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

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?

• Hi Yash! Welcome to Physics StackExchange :) Kindly consider including your equations the body of text using some beautiful MathJax :) – Abhay Hegde Apr 13 '19 at 17:36

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 $$I_1 = I_2 = I_3 = I$$.

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.

• Is the single rotation axis in general determined by the direction of $\omega$ or the direction of $L$? – BioPhysicist Apr 15 '19 at 22:12
• The direction of rotation is given by the vector $\vec{\omega}$. See edit to the answer. – John Alexiou Apr 16 '19 at 0:14
• PS. If the center of mass has a velocity vector $\vec{v}$, then the rotation axis is located at $$\vec{r}_{\rm rot} = \frac { \vec{\omega} \times \vec{v} }{ \omega^2 }$$ relative to the COM. – John Alexiou Apr 16 '19 at 0:17
• Can we express angular momentum about O in terms of vector also? – Yash Mittal Apr 16 '19 at 18:27
• @YashMittal, yes that would be $$\vec{L}_0 = \vec{L} + \vec{r}_{\rm com} \times \vec{p}$$ – John Alexiou Apr 17 '19 at 1:05

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.

• Why its impossible for a body to rotate about two axes? – Yash Mittal Apr 15 '19 at 12:46
• For example, The wheel of a car rotates about horizontal axis, and when the driver rotates the axle to turn the car, wheel rotates about vertical axis also – Yash Mittal Apr 15 '19 at 12:47
• Yes, but the car is no longer stationary. Your problem requires the sphere to be stationary. – Sidhant Rajadnya Apr 16 '19 at 17:31
• To be more precise, consider a car lifted up on a jack and its front tire are rotating, Now driver turns the wheel to turn the car. Now, car is not moving and wheel has two rotations :) – Yash Mittal Apr 16 '19 at 18:24