Inertia tensor

Inertia Tensor in 2D

We can't have rotations in one dimension so two is the lowest number of dimensions to which we can apply this. In two dimensions the inertia tensor is a scalar quantity known as the second moment of mass.

I = ∫(r²)dm

In other words, we don't just sum the mass, but mass further from the rotation axis has more effect. Each element of mass is multiplied by the square of its distance from the centre of rotation.

Inertia Tensor in 3D

In 3 dimensions the inertia tensor is grade 2, which is a matrix. This makes things a bit more complicated as it means that a torque, say around the 'x' axis, can cause acceleration around the other axes.

[I] =

∫(rz²+ry²)dm -∫rx*rydm -∫rx*rzdm -∫ry*rxdm ∫(rz²+rx²)dm -∫ry*rzdm -∫rz*rxdm -∫rz*rydm ∫(rx²+ry²)dm

The inertia tensor is derived on this page.

This matrix is symmetrical around the leading diagonal and it has the property that it can be factored into rotational and diagonal parts as follows:

[I] = [R][D][Rt]

where:

• [I] = the inertia tensor matrix
• [R] = rotation matrix made up from the eigenvectors of [I]
• [D] = a diagonal matrix with the diagonal terms made up from the eigenvalues of [I] and the non-diagonal terms are zero.
• [Rt] = the transpose of [R]

Factoring matrices in this way is discussed on this page. What this means, in practical terms, is that we can always rotate the local coordinates of the solid object in such a way that [I] becomes a diagonal matrix. Which means that, in this axis, the acceleration will be around the same axis as the torque.

Another way of explaining this is that, there are certain symmetries of any solid object around which it will rotate without wobbling into a different axis. These symmetrical axes are the eigenvectors of [I].

The inertia matrix is often described as the inertia tensor, which consists of the same moments of inertia and products of inertia about the three coordinate axes.[6][23] The inertia tensor is constructed from the nine component tensors, (the symbol {\displaystyle \otimes } is the tensor product)

{\displaystyle \mathbf {e} _{i}\otimes \mathbf {e} _{j},\quad i,j=1,2,3,}

where {\displaystyle \mathbf {e} _{i},i=1,2,3} are the three orthogonal unit vectors defining the inertial frame in which the body moves. Using this basis the inertia tensor is given by

{\displaystyle \mathbf {I} =\sum _{i=1}^{3}\sum _{j=1}^{3}I_{ij}\mathbf {e} _{i}\otimes \mathbf {e} _{j}.}

This tensor is of degree two because the component tensors are each constructed from two basis vectors. In this form the inertia tensor is also called the inertia binor.

For a rigid system of particles {\displaystyle P_{k},k=1,...,N} each of mass {\displaystyle m_{k}} with position coordinates {\displaystyle \mathbf {r} _{k}=(x_{k},y_{k},z_{k})}, the inertia tensor is given by

{\displaystyle \mathbf {I} =\sum _{k=1}^{N}m_{k}\left(\left(\mathbf {r} _{k}\cdot \mathbf {r} _{k}\right)\mathbf {E} -\mathbf {r} _{k}\otimes \mathbf {r} _{k}\right),}

where {\displaystyle \mathbf {E} } is the identity tensor

{\displaystyle \mathbf {E} =\mathbf {e} _{1}\otimes \mathbf {e} _{1}+\mathbf {e} _{2}\otimes \mathbf {e} _{2}+\mathbf {e} _{3}\otimes \mathbf {e} _{3}.}

In this case, the components of the inertia tensor are given by

{\displaystyle {\begin{aligned}I_{11}=I_{xx}&=\sum _{k=1}^{N}m_{k}\left(y_{k}^{2}+z_{k}^{2}\right),\\I_{22}=I_{yy}&=\sum _{k=1}^{N}m_{k}\left(x_{k}^{2}+z_{k}^{2}\right),\\I_{33}=I_{zz}&=\sum _{k=1}^{N}m_{k}\left(x_{k}^{2}+y_{k}^{2}\right),\\I_{12}=I_{21}=I_{xy}&=-\sum _{k=1}^{N}m_{k}x_{k}y_{k},\\I_{13}=I_{31}=I_{xz}&=-\sum _{k=1}^{N}m_{k}x_{k}z_{k},\\I_{23}=I_{32}=I_{yz}&=-\sum _{k=1}^{N}m_{k}y_{k}z_{k}.\end{aligned}}}

The inertia tensor for a continuous body is given by

{\displaystyle \mathbf {I} =\iiint \limits _{Q}\rho (\mathbf {r} )\left(\left(\mathbf {r} \cdot \mathbf {r} \right)\mathbf {E} -\mathbf {r} \otimes \mathbf {r} \right)\,\mathrm {d} V,}

where {\displaystyle \mathbf {r} } defines the coordinates of a point in the body and {\displaystyle \rho (\mathbf {r} )} is the mass density at that point. The integral is taken over the volume {\displaystyle V} of the body. The inertia tensor is symmetric because {\displaystyle I_{ij}=I_{ji}}.

Alternatively it can also be written in terms of the angular momentum operator {\displaystyle [\mathbf {r} ]\mathbf {x} =\mathbf {r} \times \mathbf {x} }:

{\displaystyle \mathbf {I} =\iiint \limits _{Q}\rho (\mathbf {r} )[\mathbf {r} ]^{T}[\mathbf {r} ]\,\mathrm {d} V=-\iiint \limits _{Q}\rho (\mathbf {r} )[\mathbf {r} ]^{2}\,\mathrm {d} V}

The inertia tensor can be used in the same way as the inertia matrix to compute the scalar moment of inertia about an arbitrary axis in the direction {\displaystyle \mathbf {n} },

{\displaystyle I_{n}=\mathbf {n} \cdot \mathbf {I} \cdot \mathbf {n} ,}

where the dot product is taken with the corresponding elements in the component tensors. A product of inertia term such as {\displaystyle I_{12}} is obtained by the computation

{\displaystyle I_{12}=\mathbf {e} _{1}\cdot \mathbf {I} \cdot \mathbf {e} _{2},}

and can be interpreted as the moment of inertia around the {\displaystyle x}-axis when the object rotates around the {\displaystyle y}-axis.

The components of tensors of degree two can be assembled into a matrix. For the inertia tensor this matrix is given by,

{\displaystyle \mathbf {I} ={\begin{bmatrix}I_{11}&I_{12}&I_{13}\\I_{21}&I_{22}&I_{23}\\I_{31}&I_{32}&I_{33}\end{bmatrix}}={\begin{bmatrix}I_{xx}&I_{xy}&I_{xz}\\I_{yx}&I_{yy}&I_{yz}\\I_{zx}&I_{zy}&I_{zz}\end{bmatrix}}.}

It is common in rigid body mechanics to use notation that explicitly identifies the {\displaystyle x}, {\displaystyle y}, and {\displaystyle z}-axes, such as {\displaystyle I_{xx}} and {\displaystyle I_{xy}}, for the components of the inertia tensor.