Teaching Computer Programming to High School students: Basic Mechanics

Here are Newton's 3 laws of motion (http://en.wikipedia.org/wiki/Newton's_laws_of_motion).

Newton's 2nd law says that to accelerate a body, you must apply a force.

F = ma	; force = mass * acceleration

Let's say we have a 1kg object. To accelerate this mass by 1m/sec², we need a force of 1N (newton).

Instead of supplying a force ourselves, we can let gravity do it for us. The acceleration due to gravity is 9.8m/sec². The symbol for acceleration due to gravity is g. The value of g should be committed to heart by everyone who lives on the surface of the earth. If we let this 1kg mass drop, then from it's acceleration, we know that it's being acted on by a force of 9.8N. Here's the formula for the force due to gravity acting on an object of mass m.

F = mg

If we allowed a 2kg mass to drop, what force would it be accelerated by ^[1] ? The acceleration by gravity is the same no matter what the mass. This was proven by Galileo (http://en.wikipedia.org/wiki/Galileo_Galilei) who dropped unequal masses from the Tower of Pisa, both masses hitting the ground at the same time. (Apparently this story is apochrephal). Why don't a feather and a 1kg weight drop to the floor at the same speed ^[2] ?

	Note
	You can get the same effect in air by using a massive object to push the air out of the way of the feather (instead of a feather, you can use a small piece of paper). If you orient a calendar or newspaper horizontally and drop it, a piece of paper sitting on top will drop as fast as the calendar.

As a kid, I remember seeing a movie of a person in a vacuum chamber (and in a pressure suit) dropping different weights (and probably a feather) to show the same thing. Here is the experiment done with a Feather and Ball Bearing Dropped in Vacuum/Air (http://www.youtube.com/watch?v=_XJcZ-KoL9o). The experiment was repeated on the moon at by Apollo 15 Commander David Scott, who dropped a geologic hammer and a feather on the moon (http://www.youtube.com/watch?v=5C5_dOEyAfk). (Notice how much slower objects drop on the moon compared to earth.)

Work (energy) is force*distance.

W = F*s   ; s is the symbol for distance	

	Note
	energy is from the Greek - "work within" (see heat work energy http://www.engineeringtoolbox.com/heat-work-energy-d_292.html). In the cgs (centimeter, gram, second) system, the unit of work is the erg. 1 erg is the work done moving a force of 1 dyne (accelerating 1 g by 1cm/sec2) through 1cm. An erg is not much; 107ergs = 1 Joule.

Work must be done to lift massive objects against gravity. If we lift the 1kg mass, 1m against gravity, we will do 9.8*1=9.8 Joules of work. If we drop the 1kg mass, 1m and let it land on our foot, it will release 9.8J of work into our foot (this energy will be used to rearrange our foot).

Power is work/time.

P = W/t	

If we lift the 1kg mass, 1m against gravity in 1 sec, we work at a rate of 9.8J/sec=9.8w.

The Jin Mao Tower (Golden Prosperity Building) in Shanghai has elevators(lifts) that travel at 9.1m/sec, taking 45secs to reach the top (88th floor). (If the elevators are all glass, this would be a great ride.) Assume the elevators weigh 1 Tonne (1 Mg) (I have no idea, I just made it up). What power is needed to run the elevator ^[3] ?

	Note
	1HP (horsepower) = 746w. The elevator needs a 120HP motor.

Standard voltages for motors are 110V, 240V, 415V and 1.5KV. If you were limited to 100A cables (any more current and the cables would be too thick), what voltage would you choose for your elevator motors ^[4] ?

Example: a hiker, all up weight (i.e. with pack) of 70kg ascends 1000m in 3hrs. What power level is the hiker operating at ^[5] ? The hiker is putting out energy at a rate that would light a small household light bulb. Assuming that the hiker returns that day and does the same amount of work descending (in terms of conservation of energy, the energy that the hiker expended gaining altitude is released on descending. However most hikers think that it's just as much work going down as going up), what is the total amount of energy expended by the hiker during the day ^[6] ? How much extra food (calories) should the hiker pack to fuel the day's hike ^[7] ? Alternately you could use the conversion between calories and joules: calorie (http://en.wikipedia.org/wiki/Calorie) which I used to calculate the 100W number in the first place. 1 calorie = 4.2kj.

total energy = 2 *686kJ
             = 2 * 686/4.2 calories
             = 326 calories

Going to the pantry, you find that this amount of energy is in a cup (100g) of raisins (raisins are mostly water, so say this is 30-50g of sugar). The hiker should pack about 15% more food than they use just hanging around the house.

Note

This would be a reasonable day's exercise for most hikers. However there's not much energy used in ascending, and the hiker doesn't need to take any more food than they would need just to stay at home for the day. Most of our food goes to maintain body temperature. Admittedly on a hike more energy is used for other things than ascending. You have to balance your body (and pack) and move it up and down a few cm just to walk horizontally.

Power * time = energy. A unit of energy is the kW-hr - the amount of energy supplied to a device that uses 1kW for an hour.

1 kw-hr = 1kj/sec * 3600secs = 3600kJ

Last time I looked, 1kw-hr of electricity cost 6c. What would it have cost the hiker to do the hike using electricity, rather than food and muscle, for the day's work ^[8] ? In How much coal is required to run a 100-watt light bulb 24 hours a day for a year? (http://www.howstuffworks.com/question481.htm) we find that a burning a ton of coal produces 2,460 kWh of electricity. Assume an imperial ton = metric tonne = 1Mg. What weight of coal would been burned to generate the above amount of electricity ^[9] ? This is comparable to the weight of raisins eaten. With the heat of combustion of anthracite =50kj/g, sugar 5calories/g = 21kj/g, I would have expected the weight ratio coal/sugar=0.5, i.e. 15g coal. However, by the wonders of modern economics, getting your energy from coal is cheaper than getting from the sun via food. The amount a work a human can do pales by comparison to what can be done by burning tons of coal, that took millions of years to produce.

Look at this derivation

P = W/t
  = F*s/t
  = F*v 	; power = force * velocity

Let's try this on the Jin Mao elevators. What power are they putting out?

P = F * v
  = m * g * v
  = 10^3kg * 9.8m/sec^2 * 9.1m/sec
  = 90kW

... the same answer we got earlier

Let's do some phyisical work. Requirements: set of scales, weights, watch with stopwatch function.

How powerful is a programmer? Find a set of stairs about 5m in height. Strap on a pack with weights (10-15kg should do it - to make you do some work and reduce any advantage due to agility); weigh yourself and the pack. Time yourself running up the stairs. Calculate your power output.

	Note
	I did this in 8th grade Physics. You can't use any advantage from agility, so you can't start back from the stairs to give you a running start, and you can't do the stairs 2 at a time. Presumably you could try variations on this experiment to see if these restrictions make any difference.

Here's the results from our class. Height was 18 stairs of 18.75cm = 3.375m.

person   all up weight,kg    best time,sec     work,J      power,W   power,HP
Z                   83.75             5.44       2770          509       0.68
J                   77.75             5.94       2571          433       0.58

The SI (System Internationale) is a scheme for describing physical quantities. The units are

Name     symbol Quantity                   Symbol
metre 	 m      length 	                   l
kilogram kg 	mass                       m
second 	 s 	time 	                   t
ampere 	 A 	electric current 	   I
kelvin 	 K 	thermodynamic temperature  T
candela  cd 	luminous intensity 	   Iv
mole     mol 	amount of substance 	   n

The scheme is dimensionally consistent (see dimensional analysis http://en.wikipedia.org/wiki/Dimensional_analysis). "dimensionally consistent" will be explained below by examples, but for the moment every physical quantity is some combination of these dimensions

If we have an equation, not only do the quantities and units have to agree on both side of the "=" sign, but Newton realised that the dimensions must agree as well (i.e. you can't add a kilogram to a metre). Fourier realised that physical laws, like F = ma, should be independent of the units employed to measure the physical variables.

We're all familiar with dimensions in an informal sense. We know that if our car uses fuel at x miles/gallon and we've done y miles, we expect the y/x will give us a result in gallons.

Let's look at the formula for velocity

v = d / t
dimensions of v = l.t^-1

Velocity has the dimensions l.t^-1 (l/t).

What are the dimensions of force? We find

F = ma

The dimensions of F then are m.l/t² or l.m.t^-2.

Let's look at the formula for power.

P = F * d / t

From this we know that the definition of power must have the units of (newton*metre/sec). The dimensions of metre are l, the dimensions of sec are t, Using the dimensions of force that we just found,

P =  m.l^2.t^-3

the dimensions of P are m.l.t^-2.l/t = m.l².t^-3.

A little earlier we did this calculation. Let's check it for dimensional correctness.

weight of coal burned  = 0.38kw-hr * 1 tonne / 2460 kw-hr
                       = 154g

LHS required dimension = m
dimensions of RHS = power.t.m/power.t

On the RHS, it doesn't matter what the dimension of power is, 
as both the numerator and denominator have it and these two terms will cancel.
As well the numerator and denominator both have t (which cancel).

dimension of RHS = m

The equation is dimensionally correct.

Let's use dimensional analsys to determine the quantities that affect the period of a pendulum. From observation we know that the period of a pendulum is not dependant on the amplitude of the swing (as long as it is "small"), it's not affected by the temperature (as long as the length of the pendulum doesn't vary) or the length of time you observe it (i.e. we can experimentally eliminate the angle/amplitude, temperature and time as variables).

Galileo (http://galileo.rice.edu/sci/instruments/pendulum.html) determined that the period of a pendulum was independant of the amplitude of the swing. According to the story I heard when I was a kid, young Galileo was watching a chandelier in the cathedral in Pisa (of leaning tower fame, and where Galileo was born and lived all his life). Presumably he was in the cathedral to pray, but was distracted and instead timed the period of the chandelier with his pulse (there were no accurate time pieces then, portable or otherwise). He then poked the chandelier (how I don't remember, but he would have had to stop pretending he was praying and stand on something), to find that with its new, larger amplitude, the period of the pendulum was unchanged. (the story I heard gets some confirmation here http://physics-history.suite101.com/article.cfm/galileos_pendulum and here, except that the wind changed the amplitude http://www.ehow.com/about_5292790_galileos-pendulum.html )

Galileo, later in life (by then he was blind), went on to develope the pendulum clock, a device accurate enough for astronomical timekeeping (and eventually solving the longitude problem).

Let's use dimensions to find the dependance of the period of a pendulum on

The formula for the period of a pendulum will thus be

t = f(weight of bob, length of pendulum, accel due to gravity)
ie
t = f(m,l,g)

where f() is shorthand for a function to be determined. If f() is the sum of a bunch of terms, dimensional analysis says that each term must have the dimension t. Let's lump all these terms together; summing them will result in a dimensionless constant in front of a single function, whose dimension is t.

	Note
	We're using the same symbols for mass, length, time and for their dimension(m,l,t). I'll try to differentiate the representation.

After combining the sums, the only possibilty left is to multiply terms (actually powers of terms) in m,l,g. Mutliplying (powers of) the three terms above gives the period of a pendulum as

mutliplying the three terms above
period = m^x.l^y.a^z 

where x,y,z are to be determined. (If x=-1, then you'll be dividing, rather than multiplying by m). Dimensional analysis says that the dimensions on both sides of the equation must be the same. Checking dimensions for the period of the pendulum

period = m^x.l^y.a^z 
the dimension of "a" are l.t^-2

thus writing out the dimensions

m^0.l^0.t^1 = m^x.l^y.(l^z.t^-2z)
            = m^x.l^(y+z).t^-2z

separating dimensions (they're independant remember)
LHS      RHS  

m^0      m^x
l^0      l^(y+z)
t^1      t^-2z

this gives us 3 simultaneous equations

x        =  0 (1)
   y + z =  0 (2)
      2z = -1 (3)

from (3)
       z = -0.5 (4)
substituting (4) in (2)
       y =  0.5
from (1)
       x = 0

the dimensional check shows that the period of a pendulum

Numbers are dimensionless, so dimensional analysis doesn't allow us to determine any constants in the formula for the period of a pendulum. The formula for the period of a pendulum then is (with k a constant whose value must be determined by other means)

t = k * sqrt(l/g)

As it turns out k=2π and thus t = 2π√(l/g)

Let's check: A pendulum with a period of 1sec: what's the period of the same pendulum

In studying waves and periodic functions (like alternating current, or electromagnetic radiation), a useful unit is the Hz which is the number of oscillations/sec (also called the frequency). See oscillations (http://en.wikipedia.org/wiki/Oscillation). Here's the frequency of some common oscillations

type of oscillation                            frequency

ringing of the earth following earthquake      1/54mins 
heart rate                                     40-240beats/min
audible sound                                  20-20kHz
middle C (physicist's scale)                   256Hz
A' (musician's scale)                          440Hz
electromagnetic radiation                      1Hz->10^23Hz (1)
BC radio (AM)                                  550-1630kHz
FM band                                        88-108MHz
TV                                             50-800MHz (approx)
cell phones                                    800-1900MHz (only slices thereof)
wifi, household remote phones, microwave ovens 2.5GHz

There isn't a symbol for frequency in electrical engineering, but in wave mechanics the symbol is ν ("nu", pronounced "new") What is the dimension of frequency (the number of oscillations is a number; a number is dimensionless) ^[12] ?

Another useful parameter to characterise waves is the wavelength (symbol λ); the distance between successive peaks in the oscillations. You can see the distance between crests in ocean waves or ripples in a pond. However most wave motions are not detected as such by human senses. Here's some wavelengths of common waves

wave                           wavelength
ocean waves                    100m-many km
middle C (physicist's scale)   4' (1.3m)
BC radio                       200-600m
SW radio                       100-10m
visible light                  390-700nm
Xrays                          0.1-1nm
gamma rays                     1pm

What's the dimension of wavelength ^[13] ?

The conversion between frequency and wavelength involves another simple physical quantity with which you will be familiar. Using dimensions, find (or confirm) this physical quantity ^[14] .

Dimensional analysis doesn't say anything about numbers (these are dimensionless), so the actual formula relating v,f,λ could have a constant in it (i.e. the formula could be v=5.4*f.λ). A simple thought experiment will show you that the constant is 1. The formula relating these three quantities is v = ν.λ.

For convenience, for simple illustrations, physicists use the velocity of sound as 1024'/sec and the frequency of middle C as 256Hz. What is the wavelength of middle C ^[15] ? The velocity of sound in water is 4.3 times higher and in iron is 5 times higher (speed of sound http://en.wikipedia.org/wiki/Speed_of_sound). What is the wavelength of middle C in water and in iron ^[16] ?

Transverse (i.e. moving sideways) waves in stretched strings (i.e. strings under tension) are well known; e.g. guitar, piano, violin. The frequency of the note produced is affected by the tension (force) in, the weight and the length of the string.

Note

The tension is the force you'd have to pull on a string. Here's one way of setting the tension if you want to test its effect. One end of the string is fixed, the other runs over a smooth surface (wheel, or polished curved metal). The end of the string holds a weight string ----------------------------------------------\ ^ O| _____________________________________________ | | sound board | |xxx| weight

The tension in a piano frame (plate) can exceed 20 tons (see piano http://en.wikipedia.org/wiki/Piano). Tuning a piano is a more difficult than a violin or guitar (where the instrument is stiff and you can tune each string independantly), in a piano, a stiff plate would be too heavy. In a piano, if you let off the tension in a string, the plate expands, raising the pitch of the other strings (but only a little bit). In a piano you can only adjust the strings a small amount at a time.

Using dimensional analysis, determine the effect on the frequency of a string, of the tension, the mass and the length of the string ^[17] . Does this equation agree with your prediction on the effect of mass, length and tension on the pitch of a string?

While musicians are interested in the pitch of a musical note, physicists deal with the propagation of waves in media of infinite extent (infinite at least in principle), so they're more interested in the velocity of the wave. You could rerun this derivation using velocity on the left hand side. Instead use your previous result v= ν.λ to determine the formula for the velocity of a transverse wave in a string (you should get the same result).

	Note
	What's the wavelength of the wave in a string? Assume that you are only concerned with the fundamental mode of oscillation i.e. the length of the string is half the wavelength. (see the λ/2 wave at standing wave http://en.wikipedia.org/wiki/Standing_wave). The wavelength then is 2*l.

Since we're only concerned with dimensions, the constant "2" here is irrelevant. Here's the result ^[18] . The constant "k" here is 1. So v=1/sqrt(T.M). Check the formula to show that changing the tension and mass/unit length of string have the expected effect on the velocity of the wave.

	Note
	Did we use the value of the wavelength (2l)? No; the dimension of l is the same no matter how long it is.

If you release a ball on top of one of the sides of a bathtub, it speeds up as it looses height. After crossing the bottom of the bathtub, it looses speed as it goes back up the other side, until (at least in a frictionless environment) it returns to the same height on the other side. The same happens to a pendulum or swing.

To lift the mass from the bottom to the top of the side of the bathtub, work had to be done against gravity. By the law of conservation of energy, the mass at height can now do the same amount of work descending to the bottom of the bathtub. The mass at height can potentially (i.e. if you let it go) do work. According to the law of conservation of energy, the mass has potential energy (P.E.).

What is a "law"?

When an observation (e.g. "energy is conserved") is repeatable and useful, but has no explanation and you aren't particularly bothered by the lack of explanation, then to get on with the other things you're more interested in, and get the unexplained observation out of your way, you call the observation a "law". Well known laws are "things fall down; opposites attract". That things fall has been known for so long and observed so consistently, that no-one questions its truth.

A "law" is not like an axiom in mathematics, some unfathomable fundamental property of the physical universe, for which there is no hope of further understanding. A law is just something we don't understand, but can describe well (possibly very accurately). As for assigning fundamental properties to the universe; we don't know enough about the universe to be doing this just yet. We used to think matter was earth, air, fire and water; but later came atoms; and then subatomic particles. You don't think we're at the end yet do you?

