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Calculation of angular acceleration (two rotating frames)

  1. Aug 5, 2010 #1
    Please see the attached figure.
     

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    Last edited: Aug 6, 2010
  2. jcsd
  3. Aug 6, 2010 #2
    Rotations don't commute. In other words a rotation by w1 followed by a rotation by w2 is different than w2 followed by w1, and in any case is NOT w1 + w2.
     
  4. Aug 6, 2010 #3
    It's true that finite rotations don't commute. But here I'm considering instantaneous angular velocities of two different reference frames with one rotating with respect to the other. Isn't the absolute angular velocity of the second frame (please see the figure) w1+w2?

     
  5. Aug 6, 2010 #4
    Not sure, maybe w1 (and w2 ) is different in expression 3 than 4, cause it is expressed in different coordinate system.
     
  6. Aug 6, 2010 #5
    I think you are misunderstanding the usual meaning of "finite" rotations. This doesn't mean "independent from time" or "instantaneous". It means rotations "by an angle big enough".
    If w1 and w2 are small, then w = w1 + w2 is valid only up to first order (in w1 and w2).
    Now note that the difference betwenn (3) and (4) in your picture is (w1 x w2), that is zero only if

    (1) w1 or w2 are parallel (rotations about the same axis commute), or
    (2) you can neglect second order terms
     
  7. Aug 6, 2010 #6

    D H

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    Staff Emeritus
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    Yes and no. Those two vectors ω1 and ω2 are not expressed in the same reference frames. One of them has to be transformed so that they can be added. Angular velocities are typically expressed in rotating frame coordinates. Three reasons:
    1. The inertia tensor of a rigid body is constant in body frame coordinates. It is a nasty, time varying beast when expressed in inertial coordinates.
    2. One standard way to measure the angular velocity of a rotating body is to attach a set gyros to that body. Those measurements will naturally be in rotating frame coordinates.
    3. Last but not least, expressing angular velocity in rotating frame coordinates is the convention.

    So, rather than saying

    [tex]
    \boldsymbol{\omega} = \boldsymbol{\omega}_1 + \boldsymbol{\omega}_2
    [/tex]

    it is better to say

    [tex]
    \boldsymbol{\omega} =
    \boldsymbol T_{1:2}\,\boldsymbol{\omega}_1 +
    \boldsymbol{\omega}_2
    [/tex]

    Here [itex]\boldsymbol T_{1:2}[/itex] is the transformation matrix from frame 1 to frame 2. Differentiating the above yields

    [tex]
    \dot{\boldsymbol{\omega}} =
    \boldsymbol T_{1:2}\,\dot{\boldsymbol{\omega}}_1 +
    \dot {\boldsymbol T}_{1:2}\,\boldsymbol{\omega}_1 +
    \dot{\boldsymbol{\omega}}_2
    [/tex]


    If you want to go against convention and represent ω1 in inertial coordinates and ω2 in frame 1 rotating coordinates, the angular velocity of frame 2 with respect to inertial is instead

    [tex]
    \boldsymbol{\omega} =
    \boldsymbol{\omega}_1 +
    \boldsymbol T_{1:I}\,\boldsymbol{\omega}_2
    [/tex]

    Differentiating will once again yield a term that involves the time derivative of a transformation matrix. So, what is the time derivative of a transformation matrix? You already have a big clue as to what this is in your expression (1),

    [tex]
    \dot{\mathbf v} = \dot{\mathbf v}' + \boldsymbol{\omega}\times\mathbf v
    [/tex]

    At the undergraduate level the derivation of the above is typically involves some rather kludgy hand waving. Let's do this with less hand waving (there still will be some hand waving at the end). Suppose q is some vector quantity. This can be expressed in two different coordinate systems, call them A and B. The relationship between these different representations of this vector is

    [tex]
    \mathbf q_A = \boldsymbol T_{B:A}\, \mathbf q_B
    [/tex]

    Using the chain rule to differentiating with respect to time,

    [tex]
    \dot{\mathbf q}_A =
    \boldsymbol T_{B:A}\, \dot{\mathbf q}_B +
    \dot {\boldsymbol T}_{B:A}\, \mathbf q_B
    [/tex]

    The components of the time derivatives [itex]\dot{\mathbf q}_A[/itex] and [itex]\dot{\mathbf q}_B[/itex] are the component-by-component time derivatives of the vector q expressed in frames A and B. In other words, these are time derivatives of the vector q as observed in and expressed in frames A and B respectively. All we are left with is the time derivative of the transformation matrix.

