Predicting Change in multiple dimensions: Difference between revisions
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Given the force: | Given the force: | ||
<math>\ | <math>\vec{F} = (F_x,F_y,F_z) </math> | ||
And change in time: | And change in time: | ||
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<math>\Delta t</math> | <math>\Delta t</math> | ||
<math>\Delta p = \ | <math>\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>\ | <math>\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: | This can also be expressed as: | ||
<math> \ | <math> \vec{F}_{net} = \frac {\text{d}\vec{p}} {\text{d}t}</math> | ||
or: | or: | ||
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====Multiple Particles==== | ====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'''. | |||
If | |||
Center of Mass: | Center of Mass: | ||
In summation notation: | |||
<math> \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 [http://www.columbia.edu/itc/sipa/math/summation.html 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> (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> | |||
<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> m, \vec{v}, </math> and <math> r </math>, we use <math> m_{total} = \sum_i^n m_i </math>,<math> \vec{v}_{cm} </math>, and <math> \vec{r}_{cm} </math>. | ||
===A Computational Model=== | ===A Computational Model=== | ||
Revision as of 15:07, 28 June 2019
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
Below are models that use change in momentum to predict how particles move (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
An electron and proton
We can also use momentum to model the path of more complex models, like a proton and electron near each other:
An electron and proton with non-zero velocities with electric force included
Elastic Collision
Two objects undergo perfectly elastic collision
Two objects collide perfectly elastically
Inelastic Collision
Two objects undergo inelastic collision
Others
Examples
Simple
A ball of mass 1000 g rolls across the floor with a velocity of (0,10,0) m/s. After how much time does the ball stop? Where does it stop if it starts at the origin? Assume the coefficient of friction is 0.3.
We need to find when velocity is 0 or when final momentum is 0.
Declare known variables:
Change mass to kilograms:
[math]\displaystyle{ \mathbf{m} = \frac{1000} {1000} = 1 kg }[/math]
[math]\displaystyle{ \overrightarrow{\mathbf{v}} = (0,10,0) \frac{m} {s} }[/math]
[math]\displaystyle{ \mu = .3 }[/math]
Find the initial momentum:
[math]\displaystyle{ \overrightarrow{p}_{initial} = \mathbf{m} * \overrightarrow{\mathbf{v}} }[/math] = [math]\displaystyle{ 1 * (0,10,0) }[/math] = [math]\displaystyle{ (0,10,0) kg * \frac{m} {s} }[/math]
Find time passed:
[math]\displaystyle{ \Delta{\overrightarrow{p}} = \overrightarrow{\mathbf{F}}_{net} * \Delta{t} }[/math]
[math]\displaystyle{ \mathbf{\overrightarrow{F}}_{net} = \overrightarrow{\mathbf{F}}_{normal}*\mu }[/math]
[math]\displaystyle{ \overrightarrow{\mathbf{F}}_{normal} = (0,\mathbf{m} * \mathbf{g} = 1 kg * 9.8 \frac{m} {s^2},0) = (0,9.8,0) N }[/math]
[math]\displaystyle{ \mathbf{\overrightarrow{F}}_{net} = (0,9.8,0) N * 0.3 = (0,2.94,0) N }[/math]
[math]\displaystyle{ \Delta{\overrightarrow{p}} = \overrightarrow{p}_{initial} = (0,2.94,0) N * \Delta {t} }[/math]
[math]\displaystyle{ (0,10,0) = (0,2.94,0) N * \Delta {t} }[/math]
[math]\displaystyle{ \Delta {t} = \frac {(0,2.94,0)} {(0,10,0)} }[/math]
[math]\displaystyle{ \Delta {t} = .294 s }[/math]
Find displacement
[math]\displaystyle{ \Delta {d} = \overrightarrow{v}_{avg} * \Delta {t} = \frac {(0,10,0)} {2} \frac{m} {s} * 0.294 s = (0,1.47,0) m }[/math]
Find final position
[math]\displaystyle{ \mathbf{r}_{final} = \mathbf{r}_{initial} + \Delta {d} = (0,0,0) m + (0,1.47,0) m = (0,1.47,0) m }[/math]
Middling
You kick a 5kg ball off a 100m cliff at a velocity of (10,15,10). How long does it take for the ball to reach the ground? How far away does it land?
Declare Known Variables
[math]\displaystyle{ \mathbf{m} = 5 kg }[/math]
[math]\displaystyle{ \mathbf{h} = 100 m }[/math]
[math]\displaystyle{ \overrightarrow{\mathbf{v}}_{initial} = (10,15,10) \frac{m} {s} }[/math]
Find Initial Momentum
[math]\displaystyle{ \overrightarrow{\mathbf{p}}_{initial} = \mathbf{m} * \overrightarrow{\mathbf{v}}_{initial} }[/math]
[math]\displaystyle{ \overrightarrow{\mathbf{p}}_{initial} = 5kg * (10,15,10) \frac{m} {s} = (50,75,50) kg * \frac{m} {s} }[/math]
Find Final Momentum
Because the ball will hit the ground the final y component of the velocity and final momentum will be 0. Because gravity only affects the y component, the x and z components are unchanged.
