# Homework Help: Rotating uniform rod about an arbitrary axis

1. Nov 18, 2016

### Dustgil

1. The problem statement, all variables and given/known data
A thin uniform rod of length l and mass m is constrained to rotate with constant angular velocity omega about an axis passing through the center O of the rod and making an angle alpha with the rod.
a) Show the the angular momentum L about O is perpendicular to the rod and is of magnitude
$$\frac {ml^{2} \omega} {12} sin(\alpha)$$

b) Show that the torque vector N is perpendicular to the rod and to L and is of magnitude
$$\frac {ml^{2} \omega^{2}} {12} sin(\alpha) cos(\alpha)$$

2. Relevant equations
I think these two:

$$L = I_{x} \omega_{x} i + I_{y} \omega {y} j + I_{z} \omega {z} k$$

$$\frac {dL} {dt} = N$$

3. The attempt at a solution

This chapter has me really confused at the moment. Where do I define the axes? I tried assigning the k axis to be along the length of the rod. then L should be perpendicular to the rod because the moment of inertia is zero in the k direction (because every point on the rod is at zero distance from that axis). The moment of inertia about the i and j axes is
$$I = \frac {ml^{2}} {12}$$

This is where i get lost. I'm not sure what the component of omega is in the x and y directions. Soooo not sure if I'm doing this right.

2. Nov 18, 2016

### haruspex

Consider the plane containing both rod and axis at some instant. Which way is the instantaneous velocity of a point on the rod?

3. Nov 19, 2016

### Dustgil

the instantaneous velocity should be perpendicular to the plane containing both the rod and the axis. if the plane coincides with the computer screen, it will be into the screen if the rotation is counterclockwise and out of the screen if clockwise.

4. Nov 19, 2016

### haruspex

That is not quite right. Can you correct that?

5. Nov 19, 2016

### Dustgil

Yeah, I see. half of it is rotating into the screen and half is rotating out. the piece at the center of mass isn't translating, just rotating.

6. Nov 19, 2016

### haruspex

Good. So can you figure out the direction of the angular momentum vector?

7. Nov 19, 2016

### Dustgil

Well it's perpendicular to the rod. My guess for why is that the two halves cancel and we just consider the piece thats not translating. I'm not sure if that fully answers your question.

8. Nov 19, 2016

### haruspex

No, they don't cancel. This is angular momentum, not linear.
Try to forget the problem statement for the moment and just consider what you deduced about the instantaneous velocities. How does the linear velocity of a point on the rod vary with its distance from the centre? What axis would you think the rod was rotating about?

9. Nov 19, 2016

### Dustgil

the linear velocity grows as you increase distance from the center. Going from that statement alone it implies that the axis of rotation is through the center of mass.

10. Nov 19, 2016

### haruspex

Yes, but you can say more than that. Consider a pair of points on the rod equidistant from the centre. Taken as a pair, what axis do they appear to be rotating about, instantaneously?

11. Nov 19, 2016

### Dustgil

Ah, that's an axis at an angle alpha to the whole rod.

12. Nov 19, 2016

### haruspex

That's not what I see.
Consider two equal particles at (-1,-1) and (1,1). One is heading straight into the page and one straight out. Taken as a pair, what axis is the pair rotating about at that instant? I.e., which axis maximises the angular momentum?

13. Nov 20, 2016

### Dustgil

I've thought about this for a while. I tried to find some visualizations to help me through it but I couldn't locate any online. I suppose it has to be perpendicular, But if the path of each particle traces a circle out in space due to rotation, the center of each circle is an axis at an angle alpha to the whole rod. instantaneously, maybe it's perpendicular...

14. Nov 20, 2016

### haruspex

That's the point - the axis of rotation keeps moving. If you just look at the instantaneous velocities of points along the rod you cannot distinguish its movement from rotating about an axis perpendicular to the rod. Since that choice maximises the moment, that is what it is doing.
The rotation of the rod about the axis at angle alpha is a second order phenomenon involving accelerations.

15. Nov 20, 2016

### TSny

One approach is to recall that the angular momentum of any system of particles about a point O is a sum over the particles of the form $\vec{L} = \sum_i \vec{r}_i \times \vec{p}_i$ where $\vec{r}_i$ is the position of the ith particle relative to O. The sum is replaced by an integral for continuous mass distributions. Using this formula you can easily see the direction of $\vec{L}$.

Alternately, the answer is easily determined from your formula $\vec{L} = I_x \omega_x \hat{i} + I_y \omega_y \hat{j}+I_z \omega_z \hat{k}$ if you interpret it correctly. Have you covered the notion of moment of inertia tensor and principal axes?

Last edited: Nov 20, 2016
16. Nov 20, 2016

### Dustgil

We have, I'm just unsure of how to apply it. I'd imagine one of the principal axes is along the length of the rod, correct?

17. Nov 20, 2016

### TSny

Yes. The formula assumes that you choose your coordinate axes to align with principal axes of the rod. So, you'll need to choose one of the coordinate axes to be along the length of the rod.

It can be a little confusing because the rod is changing its orientation relative to the inertial "lab" frame. But for a specific instant of time, you can choose the coordinate axes so that one of the axes is along the length of the rod.

Last edited: Nov 20, 2016
18. Nov 20, 2016

### Dustgil

so if i choose the z axis to be along the length of the rod then will omega work out like this?

$$\omega_{x} = cos(\alpha)$$
$$\omega_{y} = sin(\alpha)$$

This was the problem I was really having with the route. I couldn't understand the geometry well enough to break omega in components.

19. Nov 20, 2016

### TSny

No. If you choose the z axis along the rod, wouldn't $\vec{\omega}$ have a nonzero z component? The x and y components of $\vec{\omega}$ would depend on your choice of orientation of the x and y axes. Choose the orientation to simplify the calculation.

20. Nov 20, 2016

### Dustgil

I guess you're suggesting that omega would go to zero along either the x or y axis. the moment of inertia around the z axis is zero so that would leave one component. I think you can choose the axes so that either x or y is orthogonal to the direction of omega. Is this the correct track?

21. Nov 20, 2016

### TSny

Yes.

One component of what?

22. Nov 20, 2016

### Dustgil

of the angular momentum.

23. Nov 20, 2016

### TSny

OK, yes.

24. Nov 20, 2016

### Dustgil

Okay. I have to go in to work now but the second part should just involve taking the derivative of the angular momentum, right? Not sure how that shakes out to both sin and cos in the term.

25. Nov 20, 2016

### TSny

Yes, that's right (the time derivative).