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Consider a rigid body performing a rotational motion around a vertical fixed axis $z$ with constant angular velocity $\vec{\Omega}$. The angular momentum vector $\vec{L}$ is not parallel to $z$ (and so to $\vec{\Omega}$) and follows a uniform precession motion.

Supposing to take as pivot point for the calculation of momenta any point $P$ on the $z$ axis. Then the axial component of $\vec{L}$ (called $L_z$) does not depend on the choose of the pivot point $P$ on the axis, while the component of $\vec{L}$ perpendicular to $z$ (called $L_n$) does depend on the choice of $P$.

Since $\vec{L}$ is not constant there must be an external torque $\vec{\tau}$ acting on the system. In particolar, since $\vec{L}$ follows a uniform precession motion

$\vec{\tau}=\frac{d\vec{L}}{dt}=\vec{\Omega} \wedge \vec{L} \implies \tau=L_n \Omega$

Does this mean that $\vec{\tau}$ depends on the choice of the pivot point $P$? And is the $\vec{\tau}$ calculated with this formula the torque with respect to the same point $P$ used for calculation of $\vec{L}$?

Example Take the following dumbell rotating around a vertical axis $z$ not passing through its center with constant angular velocity $\omega$.

enter image description here

If the pivot point (here I called it $O$ instead of $P$) is on the dumbell (situation (A)) then the resulting angular momentum is axial, and that means no external torque on the system. If the pivot point $O$ is somewhere else on the $z$ axis (situation (B)) then the angular momentum is not axial anymore and the external torque must be different from zero.

Does this mean that if I calculate the torque on the dumbell about the point $O$ (on the $z$ axis) I necessarily find:

  • zero if $O$ is at the intersection between the dumbell and the $z$ axis
  • not zero if $O$ is somewhere on the $z$ axis?
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marked as duplicate by ja72, Kyle Kanos, user36790, John Rennie, CuriousOne Apr 10 '16 at 9:25

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Both torque and angular momentum must be calculated "about a point". I'm not sure what your question about that is. $\endgroup$ – ACuriousMind Apr 1 '16 at 12:19
  • $\begingroup$ @ACuriousMind I made an example with a practical question, hope that it is clear enough $\endgroup$ – Sørën Apr 1 '16 at 17:16
  • $\begingroup$ It appears to me that while $L_1$ and $L_2$ are different, the sum of all momenta ($L$) is the same because $L_n$ cancels. Are you asking about torques to the entire system or torques to one of the weights of the dumbbell? $\endgroup$ – BowlOfRed Apr 1 '16 at 21:55