# Precession: Deriation of formula

1. Dec 15, 2007

### JolleJ

How do they get to the formula
$$\textbf{wp}$$ = $$\frac{Q}{I*w}$$

I also note that I*w = L (angular momentum).

How do they get to that equation? I've thought and though, I just don't know how. I really hope that someone can give me a link to a deriation of it, or perhaps help me themselves.

Last edited by a moderator: Apr 23, 2017
2. Dec 15, 2007

### Staff: Mentor

Try this: http://scienceworld.wolfram.com/physics/GyroscopicPrecession.html" [Broken]

Last edited by a moderator: May 3, 2017
3. Dec 15, 2007

### JolleJ

Wow! Thank you so much.

Just one thing though: In equation (4), the d0, what is that? Change in angle? And where does that equation (4) come from? Never seen it before? Why is that true? :s

4. Dec 15, 2007

### JolleJ

I think I get it now actually. It is because the change as well as angle are inifitsmally small... :) Right? :D

5. Dec 15, 2007

### Staff: Mentor

Yes, that's it. And because the torque is perpendicular to the angular momentum.

Now that I looked it over more carefully, I'm not happy with that derivation that I linked. It seems a bit sloppy. I'll post my own version in a bit.

6. Dec 15, 2007

### JolleJ

Oh thank you so much! It's deeply appreciated.

7. Dec 15, 2007

### Staff: Mentor

I was just about to post my own derivation, when I found this on hyperphysics (one of my favorite educational sites--I highly recommend it): http://hyperphysics.phy-astr.gsu.edu/hbase/top.html" [Broken]

This is almost exactly what I would have written, so it saves me the trouble! If you have questions about this derivation, let me know.

Last edited by a moderator: May 3, 2017
8. Dec 15, 2007

### JolleJ

Thank you once more. What I need is acutally a combination of the two. The first is good because it fits the precession of a bicycle wheel (which is what I need to explain), and the next is good because of the smart derivation.

I've tried to write it down, how I think I need it. But I'm not sure, if I did any errors... I would very much like if you could just quickly look it through?
At the moment it hasn't got an illustration, I will make one however.

Last edited by a moderator: Apr 23, 2017
9. Dec 15, 2007

### JolleJ

Oh, and a last question: Is this good argumentation:
If I call the angle v. Can I say this:

"It is obvious that
$$sin(v) = \frac{\Delta L}{L}$$​
For infinitsimal small angles sin(v)=v, so that:
$$dv = \frac{dL}{L}$$​
since the angle is small when delta L is small."

Last edited: Dec 16, 2007
10. Dec 16, 2007

### Staff: Mentor

It looks OK to me. Note that your "r" is the moment arm--I assume your bicycle wheel is oriented perpendicular to the vertical?

It is certainly true that for small angles $\sin\theta \approx \theta$, as long as the angle is in radians. But you can also argue directly that as $\Delta L$ becomes small it more closely approximates the arc length of a circle, thus $\Delta L/L$ becomes the radian measure of the angle.

11. Dec 16, 2007

### JolleJ

Thank you once more! :) I think I get it now. :D