[geocentrism] Re: Angular momentum
- From: Mike <mboyd@xxxxxxxxxxxx>
- To: geocentrism@xxxxxxxxxxxxx
- Date: Mon, 25 Oct 2004 03:18:50 +0100
>> Ok, I was a little quick to respond previosly, my apologies.
>
> Apologies accepted.
>
> I've allowed myself to become too heated recently. Please accept my
> apologies.
Accepted of course. We alll suffer from this from time to time. I hope
that we can both keep the conversation constructive.
> You then state, without any justification, that I have ignored the
> fact that velocity is a vector quantity. Get off your bookshelf any
> of the elementary physics textbooks that you appear to place so much
> store by and many times you will observe that velocity (and hence
> linear momentum) is treated as a scalar.
It is true that many texts dealing with collisions do appear to ignore
the vector nature of velocity but this is precisely because they are
dealing with simplified one dimensional collisions. The fact that I
completely failed to see that you were talking about these simplified
text book collisions should tell you that I am doing my own thinking and
not just relying on what I read elsewhere.
>> If both particles are moving along the x axis (for example) and in
>> the same direction, then the maths holds, I think, because the
>> directioin of both particles is the same before and after the
>> collisio so the vector arithmetic simplifies to scalar arithmetic.
>> Is that what you are saying?
>
> Yes, although they don't even have to be initially going the same way
> along that axis.
Initially I suspected that your conclusion depended upon this equating
vector sums with their magnitude sums. However, after you asked me to
work it out on the x axis I did, and realized that your result is
correct if all the velocities are along the same axis in the same
direction and therefore your result in that case is true and your
conclusion (which isn't mathematical) needs addressing. They do need to
be going in the same direction however, otherwise one of your vector
difference in
(u1 - v1) + (u2 - v2) = 0
would not equal the scalar differences by the same names in
(u1 + v1)(u1 - v1) + (u2 + v2)(u2 - v2) > 0
because at least one of the particles would have to change direction.
Here is the short version version
**********************************
Your equation (2)
(u1 - v1) + (u2 - v2) = 0
either has (u1-v1) positive and (u2-v2) negative or the other way round
depending on your choice of labelling. You assumed a choice that gave
the former and then incorrectly concluded that that this choice was
arbitrary.
Here is the long version
************************
Your result is
u1 + v1 > u2 + v2
And your conclusion is that this is preposterous. This is not, in
itself, preposterous though. Suppose that
u1=10m/s
u2=5m/s
and the coefficient of restitution is 0.5 so that
v1=6.25m/s
v2=8.75m/s
We can see that momentum is conserverd
m(u1+u2) = m(v1+v2)
15m=15m
And kinetic energy is reduceed
m/2 * u1^2 + m/2 * u2^2 = m/2(100+25) = 62.5m
m/2 * v1^2 + m/2 * v2^2 = m/2(39.0625+76.5625) = 57.8125m
The absurdity you claim is that the choice of which way round the
particles were labelled was totally arbitrary. Here is why I think your
conclusion is wrong.
Your momentum equation (2)
(u1 - v1) + (u2 - v2) = 0
means that
(u1 - v1) = -(u2 - v2)
Unless the difference in the before and after velocities is zero (in
which case we have not had an inelastic collision) this tells us that
(u1-v1) must have the opposite sign to (u2-v2).
If (u1-v1) is positive then dividing both terms in
(u1 + v1)(u1 - v1) + (u2 + v2)(u2 - v2) > 0
by the same *positive* amouts, (u1-v1) and -(u2-v2), we get
(u1 + v1) - (u2 + v2) > 0
As per your your proof.
If, however, (u1-v1) is negative then (u2-v2) is positive so dividing
(u1 + v1)(u1 - v1) + (u2 + v2)(u2 - v2) > 0
by the same *negative* amounts, (u1-v1) and -(u2-v2), we get
(u1 + v1) - (u2 + v2) < 0
The greater-than sign turns into a less-than sign because we have
negated all of the terms. Alternatively we could have multiplied both
sides of
(u1 - v1) = -(u2 - v2)
by -1 so that we're still dividing
(u1 + v1)(u1 - v1) + (u2 + v2)(u2 - v2) > 0
by positive amounts, -(u1-v1) and (u2-v2) to get
-(u1+v1) + (u2+v2) > 0
or
(u2 + v2) - (u1 + v1)> 0
rather than the result from the other choice of labelling which gave
(u1 + v1) - (u2 + v2) > 0
Your result assumed that (u1-v1) is positive and (u2-v2) is negative.
If you swap u1 and v1 for u2 and v2 in your
(u1 - v1) + (u2 - v2) = 0
you will swap the sign of (u1-v1) and (u2-v2) and your result changes to
(u1 + v1) - (u2 + v2) < 0
Because you are now dividing the two terms in
(u1 + v1)(u1 - v1) + (u2 + v2)(u2 - v2) > 0
by a negative quantity.
If I have made any mistakes in the above please just point them out and
let me address them. I am not trying to ridicule you, I believe my
arguement is correct and am simply pointing out to you the problem I see
with your conclusion.
Regards,
Mkike.
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