Re: Moment of inertia of a tilted cylinder?

On 23 Feb 2007 16:39:06 -0800, turboenthusiast@xxxxxxxxx wrote:

Is there a formula for calculating the moment of inertia of a cylinder
that is rotating around a non-principal axis? For example, imagine a
cylinder (such as an upright can of soup) that has been tilted away
from the vertical axis by angle theta. What is the moment of inertia
of the cylinder around the vertical axis? For simplicity, the center
of gravity of the cylinder lies on the vertical axis.
(I sent this reply this morning, and it has disappeared, so here it is
again).
You have to use the inertia tensor which is a diagonal matrix with
J11, J22 and J33 on the diagonals.
Then you have to use a similarity transformation which is quite
simple. If we call A a rotation matrix for theta about x, then the
diagonaizedl inertia tensor J will become J':
J' = A*JA
where A* is the matrix for -theta. J' has become complex. Now there is
no defined moment of inertia.

J' has off-axis terms so in finding momentum:
J' * omega = L (momentum)
where momentum and omega are no longer colinear. To give you some idea
J'33 = J33 cos^2 theta + J22 sin^2 theta
If you try to spin it that way you'll find off-axis torques that are a
problem without a rigid axle and bearing. It will not be rotating in a
minimum energy configuration.
John Polasek
.

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