This is the first part of my complete notes for Classical Mechanics, covering topics of Kinematics such as Velocity, Speed, Acceleration, Free Fall, Projectile Motion, Range, and more. Due to the large scale of this topic, I have had to split even the complete notes into multiple parts.
I color-coded my notes according to their meaning - for a complete reference for each type of note, see here (also available in the sidebar). All of the knowledge present in these notes has been filtered through my personal explanations for them, the result of my attempts to understand and study them from my classes and online courses. In the unlikely event there are any egregious errors, contact me at jdlacabe@gmail.com.
Summary of Classical Mechanics, Part 1: Kinematics
Table Of Contents
I. Velocity, Speed, Acceleration.
I.I Basics of Mechanics.
Physics is based on measurement of physical quantities: Having some expertise in dimensional analysis is crucial to be able to convert from one unit of measurement to another. There are several different systems of measurements which measure the same types of quantities, the two biggest systems being the Metric System and the Imperial System. For example, length is measured in the metric system in 'meters', while the imperial unit for length is 'feet'. Dimensional Analysis is conducted using conversion factors, which are just the ratio of units equivalent to what you are trying to convert to. For example,
(2 min) x (60 sec/1 min)
has (60 sec/1 min) as the conversion factor, converting from minutes to seconds. The conversion factor is the raetion being multiplied regardless of the coefficient of the original unit, as shown in how two minutes, 40 minutes, and 542 minutes will all have the same conversion factor when converting into seconds.
# P. Rule 1. 1) All Non-zero digits are significant.
2) Zeros between significant digits are significant.
3) Zeros to the left of nonzero digits are not significant.
4) Zeros at the end of a number are significant only if there is a decimal point present.
For Example:
1200 with 3 significant figures: 1.20 x 10³
0.002 with 3 significant figures: 2.00 x 10⁻³
# P. Rule 2. THE ARCANE ROUNDING RULE - If the number ends in a perfect 5, you round to the even number.
For Example:
275 with 2 Significant figures: 280 1030 x 5.1 = 5300
# A.M.U.:
An additional unit of measurement (derived from the kg) which is specific to atoms is the Atomic Mass Unit, found as the atomic mass of each element in relation to that of Carbon-12. For those familiar with stoichiometry, the amu can converted to moles and atoms easily.
# Dimensions:
┏━━━━━━━━━━━━━━━━━┳━━━━━━━━━━━━━┳━━━━━━━━━━━━━━━━━┓
┃Base Dimensions ┃ S.I. Unit ┃Imperial System ┃
┣━━━━━━━━━━━━━━━━━╋━━━━━━━━━━━━━╋━━━━━━━━━━━━━━━━━┫
┃ length ┃ meter ┃ foot ┃
┣━━━━━━━━━━━━━━━━━╋━━━━━━━━━━━━━╋━━━━━━━━━━━━━━━━━┫
┃ mass ┃ kilogram ┃ slug ┃
┣━━━━━━━━━━━━━━━━━╋━━━━━━━━━━━━━╋━━━━━━━━━━━━━━━━━┫
┃ time ┃ seconds ┃ seconds ┃
┗━━━━━━━━━━━━━━━━━┻━━━━━━━━━━━━━┻━━━━━━━━━━━━━━━━━┛
# Density:
The density d of a material is the mass per unit volume:
d = (m / V)
The majority of the time, one will see a density listed in Kilograms (or another metric weight) per cubic meter (or another length, such as cm³). The density of water, for example, is 1.00 grams per cm³.
# Vertical Scaling:
When you have a problem such as that shown below (ignore the problem itself - it is done through integration if you must know), the "vertical scaling" is just referring to what the given value (vs) is as each y-value is multiplied by a constant, creating a ratio between the new and old scales. The old scale will almost always be a scaling of 1:1, e.g. the normal graph, and the new scale (as depicted in the image) has a ratio of 2:1 with the old graph, demonstrated by how vs is four units upward, meaning each unit is two. The exact same process can be applied to the x-axis, resulting in Horizontal Scaling.
# P. Rule 3. When you have squared or cubed units, you put the exponent on the outside of the individual component, given both units of the type share the same exponent: 12 mm² → m² = 12 mm² × ((1 m) / (1000 mm))²
# Tangent Line: A straight line that touches a curve at a point but does not cross the curve.
# Final Example:
Here is a beginning problem that should be common sense to any beginning physicist well-versed in dimensional analysis:
Iron has a density of 7.87 g/cm³, and the mass of an iron atom is 9.27 × 10⁻²⁶ kg. If the atoms are spherical and tightly packed,
(a) what is the volume of an iron atom and
(b) what is the distance between the centers of adjacent atoms?
Of course, you always want all of your measurements to be in the same units, so we will first convert the grams to kilograms and cubic centimeters to cubic meters:
d = (7.87 g/cm³) × (1 kg / 1000 g) × (100 cm / 1 hr)³
d = 7.87 × 10³ kg/m³
Now lets move on to finding the volume of the iron atom. For this, we need to remember our previous knowns of density, mass, and volume: d = (m / V) Thus, our first step must be to isolate the volume since we already know the density and mass.
(9.27 × 10⁻²⁶ kg) / (7.87 × 10³ kg/m³)
This gives us a result of 1.18 × 10⁻²⁹ m³ the volume of an iron atom.
Next, for part b, we need to find the distance between the centers of adjacent atoms. Using critical thinking, we can determine that this distance is in fact the diameter of the atom, which we can reverse engineer by isolating the radius from the volume equation of a sphere:
Volume of sphere = (4/3) × π × r³.
r = ∛(3V / 4π)
Plugging in the known volume from answer a), we find that the radius is equal 1.41 × 10⁻¹⁰ meters. Since we are looking for the diameter, we multiply this distance by two and get 2.82 × 10⁻¹⁰ meters.
I.II Accuracy & Uncertainty.
# Precision: How close your observed (or measured) values are to one another.
# P. Rule 4. High Accuracy also includes taking the average of all of the instances, so if a bullseye looks like this:
it would be high accuracy.
# P. Rule 5. If there is only one measurement that can't be compared to any other measurements, precision cannot be determined.
# Error Equations:
Relative Error measures Accuracy:
Er = ((O - A) / A) × 100
A = Accepted Value
Er = Relative Error
O = Observed Error
# Estimated Uncertainty: The estimated quantifiable range which a measurement can be within. When giving the result of a measurement, state the estimated uncertainty. 8.8 ± 0.1 cm.
# Percent Uncertainty: The ratio between the estimated uncertainty and the measured value, made into a percentage: ((0.1) / (88)) × 100% ≈ 1%.
# P. Rule 6. When you want to find the approximate uncertainty of something, such as that of a circle when given the radius, use the main area formula, in this case πr², to first find the ideal area. If r = 3.8 × 10⁴ cm, then A = 4.5 × 10⁹ cm². From there, if no uncertainty number is given, assume 0.1 of the original units, 0.1 × 10⁴ cm. From there, find area again from the original for ±0.1. Find the difference in the areas, divide by 2, and use it as the uncertainty of the ideal area: (4.5 ± 0.26) × 10⁹ cm².
I.III Displacement and V.S.A.
# Displacement: The straight-line distance between initial and final points, also including direction. ∆x is the symbol for displacement and change in position, known as positionfinal - positioninitial. Can be negative, positive, in feet, meters, or anything.
Displacement is a VECTOR (see section III, on vectors). It has both x and y distance, even though it is written as ∆x ("delta-x"), which is in reference to POSITION, not just the x value.
Using d as displacement, d = ((x₂ - x₁)î + (y₂ - y₁)ĵ), using unit vectors (see Rule 22). This equation describes displacement in terms of its component vectors on the x-axis and y-axis. These component vector indicate the direction and magnitude of the displacement (also, the equation effectively uses unit vectors - see Rule 23). In order to find displacement in the graphical sense, where you have two dots on a graph (the start and endpoints) and you need to find the straight-line distance between them, this is the first step. You must find the components in the x and y directions first to find the displacement, which is the hypotenuse of the right triangle formed by the x and y component vectors (see Rules 20 & 21).
If you want to find the magnitude of the displacement, the exact straight-line distance between any two points can be found with the following equation:
|d| = √((x₂ - x₁)² + (y₂ - y₁)²)
Use this equation when you are prompted to find the distance or the magnitude the displacement - the direction can be determined by the component vectors themselves, as demonstrated in Rule 22.
