Predicting Change in multiple dimensions
Claimed by rbose7 Fall 2015
Here we analyze how to apply the same iterative methods we have hitherto used in one dimension to predict motion in multiple dimensions. In essence the situation may be broken down into three one dimensional problems, but when we are dealing with varying forces, it is necessary to generalize the probability that motion in one dimension will alter the forces in another dimension.
The Main Idea
Just as in one dimension, the linear (also known as translational) motion 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). By utilizing Newton's second law in each dimension, it is possible to predict the motion of the object in each dimension, making use of the same modelling techniques we have applied before.
A Mathematical Model
The change in momentum due to a force is expressed in mathematical terms as:
[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]
Relation to Force
Given the force [math]\displaystyle{ \vec{F} = (F_x,F_y,F_z) }[/math], and change in time [math]\displaystyle{ \Delta t }[/math], we may express the change in momentum as
[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]
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 - Two Dimensional Harmonic Motion 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
This is an example of a generic three dimensional motion simulator constructed using numpy.
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
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
- Linear Momentum
- Projectile Motion
- Kinematics
- Momentum Principle
- Fundamentals of Iterative Prediction with Varying Force
- Two Dimensional Harmonic Motion
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