Re: Question on centre of mass.


fellow wrote:
> "PD" <pdraper@xxxxxxxxx> wrote in message
> news:1112812165.514613.311870@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
> >
> > fellow wrote:
> > > This comes from Feynman's Vol1. For a mass, F = Ma, where F is
the
> > total
> > > external force applied to the body, M the mass, and d^2r/dt^2 the
> > > acceleration of the centre of mass.
> > >
> > > Now take a cylindrical rod, where the centre of mass is located
in
> > the
> > > centre.
> > > It seems to me that the centre of mass moves in a way that
depends on
> > the
> > > orientation of the rod, even though F and M haven't changed.
> > >
> > > For example, here the centre of mass moves along the axis of the
rod
> > >
> > > F---> oooooo-centre-oooooo
> > >
> > >
> > > Whereas in the next case, the centre of mass would appear not to
> > travel the
> > > same path as above.
> > >
> > > F--->o
> > > o
> > > o
> > > o
> > > o
> > > o
> > > centre
> > > o
> > > o
> > > o
> > > o
> > > o
> > >
> > > What does F=Ma really mean then?
> >
> > Actually, the *center of mass* would move *identically* in the two
> > cases, toward the right and with the same acceleration. However, in
the
> > second case, there is *also* a rotation, which is accounted for by
the
> > rotational equivalent of Newton's 2nd law. The two configurations
are
> > *not* identical in terms of torques, so we don't expect the
rotational
> > behavior to be the same.
> >
> > PD
>
> Are you sure about this? In the first case, seems to me the motion
would be
> entirely translational,
> whereas in second case it's purely rotational.

I'm sure. The second case is NOT purely rotational IF:
- the rod is not pinned anywhere
- the force continues to push to the right and does not "follow" the
tip around (because that would be another difference from the other
case, wouldn't it?)

Here's a little experimental test. Balance a pencil vertically on its
point on a smooth countertop, lightly holding the top of the pencil by
touching the eraser with your fingertip. Now, with your other hand,
give a sharp flick horizontally near the top of the pencil with your
finger. We'll take the force of friction between your finger and the
eraser to be negligibly small and the force of friction between the
lead point and the countertop to be negligibly small.
You'll see (experimentally) that the pencil both rotates AND
translates. It's best you do this near the edge of the table so you're
not confused by the bounce against the table. To test that the
translational motion is the same in the two cases, note that the pencil
lands on the floor in about the same place whether you flick it near
the top of the pencil (lots of rotation) or near the middle (little
rotation).

Once again, the F=ma law you quoted ONLY applies to the translational
motion, which is identical in the two cases.
The rotational law is given by T=I*alpha, where T is torque (instead of
force), I is the rotational inertia (instead of mass) and alpha is the
rotational acceleration (instead of translational acceleration). The
two cases are clearly not identical for rotation, because the torque
applied is dramatically different.

PD

.

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