Derivation of the moment of inertia matrix

Angular momentum of a mass

Angular momentum is a vector quantity and can be expressed as the vector product of the potential vector and the momentum:

\[\vec{L} = \vec{r} \times \vec{p} \tag{1} \]

inertial tensor (math.)

For passing through the center of mass, around any axis with angular velocity\(\omega\)The angular momentum of a rotating rigid body, for the center of mass, is defined as:

\[H_{cg} = \int(\vec{r} \times (\vec{\omega} \times \vec{r})) dm \]

r and w can be written as vectors:

\[\vec{r} = x\vec{i} + y\vec{j} + z\vec{z} \]

\[\vec{\omega} = \omega_x\vec{i} + \omega_y\vec{j} + \omega_z\vec{z} \]

Vector cross product can be written in the form of matrix and vector multiplication:

\[\vec{r} \times \vec{q} = \begin{bmatrix} r_y q_z - r_z q_y \\ r_z q_x - r_x q_z \\ r_x q_y -r_y q_x \end{bmatrix} = \begin{bmatrix} 0 & -r_z & r_y \\ r_z & 0 & -r_x \\ -r_y & r_x & 0 \end{bmatrix} \begin{bmatrix} q_x \\ q_y \\ q_z \end{bmatrix} \]

Thus unfolding the cross product in the angular momentum formula:

\[H_{cg} = \int(\vec{r} \times (\vec{\omega} \times \vec{r})) dm = \int( \begin{bmatrix} 0 & -z & y \\ z & 0 & -x \\ -y & x & 0 \end{bmatrix} \begin{bmatrix} 0 & -\omega_z & \omega_y \\ \omega_z & 0 & -\omega_x \\ -\omega_y & \omega_x & 0 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} ) dm \\ = \int( \begin{bmatrix} -z\omega_z-y\omega_y & y\omega_x & z\omega_x \\ x\omega_y & -z\omega_z-x\omega_x & z\omega_y \\ x\omega_z & y\omega_z & -y\omega_y-x\omega_x \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} ) dm \\ = \int( \begin{bmatrix} (y^2+z^2)\omega_x-xz\omega_z - xy\omega_y \\ (x^2+z^2)\omega_y -yz\omega_z-xy\omega_x \\ (x^2+y^2)\omega_z - yz\omega_y - xz\omega_x \end{bmatrix} ) dm \\ \]

Order:

\[I_{xx} = \int(y^2+z^2)dm \]

\[I_{yy} = \int(x^2+z^2)dm \]

\[I_{zz} = \int(y^2+x^2)dm \]

\[I_{xy} = I_{yx} = \int(xy)dm \]

\[I_{xz} = I_{zx} = \int(xz)dm \]

\[I_{yz} = I_{zy} = \int(yz)dm \]

The above equation can be rewritten as:

\[H_{cg} = \begin{bmatrix} I_{xx}\omega_x-I_{xz}\omega_z - I_{xy}\omega_y \\ I_{yy}\omega_y -I_{yz}\omega_z-I_{xy}\omega_x \\ I_{zz}\omega_z -I_{yz}\omega_y - I_{xz}\omega_x \end{bmatrix} \\ = \begin{bmatrix} I_{xx} & -I_{xy} & -I_{xz} \\ -I_{xy} & I_{yy} & -I_{yz} \\ -I_{xz} & -I_{yz} & I_{zz} \\ \end{bmatrix} \begin{bmatrix} \omega_x \\ \omega_y \\ \omega_z \end{bmatrix} \]

Let the matrix I:

\[I = \begin{bmatrix} I_{xx} & -I_{xy} & -I_{xz} \\ -I_{xy} & I_{yy} & -I_{yz} \\ -I_{xz} & -I_{yz} & I_{zz} \\ \end{bmatrix} \]

Eventually get:

\[H_{cg} = I\vec{\omega} \]

I is the moment of inertia.

Both sides of the above equation are derived for time, and the inertia matrix I can be considered constant at very short times:

\[\frac{d H_{cg}}{dt} = I \frac{d \vec{\omega}}{dt} = I \vec{\alpha} \]

According to the angular momentum theorem (the total external moment is equal to the time rate of change of the angular momentum of the rigid body), the left-hand side is the external moment, therefore:

\[\vec{M} = I \vec{\alpha} \Rightarrow I^{-1} \vec{M} = \vec{\alpha} \tag{6} \]

This is the kinematic formula for rotation, where the total external moment overcomes the moment of inertia of rotation, giving the object in angular acceleration.
and the translational motion formula\(\vec{F}=m\vec{a}\)The same structure and status.

Calculating the moment of inertia

\[I_{xx} = \int(y^2+z^2)dm \]

\[I_{yy} = \int(x^2+z^2)dm \]

\[I_{zz} = \int(y^2+x^2)dm \]

\[I_{xy} = I_{yx} = \int(xy)dm \]

\[I_{xz} = I_{zx} = \int(xz)dm \]

\[I_{yz} = I_{zy} = \int(yz)dm \]

\[I = \begin{bmatrix} I_{xx} & -I_{xy} & -I_{xz} \\ -I_{xy} & I_{yy} & -I_{yz} \\ -I_{xz} & -I_{yz} & I_{zz} \\ \end{bmatrix} \]

According to its definition, the position of each mass microproduct changes when the attitude of the object changes in the case of a constant coordinate system, so each quantity needs to be recalculated, which brings a large amount of computation.

However it doesn't actually need to be that much of a hassle.

Write I in another form:

\[I=\sum m_i(r_i^T r_i 1 - r_i r_i^T) \]

where 1 represents the unit matrix.

As pictured in the reference pose:

\[I_{ref}=\sum m_i(r_i^T r_i 1 - r_i r_i^T) \]

When the object is rotated, the original microproducts are located in the\(Rr_i\)at, and therefore

\[I=\sum m_i((Rr_i)^T (Rr_i) 1 - (Rr_i) (Rr_i)^T) \\ =\sum m_i(r_i^T R^T Rr_i 1 - Rr_i r_i^T R^T) \\ =\sum m_i(r_i^T r_i 1 - Rr_i r_i^T R^T) \\ =\sum m_i(R r_i^T r_i 1 R^T - Rr_i r_i^T R^T) \\ =\sum m_i R(r_i^T r_i 1 - r_i r_i^T) R^T \\ =R I_{ref} R^T \]

Finally:

\[I = R I_{ref} R^T \]