}}+z_\mathrm{shift}^2$$ (again in units of $m$). You may, as an alternative, reformulate the last integral.

• This is not "a homework question". I'm trying to finish the "physics subsystem" of my 3D simulation/game engine. Thus I'm much too busy to take time to "go back to school", hence my desire to simply ask the question and get an answer. Also, I don't immediately understand how the parallel axis theorem gets me from "cylinder" to "tapered cylinder". Obviously nobody has an obligation to answer my question in the form I need it, but I would appreciate such an answer if someone knows how and does so. May 14 '15 at 23:29
• @honestann sorry for the late reply. Saying it is a homework type of question does not necessarily mean it is really homework. It just refers to the type of question. Moreover, it is not common to give detailed answers to this type of question here, although sometimes it happens. May 25 '15 at 6:53
• @honestann Concerning the parallel axis theorem: this theorem tells you what the tensor of inertia is when you change the origin. The tensor usually is defined with respect to the center of mass. The theorem tells you what the tensor looks like if you rotate with respect to a different point. You probably know how to get the total tensor of two objects (sum). So your tapered cylinder is a cone with the tip missing. Get the tensor for the cone and subtract the tensor for the shifted tip. Should be rather simple. Of course you could alway make the full integration yourself. May 25 '15 at 6:59
• Thanks for the information. The problem is, I'm not a math wizard - to put it mildly. It took me enormous effort to learn enough linear algebra (vector and matrix math) to create this 3D engine in the first place (not to mention multi-level articulation, semi-fancy shaders, etc). So what is "simple" for you is near impossible for me. For all the other shapes I found the tensor equations in various 3D graphics books or math websites. I need the tensors for around the center of gravity (and yes, I need to know how to compute that too). The hollow cases are also important. Feb 19 '17 at 2:48
• Ok...usually this is not a calculus help site but let me give it a try. Feb 20 '17 at 11:27

DISCLAIMER: This is my best attempt at a solution. However, I may well have mistyped something and hence arrived at an answer that is pure garbage. Viewer discretion is advised.

The equation you want is

$I_{ij} = \iiint d^3\mathbf{x}\ \varrho(\mathbf{x}) (x_kx_k\delta_{ij} -x_ix_j)$

Assuming the body is made of a uniform material, this reduces to

$I_{ij} = \varrho \iiint d^3\mathbf{x}\ (x_kx_k\delta_{ij} -x_ix_j)$

If I have understood you correctly, the slopes of the (solid) shape are $a(z) = a_0 + \Delta a \frac{z}{h}$ and $b(z) = b_0 + \Delta b \frac{z}{h}$ where $a_0$ and $b_0$ are the dimensions of the base, $\Delta a = a_1 - a_0$, $\Delta b = b_1 - b_0$ if $a_1, b_1$ are the dimensions of the top ellipse.

The volume can then be parameterised in terms of $z$, $r$ and $\theta$ as follows: $\{ra(z)\cos(\theta),\ rb(z)\sin(\theta),\ z\},\ 0\le z \le h, 0 <r \le 1, 0 \le \theta < 2\pi$

Performing the integral for $I_{33}$ $$I_{33} = \iiint_{Pipe}dx\ dy\ dz\ (x^2 + y^2) = \int_0^hdz\int_{-a(z)}^{a(z)}dx\int_{-b(z)\sqrt{1-x^2/a(z)^2}}^{b(z)\sqrt{1-x^2/a(z)^2}}dy (x^2 + y^2)$$ (forgive my abuse of notation here)

Since this is bloody horrendous, I used Mathematica to get the following results for a body centered on the middle of the bottom face:

Integrate[x^2+z^2 , {z, 0, h}, {x, -a[z], a[z]},
{y, -b[z]*Sqrt[1 - x^2/a[z]^2], b[z]*Sqrt[1 - x^2/a[z]^2]}]