Note

Do not confuse being able to describe something accurately (e.g. the effect of gravity; we can build bridges that stay up; we can land spacecraft on remote objects in the solar system), with understanding it. No-one has a clue why massive objects attract each other. Similarly no-one understands why mass has inertia (resistance to accelaration) or why the amount of inertia is equal to the amount of mass (in the formula F=ma, m is actually inertia). The two properties of mass, of mutual attraction and inertia, are well known and well described, but not understood.

Laws are completely arbitary, but useful. When they prove to no longer be useful, they are discarded. There's no implication of truth in a law, only usefulness. The very existance of a law declares our complete ignorance of how things work there. At least this is how it is in the physical world. In the legal realm of the US, somehow truths become enshrined in law, indicating that the final resting place for truth is law.

Because the "law of conservation of energy" is a law, I can't justify it and I can't prove it. I'll just be using it.

The ball, after release from the side of the bathtub and rolling across the bottom of the bathtub, can by virtue of its motion, do work. The moving ball can collide with another ball of the same mass, sending the 2nd ball up to the release height on the opposite side of the bathtub. The moving ball at the bottom of the bathtub then has energy, called kinetic energy (K.E.) (http://en.wikipedia.org/wiki/Kinetic_energy).

	Note
	The adjective kinetic is from the Greek kinesis meaning motion. The same root produces the word cinema (referring to motion pictures).

By the law of conservation of energy, the P.E. at the top of the side of the tub is converted into an equal amount of K.E. at the bottom of the tub. The K.E. is then turned into to original amount of P.E. when the ball goes up the other side of the tub. At intermediate points in the ball's trip, the P.E. descreases as it descends (and increases as it ascends), while the opposite is happening to the K.E. Since energy is conserved, the total (K.E+P.E) is constant.

We know how to calculate P.E, it's the energy put into a mass by raising it through height h

P.E. = mgh

If we see a ball whizzing past us, how do we calculate its K.E.? Let's start using dimensional analysis. All we know about the ball is its mass and velocity. Derive a formula for K.E. using these two measurements. Here's my answer ^[19] . We need to find the dimensionless constant k. To do this we need to understand the physics of the situation. We know that after falling h that the mass has gained K.E.=(lost P.E.)=mgh. Since K.E. is determined by velocity, we need to know the velocity of a mass after it falls a distance h. We don't know that yet.

	Note
	From dimensional analysis we see there is no dependance of K.E. on direction i.e. the K.E. at any instant is the same whether the mass is moving up, down or sideways.

By dropping weights from different heights into a block of clay, Gravesande (http://en.wikipedia.org/wiki/Willem_%27s_Gravesande) (early 1700's) determined that their penetration depth was proportional to the square of their impact speed (see kinetic energy http://en.wikipedia.org/wiki/Kinetic_energy).

	Note
	Joe: I don't know how Gravesande timed the weight's fall (its impact velocity). Presumably he used an inclined plane because of the requirement to be able to time the weight's fall. However the velocity on impact would have been small.

In the next sections, we're going to find first the velocity as a function of time, then the distance as a function of time. From that we'll figure out velocity as a function of distance. This will allow us to determine the K.E. gained as a function of height dropped, giving us the dimensionless constant for K.E.

Let's calculate the distance s, travelled by an object moving at constant velocity u, for time t. By the definition of velocity

u = initial velocity
s = distance travelled
t = time

then
u = s/t

thus
s = ut

Let's see if there's any other way we can derive this relation. Here's the graph of the velocity of the object as a function of time.

   |
^  |
|  |
v  |.......................u
   |
   |
   |_______________________t

             time->

Here's the position (displacement) s of the object as a function of time, with the starting position s=0 at t=0.

   |              o 
   |            o 
^  |          o   
|  |        o     
s  |      o 
   |    o 
   |  o
   |o_______________
        t, secs ->

This graph is a straight line (the relationship between x and t, under constant v is linear). If the object has an initial displacement (wasn't at the origin at t=0, but instead was at x=x₀ at t=0) then the relationship is still linear (i.e. the line is of the form y=mx + b).

   |              o 
   |            o 
^  |          o   
|  |        o     
x  |      o 
   |    o 
   |  o
x0 !o
   |
   |
   |
   |_______________
        t, secs ->

What's the area under the line on the v-t graph?

   |
^  |
|  |
v  |.......................u
   |///////////////////////
   |///////////////////////
   |///////////////////////
    -----------------------t

             time->

The area is vt. What is the dimension of the area?

dimension of vt = (l/t).t
                 = l

Dimensional analysis tells us that the area under the line (the line is a straight line) is the type of quantity (distance) that we're after. The two x-t graphs immediately above the v-t area graph, only differ in the starting position of the mass, and are both described by the one v-t graph. The area under the v-t graph describes the change in position (i.e. the distance moved) during the time t. There is a dimensionless constant to determine. From the definition of velocity, the dimensionless constant is 1. Thus for constant velocity, the area under the v-t curve is the distance. (It's true for varying velocity too, but we haven't proven it.)

Let's find v=f(t) for an accelerating body.

The definition of acceleration is the change in velocity per unit time. A stationary object accelerated at 1m/sec² will have a velocity of 1m/sec after 1sec, 2m/sec after 2sec... Here's the graph of its velocity

  2 |              o 
    |            o 
^   |          o   
|   |        o     
v 1 |      o 
m/s |    o 
    |  o
  0 |o_______________
    0      1       2  
        t, secs ->

We know that a falling body is accelerating at a rate g, which allows us to calculate the velocity at time t.

v = at

a = g
v = gt

Let's see what we can find from the area under curves (here straight lines). Here's a graph of the acceleration of an object falling in a gravitational field g for t secs.

   |
^  |
|  |
a  |.......................g
   |
   |
   |_______________________t

             time->

In this graph, the acceleration (=g) is constant.

From the definition of acceleration, the velocity for the accelerated object (assumed to have v=0 at t=0) is

   |              o 
   |            o 
^  |          o   
|  |        o     
v  |      o 
   |    o 
   |  o
   |o_______________
        t, secs ->

This graph is a straight line (the relationship between v and t under constant a is linear). If the object has an initial velocity then the relationship is still linear (i.e. is of the form y = mx + b).

   |              o 
   |            o 
^  |          o   
|  |        o     
v  |      o 
   |    o 
   |  o
   |o
   |
   |
   |
   |_______________
        t, secs ->

What's the area under the line on the a-t graph?

   |
^  |
|  |
a  |.......................g
   |///////////////////////
   |///////////////////////
   |///////////////////////
    -----------------------t

             time->

The area is gt. What's the dimensions of the area ^[20] ? The area under the curve tells us something about velocity, but what? The two v-t lines immediately above the a-t area graph are described by the one a-t graph. The area under the a-t graph describes the change in velocity during the time t. This is reasonable, considering the definition of accelaration (the change in velocity/sec). From the definition of accelaration, the change in velocity is a*t, which is the area under the a-t graph. This makes the dimensionless constant 1.

Newton's Laws of Motion (http://en.wikipedia.org/wiki/Newton's_laws_of_motion) are (here's the word "law" again)

From this we see that acceleration is independant of the initial velocity. Accelaration only causes a change in velocity. The equation above is better written

v: final velocity
u: initial velocity
a: accel
t: time
v - u = change in velocity

v - u = at 
v = u + at ; the usual form of the equation

Example: you throw a ball upwards with an initial velocity of 9.8m/sec. How long does it take for the ball to stop moving?

Inputs
u = 9.8m/sec
v = 0  m/sec
a = g
t?

v = u + at
t = (v - u)/a

t = (0-9.8)/9.8 sec
  = -1 sec

Hmm, the ball stopped 1 sec before we threw it. What went wrong here? Velocity is a vector quantity - it has direction (speed is a scalar quantity; it has magnitude but no direction).

	Note
	the words vector, scalar are adjectives. They are used routinely as nouns e.g. this is a vector, this is a scalar.

Accelaration is a vector too. To use vectors, we need a frame of reference to establish a direction. The ball is moving (going up and down) in a 1-dimensional frame of reference, the vertical frame. We can arbitarily assign up as being the +ve direction, or down as being the +ve direction.

	Note
	No matter which frame of reference we use (up is +ve, or up is -ve), we will get the same answer. The initial velocity is in the opposite direction to the acceleration, so velocity and acceleration will have the opposite sign, no matter which direction is chosen as the +ve direction.

Let's assign up as the +ve direction. Here's our problem now

Inputs
u = 9.8m/sec
v = 0  m/sec
a = -g
t?

v = u + at
t = (v - u)/a

t = (0-9.8)/-9.8 sec
  = 1 sec

The ball stops moving (i.e. its velocity is 0) 1 sec after you release it. Using the eqn v = u + at, plot the velocity of the ball for the first 2 secs of its flight.

  9.8 |x
      |  x
^     |    x
|     |      x
v   0 |--------x------ 
m/s   |      1   x   2
      |            x
 -9.8 |              x
        
          t, secs ->

The ball starts (at t=0) with a +ve velocity (it's going up). The velocity decreases till at t=1 sec the ball is not moving (i.e. v=0). After that, the ball will start falling back to earth (v<0). The velocity of the ball is decreasing linearly through the whole flight.

	Note
	The speed initially decreases till the ball stops and then the speed increases again.

If we'd chosen the down direction as the +ve frame of reference, then what would have been happening to the velocity, the speed ^[21] ?

At what time would the ball have a velocity -19.6m/sec?

Inputs
u = 9.8m/sec
v = -19.6m/sec
a = -g
t?

v = u + at
-19.6 = 9.8 + -9.8t
t = (-19.6-9.8)/-9.8 sec
  = 3 sec

Here's the velocity of the ball at t=0,1,2,3secs

t,secs        v,m/sec
0             9.8
1             0
2            -9.8
3           -19.6

At what time would the ball have a velocity 19.6m/sec?

Inputs
u = 9.8m/sec
v = 19.6m/sec
a = -g
t?

v = u + at
19.6 = 9.8 + -9.8t
t = (19.6-9.8)/-9.8 sec
  = -1 sec

What does the -ve time mean? It means that 1 sec before the ball came past your hand, it was moving upward at 19.6m/sec

Note

The equation is a continuous function of t and covers t for -∞<t<+∞. Into the equation, you plugged; the velocity at t=0 and the acceleration. The equation handles the rest. You didn't specify anything special about the behaviour of the ball for t<0 and therefore the equation specifies v for all t. Equations with discontinuous properties (e.g. for t<0; v=0) are tedious to write and use for calculations. Unless you arrange it otherwise, equations are continuous in the independant variable (here t).

This equation doesn't tell us where the ball was, when it was moving upward at 19.6m/sec. We'll need an equation, to be derived below, to figure that out. Ordinary experience says that the ball would have been coming up from below.

Example: In the US, in 2009, the average new car accelerates from 0 to 60mph (http://en.wikipedia.org/wiki/0_to_60_mph) (97kmph) in 8.4sec. Assuming that the acceleration is constant (and the car is in a vacuum), what is the acceleration during this period relative to g ^[22] ?

The weight of a US car is about 1.5tonnes. What power is being used by the car when it starts to move ^[23] ? What power is being used when the car is doing 97kmph (and still accelerating at the same rate) (give answers in kW and HP; 1HP=746w) ^[24] ?

	Note
	The same car at highway speed uses about 40kW. Cars use less fuel on freeways than driving around town, where power used accelerating in traffic is lost as heat in braking. The rest of the highway fuel economy comes from the engine being designed to be most efficient at highway speeds.

In the python course, we did some calculations involving a Boeing 747-400ER taking off. This time we have competent programmers loading the plane. Here's some numbers

Calculate the acceleration in m/sec^2, and relative to g, during the takeoff run and the power output at the moment of takeoff (assume the plane accelerates at a constant rate). Here's my answer ^[25] . A diesel train engine outputs about 12Mw.

Note

takeoff is the most difficult point of the flight as far as the power budget is concerned. The plane is at maximum weight (it will land almost empty of fuel at half the weight); the plane is flying at about 1/3 of its cruising speed (the wings are supplying less lift, and have the flaps down to increase lift at the expense of increased drag). Once the plane reaches any speed at all, no runway is long enough for it to stop by braking and there is nothing (e.g. nets) that can stop it. If there is an emergency at takeoff, the plane is too heavy to land (the wings with all the fuel will tear off, the plane erupting in a fireball).

An example in two dimensions: You shoot a projectile (canon, arrow, kick a football, hit a golf ball, throw a raw egg, tomatoe) in a vacuum, with initial velocity u, at an angle ϑ (theta) to the horizontal.

Here's the initial path of the ball.

   |       / 
   |  -   / 
   |  /| / 
v  | u  /
e  |   / 
r  |  / 
t  | /
   |/theta
    ---------------
      horizontal

	Note
	The path of the ball is a parabola (ascii art parabola shown), but we won't be proving this till the next section. | | | | y | o o | o o | o o | o o --------------- x

We'll be looking at the path of the ball, both as a function both of time t and of horizontal position x. For the rest of the discussion, we will be use s (the horizontal distance) to describe the position on the x-axis. Here's the initial path of the ball as function of t.

   |       / 
   |      / 
^  |     / 
|  |    /
h  |   / 
   |  / 
   | /
   |/theta
    ---------------
        t, secs ->

Here's the initial path of the ball as a function of s (the horizontal distance)

   |       / 
   |      / 
^  |     / 
|  |    /
h  |   / 
   |  / 
   | /
   |/theta
    ---------------
        s, m ->

Note that both graphs of the initial path (with s or t on the abscissa) are similar; only the scale on the x-axis changes. You can do this because the horizontal velocity is constant (there is no horizontal force/acceleration on the ball). The scales on the two graphs are related by s=u_horizontal.t, i.e. s=const.t.

      | 
      | 
^     | 
|     |
v,hor |---------------u
m/sec |
      |
      |
       ---------------
        t, secs ->
        s, m

The ball is acted on by the force of gravity, accelerating downwards. Here's the velocity of the ball as a function of both time and distance (two graphs are combined into one since s and t are related by a constant). The ball has an initial +ve velocity (it's going up), it slows down till it stops, and then the velocity becomes -ve. The velocity decreases linearly through the flight (becoming more -ve).

Note

Thinking about the speed only confuses the matter, but here it is anyhow: the speed decreases till the top of the arc (when the ball has zero velocity), then increases again as the ball falls back to earth. While speed is useful for driving a car, or walking; if you want numbers and something is being acted on by a force, you need to think in terms of velocity, not speed. Nature thinks in terms of velocity; if you want to converse with nature, you should use its language. Speed is fine for social situations, but nowhere else.

       u |x
         |  x
^        |    x
|        |      x
v,vert 0 |--------x------ 
m/sec    |          x   
         |            x
         |              x
      -u |                x
        
          t, secs ->
          s, m

Here's the path of the ball again.

   |       / 
   |      / 
^  |     / 
|  |    /   o o
h  |   / o       o
   |  /o           o
   | /o             o 
   |/theta           o
    ---------------
        t, secs ->

At what angle ϑ, do you have to launch your projectile, to make it go the furthest horizontal distance, before hitting the ground?

First we need to find a frame of reference.

multidimensional space, degrees of freedom

We live in 3-D space.

The previous problems have been 1-D; we threw a ball up in the vertical direction, or we accelerated a car or a cargo plane in the horizontal direction. These are 1-D problems; the position of the object is described by its height (or distance along the road/runway). The object is said to have 1 degree of freedom, i.e. only 1 parameter is needed to describe it's position (or velocity).

In the current problem, the ball is launched at an angle, and is moving in 2-D space. Its motion in any direction is independant of its motion in any other direction (you can change the ball's x independantly of its y).

A counter example, where x,y of an object in 2-D space, can't be changed independantly, is a ball fixed to the circumference of a rotating disk, or to the end of of a pendulum. For a small change in x there will be only one possible new value for y, i.e.x,y cannot be changed independantly. For the case of the rotating disk or the pendulum, x,y are completely determined by the angle of rotation of the disk (or pendulum rod). This system is said to have only 1 degree of freedom, i.e. only 1 parameter (here the angle of rotation) is needed to determine the position of the ball. While the ball at the circumference of the disk moves in 2-D space, its movement only has 1 degree of freedom i.e. it's a 1 parameter problem.

The motion of the arrow/golfball is in 2-D and because movement in the x,y directions is independant, the ball has 2 degrees of freedom. We need two parameters to describe the position (or velocity) of the ball.

Note

We need two parameters because the problem has 2 degrees of freedom, not because it's a 2-D problem. We have 2 degrees of freedom because the ball is moving in 2-D space AND because the ball can move independantly in each of the two dimensions. How about a 1-D problem with 2 degrees of freedom? Imagine two train carriages, and instead of being joined by a standard coupler, are joined by a spring. The two carriages are sent down the rail line by pushing the rear carriage. The two carriages will oscillate about their mean position, by virtue of the spring, while their center of mass travels along the rail at constant velocity. You can choose two frames of reference parameter_1: the center of mass of the two carriages; parameter_2: the distance between the two carriages. parameter_1: the position of carriage_1; parameter_2: the position of carriage_2. The first frame of reference is the simplest; the problem resolves into two separate and well studied problems the center of mass travelling at constant velocity a pair of oscillating masses connected by a spring

In a 1-D problem there is only 1 possible frame of reference (well, you can offset the origin). In 2-D there are an infinite number of them. We can choose any two directions (the frame of reference). to describe the position of the arrow. Life is simpler if we choose orthogonal directions (directions at right angles), and simpler yet if we choose one of the directions parallel to the acceleration of gravity. However we could choose one direction along (or at right angles to) the direction of the ball's initial movement and the other along the ground. We can choose any frame of reference we like and so we choose the one that gets the answer the fastest (or simplest).