NOTE THAT THE DISPLACEMENT EQUATION AND SEEING DISPLACEMENT AS A VECTOR IS NOT NECESSARY IF THE OBJECT IS ONLY MOVING ALONG A SINGLE AXIS. If the position is only moving along one axis, then you only need to subtract yfinal by yinitial or xfinal by xinitial to find displacement. If the equation is in 2D, then you want to use the equation as stated, and if it is in 3D, then you should just add + (z₂ - z₁)k̂, if you are are calculating displacement in unit vectors, or + z², if you are calculating the magnitude. Useful proof.
# Important Note:
When used in different context, the 'Average Rate of Change' can mean different things:
When used with f(x): Average Rate of Change = Average Velocity = (Displacement / change in time).
When used with f'(x): Average Rate of Change = Average Acceleration = (Change in Velocity / change in time).
Using it in relation to the second derivative would mean to take the third derivative, which is frankly ridiculous.
# P. Rule 7. Cartesian Coordinates: (x, y).
Relative Coordinates: (Left, Right).
If you walk 2 meters east and moon walk back, then your displacement is 0 m, directionless, magnitude is also 0. The distance traveled will always be greater or equal to the magnitude of displacement.
# Velocity: V = (∆x / ∆t) = (change in position) / (change in time) = (xF - xi) / (tF - ti)
Thus, velocity has both magnitude and direction.
# P. Rule 8. ALWAYS convert into the standard S.I. Units unless told otherwise.
# P. Rule 9. Speed = (total distance) / (time). MAGNITUDE ONLY. Because Velocity accounts for displacement, but speed only for distance traveled, they will only share the same magnitude when moving in a straight line. The velocity is specifically in straight-line distance, so if the line is curved, the end point and the initial point are the only things that are going to be used in a velocity calculation. However, the speed of that curve would be the length of the entire line, all of distance traveled between point a and point b.
# Slope: slope = m = (rise)/(run) = (∆y / ∆x) = (change in position / change in time) = VELOCITY. See "The Truth About Velocity & Displacement" for further elaboration. The slope of a position as a function of the time graph is the velocity. The slope of a velocity vs. time graph is acceleration.
# Acceleration: a = (∆v / ∆t) = (vF - vi) / (tF - ti)
Since velocity is in meters over seconds, then acceleration is in (meters over seconds) over seconds, which is equal to meters over second². Acceleration, as a function of velocity, has both magnitude and direction.
# P. Rule 10. The object is speeding up whenever velocity and acceleration are in the same direction (e.g. they are the same sign), while it is slowing down whenever velocity and acceleration are in opposite directions/signs.
# The Truth About Velocity & Displacement:
In your reading of Velocity and Slope, you have no doubt read these two truths:
Velocity = ∆x / ∆t (Displacement / Change in Time)
Slope = ∆y / ∆x (Change in Y / Change in X)
Velocity = Slope
The '∆x' of slope is not referring to displacement, but rather the change in the x position. This is constant cause of confusion, but know that in the majority of cases unless expressly obvious otherwise, '∆x' is in reference to displacement, the change in position.
Here are the two equations shown as equivalent:
∆x1 ∆y
━━ = ━━
∆t ∆x2
Since the two equations are known to be equivalent, we known that the ∆x1 of displacement and ∆x2 of slope are not the same, as if we were to cancel out the ∆x of the right side by multiplying it by both sides, we would find that ∆x² / ∆t is equal to ∆y, which is total nonsense. The key to understanding how this statement is true (and how the ∆x is referring to two things simultaneously) is to realize that both equations are stating the same ratio, just in terms of different graphs.
The equations are both output/input, and so they produce velocity. If the graphs were from the standpoint of velocity, then they would produce acceleration (see the "Important Note").
I.IV Uniformly Accelerated Motion.
A signifier that indicates that an object is moving with an acceleration that is constant. E.g., in calculus terms, the object has a position represented by an equation with an exponent no higher than two (a quadratic) in its time graph. U.A.M. allows the physicist to use 5 special equations not allowed otherwise:
1. Velocity Final is equal to velocity initial plus the acceleration multiplied by the change in time:
VF = Vi + (a × ∆t).
2. Displacement is equal to velocity initial multiplied by the change in time, plus 1/2 multiplied by acceleration and the change in time squared:
∆x = (Vi × ∆t) + (1/2 × a × ∆t²)
3. Velocity final squared is equal to velocity initial squared plus two multiplied by the acceleration and the displacement:
VF² = Vi² + (2 × a × ∆x)
4. The change in position is equal to one half times the quantity of velocity final plus velocity initial:
∆x = (1/2) × (VF + Vi) × ∆t
VF = Velocity final
Vi = Velocity initial
a = Acceleration
∆t = Change in time
∆x = Displacement (or change in position)
YOU CAN CHANGE DISPLACEMENT TO BE CHANGE IN Y IF YOU NEED TO, SUCH AS IN FREE FALL EQUATIONS.
Additionally, there are some lame 'alternate' versions of these equations that are useful for deriving and integrating the equations:
v = v0 + at
x = x0 + v0t + (1/2) × at²
v² = v0² + 2a(x - x0)
x = x0 + (1/2) × (v + v0) × t
# P. Rule 11. There are 5 U.A.M. variables, and there are 4 equations. If you know 3 variables, you can find the other 2, leaving 1 happy physics student. The highest exponent in the position equation must be 2 or below for U.A.M. to be active. If the highest exponent is three or above, the object is not in Uniformly Accelerated Motion. These are also referred to as the kinematic equations, the equations that govern motion (when ignoring thing like drag/air resistance - see Section III.III for that). For motion in two dimensions, see Section III.II.
If you have two equations, and you know they have three of the same U.A.M. variables, you will know that the other two have to be the same as well. In the classic bullet equation, where we knew the fired bullet and dropped bullet had a Δy of -h, an acceleration of -9.81 m/s², and a Viy of 0, we could determine that the time it would tell them to hit the ground would be the same (considering how time is the same for the x-direction and y-direction - see Rule 27 and the rest of Projectile Motion for more information).
# Memorization tools for U.A.M. Equations:
<disclaimer>
I believe the best way to memorize new information is to associate it with something that we have already memorized that we subconsciously reinforce through repetition: music. Everyone has already memorized hundreds of melodies, so finding one that would match the rhythm and meter of the equation is not too terribly difficult. Of course, fast music is most of the time unfit for memorizing an equation, of which a slower recitation and melody are much easier to perfectly and quickly recall.
</disclaimer>
1. To the theme of Funiculì, Funiculà (Slower part):
Vf equals vi + a times change in t.
2. To the theme of Funiculì, Funiculà (Faster Part):
Change in, x equals, vi times change in t. plus one, half times, a, change in t squared.
vi times change in t equals vi times the change in t!
vi times change in t plus one half times a change in t, squared,
vi times change in t plus one half times a change-in-t squared.
3. To the theme of Mozart's Dies Irae:
Vf squared equals, (da da da da) vi squared plus, (da da da da) 2 times a times, change in x equals vi SQUARED plus 2a change in time.
4. To the theme of Satie's Gnossienne No. 5
Delta x, ~~~~ equals one slash two, ~~~~~~~~~~~~~~~ times velocity final plus vi, times change, in time...
# P. Rule 12. Acceleration refers to the change in meters per seconds per second, When a ball is rolling on an incline, it will start at 0 m/s in velocity and after 1 second get to 2 m/s, and with each passing second the velocity increases by 2 m/s. Finding the position for those times is just as simple as using the second U.A.M. equation, which in this case leaves you with each change in position equaling the change in time squared.
# P. Rule 13. Order of Magnitude estimates are the most simple estimates possible, with only one significant figure. If the 'order' of magnitude itself is needed, then round to have only tohe power of ten, adding one if rounding up.External Link.
# Esoteric U.A.M. Problem:
A use of the U.A.M. equations that is not immediately apparent:
Two trains, each having a speed of 30 km/h, are headed at each other on the same straight track. A bird that can fly 60 km/h flies off the front of one train when they are 60 km apart and heads directly for the other train. On reaching the other train, the (crazy) bird flies directly back to the first train, and so forth. What is the total distance the bird travels before the trains collide?
