Rules regarding the vector cross product and dot product

  • Thread starter ElDavidas
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  • #1
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hi, I'm currently doing a mechanics module at Uni. The thing is, I'm not very sure about rules regarding the vector cross product and dot product.

For example, it says in my notes for angular momentum:

"Introducing polar coordinates

[tex] \mathbf{r} = r(cos \Phi \mathbf{i} + sin \Phi \mathbf{j}) [/tex]

[tex] \mathbf{\dot{r}} = \dot{r} (cos \Phi \mathbf{i} + sin \Phi \mathbf{j}) + r \dot{\Phi}(-sin \Phi \mathbf{i} + cos \Phi \mathbf{j})[/tex]

The angular momentum about 0 is given by

[tex] \mathbf{h} = m \mathbf{r} \times \mathbf{\dot{r}} = mr^2 \dot{\Phi} \mathbf{k} [/tex]"

How do you get all those sin and cos functions to cancel out?! I mean what's going on here? How did the k suddenly appear?
 
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Answers and Replies

  • #2
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ElDavidas said:
hi, I'm currently doing a mechanics module at Uni. The thing is, I'm not very sure about rules regarding the vector cross product and dot product.
For example, it says in my notes for angular momentum:
"Introducing polar coordinates
[tex] \mathbf{r} = r(cos \Phi \mathbf{i} + sin \Phi \mathbf{j}) [/tex]
[tex] \mathbf{\dot{r}} = \dot{r} (cos \Phi \mathbf{i} + sin \Phi \mathbf{j}) + r \dot{\Phi}(-sin \Phi \mathbf{i} + cos \Phi \mathbf{j})[/tex]
The angular momentum about 0 is given by
[tex] \mathbf{h} = m \mathbf{r} \times \mathbf{\dot{r}} = mr^2 \dot{\Phi} \mathbf{k} [/tex]"
How do you get all those sin and cos functions to cancel out?! I mean what's going on here? How did the k suddenly appear?
Well, the easy way out for you is just to calculate the vector product between r and r_dot. You have been given the components, so you should be able to do this. Try it and come back with your result...i will keep an eye on you :)


regards
marlon

edit : keep in mind that those vectors are 3 D. You have a third k unit vector next to i and j but ofcourse the components in the k-direction are 0 for both vectors.

eg : for r you actually have [tex] \mathbf{r}= r(cos \Phi \mathbf{i} + sin \Phi \mathbf{j} + 0\mathbf{k}) [/tex]
 
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  • #3
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marlon said:
Well, the easy way out for you is just to calculate the vector product between r and r_dot.
Ok, so the formula for calculating the cross product is

[tex] r \times \dot{r} = sin \alpha |r| |\dot{r}| [/tex]

So

[tex] r \times \dot{r} = sin \alpha |r(cos \Phi \mathbf{i} + sin \Phi \mathbf{j})| |\dot{r} (cos \Phi \mathbf{i} + sin \Phi \mathbf{j}) + r \dot{\Phi}(-sin \Phi \mathbf{i} + cos \Phi \mathbf{j}) | [/tex]

what do you do from here?
 
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  • #4
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Well, there are actually two formula's for calculating a vector product
1) [tex] r \times \dot{r} = sin \alpha |r| |\dot{r}| \vec{k} [/tex]

the version you quoted is not entirely correct because a vector product will yield another vector perpendicular to the plane of the two vectors in the cross product. That is why i added the k vector because it is perpendicular to the (i,j)-plane.

2) determinant with in the first row the three unit vector i j k. In the second row the i,j,k-components of the first vector in the cross product. In the third row the i,j,k-components of the second vector in the cross product.

Using method 1) you need to

a) calculate the magnitude of a vector of which you have been given the components. The formula is :

[tex]| \vec{r}| = |a \vec{i} + b \vec{j} + c \vec{k}| = \sqrt{a^2 + b^2 + c^2}[/tex]

b) determine the angle between the vectors r and r_dot. ie what is the alpha ?

regards
marlon

ps : try applying both methods.
 
