Tensors and moment of inertia

Discussion in 'Physics & Math' started by arfa brane, Dec 16, 2017.

  1. arfa brane call me arf Valued Senior Member

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    So vectors and operations on them. Scalar multiplication acts on single vectors and changes the magnitude.

    Take pairs of vectors, in physics of course, these have to have the same units if you want to add them together or "take" the inner product.
    Referring back to a previous diagram

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    If the vector P is also a position vector (rotating or fixed), then it's the sum of various other pairs of coordinate vectors present in the diagram.
    For instance \( P = p^1 + p^2 \). Another way to write this is \( P = p^i \) with implied summation. If the projection \( P_1 \) were added to the dashed line from the end of P onto X1 (maybe think of this line as a projector), the sum would also equal P. By adding the coordinate line \( \Xi^2 \) perpendicular to X1 and parallel to the projected line, you then have a real component to add.

    Note also that being parallel to implies being translated from, such that a direction is left the same . . .


     
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  3. arfa brane call me arf Valued Senior Member

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    --http://hermes.ffn.ub.es/luisnavarro/nuevo_maletin/Einstein_GRelativity_1916.pdf

    The moment of inertia tensor is a scalar in two dimensions; introduce the third dimension and cross product (in two dimensions the latter has to also be scalar), and then you can analyse precession of the axis of rotation, changes in torque etc.

    --http://www.askiitians.com/forums/Mechanics/10/7775/tensor.htm

    Steiner's theorem and formula relates parallel pairs of axes (usually one is the centre of rotation, or the centre of mass). In this theorem, the two parallel axes in say, the Z direction, have projections on them from an arbitrary point in an XY plane. Make the X and Y axes parallel to each other and use the covariance of the projection vectors. That is to say, introduce a linear system of coordinates.
     
    Last edited: Dec 29, 2017
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  5. arfa brane call me arf Valued Senior Member

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    I've made a rather egregious error with \( r \times v \). This is not the angular velocity \( \omega \) because \( r \times v \) will have units of square metres per second, and \( \omega \) has units radians per second, or just \( s^{-1} \).

    That is, \( \omega = d\theta / dt \) or \( \omega = \dot\theta \). However, the tangential velocity, \( v = \omega \times r \).
    What that means is that \( v \) as a cross product has the right units, metres per second, but in the rotating frame is a pseudovector (rank 1 pseudotensor) because of the way it transforms under a reflection. The magnitude of \( v \) is \( \|\omega\|\|r\| \).

    That's the case at least, when \( r \) lies in the same plane of rotation as the point of interest.
     
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