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Wedge Symbol

  1. Aug 13, 2013 #1
    Hi guys,
    Would you please help me by telling me the meaning of the wedge symbol in the equation attached?

    ω=angular velocity.

    Since the inertia is constant, then why do we need to add the second term? I mean why isn't the differentiation of (Iω) simply equal to (Iω')?

    Attached Files:

  2. jcsd
  3. Aug 14, 2013 #2

    D H

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    Too long, didn't read answer: Abuse of notation, reference frames, and the transport theorem.

    The left hand side of that equation would be better written as ##{\frac{d\vec L}{dt}}^{(I)}##, where the superscript (I) means "the time derivative of angular momentum from the perspective of an inertial frame."

    Without proof, the transport theorem relates the time derivative of some vector quantity ##\vec q## as observed from the perspective of an inertial observer versus that of a rotating observer as
    [tex]{\frac{d\vec q}{dt}}^{(I)} = {\frac{d\vec q}{dt}}^{(R)} + \vec\omega \times \vec q[/tex]

    Use ##\vec q = \vec r## (i.e., position vector) and differentiate twice and you'll get the standard relation involving centrifugal and coriolis accelerations between accelerations as observed in a inertial frame versus that observed in a rotating frame.

    What happens if you use ##\vec q = \vec L \equiv I\vec \omega\ ##? The left hand side, ##{\frac{d\vec L}{dt}}^{(I)}##, becomes the external torque. The right hand side becomes ##{\frac{d\vec L}{dt}}^{(R)} + \vec\omega\times \vec L##.

    In any frame, ##\vec L = I\vec \omega##, but you have to beware that the inertia tensor I and the angular velocity ω are frame-dependent quantities. The inertia tensor of a rigid body is constant in a frame rotating with that body. Thus ##{\frac{d\vec L}{dt}}^{(R)} = {\frac{d}{dt}^{(R)} (I\vec \omega)}= I\frac{d\vec \omega}{dt}##. Note that I've dropped the superscript R from the final derivative because angular acceleration (time derivative of angular velocity) is the same vector in the inertial and rotating frame.

    Putting this all together yields
    \vec \tau \equiv \frac{d}{dt}^{(I)}(I\vec \omega) =
    I \frac {d\vec \omega}{dt} + \vec \omega\times(I\omega)
  4. Aug 18, 2013 #3
    Thank you for the detailed reply. In reality, I am dealing with modelling an aircraft. ω is actually the body angular velocity of the craft and it is not the angular velocity of the rotating frame. I know that ## \vec \omega\times(I\omega) ## would represent the Coriolis effect and for finding it we need to know at what constant angular velocity the rotating frame is rotating. ω here is actually the craft angular velocity in its local frame not the velocity of the rotating frame. I read the following Wiki article and got some info. about the topic too. I might be confusing between frames and which ones I am to refer to.

  5. Aug 19, 2013 #4

    D H

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    You most certainly are working with a rotating frame. It's the body frame of your aircraft. The canonical aircraft body frame has its origin of the frame at the center of mass of the aircraft, and has axes such that the +x body axis points forward, the +y axis points the right wing, and the +z axis is x cross y to complete a right hand system.

    There are two key issues with this frame. One is that the center of mass shifts as passengers and cargo are loaded or unloaded and as fuel is consumed. The other: How do you specify the location of the center of mass? Aircraft designers use a different but related frame to address these issues. The canonical aircraft structure frame has its origin a fixed distance in front of and below the nose of the aircraft. The structural +x axis points toward the rear rather than the front, parallel to the centerline of the aircraft. The structural +z axis points up rather than down, and +y is z cross x to complete a right hand system. The body and structural frames as described differ by a shift in origin and a 180 degree yaw.

    You probably want the body frame rather than the structural frame because translational and rotational motion are decoupled in a center of mass frame. You do need to know of that other frame however because that's typically the frame in which center of mass is specified.

    Regarding that ##\vec{\omega} \times (I \vec{\omega})## term, this is *not* the Coriolis effect. The Coriolis force is given by ##-2m\vec{\Omega} \times \vec v##. The Euler torque, as it is sometimes called, is given by ##-\vec{\omega} \times (I \vec{\omega})## (note the sign change). Here, ##I## is the aircraft's inertia tensor about the center of mass expressed in some body-fixed frame and ##\vec \omega## is the angular velocity of the aircraft with respect to inertial but expressed in the same frame as is the inertia tensor. That weird way of representing angular velocity is basically what your aircraft gyros report.

    Your gyros actually report their output in a third body-fixed frame, the gyro case frame. The gyros don't know if you mounted them aligned with the body frame, upside down, or at some cockeyed angle. The gyro case frame x, y, and z axes are specific to the gyro. You know (or should know) how the gyro assembly is mounted to the aircraft, so you know (or should know) the relation between the gyro case frame and the aircraft structural and body frames. You can ignore this detail if you intentionally mount the gyros so that the case frame axes correspond to the (for example) body axes. It's still good to be aware of this the concept of a case frame. For example, you (or the aircraft designers) might have intended to mount the gyros so that they are perfectly aligned with the body axes, but nothing is perfect in reality. There's always going to be some misalignment. If you want to be very precise you should model this misalignment in your dynamics.
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