Calculating the Inertia Tensor of cone with uniform density

In summary, the problem involves calculating the moments of inertia for a homogeneous cone using the parallel-axis theorem and the given geometric properties. The inertia tensor is found to have two degenerate eigenvalues and one distinct eigenvalue due to the symmetry of the cone. The calculations can be simplified by stacking circular disks of varying radius along the z-axis and using the parallel-axis theorem to refer the moments of each disk to the apex of the cone.
  • #1
Wavefunction
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Homework Statement



Calculate the moments of inertia [itex]I_1[/itex], [itex]I_2[/itex], and[itex] I_3[/itex] for a homogeneous cone of mass [itex] M [/itex] whose height
is [itex]h [/itex] and whose base has a radius [itex] R[/itex]. Choose the [itex] x_3[/itex] axis along the axis of symmetry of the cone.
Choose the origin at the apex of the cone, and calculate the elements of the inertia tensor. Then
make a transformation such that the center of mass of the cone becomes the origin, and find the
principal moments of inertia. Hint: Steiner's parallel-axis theorem might be useful for the
second part of the problem.

Homework Equations



[itex]J_{ij}=\int_{M}\rho\delta_{ij}\sum_{k}x_{k}^2-x_{i}x_{j}dM [/itex]

[itex] I_{ij}=J_{ij}-M\sum_{k}a_{k}^2-a_{i}a_{j}[/itex]

The Attempt at a Solution



The region I'll be integrating over will be: [itex] 0\leq\theta\leq 2\pi [/itex], [itex] 0\leq r\leq R[/itex], and [itex]0\leq z \leq rcot(\alpha) [/itex] such that [itex] x_1=rcos(\theta) [/itex], [itex] x_2 = rsin(\theta) [/itex], and [itex] x_3 = z [/itex]. Also [itex]\alpha[/itex] is the angle between the z-axis and the boundary of the cone.

Case 1) [itex]\delta_{ij} = 1\rightarrow J_{ii}=\rho\int_V\sum_{k}x_{k}^2-x_{i}^2dV [/itex]

[itex] J_{ii}=\rho\int_{0}^{2\pi}\int_{0}^{R}\int_{0}^{rcot(\alpha)}[\sum_{k}x_{k}^2-x_{i}^2]dzrdrd\theta [/itex]

[itex] i=1 \rightarrow J_{11} = \rho\int_{0}^{2\pi}\int_{0}^{R}\int_{0}^{rcot(\alpha)}[x_{2}^2+x_{3}^2]dzrdrd\theta[/itex]

[itex] J_{11} = \rho\int_{0}^{2\pi}\int_{0}^{R}\int_{0}^{rcot(\alpha)}[(rsin(\theta))^2+(z)^2]dzrdrd\theta = MR^2\frac{2cot^2(\alpha)+3}{10cot(\alpha)} [/itex]

[itex] i=2 \rightarrow J_{22} = \rho\int_{0}^{2\pi}\int_{0}^{R}\int_{0}^{rcot(\alpha)}[x_{1}^2+x_{3}^2]dzrdrd\theta[/itex]

[itex] J_{22} = \rho\int_{0}^{2\pi}\int_{0}^{R}\int_{0}^{rcot(\alpha)}[(rcos(\theta))^2+(z)^2]dzrdrd\theta = MR^2\frac{2cot^2(\alpha)+3}{10cot(\alpha)}[/itex]

[itex] i=3 \rightarrow J_{33} = \rho\int_{0}^{2\pi}\int_{0}^{R}\int_{0}^{rcot(\alpha)}[x_{1}^2+x_{2}^2]dzrdrd\theta[/itex]

[itex] J_{33} = \rho\int_{0}^{2\pi}\int_{0}^{R}\int_{0}^{rcot(\alpha)}[(rcos(\theta))^2+(rsin(\theta))^2]dzrdrd\theta = \rho\int_{0}^{2\pi}\int_{0}^{R}\int_{0}^{rcot(\alpha)}[(r)^2]dzrdrd\theta = \frac{3MR^2}{5}[/itex]

Case 2 [itex] \delta_{ij}=0 \rightarrow J_{ij}=\rho\int_V[-x_{i}x_{j}]dV = \rho\int_{0}^{2\pi}\int_{0}^{R}\int_{0}^{rcot(\alpha)}[-x_{i}x_{j}]dzrdrd\theta[/itex]

[itex] i=1,j=2 \rightarrow J_{12} = \rho\int_{0}^{2\pi}\int_{0}^{R}\int_{0}^{rcot(\alpha)}[-r^2cos(\theta)sin(\theta]dzrdrd\theta = 0 [/itex]

[itex] i=1,j=3 \rightarrow J_{13} = \rho\int_{0}^{2\pi}\int_{0}^{R}\int_{0}^{rcot(\alpha)}[-rcos(\theta)z]dzrdrd\theta = 0 [/itex]

[itex] i=2,j=3 \rightarrow J_{23} = \rho\int_{0}^{2\pi}\int_{0}^{R}\int_{0}^{rcot(\alpha)}[-rsin(\theta)z]dzrdrd\theta = 0 [/itex]

so the Inertia tensor is then:

