# Inertia tensor problem

1. Jun 5, 2004

### Luminous Blob

I have just started learning about inertia tensors and am trying to do the problem in the attached document. I'm not sure how to calculate the moment of inertia about each axis. I would appreciate it if anyone could explain to me how it is done.

#### Attached Files:

• ###### Moment of inertia.doc
File size:
24 KB
Views:
147
Last edited: Jun 5, 2004
2. Jun 5, 2004

### turin

OK, so I'm assuming that you have a triangle in the z = 0 plane. You can use the x, y, and z -axes. The moment of inertia about any axis is the integral of:

&rho;(r) r dr

where &rho;(r) is the quasi-linear mass density (mass per incremental distance away from the axis) and r is the distance from the axis. Alternatively, you can use the parallel axis theorem.

3. Jun 6, 2004

### arildno

In your attachment, it is clearly stated that you are to find the inertia tensor.
Hence, it is insufficient to find the moments of inertia, you must also find the products of inertia.
I will assume henceforth, that we are to find the inertial tensor with respect to the origin.

1. Spin and inertial tensor:
Let a point particle with mass $$\delta{m}$$ have distance $$\vec{r}$$ to the origin, and let the particle's velocity be $$\vec{v}$$

The origin is at rest, so that we have:
$$\vec{v}=\vec{\omega}\times\vec{r}$$

Hence, the spin of the particle about the origin is given by:
$$\delta\vec{S}=\vec{r}\times\delta{m}\vec{v}= \vec{r}\times\delta{m}(\vec{\omega}\times\vec{r})=\delta{m}((\vec{r}\cdot\vec{r})\vec{\omega}-(\vec{r}\cdot\vec{\omega})\vec{r})$$
by a common vector identity.
With index notation, we have:
$$\delta{S}_{i}=\delta{I}_{ij}\omega_{j},\delta{I}_{ij}=\delta{m}(r_{k}^{2}\delta_{ij}-r_{i}r_{j})$$

where $$\delta_{ij}$$ is the Kronecker delta, $$1\leq{i,j,k}\leq3$$, and Einstein's summing convention has been adopted.

$$\delta{I}_{ij}$$ is the inertial tensor associated with the given particle.

Since we have a rigid body, the body's inertial tensor with respect to the origin, $$I_{ij}$$ is simply the sum of the point particles' associated
inertial tensors, i.e.,
$$I_{ij}=\int\delta{I}_{ij}$$

Note in particular, that the diagonal elements i=j in the inertia tensor are simply the moments of inertia about the axes $$\vec{i}_{1},\vec{i}_{2},\vec{i}_{3}$$

It is seen that the inertial tensor is symmetric; the off-diagonal elements are called products of inertia, and are, for example, important in wobbling/instability phenomena.

Calculations follow in the next post.

4. Jun 6, 2004

### arildno

The figure is a triangle with vertices:
$$((0,0,0),(\frac{L}{2},L,0),(L,0,0))$$

Any point $$\vec{r}$$ on the triangle may be written as:

$$\vec{r}(u,v)=((-\frac{L}{2},L,0)u+(L,0,0))v, 0\leq{u,v}\leq{1}$$

Clearly, since $$z=r_{3}\equiv{0}$$, we only need to calculate:
$$I_{11},I_{22},I_{33},I_{21} (x=r_{1},y=r_{2})$$

the particle mass satisfy $$\delta{m}=\rho{dA}$$, whereas the proper area element dA is given by:

$$dA=||\frac{\partial\vec{r}}{\partial{u}}\times\frac{\partial\vec{r}}{\partial{v}}||dudv$$

This should give you the necessary info to solve the problem.

5. Jun 6, 2004

### turin

arildno,
Just out of curiosity, why are you calling the orbital angular momentum of a particle "spin?"

6. Jun 7, 2004

### arildno

Oh dear, that's what we call it in Norwegian..