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marooned

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- Thread starter marooned
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- #1

marooned

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- #2

marooned

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- #3

DaTario

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Best regards

DaTario

- #4

marooned

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- #5

Yegor

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marooned said:

I think you aren't right here. Distance to the axis doesn't remains the same for each point (it doesn't equals the distance to the centre of disc, but shortest distance to the axis!).

If you know I according to 3 orthogonal axis, then you can calculate I for each axis (through the same point) using [tex]I=I_1 (\cos{\alpha})^2+I_2 (\cos{\beta})^2+I_3 (\cos{\gamma})^2[/tex]. Here [tex]I_k[/tex] is I according to k-axis and [tex]\alpha[/tex] is angle between your new axis and k-axis.

- #6

marooned

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- #7

Crosson

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[tex]\tau = I \alpha[/tex]

We want this equation to be valid all the time, but the torque and angular acceleration vectors are not always in the same direction, so MoI cannot always be represented by a scalar. In general, MoI can be represented by a 3x3 matrix (infacta second rank tensor).

http://scienceworld.wolfram.com/physics/MomentofInertia.html

- #8

marooned

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marooned

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