$$I_{11} = \varrho\frac{1}{240} \pi h \left(b_0 \left(2 a_0+3 a_1\right) \left(3 b_1^2+4 h^2\right)+3 b_1 \left(a_0+4 a_1\right) \left(b_1^2+4 h^2\right)+3 b_0^3 \left(4 a_0+a_1\right)+3 b_1 b_0^2 \left(3 a_0+2 a_1\right)\right)$$ $$I_{22} = \varrho\frac{1}{240} \pi h \left(a_0 \left(2 b_0+3 b_1\right) \left(3 a_1^2+4 h^2\right)+3 a_1 \left(b_0+4 b_1\right) \left(a_1^2+4 h^2\right)+3 a_0^3 \left(4 b_0+b_1\right)+3 a_1 a_0^2 \left(3 b_0+2 b_1\right)\right)$$ $$I_{33} = \varrho \frac{1}{80} \pi h \left(a_0^3 \left(4 b_0+b_1\right)+a_1 a_0^2 \left(3 b_0+2 b_1\right)+a_0 \left(a_1^2 \left(2 b_0+3 b_1\right)+4 b_0^3+3 b_1 b_0^2+2 b_1^2 b_0+b_1^3\right)+a_1 \left(a_1^2 \left(b_0+4 b_1\right)+b_0^3+2 b_1 b_0^2+3 b_1^2 b_0+4 b_1^3\right)\right)$$ $$I_{12} = I_{13} = I_{23} = 0$$ Using the formula $z_{COM} = \frac{1}{M} \iiint \varrho\ z\ dV$, I got $$z_{COM} = \frac{h \left(a_0 \left(b_0+b_1\right)+a_1 \left(b_0+3 b_1\right)\right)}{2 \left(a_0 \left(2 b_0+b_1\right)+a_1 \left(b_0+2 b_1\right)\right)}$$

Now we can fix up $I_{11}$ and $I_{12}$ with the parallel axis theorem (since they are about the point (0,0,0)): $[I_{11}]_{COM} = [I_{11}]_{offset} - Md^2$

So, the final answer is $$I_{11} = \frac{\varrho}{240} \pi h \left(b_0 \left(2 a_0+3 a_1\right) \left(3 b_1^2+4 h^2\right)+3 b_1 \left(a_0+4 a_1\right) \left(b_1^2+4 h^2\right)-\frac{10 h^2 \left(b_0 \left(a_0+a_1\right)+b_1 \left(a_0+3 a_1\right)\right){}^2}{b_0 \left(2 a_0+a_1\right)+b_1 \left(a_0+2 a_1\right)}+3 b_0^3 \left(4 a_0+a_1\right)+3 b_1 b_0^2 \left(3 a_0+2 a_1\right)\right)$$ $$I_{22} = \frac{\varrho}{240} \pi h \left(a_0 \left(2 b_0+3 b_1\right) \left(3 a_1^2+4 h^2\right)+3 a_1 \left(b_0+4 b_1\right) \left(a_1^2+4 h^2\right)-\frac{10 h^2 \left(a_0 \left(b_0+b_1\right)+a_1 \left(b_0+3 b_1\right)\right){}^2}{a_0 \left(2 b_0+b_1\right)+a_1 \left(b_0+2 b_1\right)}+3 a_0^3 \left(4 b_0+b_1\right)+3 a_1 a_0^2 \left(3 b_0+2 b_1\right)\right)$$ with the same $I_{33}, I_{12}, I_{21}, I_{13}$ as before.

To get the moment of inertia of a hollow tapered cylinder is easy once you have these formulas working - you just calculate the moment of inertia of the outside cylinder as though it were solid, then calculate the moment of inertia of the missing cylinder in the middle as if it were solid. Then subtract the two entrywise, and you have your answer! $[I]_{hollow} = [I]_{outer} - [I]_{inner}$

• Yikes... way beyond me! Can you repeat this with the center around the center of gravity? Sadly, I don't know how to compute that, either! :-( Feb 19 '17 at 3:09
• It's all there - the ones labeled as 'final answer' are about the COM, the ones before that are about the base. Feb 21 '17 at 6:24
• Hope this is right, it's a lot of typing if it's wrong :/ Feb 21 '17 at 6:25
• That is totally awesome, thanks! I'll enter this into the engine and see whether it behaves in a plausible manner. Some questions. Does the above tell me where the COM is? Are your I11, I22, I33 for x,y,z or your axis order is different? Note that all shapes are symmetric around the z-axis in my engine (the z-axis runs up the middle of the length of a tube, or out the tip of a cone). Is that consistent with your stuff, or are the I11, I22, I33 axis order different? Finally: Did you attempt to check your equations by setting taper to zero and see if result is the same as a simple tube? Feb 23 '17 at 10:13
• Honestly I was too lazy at the time, I'll do that now to see if it's sensible. Feb 26 '17 at 2:10