# Decomposing angular velocity and moment of inertia

1. Apr 12, 2015

### Wminus

1. The problem statement, all variables and given/known data
See the attached image.

2. Relevant equations
$T = 1/2 \omega^2 I$

3. The attempt at a solution
So $I_1,I_2,I_3$ are just the moments of inertia of the object with regards to the 3 axes. Right? OK, then I intuitively assume that the total kinetic energy must simply be $$T=\frac{1}{2} (I_1\omega_1^2+I_2\omega_2^2+I_3\omega_3^2)$$ since the kinetic energy around axis $i$ is just $I_i\omega_i^2$. But I can't seem to prove it.. Can you guys help me?

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• ###### Skjermbilde 2015-04-12 kl. 14.24.04.png
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2. Apr 12, 2015

### robphy

Can you show what you have done thus far?

3. Apr 13, 2015

### Wminus

It's nothing of interest.. My thinking is that each of the $\omega_i^2$ with $i \in \{1,2,3\}$ multiplied by their moment of inertia $I_i$ will contribute to the rotational kinetic energy linearly, so you can just add them all up to get the total $$T=\frac{1}{2} (I_1\omega_1^2+I_2\omega_2^2+I_3\omega_3^2)$$. It makes intuitive sense, but I can't seem to prove it.

4. Apr 13, 2015

### Telemachus

Yes, thats right. Thats the kinetic energy due to rotation for the principal moments of inertia. If you consider other axis of rotations but the principals, then that expression wouldn't be valid, think that the moment of inertia is a second order tensor, and has 9 components, not three. When you compute the rotational kinetic energy with respect to other moment of inertia but the principal, the expression for the rotational kinetic energy becomes quiet cumbersome. It is reduced to that simple expression when you compute it for the principal axes of inertia, the inertia tensor is diagonal in the principal axes. I think the exercise doesn't asks you to prove that formula, just to demonstrate that the kinetic energy reduces to formula (4), you just have to replace in the formula for the kinetic energy the given values for $\omega_i$. If what you want is to proove that the kinetic energy is given by that formula, you could instead of taking a continuous body, start by thinking of a discrete (rigid) distribution of puntual masses and add the kinetic energy of rotation for all of them. And then get the expressions for the continuum. Then it will appear the expression for the moment of inertia naturally. But that formula is only valid when the moment of inertia is taken with respect to the principal axes of inertia.

Last edited: Apr 13, 2015