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I have seen 2 formulas: $\tau = I \alpha$ and $\tau = I \dot\omega + \omega \times I \omega$. Which one shall I use in which case please? Also, am I right to think that $\alpha$ from the first formula is $\dot\omega$ from the second one?

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    $\begingroup$ It feels like the first equation applies when you have a solid rotating body. I am however not able to reproduce the second equation. Where does it come from? Are you sure that it's correct? You are right usually $\alpha = \dot{\omega}$. $\endgroup$ Oct 10 '14 at 10:20
  • $\begingroup$ @StevenMathey: I got the second equation from andrew.gibiansky.com/blog/physics/quadcopter-dynamics (long article, search for "We derive the rotational equations") and en.wikipedia.org/wiki/Moment_of_inertia for the resultant torque of a system of particles (is a system of particles a rigid body?). I understand the first equation, it seems to match my naive observations. I would like to understand if there are cases where I should use the second one. $\endgroup$
    – marcv81
    Oct 10 '14 at 11:09
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    $\begingroup$ A system of particles is not necessarily a rigid body, but a rigid body is a system of particles. $\endgroup$ Oct 10 '14 at 18:48
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Yes, $\alpha = \dot{\omega}$, being the angular acceleration. The first equation is special case of the second equation.

For a general object the moment of inertia is not just a scalar (a single value) but a tensor, in that case you have to use your second equation. $\tau$ and $\omega$ are then vectors and $I$ is a 3x3 matrix.

But when you spin an object around one of its high symmetry axes (one of the eigenvectors of the inertia matrix $I$), the equation simplifies to your first equation. Proof:

If $\vec{\omega}$ is an eigenvector of $\hat{I}$ it holds that: $\hat{I} \vec{\omega} = \lambda \space \vec{\omega} = \vec{\omega} \space \lambda$ Therefore your second equation becomes: $\vec{\tau} = \hat{I}\dot{\omega} + \vec{\omega} \times \vec{\omega} \space \lambda$ and $\vec{\tau} = \hat{I}\dot{\omega} + 0$ since the crossproduct of a vector with itself is 0.

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  • $\begingroup$ Brilliant answer, thanks. Exactly what I was looking for, no more no less. $\endgroup$
    – marcv81
    Oct 10 '14 at 21:18
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I have seen 2 formulas: $\tau = I \alpha$ and $\tau = I \dot\omega + \omega \times I \omega$. Which one shall I use in which case please?

You use the first formula in introductory physics classes on specially constructed problems that avoid the issues connected with the second formula, and also in practical settings that by design and operation make the $\vec \omega \times \mathrm I \vec \omega$ term very small. For example, its a good idea to keep the the wheels on your car balanced so as to keep the issues connected with that second term to a barely detectable minimum.

You use the second formula in more advanced settings when you are reasoning from the perspective of a reference frame fixed with respect to the rotating object. This is a rotating frame, and rotating frames result in fictitious forces and fictitious torques. One way to look at that $\vec \omega \times \mathrm I \vec \omega$ term is that it is a fictitious torque.

There is a third option, which is to reason from the perspective of an inertial frame. The rotational analog of Newton's second law results in $\vec \tau = \dot {\vec L}$. Since $\vec L=\mathrm I \vec \omega$, the rotational analog of Newton's second law becomes $\tau = \mathrm I \dot {\vec \omega} + \dot {\mathrm I} \vec \omega$. The second term, $\dot {\mathrm I} \vec \omega$, arises because the inertia tensor in general is not constant from the perspective of an inertial frame. Without derivation, the time derivative of the inertia tensor from the perspective of an inertial frame is the commutator $[\operatorname{Sk}(\vec \omega), \mathrm I] = \operatorname{Sk}(\vec \omega) \mathrm I - \mathrm I \operatorname{Sk}(\vec \omega)$ where $\operatorname{Sk}(\vec \omega)$ is the skew symmetric cross product matrix generated from the vector $\vec \omega$.

This suggests that there's more to that $\vec \omega \times \mathrm I \vec \omega$ term than just a fictitious torque. Expanding the inertial frame term $\dot {\mathrm I} \vec \omega$ results in $\vec \tau = \mathrm I \dot {\vec \omega} + \operatorname{Sk}(\vec \omega) \mathrm I \vec \omega - \mathrm I \operatorname{Sk}(\vec \omega) \vec \omega$. The final term, $\mathrm I \operatorname{Sk}(\vec \omega) \vec \omega$, vanishes, leaving just $\vec \tau = \mathrm I \dot {\vec \omega} + \operatorname{Sk}(\vec \omega) \mathrm I \vec \omega = \mathrm I \dot {\vec \omega} + \omega \times \mathrm I \vec \omega$, which is identical to the rotating frame formulation.

Hardly anyone uses the inertial frame POV because having a time-varying inertia tensor is just too nasty.

So how does that rotating frame formulation arise? The answer lies in what some call the kinematics transport theorem, $$\left(\frac {d\vec q} {dt}\right)_I = \left(\frac {d\vec q} {dt}\right)_R + \vec \omega \times \vec q$$ where $\vec q$ is any vectorial quantity in $\mathbb R^3$, the subscripts $I$ and $R$ on the derivatives denote the derivatives from the perspective of an inertial frame versus a rotating frame, and $\vec \omega$ is the angular velocity of the rotating frame with respect to the inertial frame. Substituting $\vec q$ with $\vec L = \mathrm I \vec \omega$ yields $$\vec \tau = \dot {\vec L}_I = \dot {\vec L}_R + \vec \omega \times \vec L = \mathrm I \dot {\vec \omega} + \vec \omega \times \mathrm I \vec \omega$$

A side issue is how to prove this transport theorem. That's a good question. As an aside, my officemate (a former physics professor who became disgruntled with academia) and I found we were both a bit disgusted with the hand-wave derivations used in many physics and engineering textbooks. We went through a number of them, categorizing them as (1) "OMG. Physics math again." (2) "OMG! Too much math!" and (3) "OMG!! Goldilocks math!". Most of the texts we looked at fell into the first category (and that included a number of graduate level texts). A small number had a derivation as an appendix that fell into the second category. We did not find a single text that fell into the Goldilocks category.

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  • $\begingroup$ Thanks for the extra info. By the way I was looking at this for a real-world quadcopter project, and even if it can theoretically roll/pitch/yaw at the same time, measurements during a typical flight showed the $\omega \times I \omega$ is negligible. $\endgroup$
    – marcv81
    Oct 15 '14 at 8:40

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