Re: Conservation of angular momentum

On Jun 12, 8:25 am, Peter <Poakfi...@xxxxxxx> wrote:
On Jun 12, 12:57 am, PD <TheDraperFam...@xxxxxxxxx> wrote:



On Jun 11, 12:31 pm, Peter <Poakfi...@xxxxxxx> wrote:

On Jun 10, 6:10 am, PD <TheDraperFam...@xxxxxxxxx> wrote:

On Jun 9, 11:03 am, Peter <Poakfi...@xxxxxxx> wrote:

On Jun 9, 10:54 am, PD <TheDraperFam...@xxxxxxxxx> wrote:

On Jun 9, 8:41 am, Peter <Poakfi...@xxxxxxx> wrote:

On Jun 9, 6:40 am, PD <TheDraperFam...@xxxxxxxxx> wrote:

On Jun 8, 4:48 pm, Peter <Poakfi...@xxxxxxx> wrote:

On Jun 8, 4:52 pm, PD <TheDraperFam...@xxxxxxxxx> wrote:

On Jun 8, 2:45 pm, Peter <Poakfi...@xxxxxxx> wrote:

On Jun 8, 9:59 am, PD <TheDraperFam...@xxxxxxxxx> wrote:

On Jun 8, 8:47 am, Peter <Poakfi...@xxxxxxx> wrote:

On Jun 7, 5:19 pm, "Greg Neill" <gneill...@xxxxxxxxxxxxxxx> wrote:

"Peter" <Poakfi...@xxxxxxx> wrote in message

news:1181249715.104862.315590@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

I never suggested thatconservationof energy is being violated. I say
a 93% energy loss in a steel-steel collision is highly improbable.
Kinetic energy is a function of mass and velocity^2. We can assume the
mass of either object does not change in the interaction; thus, only
the value of velocity^2 is left to change. If an object with small
mass, transfers all itsmomentumto an object with a larger mass,
necessarily, the magnitude of the velocity^2 of the larger object
diminish more that its mass increased. I am not sure this is clear.
The point is thatconservationofmomentumis not compatible with
conservationof energy, if the interacting objects have different
masses.

You just don't get it. A smaller mass *cannot* transfer
*all* of its linearmomentumto a larger mass in an elastic
collision. Both objects will have some motion after the
collision.

I tell you what, why don't you write out for us the
equations governing a collision between a mass m
moving at speed v colliding with a stationary mass
of 2m. Show us that you at least have a grasp of
mathematics of what you're talking about. Feel free
to consult one of your many text books.

In a direct collision, only if the two steel balls have the same mass
will the incident ball stop on impact (if it was not rolling) with the
other ball, at rest. If it has a smaller mass, it rebounds, and if a
larger mass, it keeps going. In collisions through a lever, however,
it is possible to make the incident ball stop on impact with the lever
(if it hits the right spot on the lever arm), and the other ball
(placed appropriately at the other arm), regardless of its mass
(within reason) will acquire all theangularmomentumof the incident
ball. But then, it is impossible for kinetic energy to be conserved,
regardless of the nature of the collision. This does not imply,
though, thatconservationof energy is violated. I hope this is clear.

It appears that, after 538 posts to this thread, you have absorbed
something.
Now see? Wasn't it better to raise your hand and ask a question?

PD-

Please hear this. You know that I have found empirically that, through
a collision, it is possible to transfer all themomentumfrom a point
object (cylinder) to an extended object (lever), even if the mass of
the cylinder was just a fraction of the mass of the lever, or if it
was several times the mass of the lever. Analytically, it was found
that if the mass of the lever was equal to, or up to 2.99 times the
mass of the cylinder, kinetic energy would be conserved, but if the
lever had a mass greater than three times the mass of the lever,
kinetic energy would not be conserved. Now, I have calculated what
would happen to the kinetic energy of the lever if the cylinder had a
mass three or more times the mass of the lever. I was surprised to
find that, in all cases, kinetic energy would be conserved, regardless
of how large gets the mass of the cylinder. Don't you think this is
very strange? Please tell me what you think.

Show me your calculations.-

You know, two simutaneous equations,conservationofmomentum:
m(xr)^2w1 = (Mr^2/3)w2, where m mass of cylinder. M mass of lever, r
length of lever arm, and w1angularvelocity cylinder;conservationof
kinetic energy: (1/2)m(xr)^2 w1^2 = (1/2)(Mr^2/3)w2^2; here, w2 is
angularvelocity lever. We have two unknows x and w2.

And the solutions are?

(You didn't identify what x represents, by the way.)

You really don't know what you're doing, do you?

PD-

Now I understand what is going on. According to current theory, when a
point object collides with a lever (or vice versa), if the incident
object has the same moment of inertia as the target object, the
incident object stops on impact, and transfers all itsangular
momentumand kinetic energy to the target object. If the incident
object has less moment of inertia than the target object, it rebounds,
and if it has a greater moment of inertia, it keeps going. This is
analogous to what happens when two point objects collide elastically,
and it makes a lot of sense; don't you think so? There is only one
little problem with this picture. Do you know what it is?

