Re: moment of inertia of a cube

On Apr 3, 12:59 am, Timo Nieminen <t...@xxxxxxxxxxxxxxxxx> wrote:
On Tue, 3 Apr 2007 mme...@xxxxxxxxxxxxxxxxxx wrote:

Timo Nieminen <t...@xxxxxxxxxxxxxxxxx> writes:

Moffat next gives a very nice result: for a translating non-rotating
particle to experience a torque, the particle must be chiral ie lack
mirror symmetry. Thus, the translating cube is torque-free!

You mean "translating in fluid", I gather.

Yes, and in the Stokes limit.

It's nice to see it's true - we assumed it was an adequate approximation
for modelling the optical trapping of a cube [1], and exactly correct is
certainly adequate.

It appears to me to be "incorrect".

Uncle Al already has already as much as stated this for general flow
regimes, though he positioned himself as a defender with the
qualification "net non-zero torque over time": but the claim makes no
mention of long term time average, so any non-zero torque, however
transient, is sufficient to complete the counterexample : a
shuttle*** illustrates.

What about Stokes flows? Like relativistic motion they have the charm
of the counter-intuitive, so maybe by some magic they contrive to keep
non-chiral bodies torque-free about _any_ axis. This would require
not merely magic though, but miraculous intervention. A sufficient
counter example follows:

Consider a compound body consisting of two identical spheres separated
by a long strut of negligible cross section. The drag force is
concentrated on the spheres, in the direction of flow, and equal. So,
one might say, no torque. Or is there? As Andy Resnick implied in
reaction to the original question, moments of inertia (or torques)
must be evaluated about a particular axis or point: we can't without
qualification simply speak of the "torque".

But surely, one might say, we mean "torque about the center of mass":
that's what allows us to decompose the motion into a linear
acceleration in reaction to net force and angular acceleration in
response to torque. The center of mass is in the center of the
strut. Or is it? Nobody told us the spheres were of equal density!
By altering the masses while preserving the shape we can place the COM
anywhere along the strut without altering the forces: the body will
experience a torque about the COM and begin to rotate.

But maybe we consider altering the density cheating. Can a similar
theorem can be salvaged for bodies of uniform density? Consider again
a compound body, now consisting of spheres of masses m, m/2 and m/2.
The spheres are again connected by struts of negligible cross-section,
as shown by the (planar) diagram below:



O m
|
|
___|___ _COM_
| |
| |
| |
o o m/2, m/2

The lighter spheres are equidistant from an axis passing through the
COM, the heavier sphere at the same distance on the opposite arm, the
separations sufficient so the perturbative effect on each other's flow
fields is negligible. (All spheres are of equal density, and flow
into the page).

Stokes drag on a sphere is proportional to the area, so if we halve
the mass of a sphere the drag is multiplied by 2^(-2/3). Adding the
areas of the smaller spheres we find the total area has increased by a
factor of 1.26. So by subdividing one sphere we have increased the
drag on that side of the moment arm by ~25% without moving the COM.
The body experiences a torque about the COM.

QED


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