This is a study question I have been struggling with, I would appreciate help defining initial vectors to start the question.

We consider a thin circle shaped disk (mass m, radius R). The moments of inertia of the disk are $I_n=\frac{mR^2}{2}$ for rotation around the axis through the center of the disk and normal to the disk, and $I_p =\frac{mR^2}{4}$ for rotation around an axis through the center of the disk and in the plane of the disk. The disk is mounted on an axle through its center. This axis makes an angle $φ=\frac{\pi}{6}$ with the axis normal to the disk. The axis rotates together with the disk with a constant angular velocity $ω$.

enter image description here

I set $\omega$ to be pointing along the axis $n$, although I am not entirely sure that this is correct. I have also defined the tensor of inertia for the disk to be: $\tilde{I}=\begin{pmatrix} \frac{mR^2}{4}&0&0\\0&\frac{mR^2}{4}&0\\0&0&\frac{mR^2}{2}\end{pmatrix}$, with $\vec{L}=\tilde{I}\vec{\omega}$, with result $\vec{L}=\frac{m\omega r^2}{2}$. A simple nudge in the right direction or a confirmation that my result is correct would be ideal. Thank you!

  • $\begingroup$ $\begin{pmatrix}\frac{mR^2}{4}&0&0\\0&\frac{mR^2}{4}&0\\0&0&\frac{mR^2}{2}\end{pmatrix} \begin{pmatrix}0\\0\\\omega$, where the second matrix is $\omega^T$ $\endgroup$
    – benmcgloin
    Mar 2, 2021 at 18:44

2 Answers 2


Your MMOI matrix $\mathbf{I}_{\rm body}$ and the rotational velocity vector $\boldsymbol{\omega} = \pmatrix{0\\0\\ \omega}$ are not in the same basis vectors. So you cannot just multiply the two. You need a rotational transformation.

Consider the rotation of angle $\varphi$ about the x axis as a 3×3 rotatation matrix $$ \mathbf{R} = \pmatrix{ 1 & 0 & 0 \\ 0 & \cos \varphi & -\sin \varphi \\ 0 & \sin \varphi & \cos \varphi}$$ that transforms from body coordinates to the inertial reference frame.

  1. Find the rotational velocity vector on the body coordinates

$$ \boldsymbol{\omega}_{\rm body} = \mathbf{R}^\top \boldsymbol{\omega} $$

  1. Find the angular momentum vector on the body coordinates

$$ \boldsymbol{L}_{\rm body} = \mathbf{I}_{\rm body} \boldsymbol{\omega}_{\rm body} $$

  1. Rotate the angular veclocity vector back into the inertial reference frame

$$ \boldsymbol{L} = \mathbf{R}\,\boldsymbol{L}_{\rm body} $$

Or you can do all of the above in one step

$$ \boldsymbol{L} = (\mathbf{R} \, \mathbf{I}_{\rm body} \mathbf{R}^\top) \; \boldsymbol{\omega} $$

  • $\begingroup$ So essentially the moments of inertia specified in the question are not in the correct plane? Also why have you multiplied by $R^T$, as I understand it the multiplication $R\omega$ is defined. Thank you! $\endgroup$
    – benmcgloin
    Mar 2, 2021 at 19:00
  • $\begingroup$ $\mathbf{R}^\top$ rotates from the world coordinate system to the body coordinate system. In the world coordinate system there is only $\omega$ along the z axis, but as far as the body (disk) is concerned it sees rotation about two axis. Body MMOI components are almost aways defined along the principal axis of rotation, but in this case the shaft is at an angle to the principal axis, and thus the need for the rotation operations. $\endgroup$
    – JAlex
    Mar 2, 2021 at 19:05
  • $\begingroup$ You have really cleared this question up for me! Thank you for the explanations. $\endgroup$
    – benmcgloin
    Mar 2, 2021 at 19:10

This answer is along the same lines as the earlier answer by @JAlex. I suggest you consult a physics mechanics text, such as Classical Mechanics by Goldstein, for more details. I used Mathematica to do the calculations that follow.

There are two sets of coordinates to consider: the inertial (space) coordinates and the body coordinates fixed (and rotating) with the body. Using the Euler angles a vector can be expressed in either set of coordinates. The figure below provides the coordinates and the Euler angles using the convention in Goldstein.

enter image description here

$X_I$, $Y_I$, $Z_I$ are the inertial (space) coordinates, $X_B$, $Y_B$, and $Z_B$ are the body axes that are fixed in the body and are chosen to be the principal axes of the body. $X’$ is the line of nodes.

I will re-label the angles and axes in your figure to correspond to those in the figure above. $Z_I$ is the axle, $\theta$ is $\phi$ in your diagram, $Z_B$ is in the n direction in your diagram.

For your problem, $\theta$ is fixed , $\psi$ is zero and $\dot \phi$ is constant. In the space coordinates, $\vec \omega$ is fixed in the $Z_I$ direction: $\vec \omega = \{0,0,\omega \}$. The inertia tensor $\bf I_{body}$ in the body coordinates is as you state it. You need to transform the $\vec \omega$ from space to body coordinates.

