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Kinetic Energy due to rotational motion

  1. Aug 24, 2004 #1

    BobG

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    The kinetic energy of a body due to rotational motion is a scalar using the following equation:

    [tex]E=\frac{1}{2}I\omega^2[/tex]

    Except, the moment of inertia is nomally expressed as a matrix and the angular velocity as a vector, as in the equation for angular momentum:

    [tex]\vec{H}=\vec{\omega}[/tex]

    It's tempting to convert the moment of inertia and the angular velocity to scalar form by taking the norm of each, but this doesn't seem right. For example, a point exactly on the North pole should not contribute any inertia to the rotation of the Earth and contributes nothing to the kinetic energy.

    To convert moment of inertia to a scalar for the purposes of determining kinetic energy, would the proper thing to do be to only use the inertia about the rotational axis? In essence, if the rotating about the x axis, take the norm of the x axis components of the matrix (the moment of inertia and the xy & xz products of inertia)?

    (The norm of the angular velocity vector would still be the proper step, since the angular velocity vector always lies along the rotational axis.)
     
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  3. Aug 24, 2004 #2

    NateTG

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  4. Aug 25, 2004 #3

    pervect

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    What you want, I think is

    [tex] E =\frac{1}{2} \vec{\omega} \cdot \vec{I} \cdot \vec{\omega} [/tex]

    Writing this out as a sum, it looks like

    [tex] \sum_{i,j=1}^3 I_{ij} w_{i}w_{j} [/tex]
     
  5. Aug 25, 2004 #4

    BobG

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    Close. Moment of inertia is a 3x3 matrix and angular velocity is a 3-dimensional vector, so my calculator doesn't handle this in a nice simple step. Doing it manually with paper and pencil made it a little easier to see what I had to do.

    Multiplying the I matrix by the w vector gets me a 3 element matrix (basically, a 3-dimensional vector). I can either take the dot product between the result and the w vector or transpose the result of the matrix multiplication and multiply by the w vector. And, somewhere along the way, I have to multiply by 1/2.

    End result is to only use the matrix elements corresponding to the rotational axis, which makes sense.
     
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