# Homework Help: Moment of intertia tensor

1. May 17, 2006

### gulsen

I have an ellipse rotating around a fixed axis that passes through it's center, with a constant angular speed $$\omega$$. I'm being asked what torque would generate this motion.

It should be $$\vec{\Tau} = \frac{d(I \vec{\omega})}{dt} = \dot{I} \vec{\omega}$$ since angular speed is constant.
I'm not sure where to go. Should I try calculating inertia tensor, that is dependent to time, maybe. I tried doing so, but because it's rotating, I've quite messed it up. Any hints on calculating inertia tensor?

2. May 17, 2006

### gulsen

I hope the attachment works this time...

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3. May 17, 2006

### nrqed

I am probably missing something obvious here, but once the axis of rotation is chosen, the moment of inertia is fixed, it cannot change in time. The only way for I to change in time would be if the mass was being redistributed as a function of time (an object which is not rigid but is "morphing" into another shape!!) or if the axis of rotation changes as a function of time. Otherwise, if omega is fixed then the net torque is zero.

It would make sense to ask for the net torque required to bring the object up to that omega in a certain amount of time, but I don't see how the net torque could be anything else than zero to keep omega constant (short of the weird situations already mentioned). Unless there is some other force involved and you are asked to find the torque that must be applied in addition to keep omega constant but again, that is a different question.

4. May 17, 2006

### gulsen

Picture is still pending for approval, so let me try with words. Draw an ellipse in the x,y plane whose center is at (0,0). Now draw the line y=ax, where a is an arbitrary constant. If you image ellipse rotating around that line...
Intertia tensor would be constant around that line for some special values for a (namely $$tan(a) = n \frac{\pi}{2}$$ where n is an integer), but I'm talking about a general case.

5. May 17, 2006

### nrqed

Ok, but are you saying that *as the object is spinning*, the line is changing? What I mean is that no matter what value of "a" you pick, once you have fixed "a", the moment of inertia is fixed. Of course, the value of I depends on what value of "a" you pick, but I will not changing with respect to time if "a" is fixed.

I could understand if your question was :what is I as a function of "a". Ok, that's one question. But I don't understand the part about I changing with time. The only way to do this would be to change the axis of rotation *as the object is spinning*. If this is what you mean, fine (I don't see how to do this physically but that can be defined mathematically). But then the question is impossible to answer unless the function a(t) is provided.

Do you see what I mean? I have the weird feeling that we are talking about completely different things here.

Regards

Pat

6. May 17, 2006

### nrqed

Sorry, I realize that I was thinking about the *moment of inertia* of the object whereas you are (obviously) asking about the components of the inertia tensor. I misunderstood the question.. However, if an object is rotating at constant angular velocity, the net torque vector must be zero, I still don't see how to avoid this.

Pat

7. May 18, 2006

### gulsen

Moment of intertia at that axis is also changing, think carefully.
I have the answer (not solution), and it's non-zero.

8. May 18, 2006

### nrqed

Interesting. To me it seems that if I break up the object into small mass elements $\Delta m$ they all remain at the same distance from the axis of rotation as the object is spinning (after all I can spin myself, the observer, around the object in order to visualize the rotation of the object). So there is obviously something I am missing. I apologize for having wasted your time a bit Hope someone else will help.

Regards

Patrick

9. May 18, 2006

### arildno

1. Introduce orthonormal vectors as follows:
$$\vec{i}_{\omega}=\cos\beta\vec{i}+\sin\beta\vec{j},\vec{i}_{\eta}=-\sin\beta\vec{i}+\cos\beta\vec{j},\vec{i}_{z}=\vec{i}_{\omega}\times\vec{i}_{\eta}=\vec{k}$$
Here, $\vec{i}_{\omega}$ along the direction of the angular velocity vector with angle $\beta$ to the positive x-axis.
The motion of any particle is 2-dimensional in a plane parallell to the $\eta-z$-plane at all times.

2. We have:
$$\vec{i}=\cos\beta\vec{i}_{\omega}-\sin\beta\vec{i}_{\eta}, \vec{j}=\sin\beta\vec{i}_{\omega}+\cos\beta\vec{i}_{\eta}$$

3. At t=0, a particular particle has position as measured from the centre of the ellipse:
$$\vec{r}_{p}(0)=ar\cos\theta\vec{i}+br\sin\theta\vec{j}=(ar\cos\theta\cos\beta+br\sin\theta\sin\beta)\vec{i}_{\omega}+(br\sin\theta\cos\beta-ar\costheta\sin\beta)\vec{i}_{\eta}$$
$$0\leq{r}\leq{1},0\leq\theta\leq{2\pi}$$
Here, different r and $\theta$ values denotes distinct particles.
For a given particle, its value of r and $\theta$ does NOT change with time (body-fixed coordinates!)

4. Now, as time goes on, a particle's position component along the angular velocity vector does NOT change.
However, in the $\eta-z$ plane, its component vector is seen to go from $\vec{i}_{\eta}$ to $\vec{k}$, rotating with constant angular velocity $\omega$
Thus, we may write an arbitrary particle's position as a function of time as follows:

$$\vec{r}_{p}(t)=(ar\cos\theta\cos\beta+br\sin\theta\sin\beta)\vec{i}_{\omega}+(br\sin\theta\cos\beta-ar\costheta\sin\beta)(\cos(\omega{t}\vec{i}_{\eta}+\sin(\omega{t})\vec{k})$$

5. Now, you are ready to calculate what you're really after!
(Then both of you may find out who was correct..)

Last edited: May 18, 2006