Textbook Error in Kleppner n' Kolenkow?

In summary, Kleppner and Kolenkow made an error when they derived Euler's equations for rigid body motion, but somehow managed to get the right answer. KandK tried something different and got the right answer. I'm really confused how what appears to be a mistake doesn't affect the final result.
  • #1
ralqs
99
1
It seems that Kleppner and Kolenkow made an error when they derived Euler's equations for rigid body motion, but they somehow managed to get the right answer, so I'm a little confused.

The customary derivation is to consider the principal axes as fixed to the rigid body, and then to transfer from this non-inertial frame to the inertial frame using the standard formula.

KandK, on the other hand, tried something different. I won't be able to explain it well in a few words, so I included pages from the text: http://tinypic.com/r/167qquh/7 and http://tinypic.com/view.php?pic=b4g3s9&s=7. At any rate, the genera; strategy is something like this: The coordinate system used in the inertial reference frame coincides with the principal axes at time t. At time t + h, where h is small, they still use the same coordinate system (ie stay in the inertial reference frame). They then evaluate the small change in angular momentum about each axis, using mostly geometric arguments. However, they appear to make a mistake: They say that "...since the components of [the tensor of inertia] are constant to first order for small angular displacement about the principal axes, [tex]\Delta (I_1 \omega _1) = I_1 \Delta \omega _1[/tex]." But this isn't true! The tensor of inertia, as I have verified, is not constant to first order; only the diagonal values are.

But he gets the right answer. I'm really confused how what appears to be a mistake doesn't affect the final result.
 
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  • #2
Ah I found a way to attach the picture directly. Sorry for the self-bump.
 

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  • #3
Hi ralqs

I am not a big fan of infinitesimal reasoning as a replacement for solid math. It can help in some instances to make results more intuitive but in other case it will just confuse the heck out of you. Infinitesimal rotations for me are almost always a recipe for a mess

The textbook line of reasoning is confusing because it talks of an inertial frame when the final equation are actually written in the non-inertial frame... I am not surprised you found a flaw in their 'demonstration'.

I would definitely go with Wikipedia in this case: http://en.wikipedia.org/wiki/Euler%27s_equations_%28rigid_body_dynamics%29" ; no infinitesimal rotations just plain old derivatives and changes of frames.
 
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  • #4
Thanks
 
  • #5
There is no flaw in the derivation. There is a bit of perhaps poor wording, the standard loosey-goosey stuff that always arises with treatments that depend on infinitesimal rotations, and a *lot* of omitted mathematics left as an exercise to the reader. That is not to say that I like this derivation. I agree with dgOnPhys that derivation based on the transport theorem is much preferred.

The transport theorem states that the time derivatives of some three dimensional vector quantity q as observed by an inertial versus rotating observer are related by

[tex]\left(\frac {d\vec q}{dt}\right)_{\text{inertial}} =
\left(\frac {d\vec q}{dt}\right)_{\text{rotating}} + \vec{\omega}\times \vec q[/tex]

The wikipedia reference on Euler's equations that dgOnPhys provided assumes the transport theorem as a given. The wikipedia page on the time derivative in rotating reference frame similarly assumes the transport theorem as a given. Proving the transport theorem is not an easy task. Undergraduate classical mechanics texts (e.g., Marion) tend to use a massive hand-wave. Even the derivation in Schaub and Junkins is still a hand-wave, just a bit longer duration one.
 
  • #6
What got me confused was that they said one thing, and did another. They said that they would stick in the inertial frame of reference, but if they did, there would be contributions to the angular momentum from the terms involving the products of inertia. And, there wouldn't be terms from the other angular momentum components. Everything they did, they did as if they were in the rotating reference frame.
 
  • #7
The authors did not say they would "stick in the inertial frame of reference." If they did everything in an inertial frame they would arrive at the rotational analog of Newton's second law,

[tex]\frac{d\vec L}{dt} = \vec{\tau}_{\text{ext}}[/tex]

There is a problem here: How does this relate to changes in angular velocity? You can't just use L=Iω because angular velocity is expressed in rotating frame coordinates, and so is the moment of inertia tensor.
 
  • #8
Hi D H

thanks for the google book reference.

I re-read the textbook pages posted at #2 and to me they are really describing motion in the inertial frame since it talks about about the principal axis of inertia moving away from the inertial axis 1,2,3... so something's fishy

In addition there is no problem calculating Euler equation in the inertial frame if one takes the time to see how perform the time derivative of the inertia tensor given its definition.

After all this is a vectorial equation so it does not matter in which frame you choose to write it. Of course the derivation won't be as painless nor the component equations as simple...
 
  • #9
D H said:
The authors did not say they would "stick in the inertial frame of reference." If they did everything in an inertial frame they would arrive at the rotational analog of Newton's second law,

[tex]\frac{d\vec L}{dt} = \vec{\tau}_{\text{ext}}[/tex]

I know, which is why I said that, although they suggested they would, they actually were looking at how angular momentum changed in the rotating frame, and then they transferred back to the inertial frame.
 

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