This question has several traits that force the physicist to utilize critical thinking skills instead of plugging in numbers into formulas mindlessly, which one may get in the habit of if asked too many uninspired questions.
For one, velocity cannot replace speed in most circumstances due to the inherent differences of velocity being straight-line difference and speed being the total distance traveled, but there is one exception: If the object (or objects, in this case) travel in a straight-line, which is to say in one dimension, then the speed is equal to the velocity and they are interchangeable.
The specific parameters of the problem make it easier to solve: image the trains as being simple objects with directionless speed (as stated in the problem), for example. From the perspective of the conductor of the first train, the second train is going 60 kilometers per hour. We can replace the entire two train scenario (which has perfect symmetry and identical U.A.M. variables) and replace it with a single moving object - the trains are individually moving 30 km/hr towards eachother at a starting distance of 60 km from eachother - this is equal to a single train moving at 60 km/hr toward a brick wall 60 km away. Mathematically, these conclusions can be derived as follows:
SpeedTotal = Speed1 + Speed2
SpeedTotal = 30 km/hr + 30 km/hr
SpeedTotal = 60 km/hr
Thus, in this question, ViT and VFT are in reference to the single moving train, are all equal to 60 km/hr. Additionally, the change in x has been changed to 60 km, and considering the doubled speed of the single train, this works out perfectly with the original scenario. Since there are three known variables, we can find the other two.
For this question only, it seems more logical to remain in km/hr instead of converting into m/s as is the usual.
We will first try to get the acceleration out of the train's U.A.M. equations - each equation has acceleration, so we need it to get to what we really want - the change in time.
VF1² = Vi1² + 2 × a × ∆x
(60 km/hr)² = (60 km/hr)² + 2 × a × 60 km.
0 = 2 × a × 60 km.
a = 0
This simplifies things significantly. Thus, for the second equation, which has two parts, both with the change in time, and the second with acceleration as a multiplier, is best suited for this scenario, as it can directly reveal the change in time.
∆x = Vi1 × ∆t + (1 / 2) × a × (∆t)²
60 km = (60 km/hr) × ∆t
∆t = 1
This is also very useful. If we were to consider the flying bird, we would find that it is flying at the same speed as the train and would thus go 60 km/hr for one hour before crashing into the wall, which is clearly equal to 1 hours worth of flight and thus 60 km of flight distance.
Utilizing vector addition (rule 18) would not work, because the directions would cause the train velocities to cancel out.
# Instantaneous Velocity: The velocity at a specific point in time. The Velocityfinal and Velocityinitial from the U.A.M. equations are instantaneous velocities because they represent the object's velocity at a specific point in time.
# Average Velocity: The velocity over a time period. The velocity definition, change in position over change in time, is an average velocity itself.
# P. Rule 14. An average velocity written as V(5-10 sec) is just (x10 - x5) / (t10 - t5). This style of writing, although can be misinterpreted as meaning V-5, specifies an interval on the graph to find the velocity of.
ALL OF THE INFORMATION ABOUT DERIVATIVES WILL APPLY IN THE NEXT SECTION. MAKE SURE YOU KNOW AT LEAST THE DERIVATIVE PART OF SINGLE-VARIABLE CALCULUS BEFORE CONTINUING ONTO FREE FALL.
II. Free Fall.
I.I Intro to Free Fall.
- The only force acting on it is the Force of Gravity.
- It is not touching any other objects.
- There is no air resistance (the object is falling in a vacuum).
# P. Rule 15. ay is the acceleration in the y direction (for free fall on planet earth) is -9.81 m/s², or -g. Thus, free fall is a U.A.M. through the constant acceleration.
g = Acceleration due to Gravity. This is separate from the specific acceleration in the y direction.
gEarth = 9.81 m/s². The acceleration due to gravity on Earth is positive. From a global perspective, the g is not constant at all locations on the planet. However, from a local perspective, the acceleration due to gravity is constant.
For other planets, g is different, like gMars = 3.75 m/s².
# P. Rule 16. Additional facts about gravity:
- 9.81 m/s can also be represented as 9.81 m/s down.
- The acceleration due to gravity, g, is the same no matter the mass of the object.
# Parallax: When the position of an object appears to differ when viewed from different locations.
# P. Rule 17. Sometimes, when you are dealing with an object that changes direction, splitting your calculations into two is necessary in order to find information that serves the equation as a whole: For example, if you were to take a ball and throw it up in the air, catching it in the same y-value that you threw it, the change in position would be 0, making it impossible to find the change in time using the the entirety of the event. Thus, one has to divide the movement of the object along the critical point, with the period of the ball moving upward and then moving down being separately used in calculations. Then, values can be combined to reflect the full event (the change in time, if the ball lands in the same place where it was thrown, would be the change in time of either half of the event doubled, due to symmetry. If the ball lands somewhere elsewhere then it was thrown, then the time for each half would have to be calculated and then added together).
# Common Mistakes in Free Fall:
-When you throw a ball upward, it will not have a positive acceleration. Having such would make it shoot upward like a rocket. Gravity is a positive constant, but the acceleration in the y direction during free fall is -9.81 m/s² for any object, whether moving upward or downward.
-Objects thrown upward do not have an initial velocity of 0 - they already have a positive velocity before being thrown due to whatever is throwing them. An initial velocity of zero in the y direction will not cause an object to move upward - it must be positive.
-The force with which an object is thrown will not effect the acceleration. Regardless of whether a ball is dropped or thrown downward, the acceleration in the y direction will be -9.81 m/s², and only the position and velocity will change to reflect this force.
III. Vectors and Projectile Motion
III.I Vector Basics.
# Scalar: A quantity that has magnitude only. Examples include distance, speed, time, volume, and density.
# P. Rule 18. Introductory Tip-to-Tail Vector Addition:
Vector addition is not simply adding the magnitude of the vectors together, but using the pythagorean theorem to find the line that connects the given vectors when they are placed tail-to-tip. This tells you the resultant vector.
If the vectors are in perfect cardinal direction, as is the example below, then of course pythagorean theorem can be quickly applied. Otherwise, using vector decomposition (Rule 20) is the easiest way to create a right triangle for use in the pythagorean theorem, unless you want to use the Law of Cosines.
Here are two vectors utilizing the vector symbols:
A = 49 (mm/s) North
B = 42 (mm/s) East
a² + b² = c²
C² = A² + B²
||C|| = √A² + B²
||C|| = 65 mm/s.
Since a vector has direction, we need to find it trigonometrically:
tanθ = (O / A) = (B / A) = (42 / 49)
θ = tan⁻¹(42/49) = 40.601° = East of North (see rule 19).
Thus, the object is moving at a resultant vector of 65 mm/s East of North.
# Vector subtraction:
Vector subtraction is similar to addition. A - B = A + (-B). Thus, all that would need to be done for vector subtraction is to put B in the opposite direction and discover the resultant vector from there. This, indeed, can be a different resultant vector from vector addition - gander:
# P. Rule 19. Cardinal directions in Vectors are as follows:
Additionally, the cardinal directions at perfect 45° angles would be as follows:
# P. Rule 20. Complex vector problems oftentimes entail vector decomposition, where one may have to use trigonometry in order to find the sides of the vector triangle. Take this North of East vector for example:
d = 90.0 cm @ 32° N of E.
sinθ = (O / H) = (dy / d)
dy = dsinθ = 90sin(32) = 47.693 ≈ 48cm.
cosθ = (A / H) = (dx / d)
dx = dcosθ = 90cos(32) = 76.324 ≈ 76cm.
Decomposing a vector is also called "breaking" or "resolving" the vector. The information of the original vector can be derived from the component vectors as well, using inverse tangent and the pythagorean theorem, known as 'recomposition'. If you are having problems with re-forming the resultant vector from the component vectors, remember that the resultant vector can be represented as the diagonals of a rectangle that has the components as its sides. This will strongly help with visualization.
The COMPONENTS of a VECTOR are ALSO VECTORS. There are just in a single direction, represented in unit vectors like so: [0î + 5ĵ] or [5î + 0ĵ] (see Rule 22).
# P. Rule 21. The Associative Property of Vector Addition:
A + (B + C) = (A + B) + C.
