# Question about moment of inertia

1. Aug 3, 2004

### gogetagritz

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?

2. Aug 4, 2004

### arildno

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.

3. Aug 4, 2004

### tomkeus

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.