Re: The dual oppsosing brake rotor system in space
- From: "Greg Neill" <gneillREM@xxxxxxxxxxxxxxx>
- Date: Mon, 28 Jul 2008 22:25:53 -0400
"Spaceman" <spaceman@xxxxxxxxxxxxxxxxxxxxxxxxxx> wrote in message
news:I_idnRS7c_7HyxPVnZ2dnUVZ_tfinZ2d@xxxxxxxxxxx
Greg Neill wrote:
"Spaceman" <spaceman@xxxxxxxxxxxxxxxxxxxxxxxxxx> wrote in message
news:z6ydncxtF7MEphPVnZ2dnUVZ_uLinZ2d@xxxxxxxxxxx
The only problem is you can't tell that linear momentum
can come from circular momentum if you design it to
do such.
No, linear momentum and rotational momentum are
strictly conserved separately.
So you are saying rotational momentum can never be
converted to linear motion?
Oh no, you can get linear motion (not the same as
linear momentum) from angular momentum. But whatever
it is that ends up moving linearly must have the
requisite angular momentum with respect to the system.
Note that angular momentum does not always imply a
rotating object. A projectile passing by a fixed
location in a straight line has angular momentum with
respect to that fixed location.
I'll give you an example where rotational motion can
result in a linear motion. Two rubber wheels are
pressed together at their treads and rotating. They
are rotating with the same angular velocity (but in
opposite direction, obviously). A thin rod is
introduced between them, pulled through, and fired out
the other side in a straight line. What happened?
The wheels taken together had the same magnitude
of angular momentum but in opposite directions.
The net anular momentum of the two is zero. They
each contributed the same amount of energy to
launching the rod, so afterwards the net angular
momentum is still zero.
The rod was given linear monentum in one direction,
but the wheel pair was kicked backwards in the
opposite direction via the reaction forces. Net
linear momentum is also still zero. The rod is
travelling on a line that is aligned with the
center of mass of the wheels, so its angular
momentum with respect to the wheels is zero.
So the kinetic energy was redistributed, but
individually the angular and linear momenta are
the same before and after.
Note that the wheel pair's linear motion was, in
effect, due to expelling the rod's mass in the
opposite direction. The center of mass of the
wheels and rod taken together still remains fixed.
.
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