I am studying mechanical engineering and I've got a problem with the angular momentum of objects that have a rotation which is rather complex to describe like the following:

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The shaft rotates around the $e_2$-axis with $w_f$. On the end of the shaft is a disk that rotates around the $e_3$-axis with $w_{rel}$.

Let's get a little bit simpler for now. I will assume that the shaft doesnt exist and only the disk rotates around its center. I know that $\theta_0 = \dfrac{m r^2}{2}$ and therefor $L=\theta_0\cdot w_{rel}$

If the disk would not spin around its center but around point p (in $e_3$ direction), I would use the parallel axis theorem: $\theta = \theta_0 + m\cdot l^2$.

Now I got 2 questions:

1: How would one deal and calculate the angular momentum of the system above

2: If $w_f$ and $w_{rel}$ would be parallel, how could one calculate the angular momentum then?

I am very happy for any advice.


The direction of the rotational Motion of the disk around ist Center Points in a direction perpendicular to the shaft Rotation direction, i.e. the angular Momentum Vector for Rotation around the shaft is orthogonal to the angular Momentum Vector around the Center of the disk. In this case you can use Addition of angular Momentum vectors. The total angular Momentum reads:

$\vec L = \vec L_{shaft} + \vec L_{center}$

The contribution due to the shaft Rotation is given by $\vec L_{shaft} = (l\vec{e_1^*}) \times (mv \vec{e_3^*})$, where the Rotation Velocity $v$ has Magnitude $v = l \omega_f$.

By evaluating the cross product, you will have an angular Momentum Vector pointing to the direction $\vec{e_2^*}$. The Rotation around the Center will be parallel to the direction $\vec{e_3^*}$. The resulting angular Momentum $\vec L$ will be expressed as linear combination of two unit vectors. To determine the Magnitude, simply take its absolute value (Pythagoras Theorem for Vector lengths).

Now to your second Question: If the two angular velocities are parallel, you can proceed as the same. In this case there will be $\vec L_{shaft} \parallel \vec{e_2^*}$, while the Magnitude is the same. Resulting angular Momentum will be pointing exactly in direction of $\vec{e_2^*}$. YOu will get the same as that you would obtained by parallel axis Theorem.


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The components of the disc angular momentum $\vec{L}_D$ given in $(e^*_1\,,e^*_2\,,e^*_3)_P$ coordinate system are:


with $\vec{L}_P=\vec{r}_P\times m\,\vec{v}_P$ we obtain:

$$\vec{L}_D=\vec{r}_P\times m\,\vec{v}_P+\Theta_0\,\vec{\omega}_{rel}\tag 1$$

$\vec{v}_P$ is the velocity at point $p\quad$,$\vec{v}_P=\vec{\omega}_f\times \vec{r}_P$


$\vec{r}_P=\begin{bmatrix} -l\\ 0\\ 0\\ \end{bmatrix}$

$\vec{\omega}_{rel}=\begin{bmatrix} 0\\ 0\\ \omega_r\\ \end{bmatrix}$

$\vec{\omega}_{f}=\begin{bmatrix} 0\\ \omega_f\\ 0\\ \end{bmatrix}$

The components of the disc angular momentum equation (1)

$\vec{L}_D=\begin{bmatrix} 0\\ m\,l^2\,\omega_f\\ \Theta_0\,\omega_r\\ \end{bmatrix} $

For $\vec{\omega}_f\parallel \vec{\omega}_{rel} $

$ \Rightarrow$

$\vec{\omega}_{f}=\begin{bmatrix} 0\\ 0\\ \omega_f\\ \end{bmatrix}$

you get for the components of the disc angular momentum

$\vec{L}_D=\begin{bmatrix} 0\\ 0\\ m\,l^2\,\omega_f+\Theta_0\,\omega_r\\ \end{bmatrix} $


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