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In other words, does $\frac{dJ}{dt} =0$ imply $J \times \omega =0$?

Counterexamples or proofs would be helpful!

EDIT: This question originally asked if $\frac{dJ}{dt} =0 \Leftrightarrow J \times \omega =0$, but clearly the $\Leftarrow$ direction is not true.

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To clarify, my reading of your question is as follows: "If the net external torque on a rigid body is zero, then is its angular momentum parallel to its angular velocity?" Is this accurate? –  joshphysics Nov 4 '13 at 4:14
    
Yes, and the converse: if the angular momentum is parallel to the angular velocity, is the net torque zero? –  hwlin Nov 4 '13 at 4:30
    
@joshphysics must be writing up an answer. I will give you the short version in the meantime: no. –  NowIGetToLearnWhatAHeadIs Nov 4 '13 at 4:35
    
Ah I see the converse is obviously false. Changed the question to just one way. –  hwlin Nov 4 '13 at 4:36
    
@NowIGetToLearnWhatAHeadIs Not this time ;) Be my guest. –  joshphysics Nov 4 '13 at 4:39

1 Answer 1

up vote 1 down vote accepted

The canonical counterexample is when you throw a football, but you don't get a good spiral. The football is freely flying through the air so there is no torque, but the $\omega$ and $L$ are not colinear.

To explain this example and the general case I will go into the math now. The moment of inertia tensor can be written in the form $I=\left( \begin{array}{ccc} I_1 & 0 & 0 \\ 0 & I_2 & 0 \\ 0 & 0 & I_3 \\ \end{array} \right)$ in the appropriate coordinate system. Now let's suppose we throw the football so $\vec{\omega}$ is not aligned with the symmetry axis of our football, but instead has a component along the $y$ axis so the vector makes some finite angle $\theta$ with the $z$ axis. Then $\vec{L}$ also makes some non-zero angle with the $z$ axis, but since $I_2 \ne I_3$, this angle is different from $\theta$, and $\vec{L}$ and $\vec{\omega}$ are not parallel. So this is a counter example; there is no torque, but $\vec{\omega}$ and $\vec{L}$ are not colinear.

So what is the theorem? We know $\vec{\omega}(t) = I^{-1}(t) \vec{L}$. Taking the time derivative, we get $\dot{\vec{\omega}}(t) = \dot{I}^{-1}(t) \vec{L}$. But what is $\dot{I}^{-1}(t)$?

Well the rotation relating the object's initial orientation to its orientation at time $t$ can be given by an orthogonal matrix $R(t)$. Now at a time $t$ the object is rotating with angular velocity $\vec{\omega}$, so $R(t)$ satisfies the differential equation $\dot{R}(t) = \vec{\omega}^\times R$, where $\vec{\omega}^\times$ is the matrix defined by $\vec{\omega}^\times \vec{u} = \vec{\omega} \times \vec{u}$.

Now $I^{-1}(t) = R(t)I_0^{-1}R^{-1}(t) $ so $\dot{I}^{-1} = \vec{\omega}^\times R I_0^{-1}R^{-1} - R I_0^{-1}R^{-1}\vec{\omega}^\times = \vec{\omega}^\times I^{-1}(t) - I^{-1}(t) \vec{\omega}^\times$.

Then $\dot{\vec{\omega}} = \vec{\omega}^\times I^{-1}(t) \vec{L} - I^{-1}(t) \vec{\omega}^\times \vec{L} = \vec{\omega} \times \vec{\omega} - I^{-1}(t) \vec{\omega} \times \vec{L} = -I^{-1}(t) \vec{\omega} \times \vec{L}$. From this we conclude that $\dot{\vec{\omega}}$ is zero exactly when $\vec{\omega}$ is parallel to $\vec{L}$

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