Note

Some frames of reference make life easier. The Copernican heliocentric frame of reference is simpler than the Ptolemaic geocentric system. Using the heliocentric frame of reference, Kepler was able to determine that the orbits of the planets around the sun were elipses, a problem that for 2000yrs had confounded people using the geocentric system. (It turns out that objects attracted under the inverse square field of gravity move along conic sections, i.e. circle, elipse, parabola, hyperbola. Kepler found one of them, the elipse.) While the heliocentric frame of reference is the simplest, for objects that are earth bound (or near earth), the geocentric system is the most useful. Amateur astronomers need to know where and when to look up at night; if you want to launch a satellite, go to the moon or calculate sunrise, sunset or the tides, the assumption that the earth is the center of the universe is simplest. When you're learning, you'll be shown the frame of reference that will solve the problem quickest. You'll soon get the idea of what to look for. You may have to translate the axes, rotate the problem, use non-orthogonal axes before you find a frame of reference in which the problem becomes tractable.

From past experience, the two directions to choose for our frame of reference are parallel to the acceleration of gravity (the y axis) and at right angles to it (the x axis, where there is no acceleration and the ball has constant velocity). For the positive directions, we'll use the Cartesian convention; for the positive y direction, we'll choose up; for the positive x direction, we'll throw the ball to the right. Since movement in two orthogonal directions is independant, we've resolved the problem into two components

• the vertical movement: the ball moves under the acceleration of gravity, just like it did in our previous 1-D problems.
• the horizontal movement: the ball moves with constant velocity.

Note

I said that movement in the two orthogonal directions is independant. What is actually happening is that movement in the direction parallel to the field of gravity (where a force acts on the ball) is independant of movement at right angles to the field of gravity (where no force acts on the ball). These two directions are orthogonal. Movement in the two orthogonal directions is independant, because movement parallel and at right angles to gravity is independant, and because the two directions are orthogonal.

The problem to be solved then is our previous 1-D problem, where we threw a ball up in the air, but now as well, the ball is moving sideways. We calculate the position/velocity of the ball for each component direction and then we combine the two results to see where the ball is and how it's moving. We need to find how long it takes for the ball to return to earth in the vertical direction, and then figure out how far the ball has travelled sideways in that time.

The initial velocity u is at an angle ϑ to the horizontal. We've separated the problem into two simpler problems; the behaviour in the x direction and in the y direction. We need the initial conditions for the two separate problems. and these are the component of the initial velocity in the two directions.

Note

To be precise, we say "the component of the velocity in the x direction", rather than "the velocity in the x direction". This is to make clear that the object has velocity in other directions and that for the moment we are only looking at a component of the velocity, not the velocity. In casual conversation you can say "the velocity in the x direction" as long as it's clear that you're talking about the components.

Note

This next section, to find the components of initial velocity, may seem a little laboured. The result may be obvious already. If not, after you understand the principle, the result will be obvious and you can use it in a problem without justification (the result is obvious to everyone else, at least after it has been explained). However you need to know how to prove the result, so that you know it's true. The results are uvert=u.sin(ϑ) uhorz=u.cos(ϑ) If these results are obvious and you think you could come up with an explanation (proof) of this, then hop over the proof of this below.

First we're going to find the component of the position in each direction. Then to find the components of velocity, we'll divide the change in position, by the change in time. Imagine that the ball/arrow is at a distance l from the origin or that the ball is on the end of a stick of length l, whose base is at the origin, and which is leaning at an angle ϑ to the horizontal.

   |
   |       o 
   |      / 
^  |     / 
|  |  l /
y  |   /
   |  /
   | /
   |/theta 
    ---------------
         x ->

To find the component of the position of the end of the stick in the y direction, you move off to the right and see how high the end of the stick appears, giving you h. (the ascii art to the right is an eyeball looking at the top of the stick)

   |                 \
   |-------o-------- O
   |      /|         /
^  |     / |
|  |  l /  |
y  |   /   | h
   |  /    |
   | /     |
   |/theta |
    ---------------
         x ->

You'll need a bit of trig here.

h = l*sin(theta)
   
          o
         /|
        / |
     l /  |
      /   | h
     /    |
    /     |
   /theta |
   --------

Since the ball is moving, at time=dt later , the ball has moved a distance dl and has changed height by dh. The hypoteneuse of the small triangle, showing the movement of the ball, is also at an angle ϑ to the horizontal, so

dh = dl * sin(theta)

   |                  \
   |---------o-------- O
   |     dl /| dh     /
   |       o- 
   |      /| 
^  |     / |
|  |  l /  |
y  |   /   |    h
   |  /    |
   | /     |
   |/theta |
    ---------------
         x ->

To find the component of velocity in the y direction, we divide both sides by dt.

dh      = dl * sin(theta)

divide both sides by dt
(dh/dt) = (dl/dt) * sin(theta)
therefore
v-vert  = u.sin(theta)

The component of initial velocity in the y direction is u.sin(ϑ).

   |        o
   |       / 
   |  _   / ^
^  |  /| /  |
|  | u  /   u.sin(theta)
h  |   /    |
   |  /     |
   | /      |
   |/theta  |
    ---------------
        t, secs ->

To find the horizontal velocity, we need to know the component of the position in the x direction.

h = l*cos(theta)
   
          o
         /|
        / |
     l /  |
      /   | 
     /    |
    /     |
   /theta |
   --------
      w

Following the same process as above, we find that the component of initial velocity in the x direction is u.cos(ϑ). (Make sure you can do the derivation yourself.)

   | ----->/ u.cos(theta) 
   |  _   / 
^  |  /| / 
|  | u  /
h  |   / 
   |  /
   | /
   |/theta
    ---------------
        t, secs ->

Now we have the components of velocity in both directions and we can use the equation v=u+at to find the velocity at some other time. First find the time of the flight (hint ^[26] ), then find the horizontal distance moved in this time. This trig identity will help.

sin(A)*cos(A) = sin(2A)/2

Then maximize the distance as a function of ϑ. (sin(ϑ) has values from -1 to +1. The maximum value occurs with x=90°.)

Do the problem in 3 parts.

First analyse the problem in the direction parallel to the acceleration of gravity. Find the time it takes for the ball to return to earth ^[27] .

Next analyse the problem in the direction orthogonal to the acceleration of gravity (i.e. in the horizontal plane). Find the distance the ball moves horizontally, in the time you calculated for the ball to fall back to earth. ^[28] .

The term 2.u²/g is constant for this problem. The only variable available to maximize the distance is theta. Using the trig identity sin(A)*cos(A) = sin(2A)/2, find the distance as a function of sin(2ϑ) ^[29] . Next find the angle which maximizes the distance travelled ^[30] .

Now plug this value for ϑ into the distance formula, to get the horizontal distance travelled ^[31] .

Let's try some numerical examples. The world record speed for hitting a golfball is 328kph (golf ball http://en.wikipedia.org/wiki/Golf_ball). How far would this ball travel if hit at 45° in a vacuum ^[32] ? The longest golf drive in air is 377m ( distance of the longest driven golf ball http://hypertextbook.com/facts/2004/CrystalCuevas.shtml). It would seem then that air resistance halves the distance of a long drive.

A good golfer can hit a 200m drive (long drive http://en.wikipedia.org/wiki/Long_drive). On the moon, the acceleration due to gravity is about 1/6th that of earth Acceleration of gravity due to the Moon How far should a normal golfer be able to drive a golf ball on the moon http://hypertextbook.com/facts/2004/MichaelRobbins.shtml. ^[33] ?

Astronaut Alan Shepherd encumbered by a bulky space suit, did a one handed sandtrap shot with a 6-iron on the moon, shooting the ball a distance quoted as 800m (Moon Trivia http://home.hiwaay.net/~krcool/Astro/moon/moonglossary/moontrivia.htm). Do you believe the distance ^[34] ?

Note

This example, of lobbing a canon ball, comes from an often ignored part of the history of physics. Once physicists understood acceleration, they were in great demand to calculate artillery trajectories. Lagrange was the professor of mathematics at The Royal Artillery School in Turin (Lagrange http://en.wikipedia.org/wiki/Joseph_Louis_Lagrange). I believe Galileo received some of his support ($) for doing this. Physicists (and less so other scientists) have always been happy to receive money to find better ways to destroy things.

How do we know that the result we get after resolving the problem into the x,y components and then putting the pieces back together is the right one? What if something is lost when you do it this way?

	Note
	This is the unknown part of my argument. If the two components are really independant, then nothing is lost.

To show that resolving the problem into components is valid, a proof is needed. When I was learning this topic, I was never shown a proof and in the subsequent time, I've never seen one. It was not obvious to me whether the proof was so trivial that none was needed or whether there was no proof and mathematicians were embarrassed to admit it.

Math departments, when I went through, had no problem delivering trivial proofs: at my university (Sydney University, 1965-6), the math department delivered proofs of 1+1=2 and talked at great lengths about the nature of "numberness". In light of their unwillingness to teach us any of the multitude of useful math tools, I regarded this as a declaration that the department had no interest in teaching its students anything and that it had no intention of fulfilling its contract with the tax payers, who provided their salaries, to do their job. The math department at Sydney University regarded themselves as unaccountable to society (they were unaccountable - the state government didn't care what they did with the money). It was a sham - I dropped math at the end of the 2nd year of university. There were only two good teachers in my courses - one I did my PhD with. The other, my first year chemistry lecturer, Peter Simpson, didn't get tenure. He was invited back for a reunion 40 yrs later by the students of my generation, showing that we were capable of appreciate his work, even if it went over the heads of the the chemistry department.

The remaining possibility then is that there is no proof. It's important when persuing a goal to know the assumptions you've made, in case when you get to the end, your destination isn't the one you expected. If there is no proof, then it should be declared in your list of assumptions. If the wings fall off your plane, you'll have to go back to find what went wrong and the first place to look is the list of things you can't prove correct.

In the meantime, in the absence of a proof that using orthogonal components is valid, let's list what we know about resolving problems into components. All of these conditions are necessary, but not sufficient for the validity of using components.

• It would be difficult to find a part of the technical world where calculations aren't based on components. If there were problems with the process, it's hard to imagine how they would have escaped notice. In general there's aren't other ways of handling the problems, so it's not like we can test for agreement using answers from methods that don't resolve the problem into components. However for the whole Middle (dark) Ages, people reverted to thinking that the world was flat. So it's possible that there are unnoticed problems in resolving components.
• If we accept the Cartesian system (positions are described by x,y), then we can resolve positions into any frame of reference. If we can resolve position, then by dividing by time, we can resolve velocity and acceleration too.
• We have to be able to show that orthogonal (ortho = right angles) directions are independant. When we say that we live in 3-D space, it means that we need 3 (independant) pieces of information to specify our position in space. That we need 3 measurements; width, length and height, seems obvious. If the 3 pieces of information weren't independant, then we'd only need (say) 2.5 pieces of information, but no-one has come up with a 2.5-D space. It seems to be a fundamental part of math, that dimensions (as in 2-D and 3-D) are independant. Maybe dimensions are defined as being independant, and no proof is needed.
• The positions of an object in each dimension in 2-D (or 3-D) space are independant. You can change an object's position in one dimension without changing its position in any of the other dimensions. Maybe proving this is enough to prove the validity of resolving problems into components. If it is, then I'd like the proof stated somewhere, so that when the wings fall off my plane, I don't have to go back and find another method of calculation.

The whole matter is unclear and needs to be straightened out, at least for people like me. In the absence of better information, the rationale that I use is that resolving problems into components works. I'll admit this is not very satisfactory, but it's the best I can do.

In the previous section, I told you that you could determine the time of flight of your projectile, by finding the time it took for the (vertical) velocity of the to be the same, but of opposite sign to the initial velocity. I told you this extra piece of information, so you could get on with solving the problem. You could see that this was reasonable and likely, if not certainly true. It wasn't till I saw this section delivered in class, that I realised that you didn't have the information to show that this was true. You won't get this information till the next section, when we'll do the problem again, only this time we'll do it properly.

In your further education, you won't always get a proof. However you need to know where in the stream of teaching, that a proof is needed (whether or not you got one, and whether or not the teacher will give it to you on later asking). You're building a castle of skills and knowledge. It will be based upon a series of proofs and dexterity derived from doing a large number of practice problems. If proofs are missing, you will find your castle built upon sand. It won't make any difference to the teacher if you don't get the proofs, but it's your life. If you don't get the proofs any other way, you should go figure them out yourself.

Look for statements, like I made about the projectile's velocity, and ask for proof.

With that in mind, I'm going to explain the condition for which the ball returns to the ground with velocity of the opposite sign. (I don't know the proof.)

A conservative force (http://en.wikipedia.org/wiki/Conservative_vector_field) (also called "field" and "vector field") is one in which the work done in moving a particle between two points is independant of the path taken. A corrolary of this, is that the particle on returning to its original position, will have its original energy. A non-conservative force is one in which the difference in energy is dependant on the path taken. A particle moving in a non-conservative field will have a different energy on returning to its original position.

With gravity being a conservative force (or field), then in the vertical component of the projectile problem, the energy of the projectile is dependant only on the difference in (vertical) distance from the starting point. When the particle returns to its starting point, it will have the same kinetic energy (½m.v²). Because of the squaring, you can't tell whether the velocity has changed sign, but you know that the magnitude is the same. From experience and from v=u+at, you know that the velocity of the ball, when it returns to the ground, has the opposite sign. (In a later section we'll prove that the magnitude is the same, but here you're taking my word for it.)

Galileo explored the acceleration of gravity by rolling a ball down an inclined plane (http://www.mcm.edu/academic/galileo/ars/arshtml/mathofmotion1.html) showing that the distance covered by the ball was proportional to the square of the time (s∝t²).

Note

Galileo used an inclined plane to decrease the acceleration of the falling masses. The problem was that he didn't have a device fast enough, to time objects falling under the acceleration of gravity. Galileo hadn't invented the pendulum clock yet (which wouldn't have helped much, it didn't resolve time into small enough increments i.e. its resolution wasn't good enough) and modern clocks and stopwatches hadn't been invented. In the story I read as a kid, Galileo used his pulse to show that the period of a pendulum was independant of its amplitude. Often when you invent something new, like an inclined plane, the infrastructure to use or test it doesn't exist and you have to invent that too. For the inclined plane, Galileo was pushing the current timing technology, the klepsydra, a clock that drained water at constant rate. The weight of water collected indicated the time. (For Galileo's description of his clock see inclined plane http://galileo.rice.edu/lib/student_work/experiment95/inclined_plane.html). In one of the urls on "inclined plane", to capture the essence of Galileo's accomplishment, students use a computer to time the rolling ball. If Galileo had had a computer, he wouldn't have needed an inclined plane, with its decreased acceleration, and he could have let the balls drop under gravity.

How far does a constantly accelarating mass fall. Here (again) is the graph of acceleration of a mass falling under gravity.

   |
^  |
|  |
a  |.......................g
   |
   |
   |_______________________t

             time->

Here (again) is the velocity (assumed to have v=0 at t=0; we're using downwards as the +ve frame of reference)

   |              o 
   |            o 
^  |          o   
|  |        o     
v  |      o 
   |    o 
   |  o
   |o
    ---------------
        t, secs ->

From experience (and reading) you know that s=f(t), for a constantly accelarating body, is a parabola. This graph shows the distance fallen as a function of time for a dropped body (drawn as 3 straight lines rather than a real parabola) (the downward direction is +ve).

   |           o 
   |          o 
^  |         o   
|  |        o     
s  |       o 
   |     o 
   |   o
   |o
    ---------------
        t, secs ->

We're now going to derive the equation for s=f(t) and show that it's a parabola.

By dimensions, the area under the v=f(t) graph is (change in) displacement. For constant velocity, by the definition of velocity, the dimensionless constant is 1, so the area under the graph is the actual displacement. The definition of velocity doesn't say anything about distance when the velocity is changing, so we have to figure this out ourselves. Let's see what we can find about the distance travelled (change in displacement) for the mass whose velocity is increasing linearly with time rather than constant.

In the (linear) v=f(t) graph above, let's say that the velocity at t=0 is v=0, while at the righthand end of the graph t=2sec, v=2m/sec. Find the displacement using the time and the average velocity.

v       = final velocity
u       = initial velocity
t       = time
ave_vel = (v + 0)/2

v = 2m/sec
t = 2sec

therefore
s       = ave_vel * t
        = ((v + 0)/2)*t
        = ((2 + 0)/2)*2
        = 2m

Let's find a general formula for s=f(t) using this average velocity method.

equations:
ave_vel    = (v + 0)/2

v = u + at
(u = 0)

therefore
v = at

s = vt	#definition of velocity


deriving average velocity=f(a,t)
ave_vel    = v/2
           = at/2   #average vel = init vel + 1/2th increment due to gravity

deriving s=f(a,t)
s          = ave_vel * t
           = (1/2)at^2

	Note
	For the moment, just consider situations where the object has u=0 at t=0. In this case s=½at2. We'll derive the formula for u!=0 shortly.

Galileo's experments had u=0, so the equation for his experiments is s=½at². We've found a formula that agrees with Galileo's data; s∝t². Using the average velocity to get our result sounds OK to most people, but to a mathematically educated person, deriving a result from the two end points, while ignoring the data in the middle (even if it is a straight line) is suspect. If the two end values were close, then a mathematician would say that the result is close to correct, but when one end is zero and the other end the highest value, you need another method to check your result.

	Note
	For a linear v=f(t) the average velocity method we just used, gives the correct answer, but we haven't proven this yet. Just because a result looks right doesn't mean it is right; you have to prove it's right. For 2000yrs people thought that heavier objects fall faster than light ones, and that the earth is flat.

