Angular acceleration in rigid body rotation due to a torque

[Soren4]

For the rotation of a rigid body about a fixed axis z the following holds.

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

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

Can I interpret (1) as follows?

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

 

[Buzz Bloom]

Hi Soren4:

If the applied torque vecor τ has no component along the axis of rotation (i.e. it is completely perpendicular to it) there is no way that an angular acceleration α appears.

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.

Regards,
Buzz