Precession: Deriation of formula

[JolleJ]

Please see this link first: http://en.wikipedia.org/wiki/Precession#Classical_.28Newtonian.29"

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.

Thanks in advance.

 

[Doc Al]

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

 

[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

 

[JolleJ]

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

 

[Doc Al]

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.

 

[JolleJ]

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

 

[Doc Al]

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"

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.

 

[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?
Here's the link: http://peecee.dk/?id=85095" (danish upload site: Click "Download fil").
At the moment it hasn't got an illustration, I will make one however.

Thanks in advance, once again.

 

[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."

Thanks in advance.

 

[Doc Al]

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?

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

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."

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.

 

[JolleJ]

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