[math]\displaystyle{ \overrightarrow{\mathbf{p}}_{final} = (50,0,50) kg * \frac{m} {s} }[/math]
Find Net Force
[math]\displaystyle{ \overrightarrow{\mathbf{F}}_{net} = \overrightarrow{\mathbf{F}}_{g} = (0,\mathbf{m} * \mathbf{g},0) = (0, 5kg * 9.8 \frac{m} {s^2},0) = (0,49,0) N }[/math]
Find time passed
[math]\displaystyle{ \Delta{\overrightarrow{\mathbf{p}}} = \Delta {t} * \overrightarrow{\mathbf{F}}_{net} }[/math]
[math]\displaystyle{ (0,70,0) = \Delta {t} * (0,49,0) }[/math]
[math]\displaystyle{ \Delta {t} = 1.42857 s }[/math]
Find Displacement in X
[math]\displaystyle{ \mathbf{v}_{avg x} = \frac{\mathbf{v}_{final x} + \mathbf{v}_{initial x}} {2} = 50 \frac{m} {s} }[/math]
[math]\displaystyle{ \Delta {\mathbf{d_x}} = \Delta{t} * \mathbf{v}_{avg x} = 1.42857 s * 50 \frac{m} {s} = 71.4285 m }[/math]
Find Displacement in Z
[math]\displaystyle{ \mathbf{v}_{avg z} = \frac{\mathbf{v}_{final z} + \mathbf{v}_{initial z}} {2} = 50 \frac{m} {s} }[/math]
[math]\displaystyle{ \Delta {\mathbf{d_z}} = \Delta{t} * \mathbf{v}_{avg z} = 1.42857 s * 50 \frac{m} {s} = 71.4285 m }[/math]
[math]\displaystyle{ \Delta {\overrightarrow{\mathbf{d}}} = (71.4285,0,71.4285) m }[/math]
Find final position
[math]\displaystyle{ \mathbf{r}_{final} = \mathbf{r}_{initial} + \Delta {d} = (0,100,0) m + (71.4285,-100,71.4285) m = (71.4285,0,71.4285) m }[/math]
Difficult
You kick a ball through 5 meter high posts from 60 meters away. It takes 2.5s to travel through the posts. Answer the following questions:
a)What is the initial velocity?
b)What angle did you shoot it from?
c)What is the velocity of the ball as it crosses the posts?
d)What was the balls maximum height?
e)What was the force on the ball if the ball is .76 kg and the impact lasts for 34 ms?
Part A
Find Initial Velocity in X direction
[math]\displaystyle{ \frac{\Delta {p}_x} {\Delta {t}} = \mathbf{F}_{net} = 0 }[/math]
[math]\displaystyle{ \frac{m *\Delta {v}_x} {\Delta {t}} = 0 }[/math]
Or the x component of the velocity is constant.
[math]\displaystyle{ \overrightarrow{v}_x = \frac{\Delta {x}} {\Delta {t}} = \frac{60} {2.5} = 24 \frac{m}{s} }[/math]
Find Initial Velocity in Y direction
[math]\displaystyle{ \frac{\Delta {p}_y} {\Delta {t}} = \mathbf{F}_{net} = - \mathbf{m}\mathbf{g} = \frac{m *\Delta {v}_y} {\Delta {t}} }[/math]
[math]\displaystyle{ \frac{\Delta {v}_y} {\Delta {t}} = -\mathbf{g} }[/math]
[math]\displaystyle{ \Delta {v}_y = - \mathbf{g} * \Delta {t} = v_{final y} - v_{initial y} }[/math]
[math]\displaystyle{ v_{final y} = - \mathbf{g} * \Delta {t} + v_{initial y} }[/math]
[math]\displaystyle{ v_{avg y} = \frac{v_{final y} + v_{initial y}} {2} }[/math]
[math]\displaystyle{ v_{final y} = 2v_{avg y} - v_{initial y} }[/math]
[math]\displaystyle{ - \mathbf{g} * \Delta {t} + v_{initial y} = 2v_{avg y} - v_{initial y} }[/math]
[math]\displaystyle{ v_{avg y} = v_{initial y} - \frac{1} {2} \mathbf{g} \Delta {t} }[/math]
[math]\displaystyle{ \frac{\Delta {y}} {\Delta {t}} = v_{avg y} }[/math]
[math]\displaystyle{ \Delta {y} = v_{initial y} \Delta {t} - \frac{1} {2} \mathbf{g} \Delta {t}^2 }[/math]
[math]\displaystyle{ v_{initial y} = \frac{\Delta {y}} {\Delta {t}} + \frac{1} {2} \mathbf{g} \Delta {t} = \frac{5} {2.5} + \frac{1} {2} * 9.8 * 2.5 = 14.25 \frac {m} {s} }[/math]
Find Initial Velocity in Z direction
No change in Z direction, therefore:
[math]\displaystyle{ v_{initial z} = 0 \frac{m} {s} }[/math]
Find Initial Velocity
[math]\displaystyle{ \overrightarrow{v}_{initial} = (24,14.25,0) \frac {m} {s} }[/math]
Part B
Find the Angle
[math]\displaystyle{ \tan {\theta} = \frac{\mathbf{v_{initial y}}} {\mathbf{v_{initial x}}} = \frac{14.25} {24} = 0.59375 }[/math]
[math]\displaystyle{ \theta = \tan^{-1} 0.59375 = 0.53358 radians }[/math] or 30.6997 degrees
Part C
Find Final Velocities
Final velocity is equal to the velocity as it goes through the poles.
