# Angular momentum of rotating and revolving body (earth)

My understanding is this: angular momentum of a body about a point cannot be in two directions at once, which is why a wheel already rotating about its axle, cannot also rotate about its diameter at the same time; it tilts sideways. However this doesn't seem to be the case for points of reference outside the body?

For example, consider the earth, with a circular orbit around the sun for simplicity. Taking the sun as origin, the earth's angular momentum points perpendicular to its plane of revolution. However, the earth also rotates around itself, with the axis making some angle with the plane of rotation, so it has another angular momentum, pointing in that direction.

Can somebody explain how the second situation differs from the first?

Both situations can be described in the same way. I think your issue is that you are treating each object as one object, but in reality you should consider these objects as extended objects made up of many mass elements.

The angular momentum of a point particle about some reference point is given by

$$\mathbf L=\mathbf r\times\mathbf p$$

where $$\mathbf r$$ is the position vector that points from the reference point to where the particle is, and $$\mathbf p=m\mathbf v$$ is the momentum of the particle.

To get the total angular momentum of an extended body, we just add up the angular momentum of each particle

$$\mathbf L=\sum_i\mathbf r_i\times\mathbf p_i$$

If you know the angular velocity vector $$\boldsymbol\omega$$ of the object as well as it's moment of inertia tensor $$\hat{\mathbf I}$$(all about the same reference point / axes), then you can also determine the angular momentum of the object via matrix multiplication:

$$\mathbf L=\hat{\mathbf I}\boldsymbol\omega$$

angular momentum of a body about a point cannot be in two directions at once, which is why a wheel already rotating about its axle, cannot also rotate about its diameter at the same time; it tilts sideways.

I'm not sure I follow this. A wheel can spin about its center while also spinning about its diameter. You can still pick a reference point and find the total angular momentum of the wheel.

Taking the sun as origin, the earth's angular momentum points perpendicular to its plane of revolution. However, the earth also rotates around itself, with the axis making some angle with the plane of rotation, so it has another angular momentum, pointing in that direction.

Here, you can also pick a reference point and determine the total angular momentum of the system.

The angular momentum is a vector, thus for the components of the angular momentum we have to define a coordinate frame .

I) Wheel

the coordinate system is at the center of the wheel and the angular momentum is:

$$\vec L=I_W\,\vec\omega$$

where $$I_W$$ is the inertia tensor of the wheel. and $$\vec\omega$$ the wheel angular velocity.

all components are given in the wheel fixed coordinate system

the wheel can rotate about the x-axis with the angle $$\varphi$$ and about the z-axis (diameter) with the angle $$\psi$$ thus die rotation matrix is:

$$R=R_x(\varphi)\,R_z(\psi)$$ from hear you can obtain the angular velocity.

II) earth sun system

the coordinate system is initial system that located at the sun center of mass. the angular momentum is :

$$\vec L=\vec r\times m_E\,\vec v+I_E\vec\omega$$

all components are given in the initial frame.

where $$\vec r$$ is the distance vector between earth and the sun, $$\vec v$$ is the earth velocity about the the, $$m_E$$ is the earth mass, $$I_E$$ earth inertia tensor and $$\vec\omega$$ the earth angular velocity .

the earth rotate about the north south axis $$\vec n$$ with the rotation angle $$\psi$$, the rotation matrix is:

$$R=R(\vec n\,,\psi)$$ thus the angular velocity is:

$$\vec \omega=\vec n\,\vec{\dot{\psi}}$$

the inertia tensor must be transform to the inertial system

$$I_E\mapsto R\,I_E\,R^T$$