Question about moment of inertia

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The discussion focuses on the moment of inertia, specifically the differences between scalar and second-order tensor solutions. It establishes that the scalar equation, represented as the integral over a volume of R^2 dm, is equivalent to I(1,1) in the matrix equation toa=I * alpha when analyzing uni-axial rotation. The conversation emphasizes that while the scalar expression is suitable for uni-axial rotation, using the tensor may simplify calculations due to its symmetric properties and diagonal representation in an orthogonal basis. It concludes that I(ω) is time-dependent when the instantaneous rotation axis changes direction.

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I have seen moment of inertia equations which give scalar and second order tensor solutions. When is it appropriate to use either equations, and is the scalar equation of just the integral over a volume of R^2 dm just the same thing as I(1,1) when solving the matrix equation toa=I * alpha?
 
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In the specific case where a rigid object undergoes merely uni-axial rotation, we may disregard the inertial tensor, and retain the object's moment of inertia about the given rotation axis (going through some point).

In fact, if you describe the inertial tensor in the (non-inertial) coordinate system of the instantaneous rotation axis and two mutually orthogonal axes to this, you will have in the energy equation, for example, that the rotational energy is given by [tex]\frac{1}{2}I_{\omega}\omega^{2}[/tex] where [tex]I_{\omega}[/tex] is called the moment of inertia with respect to the instantaneous rotation axis (going through C.M).

However, since the instantaneous rotation axis may change direction, [tex]I_{\omega}[/tex] is, in general, a function of time.

When is it appropriate to use the scalar expression?
This is appropriate only if you know (from somewhere else) that the object undergoes uni-axial rotation.
However, even if you know this, the tensor may be easier to calculate all the same, since you then have the opportunity to choose a coordinate system in which the terms are optimally easy to calculate.
 
Tensor of inertia is symmetric, thus it always has eigenvectors which make orthogonal basis in which it's representation is diagonal. In this case it acts like a scalar.
 

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