Let's try another approach; find the area under the v=f(t) curve (a straight line).

   |              o 
   |            o// 
^  |          o////   
|  |        o//////     
v  |      o//////// 
   |    o////////// 
   |  o////////////
   |o//////////////
    ---------------
        t, secs ->

the graph is a triangle
area = height * base/2
height = v (at time t)
       = at
base   = t

area   = (1/2)at^2

The area has the dimension of distance. But now we have the problem of finding the dimensionless constant.

Let's try another approach. Chop the graph into small intervals. Make each interval short enough that the velocity can be regarded as constant. Then let the intervals become smaller and smaller. See if you get anything useful. First let's look at one interval. Values for the graph below might be t_end = 1sec, n = 10

   |           
   |          
^  |          
|  |       _     
v  |      |/ 
   |      // 
   |      //
   |      //
    ---------------
   0      t0      t_end
        t, secs ->

total time over which we're calculating distance = t_end

number of intervals = n

time per interval (the width of the slice) = t_end/n

time at interval shown = t₀

Now let's find the distance moved during the time slice of duration t_end/n at t = t₀.

equations:

s = vt #definition of velocity

v = at #definition of acceleration

values:

t_slice = t_end/n

velocity (regarded as constant) during interval starting at t₀

v = a.t₀ # from definition of velocity

calculation:

s = v*t_slice # velocity can be regarded as constant

      = (a*t₀)*t_slice

      = t₀(a.t_end/n)

	Note
	The distance moved in the interval at t=t0 is t0*constant.

Sanity check 1: dimensions of RHS = a.t² = l.

Sanity check 2: Check by substituting some numbers.

• t₀ has values 0.0,0.1..1.0. Let's pick something in the middle, say t₀ = 0.5.
• Let t_end = 1sec
• Let n = 10
• Let a = 1m/sec²

The average method says that the distance moved in t = t_end secs is 0.5m ((0 + 1)/2 m/sec * 1sec = 0.5m). From the equation above, the distance moved in the time slice at t = 0.5secs is s = 0.5*1*1/10 = 0.05m . Since our timeslice is 1/10 of the total time, the distance we've calculated by the time slicing method is about what we'd expect from using the average method. (We'd accept anything here unless it was out by a factor of 100 or so.)

The particle moves t₀*(a.t_end/n) metres during the time slice (interval) at t = t₀, of duration t_end/n secs. Next we want to find the distance moved during each of the time slices and add them. To do this we can add all the t₀ terms and multiply by the constant. After that we make the slices thinner and thinner and hopefully get something sensible for the total distance moved.

Here's our velocity graph showing all intervals.

   |             _| 
   |           _|// 
^  |         _|////   
|  |       _|//////     
v  |     _|//////// 
   |   _|////////// 
   | _|////////////
   ||//////////////
    ---------------
        t, secs ->

(Don't get lost here. We're just finding the area of a triangle, only we're doing it in slices.)

Summing intervals for 0<t₀<t_end

t₀ is a series of (real) numbers. For t_end = 1sec and n = 10, t₀ = 0.0,0.1..1.0.

t₀ = 0*(t_end/n),1*(t_end/n)..n*(t_end/n)

      = (0..n)*(t_end/n)

The component (0..n) is an arithmetic progression. There are n+1 terms (remember the fencepost error problem). Sum the series using Gauss's method (where we sum the series twice):

0..n
n..0
----
n..n (the sum has n+1 terms)

sum of all terms (0..n) twice = n*(n+1)
sum of all terms (0..n) once  = n*(n+1)/2  

Sum of all t0 terms:
sum of (0..n)*(t_end/n)       = n*(n+1)/2 * t_end/n
                              = (n+1)*t_end/2 
	(note: as n increases, so does the sum of the t0 terms)

duration of time interval:    = t_end/n
        (note: as n increases, the duration of the interval decreases)

sum of distances = sum of t0 terms * constant 
                 = (n+1)*t_end/2 * (a.t_end/n)
                 = (1/2)a*t_end^2 *(n+1)/n

Let n become large, when (n+1)/n becomes 1.

s = (1/2)a*t_end^2

use the general symbol t (rather than t_end) for the time at the end of the period

s = (1/2)at^2

We summed the area of a triangle in slices, finding that the area was height*base/2. We already knew the area of the triangle, so this result should be no surprise. What we didn't know was how far a constantly accelarating mass moved. The definition of distance moved only applies for constant velocity. We have no definition for distance moved for an object which is changing velocity. By slicing time into small enough intervals that the velocity doesn't change during each interval, we found the distance moved. The distance is the area of the triangle, but we couldn't prove it with the average velocity method.

The notation we used is clunky and the derivation tortuous (at least the first time). However cutting the graph into small slices is how calculus works. You have to understand what happened in this example to understand calculus. Once the result and method becomes familiar, then you can use the simpler notation of calculus to get the same result. For the moment, you should use the slice method, to become familiar with slicing, then when you get to calculus, you'll just see calculus as a change in notation, rather than some big hairy deal.

Note

In calculus, the process we just did (finding the area under a line) is done by integration. Here we found the area by summing a series. The sums of series are only known for a limited number of series, so the method we used isn't a general method for finding the area under an arbitary curve and isn't the method used in calculus. In calculus, integration relies on knowing that integration is the reverse of differentiation. Differentiation is simple and can be performed on any function, giving a table of functions and their differentials. The reverse process, of finding a function for any arbitary differential is not known. If the differential is in the table, then you can do the integration. If the differential isn't in the table, then you can't do the integration in analytic form (i.e. you can't produce an algebraic formula for it). You can always find a numerical value for the integral using numerical integration. There are no general formulae for integration; either the answer is known from differentiation or it's not known. People have been working on calculus for 400 yrs and it's almost certain that every function has been differentiated. So "not known" likely means "doesn't exist".

	Note
	from "e: The Story of a Number", Eli Maori, 1994 Princeton U. Press. ISBN 0-691-05854-7, pp62-68. Before the advent of calculus, geometric methods had found the area under polynomials (i.e.y=xn) by Fermat and under the hyperbola by Saint Vincent.

Here's what we've accomplished in this section

Again, let's plot s=f(t) for an object moving under constant accelaration (it's a parabola). (The ascii version is 3 sections, each a straight line.)

   |           o 
   |          o 
^  |         o   
|  |        o     
s  |       o 
   |     o 
   |   o
   |o
    ---------------
        t, secs ->

Example: a ball is dropped on a planet with an accelaration due to gravity of 1m/sec² (g on the Moon is 1.62m/sec²). What is the total distance moved at the end of each second and the distance moved in each particular second ^[35] ? Notice that the difference in distances moved at the end of each interval is a linear series with the ratio 1,3,5,7,9.. What's a more familiar term for the difference in distances moved at the end of each interval ^[36] ?

At what time is the ball 18m below the dropping point ^[37] ? Again we used an equation continuous in t. The position of the ball is completely determined by the velocity at t=0 and the accelaration. If the ball is moving under an accelaration of 1m/sec² and is not moving at t=0, then at both t=-6sec, t=+6sec the ball will be 18m below the dropping point.

Experience tells us the the ball will be moving up at t=-6sec and moving down at t=6sec. Calculate the velocity at both these times ^[38] .

What if the object has an initial velocity u before it starts accelarating? You sum the area under the v=f(t) graph the same way, in slices. This time the triangle is sitting on a pedestal. The area under the line (and hence the distance moved) is the sum of the area of the triangle (at the top of the diagram, filled with "/" characters) and the pedestal below (filled with "\" characters).

   |              o 
   |            o// 
^  |          o////   
|  |        o//////     
x  |      o ///////
   |    o /////////
   |  o ///////////
x0 !o--------------
   |\\\\\\\\\\\\\\\
   |\\\\\\\\\\\\\\\
   |\\\\\\\\\\\\\\\
   |_______________
        t, secs ->

You can calculate the distance moved yourself. Here's the answer.

s = ut + (1/2)at^2

Previously (launch angle for projectile) we found the angle which maximized the distance of a projectile, although I had to inject extra information for you to solve the problem. With this new equation, the extra information is built in. Find the time of flight of a projectile launched at angle=ϑ, with initial velocity u (hint ^[39] ) ^[40] . We get the same answer as before. Notice we have an extra solution, t=0 (the equation is quadratic in t). We should see this solution in any properly handled problem. (We didn't see this solution in our first attempt.)

Let's look at where we are.

We're trying to find the velocity as a function of h. Once we find that, we can find the dimensionless constant for K.E.∝mv².

We have the equations

Let's eliminate t from these equations.

v = u + at         (1)
s = ut + (1/2)at^2 (2)

from (1)
t = (v - u)/a      (3)

substituting (3) in (2)
s = (v - u)u/a + (1/2)a(v - u)^2/a^2
2as = (v - u).2u + (v - u)^2
    = (v - u).(2u + v - u)
    = (v - u).(v + u)
    = v^2 - u^2

the usual form of the equation

v^2 = u^2 + 2as

We now have the velocity of an object after accelerating through a distance s and we can calculate the dimensionless constant for K.E.

object falls through height = h 
P.E. lost = mgh

equation:
v^2 = u^2 + 2as

initial condition:
u = 0

on falling height = h
velocity = sqrt(2as) 
         = sqrt(2gh)

therefore
v^2      = 2gh

K.E. = k.mv^2	#k is dimensionless constant
     = k.m(2gh)
     = 2k.mgh

but K.E. gained = P.E. lost
so

mgh = 2k.mgh

therefore k=(1/2)

Thus K.E. = (1/2)mv^2

	Note
	FIXME - any connection between 1/2at^2 and 1/2mv^2? (both triangles?). The (1/2)mv^2 area comes from the graph y=at=v when x=t. I guess 1/2mv^2 is the area under the graph y=mv, when x=v. Where that comes from isn't obvious.

The dimensionless constant for our K.E. formula is ½. We now have the formula for K.E.=½mv²

You let a ball drop off a 10m high building. How fast is it going when it hits the ground ^[41] ?

Objects falling in air encounter aerodynamic resistance. The resistance depends on the area of the object and the co-efficient of drag. A spherical object has a relatively high co-efficient of drag, while a teardrop shaped object has a lower co-efficient of drag. A skydiver going head first presents a low area to the air, while if falling horizontally presents a high area to the air. Because air presents a resisting force to objects falling through it, they cannot accelarate indefinitely - they stop accelerating when the resisting force is equal to the force of gravity. The velocity at this point is called terminal velocity (http://en.wikipedia.org/wiki/Terminal_velocity).

Here are some terminal velocities from Fluid Friction (http://hyperphysics.phy-astr.gsu.edu/HBASE/airfri2.html)

Falling object            Mass             Area        Terminal velocity
Skydiver                   75 kg        0.7 m2         60 m/s        134 mi/hr
Baseball (3.66cm radius)  145 gm         42 cm2        33 m/s         74 mi/hr
Golf ball (2.1 cm radius)  46 gm         14 cm2        32 m/s         72 mi/hr
Hail stone (0.5 cm radius)  0.48 gm       0.79 cm2     14 m/s         31 mi/hr
Raindrop (0.2 cm radius)    0.034 gm      0.13 cm2      9 m/s         20 mi/hr

For a small enough object (water droplets in a cloud), the terminal velocity is smaller than any updrafts and the drops stay suspended indefinitely. From USGS Water Science (http://ga.water.usgs.gov/edu/watercyclecondensation.html) water droplets are about 10microns (µm) and using the calculator at the Fluid Friction site above, the terminal velocity is 0.6m/s.

How far would a ball have to drop in a vacuum to achieve the terminal speed of a skydiver in air ^[42] ? The height of the deck above the water of the (GW Bridge (http://en.wikipedia.org/wiki/George_Washington_Bridge) in the middle is 65m.

You throw a ball upwards at 14m/sec. What is its velocity 7.5m above the launch point ^[43] ? Notice that the ball 7.5m above you can be going up and can be going down. When the ball is 15m above you, what is its velocity ^[44] ? What happened here ^[45] ? What's the highest point in the trajectory ^[46] ?

Work is defined as F*s. This is just one formulation of energy/work/heat. It was the first. People wanted to measure the amount of work needed to pump water, bore canons, or grind flour and F*s gave it to them. The idea of energy (kinetic or potential) came later. The understanding that heat was energy came later yet. Since energy and work are the same, use the equation derived in this section to show that the work done to accelerate a mass in the horizontal direction, is equal to the change in kinetic energy. Assume the initial velocity is u, the final velocity is v, the particle moves a distance s while accelerating at a ^[47] .

On earth the record high jump (http://ihttp://en.wikipedia.org/wiki/High_jump) is about 2.5m. The men's pole vault record (http://en.wikipedia.org/wiki/Men's_pole_vault_world_record_progression) is about 6m. If these events were conducted on the moon (gravity 1/6th of earth's), what would the records be? First list your assumption(s) about these events (take as given that athletes perform fine in a vacuum and that air resistance doesn't affect the record). Here's my assumption(s) ^[48] . Assume that the record on earth is s, find the record on the moon. Insert numerical values the formula you get. Here's my answer ^[49] .

This is a standard question given to students who've just learned v²=u²+2as. The assumption you missed is that the s in the equation is the height of the centre of mass of the object at zero (vertical) velocity, while the record is the height of the bar cleared by the athlete. They are not the same. By agility (e.g. Fosbury Flop http://en.wikipedia.org/wiki/Fosbury_Flop), the athlete clears a bar higher than his/her centre of gravity. Assume a high jumper gets an extra 0.5m through agility, what would be the high jump record on the moon ^[50] ?

Often this question is posed as "if you can throw a ball up height x on earth, how high would it go on the moon". In this case the simple answer is correct.

On earth, the pole vaulter lands in a bed of rubber, to cushion the shock of falling from the bar at 6m. On the moon, how much greater is the shock of landing from a height of 36m ^[51] ?

Assume no agility in the high jump or pole vault. By what ratio does the time in the air (vacuum) change on moving the event to the moon ^[52] ?

If bipedal life exists on Jupiter (where the acceleration of gravity is 2.5 times that of earth), how much faster would its nervous system have to be, relative to ours, for it to be able to stay upright while walking ^[53] ?

Let's look at the 2-D projectile problem again. Find the maximum height reached as a function of launch angle ϑ (the initial vertical velocity is u_vert=u.sin(ϑ)) ^[54] . Note that the maximum height occurs for ϑ=90°. Remember that the maximum distance occurs for ϑ=45°. What is is the maximum height reached for ϑ=45° compared to the height reached for ϑ=90° ^[55] .

The P.E. is a function of height. For a launch angle of 45° at the top of the arc what is the ratio P.E./K.E. ^[56] ?

For any launch angle ϑ, is the maximum height reached proportional to the ball's initial velocity or to the ball's initial energy (remember the initial energy is all kinetic)? Derive the height as a function of the initial parameter you choose ^[57] . You could have made a reasonable guess as to which it was, without knowing any of the equations for motion under constant acceleration. How would you have made your guess ^[58] ?

We found that the distance travelled for the projectile is s=u²sin(2ϑ)/g. Is the distance travelled proportional to the initial velocity or to the initial energy of the ball? Derive the distance as a function of the appropriate parameter ^[59] . Could you have guessed whether distance was a function of initial velocity or initial energy, without knowing the equation for the distance ^[60] ?

We've derived the 3 equations which describe the motion of a particle under constant acceleration. The parameters for the equations are u,v,a,t,s. You need to remember these 3 equations.

v = u + at #doesn't involve s

s = ut + ½at² #doesn't involve v

v² = u² + 2as # doesn't involve t

Each equation leaves out one of these parameters; e.g. in the first equation, the distance s is ignored. It will have some finite value, but you don't care what it is. You can calculate it from your input data using one of the other equations.

The first equation is derived from the definition of acceleration. The 2nd equation is derived from the first, for constant acceleration, by cutting time into small enough slices, that the velocity is constant, and then summing the distance moved in each slice. The 3rd equation is derived from the first two, by eliminating t.

In a situation where one variable is allowed to vary without constraint, the variable is called an unconstrained variable. Each of the 3 equations leaves a variable unconstrainted.

The equation v = u + at describes a situation where s is unconstrained. If u = 0 and we want to acheive the same v we can vary a,t such that the product a.t = constant. What effect do you imagine (from experience) changing a,t, while keeping the product a.t constant (say halving a while doubling t), will have on s?

The answer wasn't obvious to me. I had to look at the equations. From s = ut + ½at², we can see that s will double. From v² = u² + 2as, we get the same answer; the distance will double.

Unless you know something about the system, you can expect that unconstrained variables will change when other inputs are changed. There is a branch of calculus called "Calculus of Variations", which optimises the values of unconstrained variables, by varying the constrained variables. Example: can we optimise the profit from making cookies by varying costs of ingredients and the taste, texture of the cookies?

5th equation in "a"

In the set of 3 equations above, there isn't an equation which leaves u unconstrained. There is such an equation, but no-one uses it. You can derive everything in it from the 3 equations above and the situation it describes isn't all that common. However for the exercise, derive the equation describing the relationship between v,a,t,s, which leaves u unconstrained, i.e. eliminate u ^[61] .

Example: the world record time for drag racing (driving a car in a straight line for ¼mile≅400m) is 4.428secs crossing the finish line at 541kmph. What's the acceleration (assumed constant) ^[62] ?

How is u is not involved here? Clearly in this example, we already know that u=0 (the drag racer will be disqualified if u != 0). We're after a set of circumstances where u is unconstrained.

Let's look at our example more closely. We've assumed that a is constant. This is unlikely with the power/torque curve of a gasoline/petrol engine. Let's calculate u from this information, using the other 3 equations. Here's my answer ^[63] . There are two different answers and neither is 0. Presumably the non-linearity of acceleration accounts for this.

