| Topic: |
Science > Physics |
| User: |
"" |
| Date: |
14 Apr 2006 01:03:05 AM |
| Object: |
MUTUALLY INDEPENDENT velocity vectors? |
When explaining projectile motion my book used the term "mutually
independent" vectors in reference to the x and y components of the
velocity.
Does this just mean the vectors are perpendicular? Or is there some
other meaning?
I can't see why we couldn't use two non perpendicular velocity vectors
(if we had equations for them).
Any help would be appreciated.
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| User: "CWatters" |
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| Title: Re: MUTUALLY INDEPENDENT velocity vectors? |
14 Apr 2006 02:20:28 AM |
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<lite.on.beta@gmail.com> wrote in message
news:1144994584.957203.220820@u72g2000cwu.googlegroups.com...
When explaining projectile motion my book used the term "mutually
independent" vectors in reference to the x and y components of the
velocity.
Does this just mean the vectors are perpendicular? Or is there some
other meaning?
I can't see why we couldn't use two non perpendicular velocity vectors
(if we had equations for them).
Yes you can do that. The usual way to solve vector problems is to break each
one down into orthogonal components so that the components can be summed
easily to give the result. You don't have to do it that way. With some
problems you have a mixture of linear and rotational movement.
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| User: "Sam Wormley" |
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| Title: Re: MUTUALLY INDEPENDENT velocity vectors? |
14 Apr 2006 01:20:33 AM |
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wrote:
When explaining projectile motion my book used the term "mutually
independent" vectors in reference to the x and y components of the
velocity.
Does this just mean the vectors are perpendicular?
Yes - orthogonal
http://www.rwc.uc.edu/koehler/biophys/2b.html
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| User: "Scott J. McCaughrin" |
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| Title: Re: MUTUALLY INDEPENDENT velocity vectors? |
20 Jul 2006 09:03:24 PM |
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Sam Wormley <swormley1@mchsi.com> wrote:
: wrote:
: > When explaining projectile motion my book used the term "mutually
: > independent" vectors in reference to the x and y components of the
: > velocity.
: >
: > Does this just mean the vectors are perpendicular?
: Yes - orthogonal
: http://www.rwc.uc.edu/koehler/biophys/2b.html
The x,y components of the velocity vector are indeed perpendicular, but
that is not a necessary condition for independence. Perhaps the text
meant "linearly" independent, which only requires that such vectors
cannot be expressed as linear expressions of each other.
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| User: "Sam Wormley" |
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| Title: Re: MUTUALLY INDEPENDENT velocity vectors? |
20 Jul 2006 11:33:16 PM |
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Scott J. McCaughrin wrote:
Sam Wormley <swormley1@mchsi.com> wrote:
: wrote:
: > When explaining projectile motion my book used the term "mutually
: > independent" vectors in reference to the x and y components of the
: > velocity.
: >
: > Does this just mean the vectors are perpendicular?
: Yes - orthogonal
: http://www.rwc.uc.edu/koehler/biophys/2b.html
The x,y components of the velocity vector are indeed perpendicular, but
that is not a necessary condition for independence. Perhaps the text
meant "linearly" independent, which only requires that such vectors
cannot be expressed as linear expressions of each other.
I agree.
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| User: "Hero" |
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| Title: Re: MUTUALLY INDEPENDENT velocity vectors? |
21 Jul 2006 04:06:10 PM |
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Sam Wormley schrieb:
Scott J. McCaughrin wrote:
Sam Wormley <swormley1@mchsi.com> wrote:
: wrote:
: > When explaining projectile motion my book used the term "mutually
: > independent" vectors in reference to the x and y components of the
: > velocity.
: >
: > Does this just mean the vectors are perpendicular?
: Yes - orthogonal
: http://www.rwc.uc.edu/koehler/biophys/2b.html
The x,y components of the velocity vector are indeed perpendicular, but
that is not a necessary condition for independence. Perhaps the text
meant "linearly" independent, which only requires that such vectors
cannot be expressed as linear expressions of each other.
I agree.
I disagree. Independent vectors have independent causes, the explosion
of the gunpowder
is working on the projectile giving a directed speed. Far away from
planets it will determine the path of the projectile. Adding a gravity
acceleration by firing on the moon will add another vector, the
combination of both will determine the path now. And doing this on
earth will combine the first with the gravity of earth.
Friendly greetings
Hero
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| User: "" |
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| Title: Re: MUTUALLY INDEPENDENT velocity vectors? |
14 Apr 2006 01:45:11 AM |
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In article <1144994584.957203.220820@u72g2000cwu.googlegroups.com>, writes:
When explaining projectile motion my book used the term "mutually
independent" vectors in reference to the x and y components of the
velocity.
