Predicting Change in multiple dimensions
This page discusses the use of momentum to predict change in multi-dimensions and examples of how it is used.
Claimed by rbose7
The Main Idea
Just as in one dimension, the linear momentum (also known as translational momentum) of an object is the vector quantity equal to the product of the mass and velocity of an object. Unlike in one dimension, we must now consider the vectors components in each direction (see 3-Dimensional Position and Motion for more detail). Just as in 1 dimension, momentum is conserved in closed systems - that is, a system with no external forces acting upon it - which means that two objects colliding will have the same net momentum before and after the collision. We can apply these properties to all three dimensions and use momentum to predict the path an object will follow over time by observing the change in momentum just as we did in one-dimension.
A Mathematical Model
This change in momentum is shown by the formula:
[math]\displaystyle{ \Delta \vec{p} = \vec{p}_{f}-\vec{p}_{i} = m\vec{v}_{f}-m\vec{v}_{i} }[/math]
Or by relating it to force:
[math]\displaystyle{ \Delta \vec{p} = \vec{F} \Delta t }[/math]
Relation to Velocity
Given an object with velocity [math]\displaystyle{ \vec{v} = (v_x,v_y,v_z) }[/math] and mass [math]\displaystyle{ m }[/math], the object's momentum will be
[math]\displaystyle{ \vec{p} = m\vec{v} = m(v_x,v_y,v_z) = (m v_x,m v_y,m v_z) }[/math]
Relate by Force
Given the force:
[math]\displaystyle{ \vec{F} = (F_x,F_y,F_z) }[/math]
And change in time:
[math]\displaystyle{ \Delta t }[/math]
[math]\displaystyle{ \Delta p = \vec{F} \Delta t = (F_x,F_y,F_z) \Delta t= ( F_x \Delta t,F_y \Delta t ,F_z \Delta t) }[/math]
[math]\displaystyle{ \vec{p}_f = \vec{p}_{i} + \Delta p = \vec{p}_i + (F_x \Delta t,F_y \Delta t,F_z \Delta t) }[/math]
This can also be expressed as:
[math]\displaystyle{ \vec{F}_{net} = \frac {\text{d}\vec{p}} {\text{d}t} }[/math]
or:
[math]\displaystyle{ \Delta p = \int_{t_1}^{t_2} F(t)\, dt\,. }[/math]
Multiple Particles
If our system consists of multiple particles being acted upon by a net external force, we can use the same process to predict its evolution. The only difference is that we pretend the particles are just on large particle with its center at the center of mass.
Center of Mass:
In summation notation:
[math]\displaystyle{ \vec{r}_{cm} = \frac{\sum_i^n m_i\vec{r}_i}{\sum_i^n m_i} }[/math]
If you are not familiar with summation notation, a brief recap may be found here, but it essentially consists of adding the product of each mass with its respective position, then dividing by the total mass. Accordingly, we have in expanded notation
[math]\displaystyle{ (x_{cm},y_{cm},z_{cm}) = \frac{1}{m_{total}} (\sum_i^n m_ix_i,\sum_i^n m_iy_i,\sum_i^n m_iz_i) }[/math]
Performing unit analysis, we see that the masses cancel out and this is indeed a position. With these quantities we may perform the same computations as above, but instead of using [math]\displaystyle{ m, \vec{v}, }[/math] and [math]\displaystyle{ r }[/math], we use [math]\displaystyle{ m_{total} = \sum_i^n m_i }[/math],[math]\displaystyle{ \vec{v}_{cm} }[/math], and [math]\displaystyle{ \vec{r}_{cm} }[/math].
A Computational Model
The principles of iterative prediction in multiple dimensions are exactly the same as the Fundamentals of Iterative Prediction with Varying Force, except that as per the above the process must be performed with 3 times as much information. The most important difference is that the force in one direction may depend on the position and velocity in another - Oscillators in Multiple Dimensions will consider such cases - which requires the system to update each coordinate simultaneously, rather than treating it like three separate one dimensional problems.