For the drag race record, let's look at the uncontrolled variable v in the more familiar equation s = ut + at²; assume s = 400m, u = 0, a = 25m/sec², t = 4.4sec. What is the speed (velocity) at the end of the run ^[64] ? The real answer is 150m/sec. What constant acceleration would be needed to achieve a velocity of 150m/sec in 4.4sec ^[65] and what distance would be covered ^[66] ? The distance is short of 400m. Clearly the constant acceleration model doesn't fit here. However we're interested in the unconstrained variable u. Let's see what happens to u, assuming constant acceleration, if we change the acceleration and time in the 4th equation s = vt - ½at², with the convenient values v = 128m/sec, s = 512m.

As spotted by Zander, there is a 5th equation in which a is unconstrained. Derive the 5th equation.

Here we introduce vector notation: vectors are a better bookkeeping mechanism. All students of mechanics start with the cartesian system. For simple systems (1-D where there are no vectors, or multidimensional systems resolved as components, which are assembly of 1-D systems, where again there are no vectors), the cartesian system works quite well. However you do need to keep track of your frame of reference. Using the cartesian system, you'll find that similar problems are solved with different frames of references, leading to ad hoc'ery. You'll remember formulae with opposite signs in similar problems, but you won't really be clear as to why. When keeping track of directions and values (+ve,-ve) by hand, the chances of mistakes are high. The vector system is more complicated to setup, but extends into 3-D systems easily and reduces the sources of errors.

For the scalar of work, the direction of the vectors F (force) and s (distance) do not have to be colinear (i.e. they can be at an angle). Say a particle is confined to a rail allowing it to only move left or right. A force F acts on the particle at an angle ϑ. The work done on moving the particle a distance s is F*cos(ϑ)*s.

F
.
  .
    .
theta . 
------->
       s

You can think of the work as either

The work done F*cos(ϑ)*s, (where ϑ is the angle between the two vectors) is called the dot product of two vectors and written F⋅s. The dot operator, ⋅, multiplies the magnitude of the two vectors by cos(ϑ) and returns a scalar.

What work is done by a force F operating at right angles to the movement of a particle ^[68] ?

We're familiar with a force acting in the direction of movement (accelerating a car, a jet engine pushing a plane). In this case, the engine does work on the car/plane, raising its K.E. (or P.E.). Since energy is conserved, the engine has to loose energy. We normally think of an engine as being a generator of energy. This is a fiction; it's really converting potential (chemical) energy into kinetic energy (and heat). To show that an engine is a converter of energy, rather than a generator, we can leave it running for a while and eventually it will run out of potential energy and not be able to produce any more kinetic energy.

How does an engine arrange to loose energy? If the angle ϑ=180°, between the force and the movement, then work is -ve. An engine produces a force in the opposite direction to its movement. A jet (or rocket) engine throws massive objects (molecules of gas) at high velocity out the back of the plane.

<-front  back->

   F->/ m->
      \o
       |
  <-F /\
--------------
O O        O O

<---s

This diagram shows you on a frictionless rail car, throwing massive objects (m) off the back. This requires you to exert a force F on the massive objects. Since you are wearing friction soled shoes, Newton's 3rd law of motion says that your shoes will push on the bed of the rail car with a force F in the opposite direction. The engine (you) is exerting a force in the opposite direction to the movement, meaning the engine (you) is loosing energy.

You can throw massive objects off the back of the rail car in a vacuum. It will work even better than in the atmosphere (since there's no air resistance to moving the rail car). Goddard's (http://en.wikipedia.org/wiki/Robert_H._Goddard) proposal to fly to the moon with a rocket was greeted by derision by an editorial in the NYT

It expressed disbelief that Professor Goddard actually "does not know of the relation of action to reaction, and the need to have something better than a vacuum against which to react"

While the engine (you) is loosing energy, the railcar is gaining (kinetic) energy. The soles of your shoes are pushing on the railcar, in the direction it's moving. With ϑ=0, F⋅s is +ve. The important part of an engine, as far as the railcar, car, or rocket is concerned, is the motor mounts. (The car doesn't care what's pushing on the motor mounts.)

A plane/car can loose energy by braking - the force acts in the opposite direction to the direction of movement. Usually the energy is converted to heat, but could be used to run a generator to charge a battery, or converted to kinetic energy somewhere, to be later used to accelerate the car (this isn't practical at the moment). Generators in power stations convert hot steam to cool steam (loosing energy) and accelerate electrons (producing electric power).

In the cartesian frame of reference, P.E=-mgh (remember the numerical value of g is -ve). Why the -ve sign? Let's think about the sign of P.E. of an elevated object. If a massive object is at height, it can do work when it looses height; it can raise another massive object, or it can generate sound and heat when it lands. The object at height has +ve P.E. When we raise the object to its height, we push upwards and the object moves upwards; i.e. we do +ve work on the object (using the dot product formalism, explain why is work +ve ^[69] ?) and the object at height must have more P.E. When the object falls, the force acting on the center of mass is pointing downwards, and the object moves downwards, so again ϑ=0 and the work done by the falling center of mass (converting P.E. to K.E.) is +ve.

Now that we have an idea of what's going on, let's change to the vector formalism.

force of gravity acting on center of mass F=mg. (g is a vector pointing downwards.)

In the cartesian frame of reference, the force is pointed in the -ve direction. In the everyday world we can say that the value of the force is -ve. However we can get into trouble unless we think clearly here. In the vector world, where vectors have magnitude and direction (but not sign), the force is just pointing in a particular direction. Scalars have signs, vectors have directions. We don't get anything with a sign until we have a scalar, such as work, which we get from the dot product of force with another vector (like direction). If you want to be sure of getting the sign right, you should let the vector math handle it for you.

    o
    | F=+mg
    v
    ^
    | F=-mg	

Let's say we want to raise the massive object - we'll have to push on it. We'll need a force infinitesimally larger than mg, but pointing in the opposite direction to mg (up). Our force then is F=-mg (by Newton's 3rd Law of Motion). Since we're using vector notation, we don't need to know where g is pointing. Here g happens to be pointing downwards, but it could be up, down, or at some angle. (In day-to-day conversation, we can say that the force is pointing in the -ve direction, as long as we understand that we really mean it's pointing at 180° in the frame of reference.) (At intermediate steps, we can do checks, to make sure that we're doing something sensible.)

let's check the force of gravity on the object: it's F=+mg. g points down, so F=+mg points down too.

let's check the force we push on the object with: it's F=-mg. g points down, so F=-mg points up.

What work is done raising the object through a distance h? W=F⋅h=-mg⋅h. In the vector notation, to do the dot product, the direction of the force and the direction of distance moved are not used, only the angle between them. F=-mg points up (since g points down), and h points up. The angle between the force and the distance moved is ϑ=0°, so W=F⋅h is +ve i.e. we have to do (+ve) work to raise the massive object against the force of gravity.

Let's look at the the massive object. It's exerting a force F=mg. It's being moved a distance h by a force exerted from the outside. What work is the massive object doing on the external force? W=F⋅h=mg⋅h. Since g is facing downwards, and h is facing upwards, ϑ=180°, then cos(ϑ)=-1 and the massive object has done a -ve amount of work on the system pushing against it. i.e. the pushing system has less energy now.

What's the change in energy of the system? Change in P.E. of raised object + change in E. of the pushing system = (mg⋅h) + (-mg⋅h) = 0.

	Note
	Since most of the following sections have examples where vectors are colinear, the cartesian system will be used. However you can use vectors, in which case you should use vector equations like v2 = u2 + 2a⋅s.

In the previous section we showed using the dot product, that work was scalar and used examples with P.E. Here we do the same thing with K.E. using the vector formalism and dot products.

W = K.E. = ½mv⋅v

v is a vector (it has magnitude and direction) and is always parallel to itself, so ϑ=0, and therefore v⋅v = |v|²cos(0) = v² . For K.E. you can ignore the fact that v is a vector and just square the magnitude of v (as you've always done).

A conservative (vector) field (http://en.wikipedia.org/wiki/Conservative_vector_field) is one in which the change in energy of a particle going between two points is independant of the path taken. A consequence is that the total energy of a particle is conserved, when moving under the influence of conservative forces. Energy is conserved as a loss of potential energy is converted to an equal quantity of kinetic energy or vice versa. It seems reasonable (although I can't prove it) that the reverse is also true; a body moving in a vector field whose energy is conserved, is moving in a conservative vector field. We're going to show that energy is conserved for a particle moving under the acceleration of gravity, even if this doesn't show that gravity is a conservative field (the heading for this section).

Note

Other people have shown that gravity is a conservative field. I know that (gravity_is_a_conservative_field ⇒ energy_is_conserved_in_a_gravitational_field), but I can't show that (energy_is_conserved_in_a_gravitational_field ⇒ gravity_is_a_conservative_field). Since (energy_is_conserved_in_a_gravitational_field ⇔ gravity_is_a_conservative_field) then someone has already shown that (energy_is_conserved_in_a_gravitational_field ⇒ gravity_is_a_conservative_field) but I don't know how the basis of it.

We'll first show that energy is conserved for the 1-D problem (a ball moving only in the vertical plane), then do it for the 2-D problem (the ball launched at an angle ϑ).

The potential energy (P.E.) is measured with respect to a reference position i.e. the difference in P.E. between its current position and some arbitary reference position. There is no absolute P.E. If you throw a ball up (by imparting kinetic energy (K.E.)), then the ball will ascend, converting K.E. to P.E., reaching a height when its velocity is 0. At this position it will have +ve P.E. (with respect to the throwing point, i.e. the ball can do work if it's allowed to fall through gravity back to the throwing point), but no K.E.

If instead you dropped the ball, it would gain K.E. and loose P.E. (i.e. the P.E. becomes -ve and work needs to be done to return the ball to the dropping point).

Note

(Note the presence or absence of dot product symbol ⋅ in these next formulae.) People often exchange frames of reference in mid-stream without being aware of it. The formula for P.E. is often stated (in non-vector notation) as P.E.=mgh, where h is the distance the particle can drop. Unstated is that the frame of reference is +ve downwards, automatically giving the correct answer of a +ve number. If you're being a bit more rigourous, you'll choose up as the +ve direction, and you'll use vectors. In this case P.E.=-mg⋅. Just be aware of whether you're using vectors, or are in a 1-D situation, where you're just using magnitudes and whether up or down is +ve.

For this problem, let

You'll calculate P.E. and K.E. From there you'll calculate total energy (E_tot = K.E.+ P.E.) of the ball after it leaves your hand.

Let's calculate E=f(h). After watching the ball's up and down motion, this would be a likely first approach.

For E=f(h), here's my answer for the total energy (E_tot = P.E. + K.E.) ^[74] . (With the slopes of the lines for K.E., P.E. being the -ve of each other, we should expect that the sum of the two lines is a constant.) We find that no matter what the initial velocity (whether you throw the ball up or down, or let it drop), the total energy of the ball remains the same; the kinetic energy at the moment of launch. As height changes, K.E. is converted into P.E. (and back again). Energy is conserved.

We can plot a new diagram E = f(K.E.,P.E.). We know that K.E + P.E. = ½mu², so the graph with K.E. on the x-axis and P.E. on the y-axis is a straight line of what slope, and intersecting the P.E., K.E. axes where ^[75] ?

Here's the P.E./K.E. graph (caution - this is ascii art: the slope should be -45°)

1/2.mu^2 -|o
          |  o
          |    o
       PE |      o
          |        o
          |          o
          ------------|
               KE     1/2.mu^2

What physical principle requires slope=-1 ^[76] ?

You can make (KE,PE) diagrams for any system where KE and PE are interconverted e.g. oscillating systems (pendulum, mass on a spring). Because of conservation of energy, all diagrams are the same; i.e. there is only one (K.E.,P.E.) diagram. The only difference between the diagrams is the way the particle moves along the line (linear, 2nd order or sinusoidally as a function of some parameter e.g. time, height, velocity, angle...).

From thinking (we'll do the calculation shortly), what's the trajectory (locus) of the ball along this line as a function of h? i.e. what point on this graph corresponds to h=0 (start, end), h_max = -½u²/g ^[77] ? If there was no ground to stop the ball at the end of the flight, what would be the location of the ball on this graph as it descended through the ground level ^[78] ?

Again just using thinking at the moment (we'll calculate the result shortly), what would be the speed of the locus along this line as a function of h? Is it linear with height or does it speed up in some places ^[79] ?

Mathematicians are quite happy to shift (translate) axes if it simplifies the analysis. The ball doesn't start at the origin of our graph, so if we want to describe how far its moved, we have to repeatedly say "from its original position". Let's translate the axes so that the ball's starting position is at the origin (we only need to translate the K.E. axis). Call the new coordinates (ΔKE,ΔPE) (the Δ is Delta), representing the change in K.E.,P.E. from their initial values. Here's our new graph

h=-1/2.u^2/g o          |- 1/2.mu^2
               o        |
                 o      | dPE
                   o    |
                     o  |
                 h=0   o| 
            |-----------
    -1/2.mu^2 
                dKE 

Using K.E.=f(h), P.E.=f(h), find the coordinates of the ball on the (ΔKE,ΔPE) diagram as a function of height ^[80] ? Figure out (at least in principle, we'll do the calculation shortly) how to calculate the locus of the ball on the (ΔKE,ΔPE) graph as a function of height ^[81] . The line here is straight, simplifying our analysis. If the line was curved (e.g. a parabola), we would need some method of finding the distance moved along the curved line.

Here we derive the K.E. and P.E. as a function of time. The equations for E=f(t) are more complicated than for E=f(h) (P.E. is a simple function of height).

For E=f(t), here's my answer for the total energy (E_tot = P.E. + K.E.) ^[90] .

We now have K.E. and P.E. as functions of time and can plot the (K.E.,P.E.) diagram.

This is the same diagram as was derived for E=f(h), but this time marked with the location according to time. Here's the trajectory (locus) of the ball along this line as a function of t, showing t=start, t=-u/g (half the time of the flight at the top of the parabola), t_end

1/2mu^2  -|o t=1/2
          |  o
          |    o
       PE |      o
          |        o
          |          o t=start,end
          ------------|
               KE     1/2mu^2

Since we're looking at the same diagram as for the E=f(h) section, the ball moves similarly, starting at bottom right, then moves up the line to top left (at the top of the flight) and then returns to the bottom right (at the end of the flight).

(remember it's the same diagram as we saw with E=f(h)) If there was no ground to stop the ball at the end of the flight, what would be the location of the ball on this graph as it descended through the ground level ^[91] ?

Just using thinking at the moment (we'll calculate the result shortly), what would be the speed of the locus along this line as a function of t? Is it linear with time or does it speed up in some places ^[92] ?

Let's translate axes as we did for E=f(h) so that the ball starts at the origin. Here's our new graph (now with the locus of the ball parameterised in t, rather than h).

          o t=1/2    |-  1/2.mu^2
            o        |
              o      | dPE
                o    |
                  o  |
                    o| t=start,end
         |-----------
        -1/2.mu^2 
              dKE  

Using K.E.=f(t), P.E.=f(t), find the coordinates of the ball on the (ΔKE,ΔPE) diagram as a function of time ^[93] ? Calculate the locus of the ball along the line as a function of time, using the same methods as for E=f(h).

Now let's calculate E = f(v).

The result isn't quite as informative as the equations for E=f(t), E=f(h). Let's see what we can find from this equation. For what v is P.E. = 0 ^[98] ? After our previous work, these answers are obvious, but they may not be if we'd just started characterising a new system (the 2nd answer, v = -u, wasn't obvious in our earlier attempts to characterise the problem). When is P.E. a maximum ^[99] ? Well then you may ask, when (time wise) is v² a minimum? We need one or more of our 3 equations to find this. However we do know the minimum possible value for v² (it's 0), giving P.E._max = ½mu² . In this situation it's possible to convert (all) the initial K.E. to P.E. (this isn't true in the 2-D case).

What is the name of the curve P.E. = f(v) (it has the equation P.E. = ½m(u² - v²) ) ^[100] ?

Is the parabola symmetric about the v = 0 axis ^[101] ? Use the quadratic formula to show that parabolas without any first order terms are symmetric about the y-axis ^[102] . Is the P.E. parabola concave downwards or upwards? Is the P.E. at v = 0 a maximum or minumum ^[103] ?

Here's the parabolas for P.E., K.E.

     x       o|o       x  
PE=o  x   o   |   o   x 
KE=x   xo     |     ox
       ox     |     xo 
      o   x   |   x   o
     o-------x-x-------o

              v->

Since both P.E. and K.E. parabolas have the same constant (±½m) infront of the v² term, they are mirror images through a horizontal line (what is the height of the horizontal line ^[105] ?).

Note

These graphs being parabolas, you may think you're looking at graphs of the flight of the ball. As it turns out, parabolas describe the flight path and E=f(t | h | v) because the equations are 2nd order, not because the various parabolas represent the same thing. Note that that abscissa is v not s or t and that the point v = 0 is in the centre of the graph. On this graph, the LHS has v -ve. When/where in the flight is v -ve, +ve[106] ?

In the E = f(v) graph, the ball starts at the RHS and moves to the left, hitting the ground at the mirror image spot on the LHS.

	Note
	Since we have a 1-D system, there are no such things as vectors and v is a scalar in a cartesian frame of reference.

For E=f(v), here's my answer for the total energy (E_tot = P.E. + K.E.) ^[107] . We now have K.E. and P.E. as functions of velocity and can plot the (ΔK.E.,ΔP.E.) diagram.

This is the same diagram as was derived for E=f(h),E=f(t), but here it's marked with locus of the ball as a function of v. What's the trajectory (locus) of the ball along this line as a function of v? i.e. what point on this graph corresponds to v=u (start of flight), v=0 (top of the arc), v=-u (end of flight) ^[108] ?

Just using thinking at the moment (we'll calculate the result shortly), what would be the speed of the locus along this line as a function of t? Is it linear with time or does it speed up in some places ^[109] ?

Let's translate axes as we did for E=f(h) so that the ball starts at the origin. Here's our graph (now with the locus of the ball parameterised in t, rather than h).

          o v=0      |-  1/2.mu^2
            o        |
              o      | dPE
                o    |
                  o  |
                    o| v=u,-u
         |-----------
        -1/2.mu^2 
              dKE  

Using K.E.=f(v), P.E.=f(v), find the coordinates of the ball on the (ΔKE,ΔPE) diagram as a function of time ^[110] ?

As a function of v

In the (ΔKE,ΔPE) diagram the locus moves linearly with h, but is 2nd order with t and v. There's nothing particularly deep about this. The (ΔKE,ΔPE) line is straight so the projections on the ΔKE, or ΔPE axis can be used to follow the movement of the locus of the ball on the diagram. The movement of the locus then is determined by the behaviour of PE (or equivalently KE) as a function of the parameters h,t,v. We can find the dependency of the locus on the parameters by knowing that PE∝h and from our 3 equations we have h∝t², h∝v².

Here's what we've found from the 1-D trajectory problem (equations use the dot product even though we're in 1-D).

General info:

Energy=f(h).

Energy=f(t)

Energy=f(v)

Note: E_tot always is ½mu². We've shown three different ways that gravity is a conservative field in 1-D.

Now let's do the 2-D problem. Make the following change to the 1-D problem

We can calculate E_tot = P.E. + K.E. as a function of (among other things) time and of height.

Here we calculate P.E. = f(t).

Calculate the P.E. as a function of the variable t (and the initialising constants u,ϑ) ^[114] . We see that P.E. is 2nd order with respect to t (i.e. P.E. = f(t²)).

	Note
	When we say a function is nth order with respect to a variable, this means that the highest order term in the variable is to the nth power. This statement allows you to determine how fast the function grows (lower order terms are insignificant as the variable goes to ±∞).

What is the name of the curve which is 2nd order in a parameter ^[115] ?

Note

When you get an equation you haven't seen before, you should do some sanity checks on it. The only easy check (by eyeball, i.e. without resorting to pen and paper), for the function we have, is that for large t (+ve or -ve), (when the highest order term in t, here t2, dominates), P.E. is -ve (the projectile is below the starting height). Here we go to pen and paper, to show that P.E. is +ve for the expected values of t.

For what t is P.E. = 0? ^[116] ? We know from our original problem that the time of flight of the ball is t = -2u.sin(ϑ)/g. By plugging the start and end times into our formula for P.E., we see that P.E. = 0. This is as we'd expect; we defined our reference frame to have P.E. = 0 at the start. Since the start and end of the flight is at the same height, the P.E. at the end will also be 0.

Next check the calculated value for P.E. at the top of the arc. For this you need to know t at the vertex of the parabola (instead of knowing, the time honoured method of guessing is often just as fast). Since a parabola is symmetrical about its vertex, the top of the flight will occur half way (in time and distance) through the flight, at t_top = -u.sin(ϑ)/g (remember g is -ve, so t is +ve). Use this value for t_top to find the P.E. at the top of the arc ^[117] . We find that the P.E. at the top of the arc is the vertical component of the initial K.E. (The horizontal component of the initial K.E. doesn't affect the vertical component of the problem.)

Our sanity checks show that P.E. as a function of time

This is not proof that our equation is correct (however it's a good indication). If we get into trouble further down the line, these checks indicate that we can first look elsewhere for the cause.

The vertex of the parabola is offset from the y-axis (the vertex occurs at +ve time, rather than at t = 0), making the equation for the P.E. a little complicated.

	Note
	You can characterise the parabola using calculus. Still you need to be able to do it the simple way too (like we do here). Later we'll use calculus to find out more about the flight of the projectile.

You can simplify the equation of the parabola by translating the axes (in this case, we'll shift the x-axis). You need to know the value of t at the vertex (the top of the flight), which we've guessed occurs half way, at t_top = -u.sin(ϑ)/g. To put the vertex on the y-axis (i.e. at x=0), translate (offset) the x-axis by guessed_center = -u.sin(ϑ)/g (remember this has a +ve value). To do the offset, create a new time variable T and make the following substitution:

T = t - guessed_center = t + u.sin(ϑ)/g

∴ t = T - u.sin(ϑ)/g.

Substitute this formula for t into the formula for P.E., giving us a parabola with the new independant variable (the translated version of time) T, rather than the original t ^[118] .

Do a sanity check; check that all terms have the dimensions of energy. This parabola is of the form y + a = bx² (actually P.E. + a = bT²). With a -ve sign infront of the T² term, is the parabola concave down or up ^[119] ? The center of the arc (top of the flight) occurs for what T ^[120] ? Substitute this value of T into our formula for P.E., to check that we get the same result for the P.E. at the top of the arc, as we did before we translated the x-axis. Find the value of T that gives P.E. = 0 ^[121] and confirm that it's some value you expect (and check its dimensions).

If we wanted to simplify the parabola further, we could translate the y-axis too. (This doesn't add much clarity in our case, but in other situations it could be useful.) Pick a new energy scale for the y-axis and call it J.E., with a value

J.E. = P.E. + ½mu²sin²(ϑ)

i.e. P.E. = J.E. - ½mu²sin²(ϑ)

	Note
	J.E. and P.E. go in the same direction (both have constants with the same sign), so the direction of the parabola (concave downwards) doesn't change.

Substitute our new value for P.E. into the parabola ^[122] . Now we have a standard y=ax² parabola. Confirm that terms on both sides have the dimensions of energy (you should be able to do this by inspection). With the -ve sign infront of the T² term, the parabola is still concave downwards.

Note

We could further simplify the parabola by defining a new variable Y = J.E./½mg2T2 (or even Y = -J.E./½mg2T2). The numbers on the y-axis now are dimensionless. Mathematicians/physicists like using scaled/tranlated forms of equation(s), as the problem can be studied in a form that works for any initial conditions (the details for any particular flight are ignored). The scaling/translation factors are kept somewhere else (on another piece of paper, or in a struct in a program) and only consulted when you need the real world numbers for your answer. You have more chances of making mistakes if you change the dimensions of a problem. In the case here, it's likely that you would use a form that has dimensions of energy (rather than dimensionless).

We've found the formula for P.E. as a function of time. Now let's find the K.E. for the projectile as a function of time. We'll find the two components.

Find the vertical component of the K.E. as a function of time ^[123] .

Find the horizontal component of K.E. as a function of time ^[124] .

Find the total K.E. as a function of time ^[125] . Let's do some sanity checks on our formula for K.E. K.E. is ½mv² so it should always be 0 or +ve. Thus K.E. should always be non -ve. Whether this is true is not obvious from the formula (remember g is -ve). We could assume that we've got it right and press on regardless and return here if we find something inexplicable further down the road. In general this is not a good idea. Since we're learning and we mightn't recognize an inexplicable equation (there are lots of ways of things going wrong, not all of which you'll pick when they happen), let's check before going on. The K.E. is 2nd order in t (it's a parabola). By inspection you can see that at large t (+ve or -ve), when the terms in t² dominate, that K.E. is +ve. If the curve is always non -ve, then we should check that it doesn't cross the x-axis for some value of t.

How do we check that a parabola doesn't cross the x-axis ^[126] ? Do the calculation to tell if the parabola crosses the x-axis ^[127] . The maximum value for sin²(ϑ) is 1 (for ϑ=90°). The value for our term (inside the √ in the formula for the roots of the quadratic equation) then is either -ve or 0. The 0 value indicates two roots at the same point on the x-axis, i.e. the parabola is concave upwards and touches the x-axis at one point only (and never has a -ve value). At this value of ϑ (ϑ=90°) the projectile is fired directly upwards. The projectile reaches v = 0, with K.E = 0 and P.E. at its maximum. For any other ϑ, there are no real roots and the K.E. is +ve throughout the whole flight (i.e. the projectile is always moving). Our sanity check shows that K.E.=0 only for ϑ=90°.

We've shown that the P.E. parabola is concave downwards, while the K.E. parabola is concave upwards. Next we're going to add P.E.+K.E. What does the law of conservation of energy predict for the sum of these two energies (parabolas) as a function of t ^[128] ?

Add P.E.+K.E. (as functions of time) ^[129] . Throughout the whole flight, the energy of the projectile is constant at the initial K.E. This is both blindingly obvious (if you know about the conservation of energy) and profound at the same time. Nowhere in our equations (1-D, 2-D) did we build in the conservation of energy; we just allowed our ball to be accelarated by gravity. By doing so, we found that gravity is a conservative field.

Compare K.E._tot for arbitary ϑ with ϑ=90° i.e. find K.E._{tot, ϑ} - K.E._tot,ϑ=90° ^[130] . For 0≤ϑ<90°, this is +ve. This result shows that for flights with a non-vertical launch that the ball has a higher K.E. as a function of time. (When ϑ=90° at the top of the flight, the ball has no K.E.; When ϑ=45° at the peak the K.E. is ½E_tot.) Confirm that you get an equivalent result for P.E._tot,ϑ=90° - P.E._{tot, ϑ} ^[131] P.E. decreases with decreasing ϑ. The difference in P.E. as a function of ϑ is mg(1-sin(ϑ)ut #g is -ve

The locus of the ball in the (ΔKE,ΔPE) diagram starts at the origin. For ϑ=90° the (ΔK.E.,ΔP.E.) diagram is the same as for the 1-D case. For ϑ<90° the ball doesn't convert all its KE to PE at the top of the arc (it's moving horizontally at the top of the arc). For ϑ=45deg;, at the top of the arc, the ball only reaches half the height and ball has equal amounts of K.E. and P.E.. Here's the (ΔK.E.,ΔP.E.) for E=f(t) in 2-D for ϑ=45°. Note that the locus of the ball never reaches ΔPE=½mu². In the absence of ground, the ball continues to the lower right (not shown).

                     |-  1/2.mu^2
                     |
                     | dPE
           top  o    |
                  o  |
                    o| t=start,end
         |-----------
        -1/2.mu^2 
              dKE  

For ϑ<90° the (ΔKE,ΔPE) diagram extends a shorter distance into the 2nd quadrant. As well the time to reach the top of the arc is less. Do these two factors cancel, i.e. does the ball move along the line in the (ΔKE,ΔPE) diagram at the same velocity as for the vertical flight? Here we find the velocity of the ball along its locus in the (ΔK.E.,ΔP.E.) diagram as a function of time for any ϑ.

Before launching into a detailed mathematical analysis, it's often useful to calculate numbers for some convenient location(s) on the line. This may tell you in short time if you're barking up the wrong tree. From previous work we know

For ϑ=45° the height of the top of the flight is reduced by 0.5, while the time to get there is reduced by 0.707. The velocity of the locus of the ball in the (ΔKE,ΔPE) diagram is multiplied by a factor of sin(ϑ).

For any ϑ the P.E. (or ΔPE) is multiplied by sin²(ϑ). When dividing (ΔPE/t), we're left with a factor of sin(ϑ) i.e. the velocity of the locus of the ball is multiplied by sin(ϑ).

Let's find a more general derivation of this result. The velocity along the line as a function of time is

d(P.E.)/dt = -mg(u.sin(ϑ) +gt)

The velocity of the locus of the ball is smaller by -mgu(1-sin(ϑ).

In the previous section we calculated the (total) energy as a function of time. Here we do it as a function of height. We'll get the same answer (energy is conserved).

Calculate the total energy of the projectile at any h ^[136] .

	Note
	The -mgh terms cancel in P.E. and K.E.vert. In the K.E. terms, the horz and vert components of initial velocity add (via Pythagoras) to give the initial velocity.

	Note
	Doing this problem as function of height turned out to be simpler than doing it as a function of time, although I didn't realise this till after I'd done both methods. Sometimes any approach will do. Sometimes you want energy as a function of time and you'll need the time approach.

Here we calculate E=f(v).

Starting with P.E. = -mg⋅h and knowing h = f(v), derive P.E. = f(v) (you can use the 1-D version for hints) ^[137] . While this answer is correct, were asked for P.E.=f(v), not P.E.=f(v_vert). Change this result into a function of the form E=f(v) ^[138] . There's no dependence of P.E. on ϑ (this may be useful in some circumstances).

Find the components of K.E. in the x,y plane. Here's K.E._horz (it's a constant) ^[139] K.E._vert ^[140] .

What's K.E._tot=K.E._vert+K.E._horz ^[141] ? We could have written this result by inspection (i.e. without finding the components of K.E.). Again there's no dependency on ϑ. The dependency on ϑ has been subsumed into v, i.e. v=f(ϑ). (We usually don't have a lot of flexibility in how our equations work out; we might have got equations in E=f(v) that had ϑ as a variable, as we did with E=f(t), but as it turns out we didn't.)

Here's E_tot ^[142] .

Although we've gone to 2-D, we're still treating the variables (e.g. v) as scalars. This works because we're dividing the 2-D problem into component 1-D (scalar) problems and adding our results, the scalar quantities P.E., K.E. However while v_horz, v_vert are scalars, v is a vector. In 2-D, v doesn't become -ve at the top of the flight; only the angle (also confusingly called the argument) becomes -ve.

For ϑ=45°, what's v at the start of the flight? It's (x,y)=(u.½√2,u.½√2) in cartesian coodinates and (r,ϑ)=(u,π/4) in polar coordinates. What about the top of the arc? It's (x,y)=(u.½√2,0) (cartesian coordinates), or (r,ϑ)=(u.½√2,0) (polar coordinates) (being a vector along the x-axis, the cartesian and polar forms are the same).

For ϑ=45° here's the diagram of v during the flight in cartesian coordinates.

       |         o start 
       |         o 
       |         o
       |         o
       |         o
v_vert  ---------o-top-----
       | v_horz  o 
       |         o
       |         o
       |         o
       |         o end
                 u.cos(theta)

You can see the polar (r,ϑ) coordinates, by drawing a vector from the origin to each of these points. To determine the (r,ϑ), people usually eyeball the cartesian vector diagram. If you wanted to plot the (r,ϑ) diagram, this is one way of doing it.

      pi/2   | 
theta        |     
      pi/4   |start o |v|=u 
             | 
              ----o----------
             |          r
     -pi/4   |  end o
             |            o
     -pi/2   |

The intercept on the r-axis is ½√2.u The angle never reaches -π/2 (i.e. the ball never goes directly downwards).

To plot P.E.=f(v), we need a 3rd dimension. To plot K.E. and P.E. on separate axes to look for conservation of energy, we'd need a 4th dimension. Trying to visualize these functions will be difficult, and I can't see that any great insight will come from it, so I'm not going to attempt it. (Admittedly, the cartesian graph of the locus of v is a straight line, so you could plot K.E. and P.E. in 3-D.)

We've done E=f(t). Now let's do E=f(d) (horizontal distance).

Rather than deriving the equations from scratch, let's use d=u.cos(ϑ).t to derive our equations for E=f(d) from the equations in E=f(t). Here's h=f(t).

h = ut.sin(ϑ) + ½gt²

Substituting t = d/(u.cos(ϑ)) we get

h = d.tan(ϑ) + ½g(d/u.cos(ϑ))²

Do a sanity check. The throw is -2u²sin(ϑ)cos(ϑ)/g . The distance to maximum height is half this at d = -u²sin(ϑ)cos(ϑ)/g . Check that you get the expected value for h at d=0 ^[143] , d for max height ^[144] , and d=end ^[145] .

As another sanity check, we should expect some strange behaviour in h as a function of ϑ for ϑ→90°, when tan(ϑ)→∞, cos(ϑ)→0 and d→0. Try substituting these values into the h=f(d) ^[146] .

In the equation above for h=f(d), there are terms in g.d. Both of g,d are vectors and are orthogonal. The dot product of two orthogonal vectors is 0. Quite what is happening there I don't understand (you can't scalar multiply two vectors).

Here's K.E._tot=f(t)

K.E._tot = ½m(u² + 2u.sin(ϑ).gt + (gt)²)

Find K.E._tot=f(d) ^[147] . Do a dimensional check on any terms you don't recognise as being energy. Check K.E._tot at the start ^[148] , top ^[149] and end of the flight ^[150] .

Let's explore around the problem, to expand the radius of problem space that we understand. Mathematicians like doing this and you'll get pats on the back for finding some new result.

What is the energy imparted to the ground by the projectile at impact (compared to the initial energy) ^[151] ? Why is the P.E. the same at the beginning and end of the flight ^[152] ?.

For what ϑ is the time of flight longest (you've already done the calculation which finds this, to find the angle for the furthest throw, but you didn't record the time). ^[153] ? What is the time of flight for the trajectory with maximum throw, relative to the longest flight. ^[154] ?

Here's the parabolic flight of our projectile for ϑ=45° (maximum throw), with the x-axis showing distance. The y-axis shows P.E. For didactic purposes, I've shown the flight before the moment of throwing and after the end of the flight, allowing the projectile to pass effortlessly through the ground. The P.E. reaches half the initial K.E. The K.E. drops to half the initial K.E. (the vertices of the two parabolas touch; which isn't clear in the ascii art). The total energy (P.E+K.E.) is shown by e.

    x                   x   
    exeeeeeeeeeeeeeeeeexe
     |x               x
KE=x | x             x
     |  x           x
     |    x       x
     |       x x 
     |       o o
PE=o |    o       o
     |  o           o
     | o             o 
     |o 45deg         o
     o-----------------o
    o                   o
             x ->

Here's the diagram for the projectile shot more vertically (say ϑ=60°); the P.E. parabola extends up further (the ball goes higher), while the K.E. parabola goes down deeper (the ball goes slower) (not a great pair of parabolas I'll admit).

     x      x   
    exeeeeeexe
     |x    x
KE=x |x oo x
     |xo  ox
     | x  x
     | x  x 
     | o  o
PE=o | o  o
     |ox  xo
     |o xx o 
     |o    o
     o------o----------o
     o      o 
             x ->

Draw a similar diagram for the ball shot more horizontally (say ϑ=30°) ^[155] .