  • #5
matt grime
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ElDavidas said:
Ok, so the formula for calculating the cross product is
[tex] r \times \dot{r} = sin \alpha |r| |\dot{r}| [/tex]
No, the LHS is a vector and the RHS is a scalar, so try to have a think about what you really mean.
 
  • #6
matt grime said:
No, the LHS is a vector and the RHS is a scalar, so try to have a think about what you really mean.
r x r' = ( |r| |r'| sin a )K
Where K is a perpendicular unit vector? (Perpendicular to r and r'?)
 
  • #7
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Pseudo Statistic,

This question is already answered in the fourth post of this thread.

marlon
 
  • #8
marlon said:
Pseudo Statistic,
This question is already answered in the fourth post of this thread.
marlon
I know, but I wanted to see exactly. what matt grime was hinting at about the vector and scalar forms.
 
  • #9
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Pseudo Statistic said:
I know, but I wanted to see exactly. what matt grime was hinting at about the vector and scalar forms.
Well, just the same. If you do not add the unitvector k (as the OP didn't) than you are not describing a vector but an ordinary number (ie the scalar). A vector consists out of a number (the component) and a unitvector.

In the case of a cross product, this unit vector in the RHS of the equation must be directed perpendicular to the plane of r and r_dot.

marlon
 
  • #10
matt grime
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Pseudo Statistic said:
I know, but I wanted to see exactly. what matt grime was hinting at about the vector and scalar forms.

hinting at? i was hinting at nothing, but merely observing that there was a mistake, one that marlon has adequately explained in my opinion.
 
  • #11
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marlon said:
Well, there are actually two formula's for calculating a vector product
1) [tex] r \times \dot{r} = sin \alpha |r| |\dot{r}| \vec{k} [/tex]
Ok, so I follow the above formula I get:

[tex] |r| = r [/tex]

which seems right.

It then gets a bit messy

[tex] |\dot{r}| = \sqrt {\dot{r}^2( cos^2\Phi + sin^2\Phi) + r^2 \dot{\Phi}^2 (-sin^2\Phi + cos^2\Phi})} [/tex]
 
  • #12
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ElDavidas said:
Ok, so I follow the above formula I get:
[tex] |r| = r [/tex]
which seems right.
correct
It then gets a bit messy
[tex] |\dot{r}| = \sqrt {\dot{r}^2( cos^2\Phi + sin^2\Phi) + r^2 \dot{\Phi}^2 (-sin^2\Phi + cos^2\Phi})} [/tex]
This is incorrect.

The i component is [tex]\dot{r} cos \Phi - r \dot{\Phi} sin \Phi[/tex]
The j component is [tex]\dot{r} sin \Phi + r \dot{\Phi} cos \Phi[/tex]

Now, what is the magnitude ?

marlon
 
  • #13
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marlon said:
Now, what is the magnitude ?
marlon
Ok, I get:

[tex] |\dot{r}| = \sqrt {\dot{r}^2 + r^2 \Phi^2}} [/tex]
 
  • #14
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ElDavidas said:
Ok, I get:
[tex] |\dot{r}| = \sqrt {\dot{r}^2 + r^2 \Phi^2}} [/tex]
that is correct

now calculate the angle alpha.

It is easier to work with determinants though


regards
marlon
 
  • #15
BobG
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marlon is right. The determinant method for finding the cross product is the best way to solve this. You're just killing yourself doing it your way.
[tex]\vec{h} = \left(\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k}\\r cos \phi & r sin \phi & 0\\ \dot{r}cos \phi - r \dot{\phi} sin \phi & \dot{r} sin \phi + r \dot{\phi} cos \phi & 0\end{array}\right)[/tex]
 
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  • #16
BobG
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marlon said:
regards
marlon
Dang, you're fast! :rofl: Not only was your correction faster than mine, the correction was gone by time I could check how much you beat my correction by.
 
  • #17
Astronuc
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It would be worthwhile to point out that i x j = - j x i = k, and i x i = j x j = k x k = 0.
 

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