This property, in conjunction with vector decomposition, is very useful when doing vector problems such as the following:
# P. Rule 22. UNIT VECTORS:
Unit vectors are vectors with a magnitude of 1 unit in an absolute direction, without any decomposable angle. They are written differently than normal vectors, with a roof symbol on top rather than an arrow:
î = x̂ = 1 unit in the x direction.
ĵ = ŷ = 1 unit in the y direction.
k̂ = ẑ = 1 unit in the z direction.
Vix ≈ [31.7î + 14.8ĵ] m/s ≈ 35.0 m/s @ 25.0° above horizontal.
This way of writing is simply the pre-decomposed way of writing a vector.
If you have to go from an already decomposed roof form to the exact same form as an answer, it is easy as hell:
A = [2.0î + 2.0ĵ]m
B = [1.0î - 3.0ĵ]m
C = [4.0î + 4.0ĵ]m
R = A + B + C = [7.0î + 3.0ĵ]m
Note that given the component vectors, reconstruction of a direction for the whole vector is possible (obviously), though unless the angle is a perfect 45°, you will only get a general idea of where it is pointing, since 25° North of East is the same as 65° East of North.
# P. Rule 23. When the questions asks to find the distance between two vectors, you can use a simple mechanic:
Create your own vector that has its tail on the tip of the first vector and its tip on the tip of the second vector, call it ∆r.
r1 + ∆r = r2
∆r = r2 - r1
You do not have to move any of the given vectors, you can simply create one that serves the middle an already connected vector would have served in its place. Then, find the displacement (length) of that vector by using the strategies taught in rules 19 and 20. The vectors may look similar to this:
From this point, you can either find the answer in terms of unit vectors or in terms of magnitude in direction. Of course, unit vectors will always be faster and simpler. Here is an example problem carried out to fruition using unit vectors:
r1 = 15.0m @ 55.0° E of N
r2 = 25.0m @ 45.0° N of E
r1 = [r1sinθ1î + r1cosθ1ĵ]
= [(15)sin(55)î + (15)cos(55)ĵ]m
= [12.2872î + 8.60365ĵ]m
r2 = [r2cosθ2î + r2sinθ2ĵ]
= [(25)cos(45)î + (25)sin(45)ĵ]m
= [17.6777î + 17.6777ĵ]m
Δr = r2 − r1 = [12.2872î + 8.60365ĵ] − [17.6777î + 17.6777ĵ]
= Δr = [5.39040î + 9.07405ĵ]m
Because change in r is also change in position, this vector is also the displacement between the two original vectors:
Δr ≈ [5.39î + 9.07ĵ]m
||Δr|| = √(5.39040)² + (9.07405)²
||Δr|| = 10.5544 ≈ 10.6m
# P. Rule 24. The r position vector, r, is used to identify the location of an object in multiple dimensions.
r = xî + yĵ + zk̂ identifies the location of an object in 3-dimensional space.
Δr = rf - ri is the displacement of an object in 3-dimensional space.
Δr/Δt = vaverage & dr/dt = vinstantaneous
Δv/Δt = aaverage & dv/dt = ainstantaneous
III.II Projectile Motion.
- The only force acting on it is the Force of Gravity.
- It is not touching any other objects.
- There is no air resistance (the object is falling in a vacuum).
- The object has Uniformly Accelerated Motion, and thus can have all of the U.A.M. equations applied.
When solving a problem like this, you would need to separate your known variables in the x and y directions. The equations that describe the motions of an object in projectile motion are different in the x direction and the y direction. Various things placed in their appropriate columns include:
X-direction ┃ Y-Direction
━━━━━━━━━━━━━━━━━━━╋━━━━━━━━━━━━━━━━━━━━━━━━
ax = 0 ┃ Free-Fall
Constant Velocity ┃ ay = -g = -9.81(m/s²)
Vx = (∆x / ∆t) ┃ U.A.M.
┃
(Needs 2 Var.) (Needs 3 Var.)
In projectile motion, the object will not accelerate to the left or to the right, it will have a constant velocity only. The localized gravity here on Earth makes it only give the y direction of a falling object acceleration.
If you were to drop an object straight downward, the constant velocity would be zero as it would not move in the x or y direction whatsoever. It would also not be in projectile motion, as it is only moving in one dimension.
If the object were to be thrown however, the object would have a non-zero constant velocity and would be in projectile motion.
# P. Rule 26. - Differing equations for free fall in the X-direction and Y-directions:
X-DIRECTION:
Since the acceleration for projectile motion in the x-direction is always zero, and every U.A.M. equation has acceleration, after plugging in zero for acceleration for each one we are left with only two equations:1. VF = Vi
The first equation simplifies the velocity of x in projectile motion to a single variable, Vx, which we already intuitively knew.
2. ∆x = Vi × ∆t
Substituting in Vx, as discovered in the first equation, gives us the following:
∆x = Vx × ∆t
Upon further inspection, one should be able to identify this equation as merely being the standard velocity equation: Vx = (∆x / ∆t)
Thus, the velocity equation itself must be the one used when calculating movement in the x direction during projectile motion. Since there are only 3 variables, you would need to know 2 variables to discover the other 1.
Y-DIRECTION:
For the y direction, the equations are the same as they have always been - all four U.A.M. equations are in play with acceleration as -9.81 m/s², and you need three variables to find the other two. This simplifies the process.# P. Rule 27. The only U.A.M. variable that is a scalar is change in time (∆t). If you solve for the change in time in the y or x direction, you can use it for the other direction. This is because ∆t is independent of direction as a scalar.
# P. Rule 28. When you need to find the final velocity of an object in projectile motion, you need to find the velocity in the y direction and the velocity in the x direction. Then, you must realize that since you have the x and y components of an objects movement, that you have the decomposed vector and now need to find the vector itself, the hypotenuse of the triangle through the pythagorean theorem.
NEVER FORGET to find the direction at the end - you just have to do the sin inverse (or cosine inverse depending on which theta you need, though the theta should always be on the tail of the vector/hypotenuse) using the x and y decomposed vectors. This will give you the angle that, if you don't have any particular directions to work off at the beginning, can just be said to 'in front of the negative y-axis' if the object is moving rightward and downward, changing this as needed.
Remember that angles also must follow sig. fig. rules no matter what.
Vectors always require direction in addition to magnitude, a + or - will not do.
# P. Rule 29. As an object falls, the magnitude of the velocity increases and the direction of the velocity points farther and farther down. The velocity vector arrows increase in length as the object falls, because the magnitude of the object's velocity keeps increasing.
# P. Rule 30. Any U.A.M. variable that is a vector (everything other than time) can be decomposed to find its x and y-direction component vectors. This is most useful when you need to find the movement of an object in both directions during projectile motion. For example, if an object were to be launched with an initial speed of 3.25 m/s at an angle of 61.7°, you can easily find the initial velocity in the x-direction and y-direction by using the decomposition tricks of Rule 20.
# P. Rule 31. When doing calculations with unit vectors, as could be done in the previous question, the 2nd U.A.M. equation (and all others, for that matter) can be easily converted into a more useful form:
rF = ri + Vi × t + (1/2) × a × t²
Of course, this equation was created by recognizing r as being the displacement (∆x), and splitting it into its final and initial forms and moving the initial to the right side, as can be done for ∆x or acceleration at any time. Whereas before we were not writing the vector symbols in the U.A.M. equations, we are taking specific note of their vector status now.
'r' is a displacement vector, representing the change in position of the object - see Rule 24 for more uses of it. ri and rF are position vectors, in that they describe the object's location at specific points in time, each being the sum of its components in the x and y directions (and potentially z direction in 3D space). ri and rF are of course representative of the initial and final positions of the object, respectively. The displacement vector r indicates the direct path and distance between these two positions.
Recall that when you are working in unit vectors, U.A.M. equations are done with the x and y component vectors of each vector. For example, Vi can have the value [1.32î + 4.24ĵ] (or anything of the sort) plugged in to any equation. These form a more fun equation to solve, because you are plugging in the x and y forms simultaneously into the equation (for an example and a neat trick, see Rule 32).
