# I Angular acceleration in rigid body rotation due to a torque

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1. Apr 10, 2016

### 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.

2. Apr 10, 2016

### Buzz Bloom

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.

Regards,
Buzz