No change in x velocity:
[math]\displaystyle{ v_{final x} = 24 \frac{m} {s} }[/math]
From part A:
[math]\displaystyle{ v_{final y} = - \mathbf{g} * \Delta {t} + v_{initial y} }[/math]
[math]\displaystyle{ v_{final y} = - 9.8 * 2.5 + 14.25 }[/math]
[math]\displaystyle{ v_{final y} = - 10.25 \frac{m} {s} }[/math]
[math]\displaystyle{ \overrightarrow{v}_{final} = (24, -10.25, 0) \frac{m} {s} }[/math]
Part D
Find Max Time
The velocity at the max point is 0. Use equation from part A:
[math]\displaystyle{ v_{max y} = v_{initial y} - g \Delta{t_{max}} = 0 }[/math]
[math]\displaystyle{ v_{initial y} = g \Delta{t_{max}} }[/math]
[math]\displaystyle{ \Delta{t_{max}} = \frac{v_{initial y}} {g} }[/math]
Find Max Height
[math]\displaystyle{ \frac{\Delta {y}} {\Delta{t}} = v_{initial y} - \frac{1}{2} g \Delta{t_{max}} }[/math]
[math]\displaystyle{ y_{max} - y_{initial} = v_{initial y} \Delta{t_{max}} - \frac{1}{2} g \Delta{t_{max}}^2 }[/math]
[math]\displaystyle{ y_{max} = \frac{v_{initial y}^2} {g} - \frac{1}{2} {(\frac{v_{initial y}^2} {g})} }[/math]
[math]\displaystyle{ y_{max} = \frac{1}{2} {(\frac{v_{initial y}^2} {g})} = \frac{1}{2} {(\frac{14.25^2} {9.8})} = 10.3603 m }[/math]
Part E
State Known Variables
[math]\displaystyle{ \mathbf{m} = 0.76 kg }[/math]
[math]\displaystyle{ \Delta{t} = .034 s }[/math]
Find Change in Momentum
[math]\displaystyle{ \overrightarrow{\mathbf{p}}_{initial} = 0 }[/math]
[math]\displaystyle{ \overrightarrow{\mathbf{p}}_{final} = m * \overrightarrow{v}_{initial} }[/math]
Use V Initial from Part A:
[math]\displaystyle{ \Delta{\overrightarrow{\mathbf{p}}} = \overrightarrow{\mathbf{p}}_{final} - \overrightarrow{\mathbf{p}}_{initial} = m * \overrightarrow{v}_{initial} - 0 = 0.76 kg * (24,14.25,0) \frac{m} {s} = (18.24,10.83,0) kg \frac{m} {s} }[/math]
[math]\displaystyle{ |\Delta{\mathbf{p}}| = \sqrt{18.24^2 + 10.83^2} = 21.213 kg \frac{m} {s} }[/math]
Find Net Force
[math]\displaystyle{ |\Delta{\mathbf{p}}| = |\mathbf{F}_{net}| * \Delta{t} }[/math]
[math]\displaystyle{ |\mathbf{F}_{net}| = \frac{|\Delta{\mathbf{p}}|} {\Delta{t}} = \frac{21.213} {.034} = 623.908 N }[/math]
Connectedness
Momentum is what some physicists call the fundamental principle. Almost everything in physics is based off this principle. This is why it is one of the most interesting parts of physics. From getting a few measurements, we have shown how we can predict the final destination of any particle. We can create models for real life examples, like finding how hard we have to kick a ball in a sport, where a ball will stop rolling on the ground and for many instances of projectile motion. It basically allows us to create mathematical models and predict the final destination of an object for almost any situation.
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
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