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I Angular acceleration in rigid body rotation due to a torque

  1. Apr 10, 2016 #1
    For the rotation of a rigid body about a fixed axis [itex]z[/itex] the following holds.

    $$\vec{\tau_z}=\frac{d\vec{L_z}}{dt}= I_z \vec{\alpha} \tag{1}$$

    Where [itex]\vec{\tau_z}[/itex] is the component parallel to the axis [itex]z[/itex] of a torque [itex]\vec{\tau}[/itex] exerted in the body; [itex]\vec{L_z}[/itex] is the component parallel to the rotation axis [itex]z[/itex] of the angular momentum and [itex] \vec{\alpha}[/itex] is the angular acceleration.

    Can I interpret (1) as follows?

    If there is an angular acceleration [itex] \vec{\alpha}[/itex] there must be an exerted torque [itex]\vec{\tau}[/itex] with non zero component [itex]\vec{\tau_z}[/itex] along the axis of rotation [itex]z[/itex]: this last mentioned component [itex]\vec{\tau_z}[/itex] is the only one responsible for the present angular acceleration [itex] \vec{\alpha}[/itex]. If the applied torque [itex]\vec{\tau}[/itex] has no component along the axis of rotation [itex]z[/itex] (i.e. it is completely perpendicular to it) there is no way that an angular acceleration [itex] \vec{\alpha}[/itex] appears.
  2. jcsd
  3. Apr 10, 2016 #2

    Buzz Bloom

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    Gold Member

    Hi Soren4:
    I may be mistaken about all the implications, but I understand then when a constant torque is applied to a spinning body, and the angle of the torque vector is perpendicular to the spin vector, the the body experiences precession, and the spin vector direction will follow a circular motion. I am not exactly sure how to describe this angular motion of the spin axis in terms of an acceleration vector. I am guessing it would analogous to a centripetal force vector corresponding to the acceleration of body in a circular orbit about a central mass.

    Hope this helps.

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