Does this just mean the vectors are perpendicular? Or is there some
other meaning?
A set of vectors is independent if none of them can be expressed as a
linear combination of the other ones. In the simple case of two
vectors, "mutually independent" simply means that neither is a scalar
multiple of the other one. So, while it is true that perpendicular
vectors are mutually independent, they don't need to be perpendicular
in order to be independent. They just need not to be parallel.
I can't see why we couldn't use two non perpendicular velocity vectors
(if we had equations for them).
The answer is that we certainly could and, at times, do. However, the
arithmetic is usually simpler with perpendicular vectors, so by
default these will be used unless there is a good reason to do
otherwise.
Mati Meron | "When you argue with a fool,
meron@cars.uchicago.edu | chances are he is doing just the same"
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| User: "tadchem" |
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| Title: Re: MUTUALLY INDEPENDENT velocity vectors? |
14 Apr 2006 04:58:36 AM |
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It means the x-axis doesn't know what the y-axis is doing, and vice
versa. Your dynamics problem is divisible into two one-dimensional
problems (which makes for easier solutions).
Tom Davidson
Richmond, VA
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| User: "" |
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| Title: Re: MUTUALLY INDEPENDENT velocity vectors? |
14 Apr 2006 06:08:05 AM |
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In article <1145007688.382983.236630@t31g2000cwb.googlegroups.com>,
"tadchem" <tadchem@comcast.net> wrote:
It means the x-axis doesn't know what the y-axis is doing, and vice
versa. Your dynamics problem is divisible into two one-dimensional
problems (which makes for easier solutions).
Warning: I think this may be a stupid question. If so, just tell
me.
Does the x-axis and y-axis have to intersect? hmmm.. If yes,
do they have to intersect at single point?
/BAH
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| User: "CWatters" |
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| Title: Re: MUTUALLY INDEPENDENT velocity vectors? |
14 Apr 2006 01:32:16 PM |
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<jmfbahciv@aol.com> wrote in message
news:e1nvql$8qk_002@s1001.apx1.sbo.ma.dialup.rcn.com...
In article <1145007688.382983.236630@t31g2000cwb.googlegroups.com>,
"tadchem" <tadchem@comcast.net> wrote:
It means the x-axis doesn't know what the y-axis is doing, and vice
versa. Your dynamics problem is divisible into two one-dimensional
problems (which makes for easier solutions).
Warning: I think this may be a stupid question. If so, just tell
me.
Does the x-axis and y-axis have to intersect? hmmm.. If yes,
do they have to intersect at single point?
You can defibe your co-ordinate system anyway you want. Sometimes that can
be very helpful. For example it can sometimes be easier to solve a problem
using an unusual co-ordinate system then convert the answer to a more
conventional system.
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| User: "" |
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| Title: Re: MUTUALLY INDEPENDENT velocity vectors? |
14 Apr 2006 07:58:00 AM |
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In article <e1nvql$8qk_002@s1001.apx1.sbo.ma.dialup.rcn.com>, writes:
In article <1145007688.382983.236630@t31g2000cwb.googlegroups.com>,
"tadchem" <tadchem@comcast.net> wrote:
It means the x-axis doesn't know what the y-axis is doing, and vice
versa. Your dynamics problem is divisible into two one-dimensional
problems (which makes for easier solutions).
Warning: I think this may be a stupid question. If so, just tell
me.
Does the x-axis and y-axis have to intersect? hmmm.. If yes,
do they have to intersect at single point?
In the ordinary euclidean real plane the answers are yes and yes.
The x axis is the line defined by y = 0
The y axis is the line defined by x = 0
Those lines clearly intersect at and only at the point (0,0).
It is far from clear what insight you expect this answer to grant.
The notion of "orthogonal" is not intimately tied up in whether you
visualize the coordinate axes in the plane or outside it. It is
even generalizable to polar coordinates where may not be a theta
axis as such.
It is the notion of "orthogonal" that tadchem appears to be invoking
when he talks about the axes knowing what each other are doing.
In this context, "orthogonal" means that the equations can be solved for
the x component without regard to the y component. And vice versa.
In this sense, the x component does not care what the y component
is. Or, in tadchem's words, the x axis does not know what the
y axis is doing.
Now to maybe relate this back to your question...
If x is orthogonal to y and if x=0 is part of a solution set with respect
to x and if y=0 is part of a solution set with respect to y it follows
that x=y=0 is part of a solution set with respect to both.
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