Re: Conservation of angular momentum

On 2007-05-28, Peter <Poakfield@xxxxxxx> wrote:
On May 28, 2:16 pm, "Greg Neill" <gneill...@xxxxxxxxxxxxxxx> wrote:
[...]
But it seems that you could simply declare the lever to be
a non rotating rod initially moving inertially in space and
unattached to anything. The rod is struck in a perfectly
elastic collision at a point some given distance from one
end, where the projectile's trajectory was perpendicular
to the long axis of the rod. Choose the frame of reference
to be the inertial frame in which the rod existed prior to
the impact.

Wouldn't it be too difficult to know what a rod without a pivot would
do after it is struck by a point object off center?

No, it's quite easy. You get an acceleration on the line between the
contact point and the centre of mass, and a rotation about the centre of
mass superposed on one another.

Only if the point object strikes the center of percussion of the rod,
one would know what happens. Of course, this is an interesting
possibility.

Centre of percussion implies a pivot not through the centre of mass. If
you put a pivot near one end of the rod then there's a sweet spot where
the point object can hit it such that there will be a pure rotation
around the pivot and no transfer of linear momentum to the pivot. I
don't remember whether you call the pivot axis or the contact point the
"centre of percussion", although it's probably the contact point. For
every pivot axis you choose, there is a corresponding contact point at
which you get torque around the pivot and no force.

So I don't think you can say "centre of percussion of the rod" unless
you also specify a pivot axis for the rod.

A typical example is a baseball bat. The position of the handle defines
the pivot axis. The mass distribution of the bat and the position of the
pivot define a sweet spot. If the ball hits the sweet spot, the bat will
just rotate in the batter's hands and not jump forwards or smash back
into them causing uncomfortable jarring.
.

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