How the cross product representation is !!

[kalaianand]

Hi friends
suppose there a circular disc of radious R, rotating about its axes at the center. the relationship between angular velocity and linear velocity is given by
There is a formula V=ω x R (cross product)
where V-linear acceleration
ω- angular velocity
R-Radious

This expression shows that ω and R are perpendicular, and V is perpendiculat to both ω and R.
why there are expressing V as ωxR, why not Rxω ??????
as we know axb≠bxa.
On what assumption we are denoting ω in the first and R as second term ?
simillarly
angular momentum L=R x P
=Rx mv
P-linear momentum
L-angular momentum

Torque T=R x F
F-Force



Can anyone please clarify my doubts....:?::?::?:

 

[mrflibble]

why there are expressing V as ωxR, why not Rxω ??????

Which one to use is in principle a matter of convention. Pick one, and stick with it.

incidentally, ωxR = -Rxω

---------- Post added at 23:16 ---------- Previous post was at 22:46 ----------

Just noticed ...

This expression shows that ω and R are perpendicular, and V is perpendiculat to both ω and R.

If for the sake of simplicity ω, R and V are all 3 dimensional vectors, and x is the cross product/outer product, then no.

"V is perpendicular to both ω and R.", yes.

"ω and R are perpendicular", not necessarily. They could for example be parallel vectors, and then your cross product (V) would be zero.

But you are correct that for non-zero values of V, V will be perpendicular to both ω and R

Hope that helps somewhat...

 

[WimRFP]

I agree with mrflibble, it is just convention. I would recommend you to stay with that (to avoid minus signs).

You have similar thing with the Poynting vector:

S = E х H.

If you change E with H, S doesn't point in the direction of energy flow (given used definitions of E and H).

In your case, when omega points in positive Z direction, the disc rotates counter clockwise when you look at the xy plane (z points towards you). If you make a picture when the red dot on the rotating disc is at the x axis, v has an y-component only. This follows also from

v =ω x R

(as you noted already)

for imagination of the cross product: I use

X x Y = Z (all unit vectors, Z pointing towards you).

As you already noted, interchanging x and y will result in -Z.

As you can decompose any vector in its XYZ components, you can apply this to every orientation. If problems, make yourself a three dimensional model from metal wire.

 

[rtalcott]

We typically work in a right handed system....defined by i cross j equals k....{i,j,k} being the basis vectors (or a set of basis vectors)...to convince yours elf that this is consistent draw a diagram...a point on the x axis say at x=r...rotate that point about the z axis in the xy plane with angular frequency w....consider the angular rotation represented by the vector W = w*k (k the unit vector in the plus z direction)...now look at the motion with algebra and vectors using the cross product...

The first step is almost always to draw a diagram that represents the system...

rt

 

[kalaianand]

hi friends
Thank for your suggesions......
By referring some websites documents presented, it was found that. The assumption made is based on the Right handed rule(with curling fingers). These following we links giving some clear picture on how to make assumptions. It was really helpfull...

**broken link removed**

**broken link removed**

Thank you friends

 

[mrflibble]

Going with {i, j, k} as unit vectors, you can also think of the cross product as the determinant of the following:

a x b = det ([i, j, k ; a1, a2, a3 ; b1, b2, b3])

where a = [a1, a2, a3], and b = [b1, b2, b3]

 

[rtalcott]

Yes....re: the determinant...and that gives you trivially the i cross j = - j cross i

And...how do you do bold here...html?
rt

---------- Post added at 17:26 ---------- Previous post was at 17:24 ----------

another nice link here:
Cross product - Wikipedia, the free encyclopedia

dot product here:
Dot product - Wikipedia, the free encyclopedia
rt

---------- Post added at 17:35 ---------- Previous post was at 17:26 ----------

another nice discussion:
https://www.cs.berkeley.edu/~wkahan/MathH110/Cross.pdf
rt

---------- Post added at 17:38 ---------- Previous post was at 17:35 ----------

Long list of good stuff here...short pdf's
Prof. W. Kahan's Notes for Math. H110
rt

 

[mrflibble]

If you are interested in that sort of thing you might also want to look in on geometric algebra. That does a nice job of generalizing things.


Geometric algebra - Wikipedia, the free encyclopedia

Geometric Algebra For Computer Science

Or google on "Leo Dorst geometric algebra". He's been doing useful GA stuff for years now.


Oh and you can do bold with:

this is bold

 

[kalaianand]

mrfibble,
yes these informations are also usefull
Thank you friend......

 

[kalaianand]

Hello friends
i have an another doubt. If c=axb (cross product). Hoe to represent a with respect to b and c?

 

[mrflibble]

Short answer: you can't.

Longer answer: you cannot do this, without knowing more about the angle between vectors a and b.

Given the relation c = axb, then for fixed vectors c and a there are a whole lot of b's that satisfy this.

 

[rtalcott]

Look at the geometry....
rt

 

[_Eduardo_]

Hello friends
i have an another doubt. If c=axb (cross product). Hoe to represent a with respect to b and c?

1st: b must be normal to c.

2nd: No unique solution:
a' = b×c/(b•b) + α b ; α : any real number
verify c = a'×b = a×b

 

[Julius Madrid]

Hi friends
suppose there a circular disc of radious R, rotating about its axes at the center. the relationship between angular velocity and linear velocity is given by
There is a formula V=ω x R (cross product)
where V-linear acceleration
ω- angular velocity
R-Radious

This expression shows that ω and R are perpendicular, and V is perpendiculat to both ω and R.
why there are expressing V as ωxR, why not Rxω ??????
as we know axb≠bxa.
On what assumption we are denoting ω in the first and R as second term ?
simillarly
angular momentum L=R x P
=Rx mv
P-linear momentum
L-angular momentum

Torque T=R x F
F-Force



Can anyone please clarify my doubts....:?::?::?:

AxB = -BxA
The cross product is anticommutative. ;-)