Angular Momentum Compared to Linear Momentum: Difference between revisions
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Clindbeck3 (talk | contribs) (→Linear Momentum vs Angular Momentum Equations: Created a chart comparing the basic variables and formulas between linear and angular momentum) |
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Linear Kinetic Energy = | Linear Kinetic Energy = | ||
Rotational Kinetic Energy = | Rotational Kinetic Energy = | ||
{| {{Table}} | |||
! Linear Motion !! Angular (Rotational) Motion | |||
|- | |||
| Linear Position: <math>x</math> (m) | |||
| Angular Position: <math>θ</math> (rad) | |||
|- | |||
|Linear Velocity: <math>v</math> (m/s) | |||
|Angular Velocity:<math>ω</math> (rad/s) | |||
|- | |||
|Linear Motion: <math>x=\vec{v}t</math> | |||
|Angular Rotation: <math>θ=ωt</math> | |||
|- | |||
|Mass: <math>m</math> (kg) | |||
|Moment of Inertia: <math>I</math> (kg m^2) | |||
|- | |||
|Momentum: <math>p=m\vec{v}</math> | |||
|Angular Momentum: <math>\vec{L}=I\vec{ω}</math> | |||
|- | |||
|Linear Kinetic Energy: <math>\frac{1}{2} m v^2</math> | |||
|Rotational Kinetic Energy: <math>\frac{1}{2} I ω^2</math> | |||
|} | |||
===A Comparison of Variables=== | ===A Comparison of Variables=== | ||
Revision as of 20:49, 18 April 2018
Claimed by clindbeck3 (Christopher Lindbeck) for Spring 2018 Is currently a work in progress, in heavy flux April 18th 2018.
The Main Idea
Linear momentum tends to behave fairly intuitively, but angular momentum has some more obtuse properties that can make it behave in confounding ways. While many of the equations describing angular momentum look almost identical to those describing linear momentum, variables that were fixed in linear momentum can change based on multiple factors, this can lead to less intuitive behavior. In addition, angular momentum has two separate components, one of which is relative to a point of reference.
Linear momentum is relatively simple, it depends upon the mass and velocity of an object, and to an extent the equation for angular momentum mimics the equation for linear momentum, but angular momentum draws a relationship between mass, velocity, AND the radius from the axis to the mass about which it is rotating. Now that your motion is relative to an axis, all of a sudden WHERE your mass is in relation to that axis MATTERS, and so the radius wriggles its way into the equations for linear motion, which makes these new equations have similar form but function in a completely different way.
A Mathematical Model
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Things are neat on tuesdays
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In a linearly moving system, there is only one type of movement (translational motion), and so it is relatively simple to calculate the momentum of single objects or the momentum of a system which would just be the addition of the momentum for each part.
Linear Momentum vs Angular Momentum Equations
We can see that a number of equations look very similar
Linear momentum: [math]\displaystyle{ \vec{p} = m\vec{v} }[/math]
Rotational Angular momentum: [math]\displaystyle{ \vec{L_{rot}} = I\vec{ω} }[/math]
Linear Kinetic Energy =
Rotational Kinetic Energy =
| Linear Motion | Angular (Rotational) Motion |
|---|---|
| Linear Position: [math]\displaystyle{ x }[/math] (m) | Angular Position: [math]\displaystyle{ θ }[/math] (rad) |
| Linear Velocity: [math]\displaystyle{ v }[/math] (m/s) | Angular Velocity:[math]\displaystyle{ ω }[/math] (rad/s) |
| Linear Motion: [math]\displaystyle{ x=\vec{v}t }[/math] | Angular Rotation: [math]\displaystyle{ θ=ωt }[/math] |
| Mass: [math]\displaystyle{ m }[/math] (kg) | Moment of Inertia: [math]\displaystyle{ I }[/math] (kg m^2) |
| Momentum: [math]\displaystyle{ p=m\vec{v} }[/math] | Angular Momentum: [math]\displaystyle{ \vec{L}=I\vec{ω} }[/math] |
| Linear Kinetic Energy: [math]\displaystyle{ \frac{1}{2} m v^2 }[/math] | Rotational Kinetic Energy: [math]\displaystyle{ \frac{1}{2} I ω^2 }[/math] |
A Comparison of Variables
[math]\displaystyle{ \vec{v} }[/math] vs [math]\displaystyle{ \vec{ω} }[/math]
Direction And the Right Hand Rule
While the direction of linear momentum is rarely difficult to deduce, the direction of angular momentum can often be less intuitive than simply the direction of motion. We use the right hand rule and the cross product with the position vector to figure out what direction our angular momentum vectors should be pointing.