[itex]
J = MR^2\begin{pmatrix}
\frac{2cot^2(\alpha)+3}{10cot(\alpha)} & 0 & 0 \\
0 & \frac{2cot^2(\alpha)+3}{10cot(\alpha)} & 0 \\
0 & 0 & \frac{3}{5}
\end{pmatrix}
[/itex]

Now using Steiner's parallel axis thereom:

[itex] I_{ij}=J_{ij}-M\delta_{ij}\sum_{k}a_{k}^2-a_{i}a_{j}[/itex] I'll need the components of [itex] \vec{a} [/itex] which is the vector that points from the center of mass to the origin. So I'll need to find the center of mass of the cone using: [itex] R_{i}=\frac{1}{M}\int_{M}[x_{i}]dM[/itex]. By the symmetry of the cone I already know the center of mass will have have coordinates [itex](0,0,b)[/itex]. So I just have to find the value for [itex] x_3 [/itex]. First the total mass of the cone [itex] M [/itex] is:
[itex]M=\rho\int_{0}^{2\pi}\int_{0}^{R}\int_{0}^{rcot(\alpha)}[1]dzrdrd\theta = \frac{2R^3\pi}{3}cot(\alpha)[/itex]

[itex] \int_{M}[x_{i}]dM = \rho\int_{0}^{2\pi}\int_{0}^{R}\int_{0}^{rcot(\alpha)}[z]dzrdrd\theta = \frac{R^4\pi}{4}cot^2(\alpha) [/itex] so the center of mass is located at [itex] (0,0,\frac{3}{8}Rcot(\alpha)) [/itex]

Case 1) [itex] \delta_{ij}=1 \rightarrow I_{ii} = J_{ii} -M\sum_{k}a_{k}^2-a_{i}^2 [/itex]

[itex] i=1 \rightarrow I_{11}=J_{11} -M[a_{2}^2+a_{3}^2] \rightarrow I_{11}=J_{11}-M\frac{9}{64}R^2cot^2(\alpha) [/itex]

[itex] i=2 \rightarrow I{22}=J_{22} -M[a_{1}^2+a_{3}^2] \rightarrow I_{22}=J_{22}-M\frac{9}{64}R^2cot^2(\alpha) [/itex]

[itex] i=3 \rightarrow I{33}=J_{33} -M[a_{1}^2+a_{2}^2] \rightarrow I_{33}=J_{33}-M[0]^2 [/itex]

Case 2) [itex] \delta_{ij}=0 \rightarrow I_{ij} = J_{ij} -M[-a_{i}a_{j}][/itex] . Since the only non-zero component is [itex]a_3 M[a_{i}a_{j}]=0[/itex] so [itex]I_{ij}=J_{ij}[/itex] for[itex] j≠i [/itex]. Then the inertia tensor for the center of mass is:

[itex]
I = MR^2\begin{pmatrix}
\frac{2cot^2(\alpha)+3}{10cot(\alpha)}-\frac{9}{64}cot^2(\alpha) & 0 & 0 \\
0 & \frac{2cot^2(\alpha)+3}{10cot(\alpha)}-\frac{9}{64}cot^2(\alpha) & 0 \\
0 & 0 & \frac{3}{5}
\end{pmatrix}
[/itex]

Finally the inertia tensor has two degenerate eigenvalues (principal moments of inertia) and one distinct eigenvalue:

[itex] I_1=I_2 = [\frac{2cot^2(\alpha)+3}{10cot(\alpha)}-\frac{9}{64}cot^2(\alpha)]MR^2 [/itex]

[itex] I_3 = \frac{3}{5}MR^2[/itex]

My result checks out with everything taught in lecture as far as getting the two degenerate eigenvalues and one distinct one due to the symmetry of the cone, but this is the first type of problem like this that I have done so I would appreciate any help ya'll can give me (: thanks.
 
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  • #2
Boy, that looks like a lot of work!

There are some things you could have done to make the calculations easier. Writing out and evaluating volume integrals at each step is a bit much. An easier approach, I think, would be to stack circular disks of varying radius along the z-axis. Once you calculate the MOI of a disk of radius r and thickness dz, that eliminates much of your calculation. You can use the parallel axis theorem to refer the moments of each disk to the apex of the cone.

I think instead of expressing the elements of the inertia tensor in terms of the angle α, you should use the given geometric properties of the cone, like the radius of the base, r, and the height, h. That way, you can check your results against tables of moments of inertia:

http://en.wikipedia.org/wiki/List_of_moments_of_inertia#List_of_3D_inertia_tensors
 
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  • #3
SteamKing said:
Boy, that looks like a lot of work!

There are some things you could have done to make the calculations easier. Writing out and evaluating volume integrals at each step is a bit much. An easier approach, I think, would be to stack circular disks of varying radius along the z-axis. Once you calculate the MOI of a disk of radius r and thickness dz, that eliminates much of your calculation. You can use the parallel axis theorem to refer the moments of each disk to the apex of the cone.

