1. May 10, 2015

### haruspex

Last edited: Dec 25, 2016
2. May 10, 2015

### Staff: Mentor

Nice post, Haruspex. I'll have to come back to this when I start my physics and mechanics classes.

3. May 10, 2015

### Greg Bernhardt

Nice work haruspex!

4. Jul 30, 2015

### andrewkirk

Having read this, my future calcs involving torques and moments of inertia will be much faster and easier.

5. Jun 15, 2016

### wrobel

Nice article. I just want to add some comment on the equation of momentum for rigid body and on some very frequent errors that arise in this regard.
The equation of momentum is
$$J_A\dot{\boldsymbol \omega}+\boldsymbol\omega\times J_A\boldsymbol\omega=\boldsymbol M_A.\qquad (*)$$ Here $J_A,\boldsymbol\omega$ are the operator of inertia about the point $A$ and the angular velocity of the rigid body respectively; $\boldsymbol M_A$ is the torque about the point $A$ applied to the rigid body.

But what is the point $A$? If $A$ is a stationary point of the rigid body or its center of mass then equation (*) is correct.
In general, it is incorrect to use formula (*) for $A$ to be instantaneous centre of rotation; it is incorrect even for planar problems.

Another frequent error is concerned to the term $\boldsymbol\omega\times J_A\boldsymbol\omega$. This term is equal to zero identically in planar problems. But one can not forget it in essentially 3D problems.

Last edited: Jun 15, 2016
6. Jun 15, 2016

### vanhees71

One should also note that the tensor of inertia, $J_A$, must refer to the body-fixed point $A$.

7. Jun 29, 2016

### wrobel

sure

accidentally I came across an article
https://www.jstor.org/stable/2973359?seq=1#page_scan_tab_contents
perhaps It should be noted about a general formula. Let a point $A$ be any point of the rigid body. Then
$$J_A\dot{\boldsymbol\omega}+\boldsymbol\omega\times J_A\boldsymbol\omega+m\boldsymbol{AS}\times \boldsymbol a_A=\boldsymbol M_A;$$
where $S$ is the center of mass, $m$ is the mass of the rigid body, $\boldsymbol a_A$ is the acceleration of the point $A$.