    The derivation of that is a two or three page mathout (being blinded by too much whiteness is called whiteout; being blinded by too much math is called mathout). So, without derivation,

    [tex]
    \begin{aligned}
    \dot T_{B:A}
    &= \boldsymbol T_{B:A} \boldsymbol S(\boldsymbol{\omega}_{A:B,B}) \\
    &= \boldsymbol S(\boldsymbol{\omega}_{A:B,A}) \boldsymbol T_{B:A} \\
    \dot T_{A:B}
    &= -\,\boldsymbol T_{A:B} \boldsymbol S(\boldsymbol{\omega}_{A:B,A}) \\
    &= -\boldsymbol S(\boldsymbol{\omega}_{A:B,B}) \boldsymbol T_{A:B}
    \end{aligned}
    [/tex]

    Here, S(ω) is the skew symmetric cross product matrix generated from ω. This matrix lets one write the cross product in matrix form:

    [tex]
    \boldsymbol{\omega} \times \mathbf q =
    \boldsymbol S(\boldsymbol{\omega}) \mathbf q
    [/tex]

    I was very explicit with coordinate systems in the above. For example, [itex]\boldsymbol{\omega}_{A:B,A}[/itex] is the angular velocity of frame B with respect to frame A, expressed in frame A coordinates.

    Going against convention and expressing angular velocity in parent frame coordinates leads to (being sloppy with coordinates here)

    [tex]
    \dot{\boldsymbol{\omega}} =
    \dot{\boldsymbol{\omega}}_1 +
    \dot{\boldsymbol{\omega}}_2 +
    \boldsymbol{\omega}_1 \times \boldsymbol{\omega}_2
    [/tex]

    This is equivalent your equation (3). Going with convention will lead to

    [tex]
    \dot{\boldsymbol{\omega}} =
    \dot{\boldsymbol{\omega}}_1 +
    \dot{\boldsymbol{\omega}}_2 -
    \boldsymbol{\omega}_2 \times \boldsymbol{\omega}_1
    [/tex]

    This is what your equation (4) should have been. Note that this result is consistent with the above.
     
  8. Aug 6, 2010 #7
    Why we would expect (3) and (4) to be equal anyway??? They represent the same vector but in different coordinate systems, this means they dont necessarily have to be equal do they?
     
  9. Aug 6, 2010 #8

    D H

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    Staff Emeritus
    Science Advisor

    There is a world of difference between what a vector represents and how it is represented. The displacement vector from point A to point B is conceptually the same vector regardless of representation. Whether this vector is represented in frame 1 coordinates or frame 2 coordinates or whatever is irrelevant; it is still the same vector.

    Taking the time derivative of some vector adds a little twist. The derivative of some vector q as observed by two different observers A and B are conceptually two different vectors. Hold an arm straight out in front of you. Whether you are standing still, walking, or spinning around, the velocity of your fist as observed by you is identically zero. The derivative of your fist as observed by someone who is standing still is non-zero. Computing the as observed by velocity is more than just a simple transformation. When you are spinning around, the velocity of your fist as observed by someone who isn't spinning around involves a material derivative (aka substantial derivative, substantive derivative, along with other names).

    In this case, however, we are talking about two vectors that are materially equivalent. They should be the same vector.
     
  10. Aug 6, 2010 #9
    I think i am lost here but one last question:

    When two vectors are materially equivalent ?
     
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