1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Calculating the Inertia Tensor of cone with uniform density

  1. Apr 20, 2014 #1
    1. The problem statement, all variables and given/known data

    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.


    2. Relevant 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]

    3. 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.
     
  2. jcsd
  3. Apr 20, 2014 #2

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    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
     
  4. Apr 20, 2014 #3
    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.
     
  5. Apr 20, 2014 #4

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    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.
     
  6. Apr 20, 2014 #5
    Well I guess thats 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.
     
  7. Apr 20, 2014 #6

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    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.
     
  8. Apr 20, 2014 #7
    Okay well lets 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].
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Calculating the Inertia Tensor of cone with uniform density
  1. Inertia tensor (Replies: 2)

  2. Inertia tensor (Replies: 2)

Loading...