Torque about center of mass due to Coriolis force, is it 0?

[Hiero]

It’s essentially a question about cross product identities.

I want to analyze a problem in a frame of reference which is rotating with angular velocity $\vec \Omega$ relative to an inertial frame. In this non-inertial frame, we have a rigid body rotating with constant angular velocity $\vec \omega$ (which is neither parallel nor perpendicular to $\vec \Omega$).

Now, I wanted to find the torque about the center of mass in this frame (as an alternative to dealing with the inertia tensor in the inertial frame).

I’m not sure how to deal with torque due to the coriolis force though.

The net coriolis force is zero,
$$\vec F_c=\int d \vec F_c=-2\int (\vec \Omega × \vec v )dm = -2\int \vec \Omega × ( \vec \omega × \vec r)dm = \vec \Omega × ( \vec \omega × \int \vec r dm)= 0$$
because that last integral is zero in the center of mass frame.

But I’m at a loss as to how to simplify the torque:
$$\vec \tau _c = \int \vec r ×d\vec F_c=-2\int (\vec r×( \vec \Omega × ( \vec \omega × \vec r))dm$$
Sure would be nice if the cross product was associative!

Any help in simplifying this is appreciated.

 

[A.T.]

I don't think it is zero. Consider a hoop spinning around its symmetry axis, which is tilted from the frame rotation axis The Coriolis forces at opposite hoop points are not co-linear, because they are in the plane of the frame rotation, not in the plane of the hoop. So they will produce a torque around the center of mass.

 

[Hiero]

I don't think it is zero.

I also don’t think it’s zero for the simple reason that neglecting it gives a different answer than the inertial-frame analysis.

I’m still curious though, is there any clever usage of cross product identities that simplify this expression to a form I might be able to evaluate?

how to simplify the torque:
$$\vec \tau _c = \int \vec r ×d\vec F_c=-2\int (\vec r×( \vec \Omega × ( \vec \omega × \vec r))dm$$