Suppose you try to tell me.-

The problem is that it does not agree with reality. There is
definitely something wrong with the underlying theory. I'll try to
make a chart with the theoretical and experimental values, side by
side.

That would be good. It will help elucidate the assumptions you're
making.-

When a point object (cylinder) collides perpendicularly with the right
arm (of length r) of a lever (of length 2r), pivoted at its
longitudinal center, it is found empirically that it stops on impact
only if it hits a very specific point, otherwise, it either rebounds
or keeps going. If the cylinder stops on impact, it is assumed that
itsangularmomentumis transferred to the lever,

I understand what you're doing up to here.

and if there are no
energy losses, according to current theory, it is assumed that its
kinetic energy is also transferred to the lever.
Under the same
assumptions, it is also possible to find analytically the points of
impact where the cylinder stops..

Following are the points of impact, both experimental and
theoretical, in collisions of steel cylinders, of mass m and m/2, with
levers of mass m, 2m, and 6m. The three levers have the same
dimensions: ¼ x ¼ x 12 inches, and are made of acrylic, aluminum, and
steel, respectively. The lever of steel has a mass of about 100 grams.
=================================================
Mass of cylinder Mass of lever Point of

impact> > Experimental Theoretical
____________ __________ ___________
m m 0.33r 0.58r

m 2m 0.41r 0.82r

m 6m 0.51r 1.4r*
m/2 m 0.43r 0.82r
m/2 2m 0.51r 1.15r*
m/2 6m 0.70r 2.0r*
___________________
*This value is impossible: the length of the lever arm is only r.
===================================================
You can see that the disagreement between experimental and theoretical
results is huge.

Whoops. There's the error in your calculation. Even without losses,
this is a mistake. There IS NO theory that says that the kinetic
energy will get transferred to the rotation of the lever.

Please redo your calculations.

Is not energy (kinetic or not) supposed to be conserved in closed
systems?

No, kinetic energy is NOT supposed to be conserved in closed systems.
If you thought so, then this is your fundamental mistake. Kinetic
energy is ONLY supposed to be conserved in closed systems where there
are *elastic* collisions. You do not have elastic collisions here.
Inelastic collisions also occur in closed, lossless systems.

Keep in mind that in your calculations, you are ONLY keeping track of
kinetic energy and not any other buckets of energy, and you are
assuming that kinetic energy here must be deposited into *kinetic*
energy there. That is incorrect.

If my calculations are wrong, how would you say they should
be done?

Well, first of all, you didn't write down the formulas you used to
calculate the numbers in your table. To do this completely, and in a
way that would elicit productive comments, you should do the following
things:
a) Write in *words* the principles you are invoking. (This is what you
did, and which I commented on.)
b) Define the symbols that you are going to use for various
quantities. (v, m, M, L, etc.)
c) Write in symbols the principles you are invoking. This will be a
set of equations (one for each principle) that will be recognizable.
(As conservation of momentum, say.)
d) Write down the steps of the algebra to solve for the symbol whose
value you are going to write in a table. Put that symbol on the left
side of an equation, and the other symbols on the right hand side.
e) Write down your table of the values of the symbols on the right-
hand side of your equation and the calculated values of the symbol on
the left side of the equation.

You did (a) and (e) and left out all the steps in between. If you will
do all steps (a)-(e), then we can point out exactly where you went
wrong in your calculations. I have a pretty good idea of that already,
as I've already pointed out to you, but it would be useful to be
concrete.

PD

.

Relevant Pages

  • Re: Conservation of angular momentum
    ... a 93% energy loss in a steel-steel collision is highly improbable. ... Kinetic energy is a function of mass and velocity^2. ... In collisions through a lever, however, ... But then, it is impossible for kinetic energy to be conserved, ...
    (sci.physics)
  • Re: Conservation of angular momentum
    ... a 93% energy loss in a steel-steel collision is highly improbable. ... Kinetic energy is a function of mass and velocity^2. ... In collisions through a lever, however, ... But then, it is impossible for kinetic energy to be conserved, ...
    (sci.physics)
  • Re: Conservation of angular momentum
    ... a 93% energy loss in a steel-steel collision is highly improbable. ... Kinetic energy is a function of mass and velocity^2.. ... In collisions through a lever, however, ... But then, it is impossible for kinetic energy to be conserved, ...
    (sci.physics)
  • Re: Conservation of angular momentum
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  • Re: Conservation of angular momentum
    ... another object of different mass, then it is not an elastic collision, ... If you have a collision that you *know* to be elastic, ... the final kinetic energy of the object of mass m ... the two balls can differ on whether they conserve kinetic ...
    (sci.physics)