- #1

Wavefunction

- 99

- 4

## 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.