# Newton's second law for rotation

1. Can the second law of motion for rotation, $\vec{\tau}=I \vec{\alpha}$, be used for any axis?

2. Is there any case that acceleration $\vec{\alpha}$ is not in the direction of applied torque $\vec{\tau}$?

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The correct law is $$\sum \vec{\tau} = \frac{{\rm d}}{{\rm d} t}( I \vec{\omega}) = I \vec{\alpha} + \vec{\omega}\times I \vec{\omega}$$ and it is used for any $\vec{\tau}$ an $\vec{\alpha}$. – ja72 Dec 22 '13 at 0:26

It can indeed be used for any axis. But keep in mind that $I$ is a 3x3 matrix. Using different axes requires you to transform this matrix, so it corresponds to your set of choice. This can sometimes prove to be quite difficult as $I$ can be rather complex.
The acceleration will always be in the direction of the total torque acting on your object. The moment of inertia cannot be negative as it is calculated as follows: $$I = \rho(r)r^2dV$$
With $\rho(r)$ being the density of the object and $r$ the distance to the pivot point. These are all positive. Now we can conclude that in no situation would it be possible to have a negative relation between $\vec{\alpha}$ and $\vec{\tau}$. Of course, the angular velocity does not need to be in the direction of the total torque.
But as you said $I$ is a matrix so It could change the direction of $\vec{\alpha}$ when acting on it. So direction of $\vec{\tau}$ could be different! – richard Dec 21 '13 at 14:05
I implicitly assumed that the axes used are the principal axes of the system, in which case the non-diagonal elements of $I$ are zero. So that was an error on my part, my apologies. – Tom Dec 21 '13 at 19:30
If we would calculate the first element of $\vec{\tau}$ we would get $$\tau_1 = I_{11}.\alpha_1 + I_{12}.\alpha_2 + I_{13}.\alpha_3$$ Repeating this for the other elements it seems obvious to me that $\vec{\alpha}$ would not be in the same direction of the torque, which seems incredibly counter-intuitive to me. Right now, I'm actually convinced I'm missing something myself. – Tom Dec 21 '13 at 19:39
If the 3×3 mass moment of inertia is defined in body fixed coordinates as $I_{body}$ and the 3×3 rotation matrix for the body is $R$ then the MMOI in world coordinates is $$I = R I_{body} R^\top$$ This is interpreted as transform to local coordinates, apply the MMOI and transform back to world coordinates. – ja72 Dec 22 '13 at 0:35