To transform a vector $\vec V$ from space to body coordinates $\vec V_{body} = {\bf A} \vec V_{inertial}$ where $\bf A$ is the transformation matrix given in Goldstein. (To transform a vector from body to space coordinates $\vec V_{inertial} = {\bf A^T} \vec V_{body}$, where ${\bf A^T}$ is the transposed matrix.) For your case, with $\psi$ of zero ${\bf A}$ is

$\left( \begin{array}{ccc} \cos (\phi ) & \sin (\phi ) & 0 \\ -\cos (\theta ) \sin (\phi ) & \cos (\theta ) \cos (\phi ) & \sin (\theta ) \\ \sin (\theta ) \sin (\phi ) & \sin (\theta ) (-\cos (\phi )) & \cos (\theta ) \\ \end{array} \right) $

$\vec \omega_{body} = {\bf A} \vec \omega_{inertial}$ = $\{0,\omega \sin (\theta ),\omega \cos (\theta )\}$

$\vec L_{body} = {\bf I_{body}} \vec \omega_{body}$ = $\left\{0,\frac{1}{4} m R^2 \omega \sin (\theta ),\frac{1}{2} m R^2 \omega \cos (\theta )\right\}$

$\vec L_{space} = {\bf A^T} \vec L_{body}$ =$\left\{\frac{1}{4} m R^2 \omega \sin (\theta ) \cos (\theta ) \sin (\phi ),-\frac{1}{4} m R^2 \omega \sin (\theta ) \cos (\theta ) \cos (\phi ),\frac{1}{4} m R^2 \omega \sin ^2(\theta )+\frac{1}{2} m R^2 \omega \cos ^2(\theta )\right\}$

Since the object is rotating about an axis $Z_I$ that is not a principal axis, torque is required to maintain $\vec \omega$ fixed along $Z_I$.

The torque $\vec N$ is given by the Euler equations (see Goldstein).

For constant $\vec \omega$, $\vec\omega \times ({\bf I} \cdot \vec \omega) = \vec N$. $\vec N_{body} = \left\{\frac{1}{4} m R^2 \omega ^2 \sin (\theta ) \cos (\theta ),0,0\right\}$.

$\vec N_{space} = \left\{\frac{1}{4} m R^2 \omega ^2 \sin (\theta ) \cos (\theta ) \cos (\phi ),\frac{1}{4} m R^2 \omega ^2 \sin (\theta ) \cos (\theta ) \sin (\phi ),0\right\}$

You can also transform the inertia tensor between body and inertial coordinates. $\bf I_{inertial} = \bf A^T \cdot I_{body} \cdot A =$

$\tiny \left( \begin{array}{ccc} \frac{1}{2} m R^2 \sin ^2(\theta ) \sin ^2(\phi )+\frac{1}{4} m R^2 \cos ^2(\theta ) \sin ^2(\phi )+\frac{1}{4} m R^2 \cos ^2(\phi ) & -\frac{1}{4} m R^2 \cos ^2(\theta ) \sin (\phi ) \cos (\phi )-\frac{1}{2} m R^2 \sin ^2(\theta ) \sin (\phi ) \cos (\phi )+\frac{1}{4} m R^2 \sin (\phi ) \cos (\phi ) & \frac{1}{4} m R^2 \sin (\theta ) \cos (\theta ) \sin (\phi ) \\ -\frac{1}{4} m R^2 \cos ^2(\theta ) \sin (\phi ) \cos (\phi )-\frac{1}{2} m R^2 \sin ^2(\theta ) \sin (\phi ) \cos (\phi )+\frac{1}{4} m R^2 \sin (\phi ) \cos (\phi ) & \frac{1}{4} m R^2 \cos ^2(\theta ) \cos ^2(\phi )+\frac{1}{2} m R^2 \sin ^2(\theta ) \cos ^2(\phi )+\frac{1}{4} m R^2 \sin ^2(\phi ) & -\frac{1}{4} m R^2 \sin (\theta ) \cos (\theta ) \cos (\phi ) \\ \frac{1}{4} m R^2 \sin (\theta ) \cos (\theta ) \sin (\phi ) & -\frac{1}{4} m R^2 \sin (\theta ) \cos (\theta ) \cos (\phi ) & \frac{1}{4} m R^2 \sin ^2(\theta )+\frac{1}{2} m R^2 \cos ^2(\theta ) \\ \end{array} \right)$

$\bf I_{body} = \bf A \cdot I_{inertial} \cdot A^T = \left( \begin{array}{ccc} \frac{m R^2}{4} & 0 & 0 \\ 0 & \frac{m R^2}{4} & 0 \\ 0 & 0 & \frac{m R^2}{2} \\ \end{array} \right)$, your original inertia tensor in body coordinates using the principal axes.

  • $\begingroup$ Thank you for the response! I had not considered the situation fully but your explanation clears it up wonderfully. I solved the question fully and now understand the physics behind it. Thank you again🤝 $\endgroup$
    – benmcgloin
    Mar 4, 2021 at 19:06
  • 1
    $\begingroup$ You are welcome. $\endgroup$
    – John Darby
    Mar 4, 2021 at 19:08
  • $\begingroup$ I feel like an astronaut in the ocean aye! $\endgroup$ Mar 4, 2021 at 19:52
  • $\begingroup$ I added how to transform the inertia tensor between body and inertial coordinates. $\endgroup$
    – John Darby
    Mar 4, 2021 at 22:17

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