Angular momentum direction for a spinning top

In summary: On edit: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.
  • #1
jl12
8
2

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

 

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  • #2
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.
 
  • #3
kuruman said:
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?
 
  • #4
jl12 said:
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.
 
  • #5
kuruman said:
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?
 
  • #6
jl12 said:
'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.
jl12 said:
... 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
 
  • #7
jl12 said:
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.
 
  • #8
kuruman said:
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.
 
  • #9
haruspex said:
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.
 
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  • #10
kuruman said:
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.
 
  • #11
Ah, yes.
 

1. What is angular momentum?

Angular momentum is a physical quantity that measures the rotational motion of an object. It is a vector quantity that describes the amount of rotational motion an object possesses, as well as the direction in which it is rotating.

2. How is angular momentum direction determined for a spinning top?

The direction of angular momentum for a spinning top is determined by the right-hand rule. This rule states that if the fingers of your right hand curl in the direction of rotation, then your thumb will point in the direction of the angular momentum vector.

3. What factors affect the angular momentum direction for a spinning top?

The angular momentum direction for a spinning top is affected by the direction of the top's spin, the speed of the spin, and the point of rotation. These factors all contribute to the overall direction of the angular momentum vector.

4. Can the angular momentum direction of a spinning top change?

Yes, the angular momentum direction of a spinning top can change. This can happen if an external force is applied to the top or if there is a change in the top's spin or point of rotation. However, the total angular momentum of a closed system will remain constant.

5. Why is understanding angular momentum direction important?

Understanding angular momentum direction is important in many areas of physics, such as in the study of rotational motion and conservation of angular momentum. It is also important in engineering and design, as it can help predict the behavior of spinning objects and systems.

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