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Rules regarding the vector cross product and dot product

  1. Dec 26, 2005 #1
    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?
     
    Last edited: Dec 26, 2005
  2. jcsd
  3. Dec 26, 2005 #2
    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]
     
    Last edited: Dec 26, 2005
  4. Dec 26, 2005 #3
    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?
     
    Last edited: Dec 26, 2005
  5. Dec 26, 2005 #4
    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.
     
  6. Dec 26, 2005 #5

    matt grime

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    No, the LHS is a vector and the RHS is a scalar, so try to have a think about what you really mean.
     
  7. Dec 26, 2005 #6
    r x r' = ( |r| |r'| sin a )K
    Where K is a perpendicular unit vector? (Perpendicular to r and r'?)
     
  8. Dec 26, 2005 #7
    Pseudo Statistic,

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

    marlon
     
  9. Dec 26, 2005 #8
    I know, but I wanted to see exactly. what matt grime was hinting at about the vector and scalar forms.
     
  10. Dec 26, 2005 #9
    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
     
  11. Dec 26, 2005 #10

    matt grime

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    hinting at? i was hinting at nothing, but merely observing that there was a mistake, one that marlon has adequately explained in my opinion.
     
  12. Dec 27, 2005 #11
    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]
     
  13. Dec 27, 2005 #12
    correct
    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
     
  14. Dec 27, 2005 #13
    Ok, I get:

    [tex] |\dot{r}| = \sqrt {\dot{r}^2 + r^2 \Phi^2}} [/tex]
     
  15. Dec 27, 2005 #14
    that is correct

    now calculate the angle alpha.

    It is easier to work with determinants though


    regards
    marlon
     
  16. Dec 27, 2005 #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]
     
    Last edited: Dec 27, 2005
  17. Dec 27, 2005 #16

    BobG

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    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.
     
  18. Dec 27, 2005 #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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