EXTREMELY IMPORTANT, READ THIS:
You can easily get very confused when you draw the curved trajectory of an object in projectile motion, and then you draw a vector with a straight line. No, the vector you drew for whichever variable is not saying that the object moved in that same straight line. That is not what vectors mean - vectors represent direction and magnitude, not the actual path of motion. If you have the initial velocity for an object and you draw the vector to find the x and y components, you are only drawing a visualization of a position vector, (as described above), of how much the velocity or whatever is directed horizontally or vertically.
The vector right triangle is not the actual trajectory of the object on a graph.
The vector right triangle is not the actual trajectory of the object on a graph.
# P. Rule 32. There is something very beautiful that is the result of the independence of the x and y-directions - Say you have plugged in the unit vectors for each of the variables in the 2nd U.A.M. equation (in the same fashion as Rule 31), and you get the following:
[0.93î + yĵ] = [0î + 0ĵ] + [1.54079î + 2.86155ĵ]t + (1/2) × [-9.81ĵ]t²
0.93î + yĵ = 1.54079tî + 2.86155tĵ - 4.905t²ĵ
What I will now do, will be one of the miraculous things you have ever seen in your life, and something you shall remember until your dying days: Because x-direction and y-direction are independent, we can isolate the x-direction by removing everything else:
0.93 = 1.54079t
From here, the process is simple, isolating time and whatnot (which, as we learned in the revelations of Rule 27, is a directionless scalar quantity that can be used in both the x and y directions). Of course, we can repeat this process by isolating the y-direction, where we can now plug in the newly found time.
# P. Rule 33. Some general tricks that keep one's ducks in line when doing physakz: When you discover the value of a new variable, always immediately check whether the sign is correct.
1. Your object is falling? The velocity is going to be negative.
2. You found negative time? Time travel doesn't exist yet. Go back and find your error somewhere, which probably has to do with possibility 1.
Additionally, sometimes, specifically when you are finding time in the 2nd U.A.M. equation (where time is squared), you will need to use the quadratic formula. Try to figure any possible way in which you avoid using it, whether using a different formula or whatever. If you have to, you have to, but use your common sense in determining which answer applies to your problem, if only one.
III.III Rotation & Multiplication.
# Vector Rotation:
The coordinate system you are familiar with has the x and y axes parallel to the sides of the page. When you have a vector (lets name it a) on this natural coordinate system, the components of the vectors are going to parallel to the sides of the page as well.
If you were to rotate the axes (but not the vector a) through an angle ϕ as shown below, the components would have new values, a'x and a'y. Since there are an infinite number of choices of ϕ, there are an infinite number of different pairs of components for a:
||a|| = √(aₓ)² + (aᵧ)²
||a|| = √(a'ₓ)² + (a'ᵧ)²
Also,
θ = θ′ + ϕ
The relations among vector do not depend on the location of the origin nor on the orientation of the axes.
# ==Multiplying Vectors==
There are three types of vector muliplication. The first type is multiplying a vector by a scalar, which gives a vector. The other two multiply a vector by a vector, of which you have two choices: Taking the 'dot product' of two vectors gives a scalar answer, while taking the 'cross product' gives a new vector that is perpendicular to the original two, the direction of which can be determined using the right-hand rule.The best way to understand the difference between dot product and cross product beyond their scalar/vector nature is to consider when their products are minimized or maximized (described in each section). As a simple memorization tool, always remember that a · a = (|a|)², or the magnitude of a squared, while a × a = 0. The reasons for this are obvious once you analyze each multiplication tool in full.
# Multiplying a Vector by a Scalar:
The product of the scalar s and a vector v is a new vector whose magnitude is sv, and whose direction is the same as that of v if s is positive, and opposite that of v if s is negative. To divide v by s, multiply v by 1/s.
# Multiplying a Vector by a Vector:
# DOT PRODUCT:
Simply put, in only two dimensions,
a · b = abcosϕ
a and b being the magnitudes of a and b, respectively.
ϕ represents the angle between a and b.
The value given by Dot Product is a scalar, and you can see this through how each term on the right side of the equation is a directionless scalar value. If the angle ϕ between two vectors is 0°, the component of one vector along the other is maximum, and so also is the dot product of the vectors. If, instead, ϕ is 90°, the component of one vector along the other is zero, and so is the dot product. READ: If the vectors are parallel, dot product is maximized, and if vectors are perpendicular, the result is zero. Cross product is the opposite.
VERY IMPORTANT: The commutative law applies to the dot product. Thus,
a · b = b · a
This is not true with cross product.
Triple Vectors - If either or both of the vectors are 3-dimensional, you must utilize unit-vector notation. In this state, their dot product would be
a · b = (aₓî + aᵧĵ + azk̂) · (bₓî + bᵧĵ + bzk̂)
a · b = aₓbₓ + aᵧbᵧ + azbz
This will give THE EXACT SAME ANSWER as the original equation if you are only using two-dimensions, and will save you a lot of time if you are very lazy. Ignore the fact that we are not really 'distributing' the terms as you normally do with the distributive law. Unit Vectors are unique.
# CROSS PRODUCT:
Determining the magnitude of the resultant vector is easy - determining the angle/direction of the resultant vector is easy with extra steps.
Finding the magnitude for vector product is similar to that of dot product:
a × b = c
c = absinϕ
Of course, a and b are the magnitudes of a and b, respectively.
ϕ represents the angle between a and b.
The value given by Cross Product is a vector, so the direction must be calculated in addition to the magnitude. If a and b are parallel or antiparallel, a × b = 0. The magnitude of a × b, which can be written as |a × b|, is maximum when a and b are perpendicular to each other. READ: If the vector is parallel, the resultant magnitude is 0, and if it is perpendicular, the result is maximized. Dot product is the opposite.
VERY IMPORTANT: The commutative law DOES NOT apply to the scalar product. Thus,
a × b ≠ b × a
Instead,
a × b = -(b × a) This is the opposite of the dot product.
The direction of cross product is pointing in the n̂ direction, where n̂ ⊥ â and n̂ ⊥ b̂. You can easily determine the direction of a cross product, perpendicular to the two vectors, using the "Right-hand Rule":
The easiest way to think about it is this: Use the positioning of the right hand as shown in the images below and above. Do not forget that the cross product is non-commutative, and so it 100% matters which vectors you assign your fingers. Middle finger is A, pointer finger is B, thumb is C. If you were to do this same thing on your left hand, you would get B × A instead of A × B - it just works backwards.
Twist your hand so that your fingers are pointing in the correct direction of the vectors - this does not mean that your fingers have to look exactly like the graph, only that they must be in the correct direction as their respective vectors. Your thumb, resultant vector c, will be pointing in the direction of the cross product.
You can use this system for visualizing the x, y, and z directions as well. The Right-hand uses a right coordinate system, as opposed to a left coordinate system, which would necessitate using the left hand. These systems are merely different visualizations of the x-y-z plane.
Triple Vectors - If either or both of the vectors are 3-dimensional, you must utilize unit-vector notation. In this state, their cross product would be
a × b = (aₓî + aᵧĵ + azk̂) · (bₓî + bᵧĵ + bzk̂)
But we do not simplify to isolate every x, y, and z coefficient as we did with Dot Product. Instead, we are going to use a matrix ([[[[[[[[).
Starting from the unit vectors on top, the variables in green are multiplied, and those in red are subtracted from the ones in green, all within the specific unit vector they are attached to. Thus, from this matrix (WHICH SHOULD EASILY BE SET TO MEMORY FOR FUTURE REFERENCE) we can discern the following equation:
a × b = (aᵧbz - bᵧaz)î + (azbₓ - bzaₓ)ĵ + (aₓbᵧ - bₓaᵧ)k̂
III.IV Drag.
# Fluid: Anything that can flow, so generally gases and liquids. Any relative velocity between a fluid and a body (either because the body moves through the fluid or because the fluid moves past the body) will cause the body to experience a drag force.
# P. Rule 34. My Treatise on Drag (1):
Because projectile motion and free fall have explicit requirements of an object moving through a vacuum, accounting for air resistance means that the object is not in projectile motion or free fall. Thus,
ay ≠ -9.81 m/s²
Vx ≠ Constant
The force of drag always acts opposite to the direction of the velocity of the object:
There are several equations for drag, each with their own specifications. The equation known simply as the "drag equation" is most commonly used, and is as follows:
Fdrag = (1/2) × ρ × V² × D × A
ρ (rho) = Density of the medium
V = Velocity of the object
D = Drag Coefficient
A = Cross Sectional Area
A: The cross-sectional area of the object is the area of the object normal to the direction of it's travel. E.g., when the object is moving directly towards you, what object do you see?