A Generic 3D Simulator
Below are models that use change in momentum to predict how particles move in specific situations (Click run to start simulation):
(If it does not work take the '?outputOnly=true' out of the url and try again)
A object with no net force on it
Below is a particle that has no net force and therefore moves at a constant velocity:
A object with no net force on it
A object with the force of gravity
Below is an object moving with gravity acting on it. Because gravity acts in the 'y' direction, the object's y component for velocity decreases:
A object with the force of gravity
Many Particles
Below are several objects moving with gravity acting on it, using calculations from center of mass (it is usually more accurate to apply calculations on each particle individually, but this is good for a big picture).
A object launched from a cliff
Below is an object launched with an initial velocity that has gravity acting on it. It loses velocity in the y direction due to gravity until it hits the ground:
A object launched from a cliff
Others
Examples
Simple
A ball of mass [math]\displaystyle{ 1 \; kg }[/math] has initial velocity [math]\displaystyle{ (3,4,0) \; m/s }[/math] in a pool of water, where the z-axis is the axis of gravity. The ball has exactly the density of water, so it buoyancy exactly cancels out gravity, and it only feels the force of drag due to the water. It feels linear drag, described by the equation [math]\displaystyle{ \vec{F}_d = -b\vec{v} }[/math], with [math]\displaystyle{ b = 0.1 \; kg/s }[/math]. Find the vector form of the drag force on the ball, and state its magnitude.
All we have to do is plug our values into the given equation, but we do need to get the vector right:
[math]\displaystyle{ \vec{F}_d = -b \vec{v} \rightarrow (F_x,F_y,F_z) = -b(v_x,v_y,v_z) }[/math]
The scalar term (the coefficient) will multiply out through the vector, so it results in
[math]\displaystyle{ (F_x,F_y,F_z) = -(0.1 \; kg/s)\cdot(3,4,0) \; m/s = (-0.3,-0.4,0) \; N }[/math]
Now, to find the magnitude we take a pythagorean sum:
[math]\displaystyle{ |\vec{F}_d| = \sqrt{F_x^2 +F_y^2+F_z^2} = \sqrt{0.25 \; N^2} = 0.5 \; N }[/math]
Note that the magnitude is strictly positive, and that the negatives are therefore coming from the direction of the vector. Although it's not asked for, note as well that it may be useful to determine the angle of the force with the x-axis, which we do by computing
[math]\displaystyle{ \theta = \arctan\biggr{(} \frac{F_y}{F_x}\biggr{)} = 53.13^\circ = 0.927 \; \text{rad} }[/math]
Middling
In the same situation as described above, compute the position (assuming the ball starts at the origin) after two time steps of [math]\displaystyle{ 0.1 \; s }[/math].
We have the first iterations force already worked out, so we compute the change in momentum:
[math]\displaystyle{ \Delta \vec{p} = \vec{F}_d \Delta t }[/math]
[math]\displaystyle{ \Delta \vec{p} = ((-0.3,-0.4,0) \; N)\cdot (0.1 \; s) = (-0.03,-0.04,0) \; \frac{kg\cdot m}{s} }[/math]
The adding this to our initial momentum gives
[math]\displaystyle{ \vec{p}_1 = (2.97,3.96,0) \frac{kg \cdot m}{s} }[/math]
[math]\displaystyle{ \vec{v}_1 = (2.97,3.96,0) \frac{m}{s} }[/math]
Then we compute the average velocity over the time step
[math]\displaystyle{ \vec{v}_{01} = ((2.97,3.96,0)+(3,4,0))/2 \; \frac{m}{s} = (2.985,3.98,0) \; \frac{m}{s} }[/math]
And so find displacement
[math]\displaystyle{ \vec{r}_1 = ((2.985,3.98,0)\; \frac{m}{s}) (0.1 \; s) = (0.2985,0.398,0) \; m }[/math]
Now we repeat the entire process:
[math]\displaystyle{ \vec{F}_d = -(0.1 \; \frac{kg}{s})((2.97,3.96,0) \; \frac{m}{s}) = (-0.297,-0.396,0) \; N }[/math]
[math]\displaystyle{ \Delta \vec{p} = ((-0.297,-0.396,0) \; N)\cdot(0.1 \; s) = (-0.0297,-0.0396,0) \frac{kg\cdot m}{s} }[/math]
[math]\displaystyle{ \vec{p}_2 = (2.9403,3.9204,0) \; \frac{kg\cdot m}{s} }[/math]
[math]\displaystyle{ \vec{v}_2 = (2.9403,3.9204,0) \; \frac{m}{s} }[/math]
[math]\displaystyle{ \vec{v}_{12} = (2.95515,3.9402,0) \; \frac{m}{s} }[/math]
[math]\displaystyle{ \vec{r}_2 = (0.594015,0.79202,0) \; m }[/math]
This is our final answer.
