This is a concept which is bothering me a lot. I'm sure there are other duplicates to this question but on seeing a few such duplicates, I haven't yet obtained an answer.

My question is, for a rigid body undergoing rotational motion, Can I find ' Total Torque about any line on the body and equate it to the moment of inertia of the body about that line multiplied by angular acceleration and say without doubt that that angular accelerarion is the angular acceleration with which the whole body rotates ? ' In other words, ' Will the CALCULATED angular acceleration about different lines be the same'. I do know that angular acceleration is the same at all points in a rigid body. However, when we proved $\tau=I\alpha$ to hold in rigid body we did such that τ and I were taken about AXIS OF ROTATION and finally after proof we said $\tau=I\alpha$ , where I and τ are calculated about AXIS OF ROTATION; and $\alpha$ is the angular accelerarion of all all the particles of the rigid body and hence it is the angular acceleration of the body. Will this $\tau=I\alpha$ equation still hold good and will it give me the same $\alpha$ if I change $\tau$ and $I$ by calculating it about a different line?

Now I considered a sphere and pure rolling on the ground with friction acting at contact along with an external force F on another part of the rigid body. Working with axis of rotation as Centre of mass or working with axis of rotation as contact point/instantaneous axis of rotation gave me the same result when I solved a problem. Now in any general case how many points can I take as axis of rotation? Or is there only one point which i can take which is the actual axis of rotation and along with it in cases of pure rolling I can take even instantaneous axis of rotation?

  • $\begingroup$ Sorry I misunderstood your question. I've deleted my post just now. $\endgroup$
    – PeaBrane
    Mar 22, 2018 at 11:09

1 Answer 1


In other words, 'Will the CALCULATED angular acceleration about different lines be the same'.

If you do it right, yes. Torque, angular velocity, and angular acceleration are free vectors for a rigid body. Compare with force, which is not a free vector. The torque that results from a force depends very much on where the force is applied.

That said, "doing it right" is a nontrivial task. The key reason people oftentimes choose to use the center of mass of a rigid body as the point of reference is that this is the one point for which the translational and rotational equations of motion decouple: $$\begin{align} \boldsymbol a &\equiv \dot {\boldsymbol v} = \frac1m \boldsymbol F_\text{net} \\ \boldsymbol \alpha &\equiv \dot {\boldsymbol \omega} = \mathrm{I}^{-1}\left(\boldsymbol \tau_\text{net} - \boldsymbol \omega \times (\mathrm{I}\,\boldsymbol \omega)\right) \end{align}$$ One must resort to the much messier Newton-Euler equations elsewhere because, in general, translational and angular acceleration couple with one another. I'm not going to bother with replicating that messiness here. The linked wikipedia article gets the equations of motion correctly.


However, when we proved $\tau = \mathrm{I}\alpha$ to hold in rigid body ...

Note well: In general, this doesn't even hold for the center of mass. This misses the Euler torque $-\boldsymbol\omega\times(\mathrm{I}\boldsymbol\omega)$ The Euler torque becomes even messier when the point of interest is somewhere other than the center of mass.

So why use anything but the center of mass? There are good reasons for doing so, particularly in robotics. Robotic joints are constrained in one or more dimensions. It's often better to stay true to those constraints (which means dealing with the Newton-Euler messiness) than it is to use the center of mass as the point of interest and somehow fudge things to reflect the fact that those constraints exist.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.