Question about moment of inertia

AI Thread Summary
Moment of inertia can be expressed as either a scalar or a second-order tensor, with the choice depending on the object's rotational dynamics. The scalar equation is suitable for rigid bodies undergoing uni-axial rotation, while the tensor approach is more versatile, allowing for optimal coordinate system selection. In cases of changing instantaneous rotation axes, the moment of inertia becomes time-dependent. The tensor representation is symmetric and can be diagonalized, simplifying calculations. Understanding when to use each form is crucial for accurate rotational energy assessments.
gogetagritz
Messages
8
Reaction score
0
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?
 
Physics news on Phys.org
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 \frac{1}{2}I_{\omega}\omega^{2} where I_{\omega} 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, I_{\omega} 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.
 
I have recently been really interested in the derivation of Hamiltons Principle. On my research I found that with the term ##m \cdot \frac{d}{dt} (\frac{dr}{dt} \cdot \delta r) = 0## (1) one may derivate ##\delta \int (T - V) dt = 0## (2). The derivation itself I understood quiet good, but what I don't understand is where the equation (1) came from, because in my research it was just given and not derived from anywhere. Does anybody know where (1) comes from or why from it the...
Back
Top