Difficult
In the situation given above, approximate the time it takes the ball to come to a rest (note that in this ideal case it will never fully stop, but Xeno's paradox is not our problem) and its position once it has stopped. It is recommended to do this with an iterative prediction program, and use a time step of width [math]\displaystyle{ 0.01 \; s }[/math] or smaller. Now, set [math]\displaystyle{ b = 0.2 \; \frac{kg}{s} }[/math] and give the new results. Finally, set the mass to [math]\displaystyle{ m = 0.3 \; kg }[/math] (keeping [math]\displaystyle{ b = 0.2 \frac{kg}{s} }[/math]), and give these results. Deduce the formula for final position from these relationships.
I used the program linked in the Computational Method section above. Plugging in the force equation and other parameters, we get a result for convergence time of [math]\displaystyle{ \Delta t \approx 40 s }[/math], and a final position [math]\displaystyle{ \vec{r}_f = (30,40,0) \; m }[/math]. Doubling the value of [math]\displaystyle{ b }[/math] gives a time of convergence of [math]\displaystyle{ \Delta t \approx 20 \; s }[/math] and [math]\displaystyle{ \vec{r}_f = (15,20,0) \; m }[/math]. Tripling the mass, we now have a convergence time of [math]\displaystyle{ \Delta t \approx 60 \; s }[/math] and [math]\displaystyle{ \vec{r}_f = (45,60,0) \; m }[/math]. The formula is therefore fairly simple: [math]\displaystyle{ \vec{r}_f = \frac{v_0\cdot m}{b} }[/math]. This result may be found analytically from a straightforward application of the separation of variables technique, sketched out in 1 dimension (since there is no dimensional dependence, it extends trivially to multiple dimensions) below:
[math]\displaystyle{ m\frac{\text{d}v}{\text{d}t} = -bv \rightarrow \ln (v) = -\frac{bt}{m} + C }[/math]
[math]\displaystyle{ v = A \exp(-\frac{bt}{m}) }[/math]
[math]\displaystyle{ v(0) = v_0 = A }[/math]
[math]\displaystyle{ x = \int_0^t v(t') \text{d} t' = -\frac{v_0 m}{b} (\exp(\frac{bt}{m}) - 1) }[/math]
For large [math]\displaystyle{ t }[/math], the above solution has the exponential goint to zero, cancelling out the negative and giving the relationship we found.
Connectedness
For my major, computer science, there are a few applications. For example, every computational model in the section above is coded using Python. A big part of computer science is modeling real life examples on a computer to mimic situations that would be otherwise impractical or impossible to set up. For example, the model we created of the electron and proton moving would take several expensive instruments to set up and visualize the two particles. For several other instances, using the principle of momentum and analyzing the change in momentum we can model several situations using coding that would otherwise not be possible.
History
Before Newton, French scientist and philosopher Descartes introduced the concept of momentum. He used the concept of momentum to describe how people moved when objects were thrown at them. He focused generally on the conservation of momentum when dealing with collisions. Newton's laws further expanded on the idea of conservation of momentum. The ideas that F = ma and the idea that for every action there is an equal and opposite reaction are the basis for many problems and concepts explained in this section. More information here:
http://galileoandeinstein.physics.virginia.edu/lectures/momentum.html
https://en.wikipedia.org/wiki/Momentum#History_of_the_concept
See also
Fundamentals of Iterative Prediction with Varying Force
External Links
https://en.wikipedia.org/wiki/Momentum
https://en.wikibooks.org/wiki/General_Mechanics/Momentum
http://hyperphysics.phy-astr.gsu.edu/hbase/mom.html
http://galileoandeinstein.physics.virginia.edu/lectures/momentum.html
References
https://en.wikipedia.org/wiki/Momentum
https://en.wikibooks.org/wiki/General_Mechanics/Momentum
http://galileoandeinstein.physics.virginia.edu/lectures/momentum.html