Angular Momentum Compared to Linear Momentum
claimed by jmcmahon9
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/31d0f9ad9e
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!
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