Re: physical pendulums

["Prof. John P. Ertel (wizard)" <jpe@NADN.NAVY.MIL>]

On Wed, 22 Sep 1999, fred brace wrote:

> > I am using a set of physical science lab materials on the pendulum.
> > One of the activities asks the students to get a washer, a bolt
> > and a long wooden dowel to swing with the same frequency.  It says
> > that if the center of masses line up, then the pendulums will have
> > the same frequency.  Is this true?


The comment is true if you can believe that the moments of inertia
of each of the objects are substantively the same (a real stretch
for me, see Aside-1).


Also, the "bolt" has a geometry that could well require treating the
moment of inertia as a tensor rather than a simple scalar.  One could hope
that this would be a small correction which would not cause a complication
in the motion.  It would also require that the "physical pendulum" formed
by the object and whatever suspension it has (fine string, etc.) must move
in the "simple pendulum form."  That is, the IAR (Instantaneous Axis of
Rotation, the suspension, and the CM (Center of Mass) must all be
co-linear.  Lastly, there must be little or no air drag.

[All of the extra requirements are to make sure that the extended objects
are irrotational wrt (with respect to) the line joining the IAR and the CM
as well as that there are NO "sub-oscillations" (twisting of the washer,
"wobbling" of the bolt, etc.) to the principally pendular motion.]


To the extent that we can treat the moment of inertia as a simple scalar
[L (angular momentum) identically parallel to w (angular velocity) for
arbitrary rotations], the proper expression for the period, T, would be

                                 1/2
                   [  I(total)  ]
        T  =  2Pi  [ ---------- ]
                   [   m g D    ]

where I(total) is the scalar moment of inertia of the object
             taken about the IAR,
      m is the mass of the object which is part of the pendulum,
      D is the distance between the IAR and the CM,


If all of this seems overly specified just to get a simplified answer,
it's because it must be.  The problem has to be very idealized to possibly
cast it in mathematics that most of our students (as well as most of us)
can understand.


There will, no doubt be several "sub-oscillations" but we could
concentrate only on the gross overall pendular motion.

+=================================+
Aside-1:

First, we define the "point mass" moment of inertia to be
                     2
        I(pm)  =  m D

where m is the mass of the object which is part of the pendulum,
      D is the distance between the IAR and the CM.

Then we define the "extended body" correction to be the moment of inertia
of the object about its own CM
        I(CM)

The total moment of inertia I(total) is of course

        I(total)  =  I(pm) + I(CM)      (the parallel axis theorem)

Now it is difficult for me to believe that we will be lucky enough to have
the CM of each object at the same distance from the IAR and also have the
moments of inertia also be the same.
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