Angular momentum of a body about a point rotating about its own axes

Question

I want to calculate angular momentum of a sphere about point O. The sphere is rotating about its two axes with angular velocities $w_1$ and $w_2$.

I know that angular momentum = $m\vec{r}\times\vec{v} + Iw$, where v is velocity of centre of mass. Here, v=0, therefore angular momentum of COM = 0. But, the body is itself rotating. Now, which angular momentum should I take? $Iw_1$ , $Iw_2$ , $Iw_1 + Iw_2$ , Components of $Iw_1$ and $Iw_2$ along r. or What?

Answer 1

A rigid body can only have one rotation axis. When the angular velocity vector has multiple non-zero components, like $$\vec{\omega} = \pmatrix{ \omega_1 & \omega_2 & 0}$$ then the magnitude of rotation is described by the length of the vector $$ \omega = \| \vec{\omega} \| = \sqrt{ \omega_1 ^2 + \omega_2 ^2 } $$

The rotation axis direction is the unit vector along $\vec{\omega}$

$$ \hat{\rm rot} = \frac{ \vec{\omega} }{ \| \vec{\omega} \|} = \pmatrix{ \frac{\omega_1}{\sqrt{ \omega_1 ^2 + \omega_2 ^2 }} \\ \frac{\omega_2}{\sqrt{ \omega_1 ^2 + \omega_2 ^2 }} \\ 0 } $$

In case of a sphere (where the mass moment of inertia is uniform with direction) the angular momentum magnitude is $$ L = {I}\, \omega ={I}\, \sqrt{ \omega_1 ^2 + \omega_2 ^2 } $$

or by component

$$ \begin{matrix} L_1 = I\, \omega_1 \\ L_2 = I\, \omega_2 \\ L_3 = 0 \end{matrix} $$

and $$ L = \sqrt{ L_1^2 + L_2^2 + L_3^2 } $$

In general, it is easier to consider the vector form of the above with a matrix/vector equation

$$ \vec{L} = \mathrm{I}\, \vec{\omega} $$ $$ \pmatrix{L_1 \\ L_2 \\ L_3 } = \begin{vmatrix} I_1 & 0 & 0 \\ 0 & I_2 & 0 \\ 0 & 0 & I_3 \end{vmatrix} \pmatrix{\omega_1 \\ \omega_2 \\ \omega_3} $$

_{For a sphere $I_1 = I_2 = I_3 = I$.}

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Even more useless information below:

Rotation of a rigid body happens along a line in space (the rotation axis). The location of this line relative to the COM is given by $$ \vec{r}_{\rm rot} = \frac { \vec{\omega} \times \vec{v} }{ \| \vec{ \omega} \|^2 } $$ where $\vec{v}$ is the velocity vector of the COM.

Corollary to this is the fact the momentum happens along a line in space (the axis of percussion), in such a way that a single impact along this line can instantaneously immobilize a rotating rigid body. This axis has direction along the linear momentum $\vec{p} = m \vec{v}$ and is located relative to the COM at $$ \vec{r}_{\rm imp} = \frac{ \vec{p} \times \vec{L} }{ \| \vec{p} \|^2}$$ where $\vec{L} = \mathrm{I}\, \vec{\omega}$ is the angular momentum vector at the COM.

Welcome to the introduction of screw theory in mechanics.

Answer 2

I think that you would need to find the vector sum of the two angular velocities and then find the total angular momentum from there.

However, it's impossible for a rigid body to rotate over more than one axis at a time. I think that you may have phrased your question incorrectly.