I think instead of expressing the elements of the inertia tensor in terms of the angle α, you should use the given geometric properties of the cone, like the radius of the base, r, and the height, h. That way, you can check your results against tables of moments of inertia:

http://en.wikipedia.org/wiki/List_of_moments_of_inertia#List_of_3D_inertia_tensors

I understand the direction you're going with this and I like it, but I can't reconcile the following:
how can I integrate [itex] \delta_{ij}\sum_{k}x_{k}^2-x_ix_j [/itex] only with respect to [itex] z [/itex]. I understand how the volume will look: [itex] V=\int_{0}^{h}[\pi(\frac{Rz}{h})^2]dz [/itex], but as I said I can't reconcile that with the definition of the inertia tensor.
 
  • #4
Wavefunction said:
I understand the direction you're going with this and I like it, but I can't reconcile the following:
how can I integrate [itex] \delta_{ij}\sum_{k}x_{k}^2-x_ix_j [/itex] only with respect to [itex] z [/itex]. I understand how the volume will look: [itex] V=\int_{0}^{h}[\pi(\frac{Rz}{h})^2]dz [/itex], but as I said I can't reconcile that with the definition of the inertia tensor.

While the notation defining the components of the inertia tensor is remarkably compact, some degree of clarity is lost when trying to compute the inertia tensor for certain geometric figures, like those having symmetry, which qualities can be exploited to reduce the overall amount of calculation.

Unless you are some sort of mathematical savant, it takes a certain amount of 'decrypting' the notation [itex] \delta_{ij}\sum_{k}x_{k}^2-x_ix_j [/itex] in order to find simplifications. Sometimes, it pays to do a little analysis of the problem before charging headlong into evaluating various integral expressions.
 
  • #5
SteamKing said:
While the notation defining the components of the inertia tensor is remarkably compact, some degree of clarity is lost when trying to compute the inertia tensor for certain geometric figures, like those having symmetry, which qualities can be exploited to reduce the overall amount of calculation.

Unless you are some sort of mathematical savant, it takes a certain amount of 'decrypting' the notation [itex] \delta_{ij}\sum_{k}x_{k}^2-x_ix_j [/itex] in order to find simplifications. Sometimes, it pays to do a little analysis of the problem before charging headlong into evaluating various integral expressions.

Well I guess that's why he gives us the problems he gives us right? which have been helping a lot along with all the help that I get here, but with this problem I think we are suppose to crank through the integrals because we might be given something that does not have nice symmetry like a cone, but I do see I shouldn't have defined [itex] \alpha [/itex] I just redid [itex] J_{11} [/itex] and I got [itex] J_{11} = \frac{3MR^3}{10}[\frac{1}{R}+\frac{2h^2}{R^3}] [/itex]. So I think that when I calculated everything out the first time I was a bit careless when plugging into the inertia tensor.
 
  • #6
Well, usually you are doing these problems for some known geometric shape, which is, after all, what most things are composed of. If you get a truly random object, which cannot be decomposed into simpler pieces, that would be remarkable itself.
 
  • #7
SteamKing said:
Well, usually you are doing these problems for some known geometric shape, which is, after all, what most things are composed of. If you get a truly random object, which cannot be decomposed into simpler pieces, that would be remarkable itself.

Okay well let's go back to what you were saying about the disk method. As I said, I understand how to get the volume with such a method and I did some thinking about it and this is what I came up with:

[itex] J_{11} = \rho\int_{V}\sum_{k}x_k^2-x_i^2dV \rightarrow J_{11} = \rho\int_{V}[x_1^2+x_2^2+x_3^2-x_1^2]dV [/itex] the volume of one thin disk is: [itex] dV=\pi[r(z)]^2dz [/itex]. Then I get [itex] J_{11}=\rho\int_{0}^{h}[x_2^2+x_3^2]\pi[r(z)]^2dz \rightarrow J_{11}=\rho\int_{0}^{h}[x_2^2+z^2]\pi[r(z)]^2dz[/itex] So I'm just kind of stuck on how to parameterize [itex] x_2 [/itex].
 

Related to Calculating the Inertia Tensor of cone with uniform density

1. What is the definition of the inertia tensor?

The inertia tensor is a mathematical representation of an object's resistance to changes in rotational motion. It takes into account the distribution of mass within an object and its orientation in space.

2. How is the inertia tensor calculated for a cone with uniform density?

The inertia tensor of a cone with uniform density can be calculated using the formula I = (1/20)mr^2, where I is the inertia tensor, m is the mass of the cone, and r is the radius of the base of the cone.

3. Can the inertia tensor of a cone with non-uniform density be calculated in the same way?

No, the formula for calculating the inertia tensor of a cone with uniform density only applies when the density is constant throughout the object. For a cone with non-uniform density, the inertia tensor must be calculated using the integral of the mass distribution over the volume of the cone.

4. What are the units of the inertia tensor?

The units of the inertia tensor are kg*m^2. This represents the moment of inertia, which is a measure of an object's rotational inertia.

5. How is the inertia tensor used in physics and engineering?

The inertia tensor is used in the study of rotational motion and dynamics, as well as in various engineering applications such as designing vehicles and aircraft. It can also be used to determine the stability and balance of an object.

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