- #1
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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.)
[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.)