Immediately note that Fdrag is a scalar. Other equations, like the linear drag equation, will result in a vector.
# Analysis:
Take the object to be a ball, for the sake of discussion.
Acircle = πr²
Additionally, let's assume that the following information about the ball:
r = 0.031835 meters
m = 0.14529 kg
Vix = 4.469444 m/s, in the first 1/100th of a second.
D: The drag coefficient is the number used to model all of the complex dependencies of drag on shape, inclination, and some flow conditions. Essentially, it is a number determined through experiment that helps determine an object's drag. Thus, the coefficient will change depending on the shape of the object, the type of fluid flow around the object, and the object's speed.
The drag coefficient for a smooth sphere is 0.5, conveniently.
ρ: The medium in which the object is traveling is of course air, and the density of air can fluctuate due to temperature and pressure differences. The generalized air density we will use is 1.275 kg/m³.
V: Since we know we will be working with the force in the x-direction, we can use Vi and VF interchangeably with VT. As the velocity of the object changes, the drag coefficient will change with it, changing the net force (as shown in the free body diagram) of the ball, and changing the ball's acceleration. Summarized,
∆V → ∆Fnet → ∆a.
↑ ↓
┗━━━━━━━━━━━━━┛
Clearly, the acceleration is not constant throughout this interaction, and so the object would not be in U.A.M.
However, the mathematician Leonhard Euler devised a trick to emulate U.A.M. by splitting the motion of the object into many short-lasting parts, such as 1/100th of a second, under which the acceleration is approximately constant. This is similar to the concept of Riemann sums being used to find the area under the curves, with the more subdivisions, the more accuracy. Let us do some preliminary work to find the net force in the x-direction before utilizing Euler's method:
Σ Fx = -Fdrag x = m × ax
ax = -((1/2) × ρair × (Vix)² × D × A)
ax = -(ρair × (Vix)² × D × (πr²)) / 2m
ax = -(1.275 × 4.4694² × 0.5 × (π × (0.031835)²)) / 2 × (0.14529)
ax = -0.13953 m/s²
There is only one force in the x-direction, the drag, which is of course going in the opposite direction of the object in the x-direction and so is negative. This is the net force of the ball, and by Newton's Second Law of Motion (see Rule 70), we know that the net force is equal to the mass of the ball times the acceleration in the x-direction. Continuing onward, we can substitute in the drag equation to be equal to mass times acceleration. From there, the algebra is simple, and you just need to plug in all of your variables to find the answer.As a result, we have found the acceleration in the x-direction, which DOES NOT EXIST in projectile motion. Since you now know acceleration, Velocity initial, and change in time (1/100 sec), you find 4.6805 m/s as your velocity final. Therefore, compared to the Velocity initial of this 0.01 second period of 4.4694, the velocity in the x-direction (which we took to be constant) has decreased by 1.4 thousandths of a meter per second, accounting for air resistance in our calculations and contrary to what the standard U.A.M. equations for projectile motion would suggest.
∆x = xF - xi = Vix × ∆t + (1/2) × ax × (∆t)²
xF = xi + Vix × ∆t + (1/2) × ax × (∆t)²
x = (0)+(4.4694) × (0.01) + (1/2) × (-0.13953) × (0.01)²
= 0.04469 meters
# TERMINAL VELOCITY:
"The speed at which a falling body will cease accelerating and begin moving at a constant speed downward, as a result of the effects of drag."
As a body falls, the drag force gradually increases in magnitude (from zero) as the speed of the body increases. If the body falls long enough, the force of drag will eventually equal to the force of gravity, ideally. In the real world, the force of gravity will never exactly equal the force of drag, of course, but the acceleration will become small enough that it will be negligible for any calculations.
Though the forces will cancel eachother out, this does not mean that the object will stop moving (see Rule 76 for the Third Law of Motion) - it only means that the object will have no acceleration, and that it will be in equilibrium, in the sense that the velocity will be constant.
From this point onward, the body will fall at a constant speed, known as the terminal velocity VT.
# LINEAR DRAG:
There is a simpler, more memorizable equation for drag that THEY don't want you to know about. It is known as the Linear Drag Equation, and it can only be used for "small objects moving at slow speeds":
Fdrag = -b × v
b = Proportionality constant
v = Velocity of the object
A "Proportionality constant" is just the ratio between two quantities, and here it has the units kg/s.
While the traditional "drag equation" as elaborated previously results in a scalar, this new drag equation results in a vector.
The negative in the equation means that the drag force has the opposite direction of the velocity of the object. Thus, when the object is moving upward on Earth, the force of drag is downward, and when the ball begins falling downward, the force of drag moves upward.
# DRAG ANALYSIS:
As an object moves upward, the acceleration will decrease to -g, because of the following sequence of events:
The object will have a downward acceleration and drag force.
Because of the downward acceleartion, the speed of the object will decrease.
Because the speed of the object is decreasing, the magnitude of the drag force will decrease as well, because the drag force is proportional to the square of the velocity.
As the magnitude of the drag force decreases, the magnitude of the net force will decrease as well (as the net force is composed of the acceleration and drag force, which are acting in the same direction here).
Thus, because the net force of the object is decreasing, the acceleration is going to decrease, and when the object is at the top of the parabolar path, it will have an acceleration of -g, and will temporarily have a speed of 0 in the y direction.
When the object moves down, of course, the drag force will grow in the opposite direction of gravity as the objects falls.
Thus, although the speed of the ball is increasing, the magnitude of the net force will be decreasing, seeing how the acceleration and force of drag will increasingly cancel out as the object picks up speed.
Finally, because the magnitude of the net force is decreasing, this means the acceleration is decreasing, and it will be continue decreasing until it reaches zero (or near-zero, at which point the object will be in Terminal Velocity.
What this means:
Because the average acceleration on the way up is lower than the average acceleration on the way down, as has just been proven, and given the object lands at the same height is was thrown (e.g. ∆y is the same going up and going down), then if the only two forces acting on the object are drag and gravity, then
The time it takes for the object to go up a distance will LESS than the time it takes to go down the same distance.
IV. Range & Relative Motion.
IV.I Basics of Range.
# P. Rule 35. Here is the 'Range' equation, giving you the horizontal displacement (which, you should note, is not written with a Δx, as that would displacement as a whole). There are specific characteristics of a question that will allow you to use the range equation, in the same manner as the U.A.M. equations. If a question explicitly says that an object in projectile motion is landing at the same height which it was thrown (or something to that effect), then the range equation can be used as the displacement in the y direction is zero.
Range = (Vi² × sin(2θi)) / g
R = Range = Δx
Vi = ||Vi|| (Magnitude of Vi)
θi = Initial Angle or Launch Angle
g = Acceleration due to gravity on Earth, POSITIVE 9.81 m/s²
Because both the magnitude and direction of the initial velocity vectors are present in the equation, the initial velocity does not need to be resolved into its components for use in the equation.
g: In projectile motion, ay = -g
g = +9.81 m/s²
ay = -9.81 m/s²
Given this information, we know that the value in the bottom of the equation is not the negative acceleration we are accustomed to, but rather the positive acceleration due to gravity on planet Earth, 9.81 m/s². Because gravity has the dimensions of m/s², the rest of the variables must use the standard meters and seconds for their dimensions, m/s for ||Vi|| and meters for the range itself.
θi: The sine of any angle can never be greater than one. Therefore, the maximum range an angle can have (given the magnitude of the velocity initial is constant, which is should considering this is the x-direction) is discoverable through simply setting the sine value of the equation equal to one:
1 = sin(2θi)
2θi = sin⁻¹(1) = 90°
θi = (90° / 2) = 45°.
Therefore, a launch angle of 45° will give us the maximum range of an object given a constant magnitude of the initial velocity. Any angle other than 45° will give a smaller range.
Furthermore, being familiar with the graph of sin(2θ), we know that any two complementary angles will have the same sine value and range, such as 30° and 60°. Mathematically, this is shown as sin(2θi) = sin(2(90 - θi)).