The diagrams above (and experience) allow us to compare the slope of the h=f(x) parabolas at the start and end, for different launch angles. For ϑ = 0°, the initial slope is 0, while at high ϑ the initial slope is highest. We say that the initial (and final) slope is a monotonic function of the launch angle.

monotonic: a function whose value is always increasing (monotonically increasing) or decreasing (monotonically decreasing); i.e. the slope of the line does not change sign. A straight line is a monotonic function. The height of the projectile as a function of time or distance is not monotonic; it increases and then decreases (it's a parabola). The length of the throw (the distance) is not a monotonic function of launch angle; starting at ϑ=0°, the throw increases till ϑ=45° and then decreases again. The initial slope of the projectile is a monotonically increasing function of ϑ; the final slope of the projectile is a monotonically decreasing function of ϑ.

Since P.E. is a constant multiple of height (P.E. = -mgh) we expect (and we'll show below) that the relative initial slopes of the P.E.=f(x) curves (as above) will be in the same order; i.e. the initial slope of the P.E.=f(x) parabola is lowest for ϑ=30° which is less than the slope for ϑ=45° which is less than the slope for the ϑ=60° shot.

We've plotted height (or P.E.) on the y-axis, with distance on the x-axis; i.e. h=f(s). For the x-axis, we could equally well have plotted t; i.e. h=f(t). For any particular flight (i.e. ϑ fixed), there's no practical difference in the two graphs. However the two families of graphs (h=f(s), h=f(t)) is different when ϑ is varied.

Here's h=f(s) for 3 different values of ϑ (say 30°, 45°, 60°). This graph is what we'd see if we were standing to the side of the thrower. The distance starts at 0 for ϑ=0°, increases till ϑ=45° and then decreases again to 0 for ϑ=90°.

    | 6
    | 6 
 ^  |6 6 44 
 |  |6 6    4
 h  | 33     4 
    |3 63     4
     --------------
          s->

The family of parabolas turns back on itself (i.e. the horizontal distance thrown becomes shorter) at ϑ=45°.

For the same flights, the h=f(t) graphs don't fold back on themselves when ϑ=45°. Here's the h=f(t) graphs for 3 different values of ϑ (say 30°, 45°, 60°). Unlike the h=f(s) parabolas, you wouldn't ever physically see this family of curves; you'd have to record the data and then plot it on a graph. The ϑ=60° is the longest flight (in time) of the three shown, but as you see from the previous graph, it has less throw than the ϑ=45° flight.

    |      6 6
    |    6     6
 ^  |  6 44      6 
 |  | 64    4     6
 h  | 33     4     6
    |3  3     4     6 
     ----------------
            t->

From inspection of the diagrams, the variation in initial slope as a function of ϑ is smaller in the h=f(t) graph. Derive the slope of the h=f(t) graph (i.e. find the first derivative) ^[156] . Sanity check; t_dh/dt=0 = -u.sin(ϑ)/g. (This is the top of the flight.)

The initial slope is the slope at t = 0 and has the value

dh/dt_t=0 = u.sin(ϑ)

Here are the distance and time formulae as a function of ϑ

Here's these two sin functions plotted (y-axis has s = distance thrown, t = time of flight)

    |    ss 
 ^  |  s    s 
 |  | s      s
 s  |     
    |s        s 
     ----------
     0  theta 90

    |         t
 ^  |     t
 |  |   t
 t  | t
    |t 
     ----------
     0  theta 90

In the s=f(ϑ) graph, there are two values of ϑ for any s. To show the two valued property of the flight, here's another way of presenting it, this time as h=f(s).

We've previously found height = f(t), as an intermediate step to finding P.E.= f(t), but we didn't explicitely note the equation. (re)Derive h = f(t) here ^[157] .

Instead of the height as a function of time, we now want the slope of the flight as a function of time. We need calculus for this, so let's do some calculus. We'll do some noodling around as well. Find the derivative of height with respect to time, i.e. dh/dt. If you know calculus, do it yourself, otherwise here's the answer ^[158] . Check dimensions (what should they be ^[159] ?). You've seen the time derivate of height before, when you looked for u_vert (they're the same thing). Then you derived the result from v = u + at. (This formula itself is the time differential of s = ut + ½at² ).

At t = 0 what is the slope of the height ^[160] ? The initial slope is the initial u_vert.

The derivative (dh/dt) is 0 at the top of the flight (why is this true ^[161] ?) We should get a familiar expression by equating dh/dt = 0. What is it ^[162] ? We've (re)found the time required to reach the top of the arc (which you'll remember is half the time of the flight).

Let's find the formula for the slope of the P.E.,x (P.E, distance) parabola. First let's find the slope of the h,x (height, distance) parabola. This parabola has an initial angle of ϑ, so by definition slope=tan(ϑ).

The formula for the P.E. parabola is P.E. = -mg(ut.sin(ϑ) + ½gt²)

We've already found, for the flight that throws the projectile the longest horizontal distance (ϑ=45°), that at the top of the flight P.E./K.E. = 1. This result is for one point only in one particular trajectory (the longest horizontal distance). On finding such a result, mathematicians like to see if they can generalise the result (to expand the radius of understanding the problem); e.g. is there a simple formula for P.E./K.E. for any point in the flight for any launch angle. It's possible that the formulas will be nasty in the general case, in which case you can always throw them at a computer to get numerical results. If you get one of these nasty formulas, the you back off and look for subsets of the problem for which there are simple answers; you might find several of those (or only one). Let's see what we can find. Write down the formula for P.E./K.E. for any ϑ,u,t. ^[163] .

This formulae for P.E./K.E. is a little unwieldy. Remember that we already know that P.E.=K.E at the top of the arc when ϑ=45°. We're looking (hoping) for something equally simple here. You could try asking a simpler question. Notice that K.E. is dependant on three variables u, ϑ, t. You should look for (simpler) answers that depend on a smaller number of variables. Remember that there's really only one parabola (after translating/scaling the axes). (Similarly there's only one straight line, one circle...). The route to a simpler question/answer isn't obvious at this stage, and may not be until you get to the end by brute force. So let's perservere and see what happens.

Here's the summary of what we've found out about the 2-D flight of a ball. (remember g is -ve).

General:

equations as a function of time

equations as a function of height

Equations as a function of velocity

Equations as a function of distance

	Note
	these are just examples of things you can calculate - you don't need to remember them. When you need them, you calculate them on the spot from these more fundamental equations, which you should remember (or do enough examples that you remember them). v = u + at s = ut + ½at2 v2 = u2 + 2as

FIXME - do length of arc integral to find length of parabola. As check do length of arc integral of straight line, semicircle

	Note
	I had fun here. This isn't central to what I'm teaching. You can read this in your own time if you like and ask questions later (see classical mechanics http://en.wikipedia.org/wiki/Classical_mechanics).

We've found that mgh=m.(l/t²).l and mv² = m.(l/t)² have the dimensions of energy. What about m.l²/t² and m.(l²/t).t^-1 both of which have the dimension of E?

I'm just playing games here (but that's where discoveries are made). Let's see if we can think of anything physical for these new quantities.

One of the laws of physics is that if something can exist, it must exist. You may not be able to guess what it looks like, or you may not know where to look for it, but that doesn't mean it doesn't exist. In this case of looking for new quantities, only the rate of change of mass (m/t) isn't in use (hasn't been discovered). If these quantities aren't known yet, you can guarantee that one day someone will find them.

Just as it takes energy to lift a hiker (and pack) or elevator against the force of gravity, energy is released (and can be used) when massive objects fall with the force of gravity. An example is hydroelectric power; falling water pushes the blades of a turbine and produces electricity. In earlier times a small dam in a river was used to power a mill to grind flour.

The tallest hydroelectric dam in the world is the 300m Nurek dam at Vakhsh, Tajikistan (http://en.wikipedia.org/wiki/Nurek_Dam) (I couldn't find a good photo.) (here's a list of high dams http://www.infoplease.com/ipa/A0113468.html).

Let's calculate the amount of work that water can do in a hydropower station (assume turbines and generators efficiently turn this energy into electricity - they do a pretty good job, I think it's >90%).

First let's calculate the amount of energy that a drop of water of mass m picks up when it free falls through a height h (let's assume the water doesn't meet any air resistance).

	Note
	The height of water, in day-to-day running of the dam, is called the head. For a physics problem, we'll use the generic term "height".

---------------------------water level
|
|
|
|

h

|
|                       .
|                       .      
|                       .      
                        o drop of falling water 

                      __x__ turbine blades     
                        x

The force acting on the mass is

F = ma	
the acceleration of gravity is g, so

F = mg

The amount of work done to accelerate the water drop is F*s

W = F*s
  = mgh

	Note
	This amount of work is seen in many problems; the amount of energy that a mass gains when falling in a gravitational field is mgh. It's easy to derive so you don't neccessarily have to remember it, but don't be surprised when you keep seeing it.

How much power do we get? Assume that mass m passes through the turbine every t seconds, then the power available from the dam is

P = W/t
  = mgh/t watts

If we want more power we can increase the mass flow (m/t) or the height of the dam h. Since in some places, the amount of water is finite, we can get more power from the same amount of water just by building a higher dam. The higher dam isn't always free. Sometimes the land around is too flat to have a deeper/higher dam; sometimes the extra weight of water causes seismic problems (earthquakes); building a higher wall is expensive and requires a good understanding of the geology near the wall. Still if you can double the height of the wall, you double the amount of power you can get from the same mass of water. (Dam operators like to keep the dams as near full as possible for this reason. This requirement conflicts with another purpose of dams; flood mitigation, where the dam level is required to be low enough to absorb any conceivable flood.)

In hydrodams, water isn't free falling at high speed into a turbine; it's waiting almost stationary at the bottom of the dam under enormous pressure at the turbine intake. Let's see how much work we can get out of a mass m of water, at the bottom of a dam of height h. Let's say the mass of water occupies a rectangular box of horizontal area A and height dh.

---------------------------water level
|
|
|
|

h

|
|
|                        ___
|                      / A  /| dh 
|                     /____/ /    
                      |____|/     

                       __x__ turbine blades     
                         x

The force acting on the area A is the (mass of water above it)*g. We know the volume of water above the box (it's A*h). To get the mass, we multiply volume*density. Density is usually represented by the symbol ρ (pronounced "rho" with units mass/volume). What is the dimensions of ρ ^[164] ?

	Note
	The density of water, by original definition (the definition of the gram was the mass of 1cc of water), is 1gm/cm3, 1kg/l, 1tonne/m3.

What is the force acting on the area A?

F = Mg  ; M = mass above rectangular volume

calculating M 
M = h.A.rho

thus
F = h.A.rho.g 

The mass of water in our rectangular box is m=vol*density=dh.A.ρ. Let's say our mass m moves a distance dh without encountering any backpressure (i.e. the only force acting on it is the pushing force described above) by moving a distance dh.

	Note
	in the falling drop calculation we put a similar restraint on the problem, that there be no backwards force on the drop as it fell.

How much work is done by the volume of water as it moves through the distance dh.

W = F*s
  = h.A.rho.g.dh
  = (A.rho.dh).g.h

the mass of the rectangular volume is m=A.rho.dh

thus

W = mgh 

This work is only done when the water moves from the pressure at the bottom of the dam at the intake of the turbine, to a region of no pressure - i.e. at the turbine outlet. The water sitting a short distance back from the intakes, with equal pressure on all sides, isn't capable of doing any work at all (and has no reason to move in any direction).

In reality the water doesn't move from high to low pressure in the distance dh. It moves through the turbine in a descreasing gradient of pressure, releasing a bit of energy at a time, till it reaches the outlet at 0 pressure, when it can do no more work. The total amount of work is determined by the difference in pressure from turbine intake to outlet.

We've found that no matter whether we let gravity accelerate the water by falling the height of the dam, or let gravity push on the water at the bottom of the dam through the mass of water above it, we get the same amount of energy from the water. It turns out that energy is energy and it doesn't matter which way we use it (to first convert to motion or to pressure), it's still the same amount of energy. If this were not true, then it would be possible to construct a perpetual motion machine, where two sets of turbines, one operating through water pressure and another set, operating through water velocity were operated back-to-back, with the difference in energy being available for work.

Let's say we want to boil a cup (250ml≅250g) of coffee in a microwave oven using hydroelectric power. How much water has to pass through the turbines to do this? To simplify the problem, assume there is no loss in the turbines, transmission lines or microwave oven (a real microwave oven is about 60% efficient). Assume that our coffee water starts at 0° (in reality we'd start at 20-25°). Let's get our power from Nurek.

Energy needed to heat 250ml water from 0-100deg
Work needed to heat water for coffee

W = m.C.dT
W = 250*4.2*100
  = 105kJ

Work available from mass m of water at Nurek 

W = mgh
  = m*9.8*300J
  = 2.94m kJ

equating these two quantities

m = 105/2.94 kg
  = 35.7kg
  = 142 cups (of 250g)

Thus using power from Nurek, to boil our cup of coffee, we'd have to let 142 cups of water flow through the turbines.

Lets say instead we used power from the 233m Hoover dam at Boulder Az. (http://en.wikipedia.org/wiki/Hoover_Dam). How much water would need to be used to boil our cup of coffee now ^[165] ?

How about using hydropower from the 168m Grand Coulee Dam on the mighty Columbia R (http://en.wikipedia.org/wiki/Grand_Coulee_Dam) ^[166] ?

Our local park Eno River (http://www.enoriver.org/eno/Shop/Journals/MillJournal/Geos.htm) has a dam once used for a mill (the water wheel is still there). I can't find the height of the dam, but let's say it's 10m (tops). How much water would have to flow through the Eno River mill to boil my cup of coffee ^[167] ?

Even in the best of circumstances (no power loss), and getting our power from Nurek, the highest dam in the world, we need 142 cups of water going through the turbines to boil a cup of water. How much is 142 cups? My kitchen faucet (tap), running full blast, fills my coffee cup in 3 secs. It takes 2mins to boil my coffee. During that time, even if I left the kitchen tap running full blast, for the time I was using the microwave oven, I would only be letting 40 cups of water go down the drain.

The US has plenty of water (maybe), but most of the world doesn't have enough water to turn on faucets full blast everytime you use electricity. Most hydroelectric dams in the US are quite low (50-100m?)

	Note
	looking up hydrodams in NC with google, I find information about the area of the lake behind the dam, but rarely is the height of the dam listed. The Apalachia Dam http://en.wikipedia.org/wiki/Apalachia_Dam is 46m high. The Idol Dam on the Yadkin R (http://www.digitalforsyth.org/photos/818) and Idol Dam (http://www.digitalforsyth.org/photos/836) looks to be no higher than 5m.

The Hoover dam was a technological marvel of its era (I remember being impressed with it in my childhood, and took the first opportunity on moving to the US to visit it). It sits astride the mighty Colorado River, which appears as a mountain of water as it flows through the Grand Canyon. However the demands on the Colorado are such that none of it reaches the ocean anymore - it's all been diverted for agriculture. The Hoover Dam was planned in an era of unprecedented rainfall (although the builders, without historical records didn't know this). The amount of water available to be flushed to boil cups of coffee is much less than anyone planned. It's possible that the lake behind the Hoover Dam will be empty in our lifetimes. Still there's lots of water in the Columbia R for the Grand Coulee dam, isn't there? Well sure, but no anadromous fish (http://en.wikipedia.org/wiki/Anadromous#Classification) live in the river anymore (they can't get past the dam to spawn) and the way of life of the native americans went with it. Putting high dams and large lakes in the middle of a river is an environmental and habitat catastrophy (see Environmental_and_cultural_consequences http://en.wikipedia.org/wiki/Grand_Coulee_Dam#Environmental_and_cultural_consequences) that might have been acceptable 100yrs ago, when the world's population was much less, but it isn't anymore.

Hydropower has worked well for Norway, which has lots of snow, and deep valleys to produce the head (height) needed to generate the power. Hydropower would be fine if it was just a matter of sticking a turbine at the outlet of a dam that was already needed for something else, but the environmental costs of dams are such that now there are many successful projects that have removed dams and allowed the fish to return to their habitat.

What height would the 250ml of water in my coffee cup have to fall to have its temperature raised 100° on impact ^[168] ?

A little side trip as to why public transport is important.

In 1977 Caesare Marchetti published an analysis of the commute times from the Neolithic to the present (FIXME). He found that humans are prepared to commute upto 1.5hrs. As modes of transport have changed (foot, horse, car) only the size of the conurbanisations have changed not the commute time. Thus the fallacy that this generation's transport gets you to work quicker than the last generation's transport: no - you wind up living further from work in a bigger city.

In america, there are people who make their living making/selling cars. With a car you have "freedom" they say.

FIXME You need 50-100yrs and stable zoning laws to redesign/build a city around a train line (cf Perth).

Example: The TGV does 500kmph and uses power at a rate of 12MW. Assuming that all tractive power goes to overcome aerodynamic resistance (a reasonable assumption), what is the aerodynamic force resisting the TGV ^[169] ? What mass, under the attraction of gravity, would need that amount of force to resist it's movement ^[170] ? The nose of the TGV must be able to withstand the weight of 6 US cars.

