Angular momentum direction for a spinning top

[jl12]

Homework Statement

Hi I'm having trouble with a question that's asking me to calculate the precession rate for a spinning top. The trouble that I'm having is that I don't understand how the angular momentum points along the axis of the spinning top (picture attached). When I use the formula $L=r \times p$ (where r and p are meant to be vectors) to find the direction I get it pointing inwards and slightly up at an angle. However, when I look online they seem to point it along the axis of the spinning top. Am i not allowed to use $L=r \times p$ for this case? If so why?
Thanks in advance

Homework Equations

$L=r \times p$

The Attempt at a Solution

 

[kuruman]

For spin angular momentum (not the precession) r is a vector from the axis of the top to the point on the rim that has velocity v (into the screen). Both $\vec r$ and $\vec v$ are in the plane of the wheel so $\vec r \times \vec p$ is perpendicular to that plane.

 

[jl12]

For spin angular momentum (not the precession) r is a vector from the axis of the top to the point on the rim that has velocity v (into the screen). Both $\vec r$ and $\vec v$ are in the plane of the wheel so $\vec r \times \vec p$ is perpendicular to that plane.

So when its been pointing along the axis of the spinning wheel its because they're considering the angular momentum of the spinning top rather than the angular momentum of the whole system?

 

[kuruman]

So when its been pointing along the axis of the spinning wheel its because they're considering the angular momentum of the spinning top rather than the angular momentum of the whole system?

Correct. If you look at the drawing, the precession frequency $\vec{\omega}_p$ and hence the precession angular momentum $\vec {L}_p$ are straight up. Imagine a point at the tip of the $\vec{L}$ vector. This point precesses in a plane perpendicular to the direction of gravity and its precession frequency is perpendicular to $that$ plane, i.e. along the vertical. It is customary to describe the motion in terms of two separate frequencies, spin and precession, because it's easier to see what's going on: The top spins at constant frequency $\vec{\omega}_s$ about a tilted axis and at the same time the axis precesses about the direction of gravity at constant frequency $\vec{\omega}_p$. The motion wouldn't be as obvious to visualize if one wrote down the time-dependent vector sum of the two angular momenta.

On edit: By the way, Welcome to PF.

 

[jl12]

Correct. If you look at the drawing, the precession frequency $\vec{\omega}_p$ and hence the precession angular momentum $\vec {L}_p$ are straight up. Imagine a point at the tip of the $\vec{L}$ vector. This point precesses in a plane perpendicular to the direction of gravity and its precession frequency is perpendicular to $that$ plane, i.e. along the vertical. It is customary to describe the motion in terms of two separate frequencies, spin and precession, because it's easier to see what's going on: The top spins at constant frequency $\vec{\omega}_s$ about a tilted axis and at the same time the axis precesses about the direction of gravity at constant frequency $\vec{\omega}_p$. The motion wouldn't be as obvious to visualize if one wrote down the time-dependent vector sum of the two angular momenta.

On edit: By the way, Welcome to PF.

I'm sorry but I don't understand why they've ignored the angular momentum of the system because if it was just a point mass rotating around in a circle and we looked at the direction of angular momentum taken around a point just below the plane of the circle then wouldn't it point inwards and slightly up?

 

[kuruman]

'm sorry but I don't understand why they've ignored the angular momentum of the system ...

What do you mean by "angular momentum of the system"? There is spin angular momentum about the symmetry axis of the top and angular momentum of precession about the vertical axis. Neither of these have been ignored.

... if it was just a point mass rotating around in a circle and we looked at the direction of angular momentum taken around a point just below the plane of the circle then wouldn't it point inwards and slightly up?

The top cannot be a point mass. It's an extended rigid body with moment of inertia $I$ about its symmetry axis. If it were a point mass, it would not have a moment of inertia and its spin angular momentum would be zero. A point mass is an incorrect way to look at the this problem. Perhaps you might find the reference below informative.
http://hyperphysics.phy-astr.gsu.edu/hbase/top.html

 

[haruspex]

I'm sorry but I don't understand why they've ignored the angular momentum of the system because if it was just a point mass rotating around in a circle and we looked at the direction of angular momentum taken around a point just below the plane of the circle then wouldn't it point inwards and slightly up?

Typically, the precession is quite slow compared with the spin about the top's axis. Yes, technically the tangential velocity of the top's mass centre about the precession axis does add a small angular momentum about that axis, but you can ignore it. In particular, it does not alter the precession rate, since the torque is dL/dt, and the addition to L is a constant.

 

[haruspex]

If it were a point mass, it would not have a moment of inertia and its spin angular momentum would be zero.

jl12 is thinking about a point mass going in a circle, and the angular momentum that has about the centre of the circle. For a precessing top, this does indeed describe what the mass centre is doing, and this adds to the angular momentum of the top's spin about its own axis.

 

[kuruman]

jl12 is thinking about a point mass going in a circle, and the angular momentum that has about the centre of the circle. For a precessing top, this does indeed describe what the mass centre is doing, and this adds to the angular momentum of the top's spin about its own axis.

Indeed. There is spin angular momentum about the center of mass $\vec L$ characterized by the spin frequency $\vec{\omega}_s$ and "orbital" angular momentum of the center of mass characterized by the precession frequency $\vec{\omega}_p$. The two together describe the motion much like a rolling wheel has angular momentum $I\omega$ about the center of mass and angular momentum $mvR$ of the center of mass relative to a point on the surface. What puzzles me is why OP thinks that the precession frequency is being ignored.

The total angular momentum is $\vec L = \vec L_s+\vec L_p$. My point is that analyzing the motion using the two-vector right side of the equation is easier to visualize than using the one-vector left side. Nothing is ignored.

 

[haruspex]

What puzzles me is why OP thinks that the precession frequency is being ignored.

Presumably because jl12 did not observe that if in the torque equation, $\vec\tau=\vec{\omega_p}\times \vec L$, we substitute $\vec L=\vec {L_s}+\vec {L_p}$ then the $\vec{\omega_p}\times\vec {L_p}$ term vanishes because those two vectors are along the same axis.

 

[kuruman]

Ah, yes.