JUST BECAUSE TWO ANGLES HAVE THE SAME RANGE, THAT DOES NOT MEAN THAT THEY HAVE THE SAME PROJECTILE PATH. An object launched at 60 degrees will go higher than one launched at 30 degrees, but they will end up in the same place.
# Memorization tool for Range:
I believe the best way to memorize new information is to associate it with something that we have already memorized that we subconsciously reinforce through repetition: music. Everyone has already memorized hundreds of melodies, so finding one that would match the rhythm and meter of the equation is not too terribly difficult.
1. To the theme of Nessun Dorma:
Velocity initial squared, times sine 2 theta, initial, divided by gravity...
IV.II Frames of Reference.
# P. Rule 36. Relative motion is done with vectors. Imagine a van and a motorcycle driving side by side: the velocity of the van relative to Earth is 24 mi/hr in the East direction. The velocity of the motorcycle relative to the Earth is 13 mi/hr in the East direction. These values would be written thusly:
VvE = 24 mi/hr E
VmE = 13 mi/hr E
In order to find the velocity of the van with respect to the motorcycle, we must use Vvm as the term. This value will be the difference between the two vectors.
VvE = VmE + Vvm
Vvm = VvE - VmE
Vvm = 24 mi/hr E - 13 mi/hr E
Vvm = 11 mi/hr E
You may notice that by taking the negative of the vector, we are in effect reversing its direction. This can be represented by switching the order of the subscripts.
-VmE = VEm
= -13 mi/hr E
= 13 mi/hr W
Therefore, from the point of view of the motorcycle, it would appear that the Earth is moving westward at 13 mi/hr. Finally, in order to solve the velocity of the motorcycle with respect to the van, we simply need to take the negative of the already known value of the motorcycle with respect to van:
Vmv = -Vvm
=-11 mi/hr E
11 mi/hr W
Thus, the van, which is moving faster than the motorcycle, would see the motorcycle as moving backwards (west) as 11 mi/hr.
# P. Rule 37. There is a trick that can be used in all relative motion subscripts: any time you sum two relative motion velocities and they share a common subscript on the inside or the outside, that subscript will cancel out in a specific way. Using the example from Rule 36, we can understand what this specifically means:
Vvm = VvE + VEm
The "E" (Earth) on the inside of the two velocities will cancel out, and because the common subscript was on the inside, the subscripts will combine forwardly into vm.
VvE = VmE + Vvm
With this equation, because the common subscript is on the outside, the resultant subscripts are presented backwardly. Of course, you can switch the order of the terms to make the common subscript be on the inside and get the same answer, but knowledge of these tricks either way will be very helpful.
# KINEMATICS POSTMORTEM:
We have now come to the end of Non-Rotational Kinematics (at least, what I will consider 'Non-Rotational Kinematics' in my notes), our studies in Vector Addition, Projectile Motion, Relative Motion, and everything else. Let us review the similarities between the vector decompositions of the various motions: The core principle behind decomposition is that when a vector is not directly in the x or y direction, break the vector in its components. Why do we do this?
Vector Addition: We create components in order to make a right triangle so we can use the Pythagorean Theorem and the Trigonometric functions.
Projectile Motion: We resolve the initial velocity into components because we use different equations in the X and Y directions for projectile motion. We use the equation for constant velocity in the X direction (because acceleration is zero), and in the Y direction we use the U.A.M. equations for free fall.
Relative Motion: Same as Vector Addition, decompose to make a right triangle.
For ever and ever, you will use vectors and vector decomposition in your daily life, because if you are not living as a physicist, are you truly living?
V. Rotational Motion.
V.I Angular & Tangential Velocity.
# P. Rule 38. The position of an object in rotational motion can be determined in Cartesian and Polar coordinates, going back and forth using trigonometry. When an object is moving along a circle with a constant radius, the object undergoes an angular displacement (with regard to the Polar coordinates (r, θ)):
Δθ = θF - θi
The curved linear distance the object would move when moving along an arc is called the arc length. It is the curved length on the side of the circle outside of the angular displacement Δθ. The equation for arc length is as follows:
s = r × Δθ
Where s is the arc length, r is the radius, and Δθ is the angular displacement. THE UNITS FOR ANGULAR DISPLACEMENT MUST BE IN RADIANS. CONVERT USING (π / 180) IF NEEDS BE.
# P. Rule 39. π = 3.1, give or take. It refers to the ratio of a circle's circumference to its diameter, and as an irrational number, it will go on forever. Since the formula for obtaining pi cancels out any dimensions in length (c/d = (86.9m)/(27.5m) ≈ 3.1415926...), the value of pi is in Radians, which is just a dimensionless placeholder unit. The abbreviation for Radians is 'rad'.
# P. Rule 40. While average linear velocity is conveyed as Vavg = (∆x / ∆t), average angular velocity is written as ωavg = (∆θ / ∆t), which can be represented in Radians per Second, Revolutions per Minute, Degrees per Millisecond, etc.
Radians per Second are most commonly used in Physics, while revolutions per minute is most used in Applicationism and "the real world".
# P. Rule 41. The symbol for angular acceleration is α, which is very convenient considering how similar it looks to a. The average angular acceleration is equal to αavg = (∆ω / ∆t). Therefore, the units for angular acceleration are radians per second squared or revolutions per minute squared.
# P. Rule 42. Just like an object can have uniformly accelerated motion, an object can also have uniformly angularly accelerated motion. This is written as U.α.M., instead of U.A.M., even though the alpha is lowercase. As with linear acceleration, the U.α.M. constants can be used when α is constant. All of the variables from U.A.M. must be converted into their angular form:
ωF, ωi, α, ∆θ, and ∆t.
Now, we can rewrite every U.A.M. equation using these variables:
ωF = ωi + (α × ∆t).
∆θ = (ωi × ∆t) + (1/2 × α × ∆t²)
ωF² = ωi² + (2 × α × ∆θ)
∆θ = (1/2) × (ωF + ωi) × ∆t
ALWAYS use radians in your calculations for U.α.M. equations. Additionally, remember that ωF and ωi are instantaneous angular velocities, as opposed to the average velocity (∆θ / ∆t).
# P. Rule 43. Since angular velocity, angular acceleration, and change in angular position are vectors, they all have direction. However, since clockwise and counterclockwise are observer-dependent, we use the right-hand rule to determine direction.
# Tangential Velocity: The linear velocity of an object moving in a circle. Given by the equation vt = r × ω. Because it is LINEAR, you use m/s as your base units. Furthermore, notice that none of the values in the equation are vectors - this equation refers purely to the magnitudes of each variable.
# P. Rule 44. The farther along you go on the radius of a circle, the linear distance that point travels when moving through a circle, which is called arc length, will increase. This is in spite of the fact that the angular velocity and angular displacement is the same, regardless of radii length, as each point on the radius covers the same number of degrees in the same amount of time.
The path traveled by each point on the radius will form an internal circle, the point moving around the circumference of this circle. Each point on the radius has a linear velocity when it is moving in the circle, known as tangential velocity, which increases proportionally as the radius and arc length (defined by the formula s = r × Δθ) increase. The tangential velocity is given by the equation vt = r × ω, which, just like the arc length equation, requires radians. The angular velocity in the equation needs to be in radians per second.
See the different internal arc lengths that each radius-length has below:
Tangential Velocity is named for how its velocity vector is tangent to the circumference of the circle, perpendicular to the radius. Tangential Acceleration is the exact same way:
# P. Rule 45. Tangential Acceleration is defined by the following equation:
at = r × α.
Like the equations for arc length and tangential velocity, you must use Radians as your angular quantity. Because it is LINEAR, you use m/s² as your base units. Furthermore, notice that none of the values in the equation are vectors - this equation refers purely to the magnitudes of each variable.
# P. Rule 46. On the nature of Tangential Velocity (and others) as a vector:
Imagine a situation in which the radius and angular velocity of a point on the circle are both constant - This just means that the circle is rotating at a constant speed.
If this were to be the case, then the linear velocity of the radius, the tangential velocity, would still not be constant, even though tangential velocity is literally equal to the radius times the angular velocity. This is because tangential velocity is a vector with both magnitude and direction. While the magnitude would be constant, the direction of the tangential velocity would be constantly changing (perpendicular to the radius as it makes it revolution around the circle), and so tangential velocity itself would not be constant. Therefore, the radius has neither a tangential acceleration nor an angular acceleration.