However in a rotational system, we have two different subcategories for angular momentum: translational and rotational. Translational angular momentum depends upon a part's component of momentum that is orthogonal to the radius from a point of reference. Rotational angular momentum depends upon the moment of inertia of each part and how fast it rotates around its own axis, so the point of reference does not play a part in rotational angular momentum (unless the object you are calculating for is coincidentily spinning around that exact point).
- Rotational angular momentum: [math]\displaystyle{ \vec{L}_{rot} = I\vec{ω} }[/math]
- Translational angular momentum: [math]\displaystyle{ \vec{L}_{trans} = |\vec{r}||\vec{p}|sin{θ} }[/math]
- for the direction you can use the right hand rule.
- Total angular momentum: [math]\displaystyle{ \vec{L}_{tot} = \vec{L}_{rot} + \vec{L}_{trans} }[/math]
A Computational Model
Glowscript Link: https://trinket.io/glowscript/239aa4fac7
In this model I have a very simple orbit of a spacecraft around Earth while the Moon also rotates around Earth. I wanted to leave the trails for the motion to see better, however the program runs at a turtle pace if I leave the trails in. But you can notice some things about what is going on in terms of angular momentum. When the spacecraft comes near to the Earth, its velocity increases, and you can see this visually. Why? Well we know that [math]\displaystyle{ \vec{v} = \vec{ω}*r }[/math], therefore [math]\displaystyle{ \vec{ω} }[/math] has increased by this logic. All of the forces on the spacecraft are from within the system and so momentum must be conserved. Now, what is the only way to increase your [math]\displaystyle{ \vec{ω} }[/math] in a system where momentum is conserved and there are no outside forces? Looking at our previous equations, the only way to do this is to decrease [math]\displaystyle{ \vec{I} }[/math]. We can visually verify this too because the spacecraft has indeed moved closer to Earth!
Examples
Be sure to show all steps in your solution and include diagrams whenever possible
===Simple===
===Middling===
===Difficult===
Connectedness
- Connection with Mathematics Curriculum
- In multivariable calculus, there is an operation called the cross product, which can be easily derived from the dot product, that can give you some strange things. If you take the magnitude you get the area of a paralellogram, but more relevant to this section if you do not take the magnitude you get a vector orthogonal to the vectors you used in the cross product. There are two vectors that can be orthogonal to two other vectors if those two other vectors span a plane, both have the same magnitude but opposite directions. Which vector you use first in the cross product determines which of these two vectors is your answer, and thus the right hand rule was formed as a shortcut for an easy answer. The operation that you do in determining the cross product actually comes from determinants of matrices from linear algebra which help in determining properties of a matrix or the area of a paralellopipid created by the vectors that make the columns of that matrix. So, this operation was originally utilized for matrices and has now intertwined with physics for angular momentum and torque.
- Interesting Applications
- What I have been really interested in is the use of the fact that you can change [math]\displaystyle{ \vec{I} }[/math] or [math]\displaystyle{ \vec{ω} }[/math] without adding an external force which you cannot do with linear momentum (because [math]\displaystyle{ m }[/math] can never change internally, but [math]\displaystyle{ I }[/math] can).
History
Kepler created the laws of planetary motion, but Newton was the one who used this information to come up with angular momentum. He used a combination of triangles to geometrically prove that the area swept out by an object is equal in all places given that the change in time is the same. The height of these triangles is the tangential velocity and the base of these triangles is the radius of orbit. So, if the area swept out by an object must stay constant with a constant change in time, and velocity or radius changes in magnitude, the other variable must counteract this change! 
See also
Are there related topics or categories in this wiki resource for the curious reader to explore? How does this topic fit into that context?
Further reading
Books, Articles or other print media on this topic
External links
- example in space
- example with wheel
- merry go round fun
References
Print Sources:
- Chabay, Sherwood. Matter and Interactions. 3rd Ed. Hoboken: Wiley, 2011.
Online Sources: