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I Angular veolcity in rotating frame

  1. Apr 28, 2017 #1
    Hello everyone,

    I have some conceptual problems understanding the rotating frame transformation.

    Take the center of the Earth as inertial frame's origin and another point in Hawaii as rotating frame's origin.
    In many lecture notes from internet, or Marion chapter 10.
    The vector describing the rotating frame is ω, which is pointing upward to the north pole of the Earth.
    However, if we think the angular velocity of rotating frame's origin as r×v. It is not directly pointing to North pole and shifted some amount.

    I have read the wiki page and it used
    ω = r×v
    v = ω×r simultaneously, which confuses me a lot !

    Can someone help me clear the concept?
    I believe the answer should be ω is point to north pole in the transforming equations
    but what is wrong with ω in ω = r×v ? This is the definition of angular veolcity but it seems does not match the ω in the frame transforming equations.

    Thanks for you help.
     
  2. jcsd
  3. Apr 28, 2017 #2
    Let's clarify something indeed.

    Theorem 1. There exists a unique vector ##\boldsymbol \omega## such that for every pair of points ##A,B## belonging to a rigid body it follows that
    $$\boldsymbol v_A=\boldsymbol v_B+\boldsymbol\omega\times \boldsymbol{BA};$$ here ##\boldsymbol v_A,\boldsymbol v_B## are the velocities of the points ##A,B##.

    Definition. The vector ##\boldsymbol\omega## is referred to as angular velocity of the rigid body.

    Remark. Since a moving coordinate frame can be considered as a rigid body we also obtained the definition of the angular velocity for a frame.
    2) The vector ##\boldsymbol\omega## is actually an axial vector.

    To compute the angular velocity in the concrete problems the following two theorems are used.

    Theorem 2. Let ##Oxyz## be a fixed Cartesian frame and another Cartesian frame ##Ox'y'z',\quad z\equiv z'## is rotating about their common axis ##z## such that
    $$\boldsymbol e_{x'}=\cos\psi \boldsymbol e_x+\sin\psi \boldsymbol e_y,\quad \boldsymbol e_{y'}=-\sin\psi\boldsymbol e_x+\cos\psi \boldsymbol e_y,$$ here ##\psi=\psi(t)## is the angle of rotation.
    Then the angular velocity of the frame ##Ox'y'z'## is given by the formula ##\boldsymbol \omega=\dot \psi \boldsymbol e_z##.

    Theorem 3. Let ##Oxyz## be a fixed Cartesian frame. There are also two moving Cartesian frames ##O'x'y'z'## and ##O''x''y''z''##. Angular velocity of the frame ##O'x'y'z'## relative the frame ##Oxyz## is equal to ##\boldsymbol \omega_1## and the angular velocity of the frame ##O''x''y''z''## relative the frame ##O'x'y'z'## is equal to ##\boldsymbol \omega_2##.
    Then the angular velocity of the frame ##O''x''y''z''## relative the frame ##Oxyz## is computed as follows
    ##\boldsymbol \omega=\boldsymbol \omega_1+\boldsymbol \omega_2##
     
  4. Apr 28, 2017 #3
    Thanks for your reply zwierz.
    I 100% follow and understand your paragraphs and it helps a lot.
    One more question,
    You mentioned that there exist a unique vector ω to transform from inertial into the rotating frame
    and it is called 'angular velocity of the frame'.
    Does this ω have anything to do with the r × v, 'the angular velocity of a particle' ?
    Or they just share similar name but actually are different things?

    I understand in the case of your thm2. The angular velocity of the frame coincide 'ω = dψ/dt ' coincide with the 'r × v of origin of rotating frame'. However the coincidence does not occur in a 3D case, say a rotating frame on Earth's surface.
     
  5. Apr 28, 2017 #4
    I just give you an informal advice (whispering) since for sure other participants will not agree with me. Forget the formula r × v because it helps to do mistakes and nothing else.
     
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