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.

  • $\begingroup$ 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? $\endgroup$ – joshphysics Nov 4 '13 at 4:14
  • $\begingroup$ Yes, and the converse: if the angular momentum is parallel to the angular velocity, is the net torque zero? $\endgroup$ – hwlin Nov 4 '13 at 4:30
  • $\begingroup$ @joshphysics must be writing up an answer. I will give you the short version in the meantime: no. $\endgroup$ – Brian Moths Nov 4 '13 at 4:35
  • $\begingroup$ Ah I see the converse is obviously false. Changed the question to just one way. $\endgroup$ – hwlin Nov 4 '13 at 4:36
  • $\begingroup$ @NowIGetToLearnWhatAHeadIs Not this time ;) Be my guest. $\endgroup$ – joshphysics Nov 4 '13 at 4:39

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 $\vec{L}$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.