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Mechanics rotation Question

  1. Jan 11, 2014 #1
    Consider two cartesian coordinate system xyz and x` y` z` that initally concide. The x` y` z` undergoes three successive counterclockwise 45 rotations about the following axes: first, about the fixed z-axis;second, about its own x`-axis( which has been now rotated); finally, about its own z`-axis ( which has now been rotated);finally, about its own z`-axis (which has also been rotated). Find the components of a unit vector X in the xyz coordinate system that points along the direction of the x`-axis in rotated x` y` z` system.



    Attempted solution

    Let matrix A,B,& C denote the 3 rotation such that A denote the rotation 45 about z-axis,B rotation 45 about its x-axis,& C denote the matrix which has been rotated about its own z` axis.

    A = [cos 45 sin 45 0]
    [-sin 45 cos45 0]
    [0 0 1 ]

    B = [ 1 0 0 ]
    [ 0 cos45 -sin45]
    [ 0 sin45 cos45]

    C = [cos45 sin45 0]
    [-sin45 cos45 0]
    [0 0 1]



    I have them everything right but in my book it has in second B matrix it has
    j . k` as -sin45 but why is that!!?
     
  2. jcsd
  3. Jan 11, 2014 #2

    D H

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    I'm confused about what you are asking. Are you asking why it's then second row in the B matrix that has the negated value rather than the third?
     
  4. Jan 11, 2014 #3
    Yes in the book its the answer for j.k` = -sin(45) instead of sin(45) even though its cos(90 - theta) = sin(theta)
    or is their something I am missing?
     
  5. Jan 11, 2014 #4

    haruspex

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    There's an isomorphism consisting of rotating the three axes, z to x, x to y, y to z. This should map A to B.
    Applying that to your A matrix gives
    B = [ 1 0 0 ]
    [ 0 cos45 sin45]
    [ 0 -sin45 cos45]

    Edit: on second thoughts, that's a rotation about the original x axis.
     
    Last edited: Jan 12, 2014
  6. Jan 11, 2014 #5
    Ye I can see that coz cos(90 - theta) = sin(theta) thank you haru
    And cos(90 + theta) = -sin(theta) so I was wondering do I have the other matrices right for A B and C? @ haru?
     
  7. Jan 11, 2014 #6

    D H

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    Talking about isomorphisms is perhaps a bit advanced for this Introductory Physics question.

    The answer lies in the right hand rule. Set up a right handed coordinate system, with positive rotations indicated by the right hand rule. Now describe what happens with rotations about the various axes.

    I'll start with rotation about the +z axis because that's equivalent to the canonical 2D rotation in the x-y plane. Here's a rotation about the z axis looks:

    handz.gif

    Notice that a positive rotation about the +z axis rotates the +x axis toward the +y axis and rotates the +y axis toward the -x axis:
    [tex]\begin{aligned}
    \hat x' &= \phantom{-}\cos \theta\,\hat x + \sin \theta\,\hat y \\
    \hat y' &= -\sin\theta\,\hat x + \cos\theta\, \hat y \\
    \hat z' &= \phantom{-\sin\theta\,\hat x + \cos\theta\,\hat y} + \hat z
    \end{aligned}[/tex]
    Writing the above in matrix form,
    [tex]
    \begin{pmatrix} \hat x' \\ \hat y' \\ \hat z' \end{pmatrix} =
    \begin{pmatrix}
    \phantom{-}\cos \theta & \sin \theta & 0 \\
    -\sin \theta & \cos \theta & 0 \\
    0 & 0 & 1
    \end{pmatrix} \,\,
    \begin{pmatrix} \hat x \\ \hat y \\ \hat z \end{pmatrix}
    [/tex]
    That matrix on the right? That's your matrix C.


    Next let's look at a rotation about the +x axis:

    handx.gif

    Here a positive rotation about the +x axis rotates the +y axis toward the +z axis and rotates the +z axis toward the -y axis:
    [tex]\begin{aligned}
    \hat y' &= \phantom{-}\cos \theta\,\hat y + \sin \theta\,\hat z \\
    \hat z' &= -\sin\theta\,\hat y + \cos\theta\, \hat z \\
    \hat x' &= \phantom{-\sin\theta\,\hat y + \cos\theta\,\hat z} + \hat x
    \end{aligned}[/tex]
    Writing the above in matrix form,
    [tex]
    \begin{pmatrix} \hat x' \\ \hat y' \\ \hat z' \end{pmatrix} =
    \begin{pmatrix}
    1 & 0 & 0 \\
    0 & \phantom{-}\cos \theta & \sin \theta \\
    0 & -\sin \theta & \cos \theta
    \end{pmatrix} \,\,
    \begin{pmatrix} \hat x \\ \hat y \\ \hat z \end{pmatrix}
    [/tex]
    That matrix on the right? That's your matrix A.


    Finally, let's look at a rotation about the +y axis:

    handy.gif

    Here a positive rotation about the +y axis rotates the +z axis toward the +x axis and rotates the +x axis toward the -z axis:
    [tex]\begin{aligned}
    \hat z' &= \phantom{-}\cos \theta\,\hat z + \sin \theta\,\hat x \\
    \hat x' &= -\sin\theta\,\hat z + \cos\theta\, \hat x \\
    \hat y' &= \phantom{-\sin\theta\,\hat z + \cos\theta\,\hat x} + \hat y
    \end{aligned}[/tex]
    Writing the above in matrix form,
    [tex]
    \begin{pmatrix} \hat x' \\ \hat y' \\ \hat z' \end{pmatrix} =
    \begin{pmatrix}
    \cos\theta & 0 & -\sin\theta \\
    0 & 1 & 0 \\
    \sin\theta & 0 & \cos \theta
    \end{pmatrix} \,\,
    \begin{pmatrix} \hat x \\ \hat y \\ \hat z \end{pmatrix}
    [/tex]
    That matrix on the right? That's your matrix B.



    I took the above images from slide 63 of a set of tutorial slides presented at Siggraph 98:
    http://www.sdsc.edu/~moreland/courses/Siggraph98/vrml97/slides/mt0000.htm
     
  8. Jan 11, 2014 #7
    Thanks alot DH that makes sense.
     
  9. Jan 12, 2014 #8
    I have a question why do we have to multiply the matrix right to left instead of left to right to get the product


    for example in my example it is :

    CBA to get the total rotated components.
     
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