Angular Momentum Compared to Linear Momentum: Difference between revisions
Clindbeck3 (talk | contribs) (→Connectedness: Removed mention of using perpetual motion machines as a power source, also references to links that do not exist.) |
Clindbeck3 (talk | contribs) (→External links: Removed links to hoax/pseudoscience perpetual motion machine videos, most of which do not appear to be actually related to angular momentum anyway) |
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==References== | ==References== |
Revision as of 13:02, 18 April 2018
Claimed by clindbeck3 (Christopher Lindbeck) for Spring 2018
The Main Idea
Linear momentum is great and all, but angular momentum is where the magic of physics kicks into high gear. See the transition of motion from translational to rotational calls for adding another variable that drastically changes the equations for linear motion that we know and love. The two that I will focus on are the equations for both angular and linear momentum.
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
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: [math]\displaystyle{ \vec{p} = m\vec{v} }[/math]
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
Connectedness
- Connected to my major?
- I am a math major and so superficially I could say that all physics holds up under mathematical scrutiny but there is a crossover when it comes to angular momentum and matrices. 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 (not sure whether it was mathematicians or physicists or both who created this elegant method). 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: