2
$\begingroup$

If we are using the equation $\tau=I\cdot \alpha$ (where $\tau$ is the torque, $I$ is the moment of inertia, $\alpha$ is the angular acceleration) in a rotating frame of reference, we have to account for the fictitious forces, so, my question is, do these forces 'act' on the center of mass of the extended object?

As far as I know, the force $-ma_{frame}$ can be thought of as acting at the center of mass effectively, it gives the correct force and torque as we would get if we were to apply the force on each infinitesimal part of the body.

What about the coreolis term, the euler term and the centripetal term of fictitious force?

Improve this question
$\endgroup$
1
  • $\begingroup$ The correct expression for rotational motion is $$\boldsymbol{\tau}_{C}={\bf I}_{C}\boldsymbol{\alpha}+\boldsymbol{\omega}\times{\bf I}_{C}\boldsymbol{\omega}$$ and not $\tau_C = I _C\alpha$. Here the subscript C denotes summing at the center of mass. $\endgroup$
    – John Alexiou
    Commented Sep 6, 2022 at 20:43

1 Answer 1

Reset to default
0
$\begingroup$

Fictitious forces need to be considered in the non-inertial frame, specifically the centrifugal, Coriolis, and Euler forces as well as the fictitious force from translational acceleration.

For a system of particles, of which a rigid body is a special case, in a non-inertial frame the translational motion of the system can be evaluated assuming all forces- real and fictitious- act on a particle with the total system mass located at the center of mass. The rotational motion about the CM cannot be evaluated assuming all the forces- real and fictitious- act at the CM.

The force of gravity is a special case for which the total force can be assumed to act at the CM. The fictitious force from translational acceleration can also be assumed to act at the CM. The centrifugal, Coriolis, and Euler forces cannot be assumed to act at the CM.

See Where does pseudo force act at? this exchange for more details.

Improve this answer
$\endgroup$

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.