What force causes precession?

[FactChecker]

Homework Statement: Take a rope attached to the ceiling, with a rod perpendicular to it. On the end of the rod there is a spinning bicycle wheel that can be treated as a disk. What force causes the wheel to precess, if any?
Relevant Equations: T = r x F
L = I*w
dL/dt = T

In class, the prof. gave the following explanation for why the wheel precesses:
Whenever the wheel is spinning, it has a angular momentum perpendicular to its face. The torque due to gravity changes the direction of the angular momentum, causing precession.
However, this does not seem like a good enough explanation for this problem, because:
1. There is not an explanation as to why the angular momentum must be perpendicular to the wheel.

Any instantaneous linear momentum of a particle on one side of the wheel is exactly balanced by an opposite instantaneous linear momentum on the other side of the wheel. So the angular momentum can not have a component in either of those directions. The only thing left, if we want to represent angular momentum as a vector, is that the angular momentum is perpendicular to the wheel.

2. It does not explain where the force that causes precession comes from.

Angular momentum is just that -- momentum. It is not a force. The force is gravity which is trying to pull one end of the wheel down, while the other end is held up. Consider a particle on the the outer edge of the wheel. It is going in one direction and inertia wants it to keep going in that direction. When gravity tries to tilt the wheel, the direction of the particle will change. In order to minimize that change, the wheel rotates in the direction of a precession. From that, you can figure it out.

 

[Orodruin]

But even in those cases the direction of $\vec{\omega}$ relative to the direction of $\vec{L}$ is a consequence of the definition $\vec{L}=\vec{r}\times\vec{p}$. Is it not?

Yes, but $\vec L = \vec r \times \vec p$ does not imply that angular momentum is parallel to angular velocity so the latter cannot be a consequence of the former.

 

[Herman Trivilino]

Yes, but $\vec{L} = \vec{r} \times \vec{p}$ does not imply that angular momentum is parallel to angular velocity so the latter cannot be a consequence of the former.

$\vec{L} = \vec{r} \times \vec{p}$ does imply the relationship between the directions of $\vec{L}$ and $\vec{\omega}$. So the latter is a consequence the former.

 

[kuruman]

Euler's equations for rigid body dynamics allow torque-free precession. An often-seen problem in intermediate Classical Mechanics is to show that when a rigid body with three different principal moments of inertia is started spinning along each one of the axes in free space, the motion will be unstable along the axis with the intermediate moment of inertia and stable along the other two axes.

See Tennis racket theorem.

 

[Orodruin]

$\vec{L} = \vec{r} \times \vec{p}$ does imply the relationship between the directions of $\vec{L}$ and $\vec{\omega}$. So the latter is a consequence the former.

Yes, but not that they are parallel, which was the original statement.

 

[Filip Larsen]

What I am asking, is why the spindle has to point toward the angular momentum vector.

I am late to this thread and may not have a full understanding of the current context of the discussion, but for a general rigid object the relation $L = I\omega$ implies that $L$ only is parallel with $\omega$ when both are parallel with an eigenvector for the moment of inertia tensor $I$. Every rigid body has at least one such set of 3 orthogonal directions (relative to the body) that, as Orodruin also points out, is called the principal axes of the body. If the body posses inertial symmetries, then the body may have many of such sets and it is also then very easy to identify which body axis is a principle axis, which is why people prefer to work with these simpler body geometries.

So, to answer your question, for a disc or a top spinning around one of its (inertial) symmetry axes the equation $L = I\omega$ do imply that $L$ has same direction as $\omega$ because the rotation axis is a principal axis.