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Consider a body pivoted at the origin and now apply a force at the origin. The body experiences zero torque but can still rotate with constant angular velocity $\vec\omega$ where $\hat {\omega}$ is a fixed principal axis.

(The above thing is true for any arbitrary origin. The only necessary conditions are that the object is pivoted there and the force is applied on this pivot.)

I have two questions regarding this:

  1. What is the intuitive explanation for the fact that the body can start rotating even in the absence of torque?

  2. Why does it not work for non-principal axes?

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  • $\begingroup$ Is the center of mass fixed at the origin? Or is some other point of the body fixed at the origin? Or is the body not fixed at the origin and some point coincides with the origin? Or is the origin placed at the center of mass with a force through it? Please clarify the situation and all that! $\endgroup$
    – JAlex
    Commented Oct 26, 2021 at 13:28
  • $\begingroup$ The origin can be chosen arbitrarily anywhere. But wherever it is chosen, the object is pivoted there and the force acts on the pivot. $\endgroup$
    – Lost
    Commented Oct 26, 2021 at 13:41
  • $\begingroup$ A force acting on a fixed pivot is meaningless as there will be an equal and opposite reaction to it canceling it out. The end effect is the same as no force applied. $\endgroup$
    – JAlex
    Commented Oct 26, 2021 at 13:44

1 Answer 1

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  1. Is not true. A body cannot start rotating in the absence of torque about the center of mass. The rotational law of motion makes this clear when $\vec{\omega} =0 $, you need torque $\vec{\tau}$ to affect rotational acceleration $\dot{\vec{\omega}}$. $$\vec{\tau} = \mathrm{I}\, \dot{\vec{\omega}} + \vec{\omega} \times \mathrm{I}\, \vec{\omega} $$

  2. For a rotating body, in the absence of torque, angular momentum vector $\vec{L} = \mathrm{I}\,\vec{\omega}$ is preserved in magnitude and direction. So the second term in the equation above is zero only when angular momentum and rotational velocity are parallel to each other. This happens along the so called principal axis of rotation. In all other cases, when $\vec{\tau}=0$, if the second term is not zero, then first term must be non-zero also in order to cancel out and produce zero net torque.

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  • $\begingroup$ "For a rotation around a principal axis ωˆ , the lack of need for any torque means that if the object is pivoted at the origin, and if the origin is the only place where any force is applied (which implies that there is zero torque around it), then the object can undergo rotation with constant angular velocity ω. If you try to set up this scenario with a nonprincipal axis, it won’t work." I found this statement in Morin's book. Maybe I misunderstood it. $\endgroup$
    – Lost
    Commented Oct 26, 2021 at 13:44
  • $\begingroup$ What is this statement trying to say? $\endgroup$
    – Lost
    Commented Oct 26, 2021 at 13:44
  • $\begingroup$ Look at the equation of motion above. Constant rotation means the $\dot {\omega}$ term is zero. As I said this can only happen when the 2nd term $\omega \times I \omega$ is zero also. $\endgroup$
    – JAlex
    Commented Oct 26, 2021 at 16:57
  • $\begingroup$ @Lost The bit about the force isn't telling you what happens when you push at the origin. It's telling you that as long as you don't push anywhere else, the rotation around the principal axis is constant. Forces at the origin (zero torque) are irrelevant here. The rotation was happening before any such force was applied and continues afterward. $\endgroup$
    – BowlOfRed
    Commented Oct 26, 2021 at 17:14

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