# Inertia Tensor of a cylinder at a distance

1. Jun 26, 2007

### pinodk

I have a cylinder, for which i want to find the inertia tensor.

Where the rotational axis are either the x (red) or y (green).

I know that the inertia tensor for a cylinder is of the form

Then I believe that the bottom right element stays the same, since this describes the rotation around the z-axis.
The tricky part for me is the rest of the matrix. I am no expert, and do not understand inertia tensors fully, so I would like some pointers.

My immediate idea is that the matrix should remain in its diagonal form, the zeros will remain zeros, is this correct?

I know that for complex forms i can split up the moments of inertia, so i have the moment of inertia for the blank space d, which is 0. and then i can add the moment of inertia of the cylinder, but how do i calculate this, when the rotational axis is x-axis for example?

2. Jun 26, 2007

### pinodk

Oops, i should have posted this in the homework section,, dunno how to move it, so will repost there.

3. Jun 26, 2007

### smallphi

http://en.wikipedia.org/wiki/Moment_of_inertia

contains section 'Parallel axes theorem' at the bottom of the page where they show you how to obtain the tensor for any point of rotation that is offset from the center of mass once you know the tensor for origin at the center of mass.

Keep in mind the axes of the two coordinate systems, one at center of mass and one at your point of choice, must remain parallel, hence the name of the theorem.

4. Jul 8, 2007

### pinodk

I have looked at the formula listed, but i dont know quite how to use it...

Suppose i have an inertia tensor in the center of mass like this
$$\left(\begin{array}{c c c} 1&0&0\\0&1&0\\0&0&1\end{array}\right)$$
The mass m is 2 kg, and the distance vector R is
$$\left(\begin{array}{c} 1\\0\\0\end{array}\right)$$

what i thought of doing was
$$I+m\cdot R^2= \left(\begin{array}{c c c} 3&2&2\\0&1&0\\0&0&1\end{array}\right)$$
But when i look at the formula i read it as
the jk'th of I + M*((R dot R when j equals k) -(the j'th of R times the k'th of R))
for example j=1, k=1, i get
1 + 2* (1-1)=1
for j=2, k=3
0+2* (0-0)=0
and so forth. The M*(R... part always yields 0! what am i misunderstanding?

5. Jul 8, 2007

### pinodk

Never mind... i got it now... stupid me