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Lagrangian dynamics problem

  1. Nov 9, 2006 #1
    My question pertains to Example 1.2 of Schaum's Outline of Lagrangian Dynamics by Dare A. Wells, chapter 1, page 4.

    You can view the diagram and the example (1.2) by going to the following link on Amazon.com and clicking on Excerpt, and then going to page 4:

    http://www.amazon.com/gp/reader/0070692580/ref=sib_dp_pt/002-4293223-1548865#reader-link

    The goal of the example is to take the equations of motion in the inertial coordinates and find the corresponding equations of motion in the noninertial coordinates.

    I'm confused on the part that says, "Reference to the figure shows that":

    [tex]x_1 = x_2 \cos \omega t - y_2 \sin \omega t[/tex]
    [tex]y_1 = x_2 \sin \omega t + y_2 \cos \omega t[/tex]

    I don't understand how these two expressions have been derived. Can someone please explain this to me?

    Thank you.
     
  2. jcsd
  3. Nov 10, 2006 #2

    dextercioby

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    It's simply the rotation matrix

    [tex] \left(\begin{array}{cc}\cos \omega t & -\sin\omega t \\ \sin\omega t & \cos\omega t \end{array}\right) [/tex]

    applied to the vector

    [tex] \left(\begin{array}{c}x_{2} \\ x_{1}\end{array}\right) [/tex]

    Daniel.
     
  4. Nov 10, 2006 #3
    Thank you.

    But my math background isn't so solid. Could you please explain how the rotation matrix works?
     
  5. Nov 12, 2006 #4
    The rotation matrix causes a pure rotational transformation of the coordinate system through an angle theta. I would advice you to look through your Linear Algebra Book for information on it.
     
  6. Nov 12, 2006 #5

    quasar987

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    It takes a vector (x2,y2) and rotates it anti-clockwise by an angle of wt (radians):

    [tex] \left(\begin{array}{c}x_{1} \\ y_{1}\end{array}\right) = \left(\begin{array}{cc}\cos \omega t & -\sin\omega t \\ \sin\omega t & \cos\omega t \end{array}\right) \left(\begin{array}{c}x_{2} \\ y_{2}\end{array}\right) [/tex]
     
    Last edited: Nov 12, 2006
  7. Nov 12, 2006 #6

    nrqed

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    You should probably first check using simple geometry that the equations are valid before using a rotation matrix (otherwise it might always sound like something a bit mysterious to you).

    Consider the equation for x1.

    Construct a right angle triangle with x2 being the hypothenuse and the two sides parallel to the x1 and y1 axis. Then look at what the quantity [itex] x_2 cos (\omega t) [/itex] represents. It is equal to x1 plus an extra horizontal piece that sticks out at the left of the origin. You have to substract that extra piece to get x1.

    Now construct a second right angle triangle with y2 being the hypothenuse and the other two sides being parallel to the x1 and y1 axis. You should see that the extra piece we had is just the side of this new right angle triangle parallel to the x1 axis. And that piece is simply [itex] y_2 sin (\omega t) [/itex].

    So [tex] x_1 = x_2 cos (\ometa t) - y_2 sin (\omega t) [/tex].

    It's only after you have derived those equations using simple geometry that you should then learn that this can be couched down in the language of a rotation matrix. I personally think that it is inappropriate pedagogically to throw in the rotation matrix without first showing that it comes out of simple geometry. But that's just my opinion.

    Patrick
     
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