V.II Centripetal Velocity.
"Centripetal Acceleration is the linear acceleration that causes the tangential velocity to change direction."
Utterly meaningless on their surface, searching for any possible knowledge in these words is an exercise in futility. However, beyond their nonsensical face-value, an elaboration of these words will reveal their value:
The word 'Centripetal' is derived from "centrum", meaning center, and "petere", meaning to seek. Centripetal acceleration is the acceleration that causes circular motion. Centripetal acceleration is always directed inward to the circle, toward the center. It is a center-seeking linear acceleration. The force inward is what causes circular motion - see Rule 51 for more information.
The equation for Centripetal Acceleration is as follows:
ac = (Vt²) / r
Notice that none of the values in the equation are vectors: this equation refers purely to the magnitudes of each variable.
Through the substitution of the substitution of the equation for Tangential velocity, we can simply the equation further:
= ((r × ω)² / r) = ((r² × ω²) / r) = r × ω²
Because it is LINEAR, you use m/s² as your base units.
# P. Rule 48. We can extrapolate the known information about circular centripetal acceleration to apply to any curved path an object may follow, and therefore any object moving on a curved path will experience a centripetal acceleration.
When going over the top of a hill or other circular object that can have its forces summed to form the centripetal force (Rule 51), the normal force will be less than the force of gravity, since you feel less of a normal reaction force from the ground. This is why you feel weightless when you are driving over a hill!
# P. Rule 49. The whole idea of centripetal acceleration and tangential velocity and acceleration and everything else in rotational motion is that all the equations use all the same variables, and you can connect what you have to what you need by repeatedly plugging in numbers into equations. For example, to find centripetal acceleration (equation below), you tangential velocity, which you need the average angular velocity for, and so on. To really instill this into your brain, stare at the equations for rotational motion until you imagine them every time you see a rotating circle. For a refresher on all three types of acceleration, watch this videos.
# P. Rule 50. By the nature of their equations, Tangential velocity requires the object to be in acceleration (either speeding up or slowing down as it goes around the circle), while Centripetal acceleration only needs the object to be going around in the circle at a constant rate.
# P. Rule 51. Newton's Second Law (Rule 70) states that net force equals mass times acceleration, or ΣF = m × a. Since we determined in Rule 48 that every object moving on a curved path will have centripetal acceleration, we can apply centripetal acceleration to Newton's Second Law and end up with the equation for Centripetal Force:
ΣF = m × ac
There are several important rules for comprehending centripetal force:
1) Centripetal Force is not a New Force. Most other forces, such as the force of gravity, tension, friction, and the normal force (Sections V.II - V.IV), are 'independently-defined' forces, meaning they are not dependent on anything else and exist despite of what humans think is possible or impossible. Centripetal force, on the other hand, is composed out of these bedrock forces, whether by a mixture of them or just one. This is because it is the NET force in the inward direction, the net of all the forces acting in the inward direction.
2) Centripetal Force is never in a Free Body Diagram. Because centripetal force is not a new force, it never appears in a free body diagram. In order to determine the centripetal force, you need to draw out your free body diagram and then sum the forces in the inward direction.
3) In is positive, and out is negative. When you sum the forces in the inward direction, the in direction is positive, while the out direction is negative. This is almost the same for what was done before for summing forces, just now you sum them in relation to the in direction, and set them equal to centripetal acceleration instead of m × a. NEVER attempt to sum forces in the "tangential" direction - there is no such thing (you may be thinking of summing in the y or x directions). Instead, sum your torques (see Rule 126).
# P. Rule 52. Time Constant:
The time constant, τ, is a means of determining the time it takes for an exponential equation to get to a percent of its maximum value, calculated similarly to a half-life. The first time constant is 63.2%, which is a universal constant that must be remembered. It is the result of 1 - (1 / e), exactly. This number comes up with some frequency in physics.
Thus, since the first time constant is 63.2% of the time it takes to reach terminal velocity, then 2τ will be the 86.5%, because it increased by 63.2% of the percent remaining until 100%. 3τ will be 95%, and so on and so on. Every increase in the time constant will make the velocity 63.2% closer to terminal velocity. Thus, it will take 6.91τ for the velocity to be 99.9% of the terminal velocity.
Objects that encounter more air resistance and meet their terminal velocity faster, like coffee filters, will have a small time constant, while an object like a baseball will have a larger time constant.
# P. Rule 53. Rotational Right-Hand-Rule:
There is a second right-hand rule, one separate from the Vector right hand rule described in the "Multiplying Vectors" section of III.III. The 2nd R.H.R. is the Rotational Right-Hand-Rule (R.R.H.R.), which is performed somewhat differently. This rule gives you the direction of the Angular Velocity (ω) and the Angular Displacement (∆θ) of an object in centripetal motion, which are always the same and perpendicular to the plane of the object:
1. First, you need to know the direction of the revolution of the circle itself. This means whether the circle is moving clockwise or counter-clockwise. Other characteristics of the circle, such as how its placement in 3-dimensional space will effect the result, are explained in Rule 54.
2. Now that you have taken note of the movement of the circle, curl the fingers of your right hand in the direction of the movement of the circle.
To curl your hand "in the direction of the movement of the circle" means that the curve of your fingers, from the base to the fingertips, matches the movement of the circle: following the path of the your fingers, towards the fingertips, will be identical to the path of the movement of the circle.
Have no restraint in turning your hand to have the path follow correctly - this can be necessary to obtain the correct direction in Step 3.
3. Once your fingers have been curled correctly, stick out your thumb. Your thumb will be pointing in the direction of the Angular Velocity and Angular Displacement of the centripetal motion.
If you have performed this rule correctly, from your point of view (like the bird's eye view of a flat circle in centripetal motion), a circle moving clockwise will always produce a direction in the negative z-axis direction when using the R.R.H.R., and that using it on a circle moving counter-clockwise will result in the positive z-axis direction.
You may wonder, "Why bother with the R.R.H.R. anyways? Can't you just memorize the direction based on whether the circle is moving clockwise or counter-clockwise?". The issue with doing so is that "clockwise" and "counter-clockwise" are observer-dependent directions, and that they can change from whichever perspective you are looking at the circle from (and a problem can require you to look at it from several). Thus, by using the R.R.H.R., you can easily determine which direction the angular velocity & displacement are in by finding the "angular" direction, which, though it can change in positioning depending on perspective, will always be in the same direction.
For example, if you were to determine on the front side of an object in centripetal motion that it's direction is outward, and then when you move to the backside that it's direction is inward, you would have found the same direction, just from different perspectives. For all intents and purposes, it remains wise to define one view of the object in centripetal motion as being the "front" view, so that any direction inward can be labeled as being on the negative z-axis and any direction outward as being on the positive z-axis.
# P. Rule 54. If you were to look at the object in centripetal motion from the side, as depicted in figure (a) below, you would perform the exact same steps as dictated by Rule 53 - curl your fingers in the direction of the movement of the circle, with the added cognizance of the positioning your hand. You must make sure to turn your hand to best reflect the motion of the object, as if the path of your fingers was on the same plane as the object, as seen in figure (b).
This enables for directions to be found in the negative and positive x and y directions.
# Conical Pendulum: A pendulum (a hung weight that can swing freely back and forth in oscillations) that swings circularly, the motion of which forms a conical shape. In these problems, the centripetal motion will be along the horizontal plane (if hanging normally), or whichever plane the problem specifies if moving against gravity.
# Uniform Circular Motion: Centripetal motion at a constant angular velocity. This motion thus has no change in angular velocity, no angular acceleration, nor any tangential acceleration. An object spinning on the y-plane, like a yo-yo, is considered to be in nonuniform circular motion, as the force of tension will always be changing as it moves through the air (weakest when at its maximum height, and strongest at the bottom), in order for the net centripetal acceleration to be conserved. The yo-yo will take longer to move through the top semicircle of the revolution than the bottom, since the tangential speed is greater at the bottom. For a conceptual refresher, watch this video proof.
# P. Rule 55. The acceleration of any object in non-uniform circular motion has two perpendicular components: tangential acceleration and centripetal acceleration. Thus, to determine the absolute magnitude of the acceleration of an object in centripetal motion at any moment, you would simply perform the Pythagorean Theorem with the two component accelerations being a and b.