Heat is energy; it has the same dimensions as energy (this took a while to understand too). Heat can be used to lift objects against gravity. The energy in objects falling from a height is converted to heat when they impact.

Temperature is a familiar concept and has been so for many millenia. In winter it's cold and water turns to ice. In summer it's hot. We all know that when you add heat to an object that it becomes hotter and when you remove heat it becomes cooler. When objects of different temperature are in contact, heat flows from the hotter to the colder object, till they are at the same temperature. There is a strong connection between heat and temperature. We all have a good idea and agree upon temperature.

But what is temperature? Temperature is a different beast altogether from heat. In SI (but not cgs, or mks), temperature has its own dimension T (I don't understand enough about temperature to know why it has its own dimension). It took till the early 1900s for anyone to understand temperature. (They were in the same state, we are now, about gravity.)

In the meantime people had invented thermometers, which gave consistent results and which indicated something (whatever it was) that agreed with what you felt as temperature by sticking your finger in it. You could produce a temperature scale (Fahrenheit, Celsius) allowing people to brew beer and control steam in boilers (without them blowing up) and until something better came along, everyone agreed that a thermometer measured temperature.

thermometer:

a device with a property that changes with temperature e.g. a volume of liquid that expands with temperature along a bore of constant cross section; a piece of metal whose resistance changes with temperature; a bimetallic strip (http://en.wikipedia.org/wiki/Bimetallic_strip).

FIXME Various temperature scales are in use

• Centigrade/Celsius: defined by the temperature of liquid and solid water in equilibrium at atmospheric pressure=0°C and the temperature of gaseous and liquid water in equilibrium at atmospheric pressure=100°C
• Kelvin;

I'm not going to tell you what temperature is just yet. First let's look at the things people discovered on the way to figuring out what temperature is.

Materials differ in the the temperature rise produced by a fixed amount of heat. Each type of material has a different heat capacity, called the specific heat capacity, or specific heat (for short). "specific" because it's for a fixed amount of material (usually a gram or a mole). If you have a large blob of the material, the heat capacity is the specific heat * mass.

The symbol for specific heat is C (from capacity? from calorific value?). The Specific heat (http://en.wikipedia.org/wiki/Specific_heat_capacity) of water is 4.2J/g.K.

(I'm just giving you practice in dimensions. There's nothing particular here.) What is the dimension of specific heat in the S.I. system ^[171] ?

In the cgs system, the unit of heat is the calorie, being the amount of heat needed to raise 1g of water by 1°C. The calorie is a useful unit, giving water an easy to remember heat capacity of 1.0. It's most unfortunate that nature didn't cooperate giving a heat capacity of 1.0 in the self consistent SI scheme. The unit of energy used for food in the US is the Calorie being the amount of heat that will raise the temperature of 1kg of water by 1°C (i.e. Calorie = 1000 calories). Since most people don't realise that case matters here, for food labelling, the "Calorie" is usually spelt "calorie". If car manufacturers described their cars as having a 3cc engine, when it had a 3l engine, they would be mocked. However when you do the same thing with food, no-one notices.

	Note
	The rest of the world uses Joules to describe energy, including energy in food.

Example: a 50kg person ingests 2000 Calories a day. If the energy is just used as heat (which it mostly is), and the person did not loss any heat (to the air), what temperature should their body increase in a day assuming their body is mostly water (which it is) ^[172] ? With your basal temperature at 37° your blood would boil in about 1.5days if you were completely insulated.

One of the ways of loosing heat is to evaporate water, either by sweating or by evaporation of water from your lungs. These mechanisms are only effective at elevated temperatures. To turn water into water vapor, the water needs to absorb 540calories/gram (the latent heat of vaporisation of water). How much water do you need to loose/evaporate to maintain body temperature (assuming there are no other mechanisms for loosing heat) ^[173] ? This is about as much water as a 50kg person could loose without having problems from dehydration. If it was 37°C and you couldn't offload 2000Calories of heat into the air, you would have to evaporate 3.7kg of water/day to maintain temperature.

Look at the specific heat of a few substances. Notice that water has one of the highest specific heats, measured in J/(g-K). If you want to raise the temperature of a pan of water by 100° you need more heat than if you were doing the same thing to a pan of aluminium.

In the 1700s and 1800s industrialists wanted to know the efficiency of their steam engines (and later electric generators and motors). How much work should they expect from the heat produced by burning coal? Until they knew the conversion between heat and work, they couldn't tell.

Earlier work by Count Rumford (1798) boring canons (http://en.wikipedia.org/wiki/An_Experimental_Enquiry_Concerning_the_Source_of_the_Heat_which_is_Excited_by_Friction)

Rumford had observed the frictional heat generated by boring cannon at the arsenal in Munich. Rumford immersed a cannon barrel in water and arranged for a specially blunted boring tool. He showed that the water could be boiled within roughly two and a half hours and that the supply of frictional heat was seemingly inexhaustible. Rumford confirmed that no physical change had taken place in the material of the cannon by comparing the specific heats of the material machined away and that remaining were the same.

Rumford argued that the seemingly indefinite generation of heat was incompatible with the caloric theory. He contended that the only thing communicated to the barrel was motion.

Rumford made no attempt to further quantify the heat generated or to measure the mechanical equivalent of heat.

	Note
	The canon boring operation was powered by horses.

Joule (http://en.wikipedia.org/wiki/James_Prescott_Joule) was an educated brewery manager from a wealthy family. He was regarded by the Royal Society (the premier scientific body in England at the time) as a provincial dillettante. The prevailing theory at the time was that heat was a property of a fluid caloric, that moved from one object to another as they changed temperature. In an exacting set of experiments (financed by and using the physical facilities of the brewery), Joule determined in multiple independant ways (chemical, mechanical, electrical, hydraulic), the conversion factor (the mechanical equivalent of heat), showing that heat was energy (and hence not caloric). His paper in 1843, read before British Association for the Advancement of Science in Cork, was greeted with silence. As it is with scientists, new ideas are not well accepted and it often takes the death of the old generation before new ideas become widespread.

Although the calorie (the amount of work needed to raise 1g of water by 1°) and the Calorie (the amount of work needed to raise 1kg of water by 1°) are no longer used (except in some holdout countries, to measure the amount of energy in food), the conversion factor is 4.1860 J.cal^-1.

Note

What is the dimension(s) of this conversion factor, the Joule constant? The factor converts between two different units of work; the joule and the calorie. It is a number; it has no dimensions. Although it is dimensionless, it does have units J.cal-1. (Note the similarities with the dimensionless factor, which converts between metres and feet = 3.2809ft/m, or the number which converts dozens to units.)

As we all know, the main benefit of travel is self education and broadening your viewpoint on the different cultures in the world. If you read the experts on travel (the travel section of magazines and newspapers), you'll find that the best education comes from eating at the destination's most expensive restaurants. Joule, on his honeymoon, didn't have the benefit of the New York Times travel section and

marital enthusiasm notwithstanding, Joule and Thomson (later Lord Kelvin) arranged to attempt an experiment a few days later to measure the temperature difference between the top and bottom of the

• Cascade de Sallanches waterfall (http://www.flickr.com/photos/sharman/253526381/)
• Cascade de Sallanches waterfall (http://www.flickr.com/photos/louistib/3567924954/)

though this subsequently proved impractical. The height of the waterfall (http://www.world-waterfalls.com/waterfall.php?num=202) is 365m top to bottom, with the largest single drop of 270m.

Looking at the photos of the Cascade de Sallanches waterfall, what sources of errors will confound an attempt to measure the increase in temperature of water falling through the waterfall. (hint: what factors, other than falling through a gravitational field, will affect the temperature of the water) ^[174] ?

We recognise Joule's attempt to measure the increase in temperature accompanying the fall of massive objects in gravity, as the same problem as calculating the energy used by a hiking climbing a mountain, a programmer running stairs with a pack on his back, lifting an elevator against gravity, or in the energy released by water falling through gravity in a hydroelectric plant (next section).

We've seen examples of the conversion of potential and kinetic energy into heat for events in our experience. Now let's recreate Joule's experiment at Cascade de Sallanches in your own home.

Prerequisites: piece of pvc pipe with end caps, thermometer, weighing scales and massive object (small bag of rice).

FIXME experimental details

The Barringer Meteor Crater (http://en.wikipedia.org/wiki/Meteor_Crater), was caused by the impact of the Canyon Diablo meteorite about 50kyA. The 50m diameter nickel-iron meteorite, weighing 150ktonnes, hit the ground at about 12.8km/sec.

	Note
	I don't know how this is determined, but it will be the starting point for our calculations.

	Note
	The area at the time was cooler and wetter than now, being grassland and open woodland. How likely was it that the impact was witnessed by humans [175] ?

100yrs ago, geologists thought that the crater was a steam vent from a volcano. Barringer thought that the crater was from a meteor and went bankrupt trying to find and excavate it for it's iron content. The origin of the crater was settled about 1960 by Gene Shoemaker who produced enough evidence to convince people of the impact, making the Barringer crater the first identified impact site on earth.

At the time (1960's) the mountains of the Moon were also thought (by many) to be volcanic in origin, despite the characteristic central peak in the craters (see impact craters http://en.wikipedia.org/wiki/Impact_crater). Here's a photo of Tycho (http://www.lpi.usra.edu/education/timeline/gallery/slide_61.html), a relatively young (160Mya) impact site, easily visible from earth with binoculars. Rays of ejecta from Tycho can be followed acros most of the visible side of the the Moon. (from Impact Crater Geology http://www.lpi.usra.edu/expmoon/orbiter/orbiter-craters.html)

The crater Tycho, 85 kilometers in diameter, is the youngest large impact crater on the Moon's nearside. Ejecta from this crater was spread across much of the nearside of the Moon and is visible in the form of bright rays at full Moon. One such ray crosses the Apollo 17 landing site, 2000 kilometers from Tycho. Arrival of this material from Tycho is believed to have triggered a landslide from the mountains surrounding the Apollo 17 landing site. Laboratory analysis of samples from this landslide suggest that Tycho's age is about 100 million years.

	Note
	Tycho (http://www.daviddarling.info/encyclopedia/T/Tycho_crater.html) features in the film and book "2001: A Space Odyssey" as the location of TMA-1 (Tycho Magnetic Anomaly 1), which turns out to be one of the enigmatic monoliths placed in the solar system, by an alien civilization. In the story, Dr. Heywood Floyd travels from Clavius Base to see the artifact.

The Apollo program was a cold war effort to intimidate the Russians. The astronauts were military pilots and science was not on the table. Gene Shoemaker recognised that on arrival at the moon, the astronauts would be seeing rocks and not communists. A photo of astronauts saluting under a US flag would be shallow reward for the efforts to get there. Gene is to be remembered for his successful efforts to use the lunar landings (and the reluctant astronauts) to advance science. (Gene's ashes are on the moon Moon Trivia http://home.hiwaay.net/~krcool/Astro/moon/moonglossary/moontrivia.htm) The Apollo program found that the craters on the moon were impacts and not volcanoes and provided much useful information on the early history of the solar system (on earth much evidence has been erased by erosion). After beating the Russians to the moon, and there being no other reason to be in space, the lunar exploration program was discontinued, revealing the manned space program for the sham it was. The best the US can do at the moment is use the ISS to rediscover the knowledge the Russians gained with Mir (http://en.wikipedia.org/wiki/Mir).

	Note
	Most of the craters on the Moon (and by inference the terrestrial planets) were created during the late heavy bombardment (http://en.wikipedia.org/wiki/Late_Heavy_Bombardment) a short period about 4Ga, when possibly a Mars sized planet was disrupted by Jupiter's gravitational field, and the shards roamed the inner solar system, being swept up by the terrestrial planets.

Why wasn't the Canyon Diablo meteorite never found? Let's look at what happened on arrival of the meteorite.

How long did it take the meteorite to traverse the atmosphere? Well, how thick is the atmosphere? The atmosphere gets exponentially thinner with altitude and thus is infinite in extent. We could take some arbitary cutoff, like the altitude below which lies 50% (or say 90%) of the mass of the atmosphere. A less arbitary, but mostly convenient, artifice is to calculate the height of an equivalent atmosphere with constant density, temperature and pressure, the same as the real atmosphere has at ground level. The top of the equivalent atmosphere is a sharp cutoff, with no atmosphere above it.

Note

This height has a name like "height equivalent" but this name is used for many things, so presumably you have to specify it further, but I don't know how. A meteor traversing this height equivalent of atmosphere (all at a pressure of 1Atm) would encounter the same number of air molecules at it would in the real atmosphere. You would use this height if you wanted to calculate the refraction of stars near the horizon, using the refractive index of air at the ground.

Here's some more info.

What's the height equivalent of the atmosphere (hint 1 ^[176] , hint 2 ^[177] ) ^[178] ? (The height of Mt Everest is 8.8km.)

At the meteorite's impact speed, how long would it have taken to traverse the height equivalent of the atmosphere (assume that deceleration did not change its velocity appreciably, we will check this assumption below) ^[179] ?

What force was acting on the meteorite while it was traversing the atmosphere? We're going to make a quick side trip into aerodynamics and fluid mechanics (thanks to Matt at work for explaining the equation we're about to derive). We need a formula for the force acting on an irregularly shaped object falling through the atmosphere. Where are we going to get it (no, we don't need a book or wikipedia) ^[180] ? Make a list of factors that could affect the force acting on an object falling through the atmosphere. Here's mine ^[181] . Setup the simultaneous equations to determine the power to which you must raise each factor to get force. Do not solve them just yet. ^[182] .

We've got 4 unknowns, w,x,y,z but only 3 equations. We don't have enough information to uniquely solve for the 4 unknowns. We need an extra piece of information. We'll use experience (we could be wrong, but we'll find out when we test the equation by experiment). From eqn (2); y = 1. Next we tackle eqn (1). Obvious values to try are w=0, z=1 or w=1, z=0. We expect from experience that the aerodynamic braking will be dependant on the density of the fluid, so the combination w=1,z=0 is out. It's possible that w=½,z=½ or other combinations of square roots or inverse square roots. However that would result in strange values for x and z; eg the force would depend on the sqrt(A) i.e. the diameter rather than the area. Let's assume then that w=0 and see what that gets us. Here's my derivation of the force acting on the meteorite ^[183] . The formula we find for the force acting on an object falling through a fluid is

F = k.A.v^2.rho

This is the answer confirmed by experiment. In fluid dynamics this formula is written F=½.ρ.v².A.C_d where C_d is the coefficient of drag (it's our missing dimensionless constant; I don't know where the ½ comes from). Fluid dynamics is a bit of a black art and AFAIK the coefficients of drag can't be calculated simply, even for a sphere. They are measured, or calculated by supercomputers. Have a look at a few values at Drag coefficient (http://en.wikipedia.org/wiki/Drag_coefficient). Notice how streamlined (aerodynamically slippery) objects have lower coefficients of drag.

Let's assume our meteor had C_d=1.0. Calculate the aerodynamic force acting on it at impact velocity ^[184] . What's the deceleration (calculate it in g if you like) ^[185] ? This is not a particularly strong deceleration. Humans can survive this for a short time. The Apollo 11 astronauts, returning from the Moon, endured 10g when they first hit the earth's atmosphere. The deceleration wouldn't cause a rocky meteorite to break up, much less a Fe/Ni meteorite.

What velocity would the meteorite have had when it hit the atmosphere (assume deceleration was constant) ^[186] ? What is the change in velocity? It's a.t. A back of the envelope calculation gives a change in velocity of a.t=104m/sec^2*0.65sec=68m/sec, a miniscule change in speed for an object doing 12.8km/sec. (Our initial estimate of the time to traverse the atmosphere assumed that the change in velocity was small. We've confirmed our assumption, thus validating the calculations so far.)

The atmosphere has little effect on the meteorite's velocity. As well only a miniscule fraction of the meteorite's energy is lost in the atmosphere (the K.E. on entering the atmosphere - the K.E. on impact).

The size and mass of the meteorite affect how much it's slowed by the atmosphere. A large enough meteorite, e.g. the Chicxulub Meteor (http://en.wikipedia.org/wiki/Chicxulub_crater), implicated in the extinction of the dinosaurs at the end of the Cretaceous, and estimated to be 10km in diameter, wouldn't even notice the ocean and would punch straight through the ocean floor to the mantle. A 1cm meteorite burns up in the atmosphere.

Even though only a miniscule fraction of the meteorite's energy is dissipated in the atmosphere, calculate the rate at which energy is dissipated (i.e. the power generated) ^[187] . Remember this power is being generated in a ball 50m in diameter.

	Note
	Peak electrical consumption over the whole US in 2007 was 0.8Tw (peak consumption is in summer). (http://www.eia.doe.gov/cneaf/electricity/epa/epa_sum.html http://www.eia.doe.gov/cneaf/electricity/epa/epa_sum.html). You could calculate the length of time this meteorite could power the US.

How much energy was produced by the meteor passing through the atmosphere ^[188] ? Below, we'll compare this to the energy produced on impact.

The Stefan-Boltzmann law (http://en.wikipedia.org/wiki/Stefan%E2%80%93Boltzmann_law) allows you to calculate the temperature of a black body from the radiance (power/area). (I'm not going to derive this. I'm just going to give you the result.)

Stefan-Boltmann constant:
sigma = 5.67*10^-8 W.m^-2.K^-4

surface area of meteor diameter 50m = 7850m^2
power radiated = 200Tw

thus
200*10^12w = 5.67 * 10^-8 W.m^-2.K^-4 * 7850m^2 *T^4
T^4 = 200*10^12/(5.67 *10^-8 * 7850)K
T = 26,000K

This temperature gives (assuming a black body, reasonable in this case) allows you to determine the color of the maximum light intensity via Wien's displacement law (http://en.wikipedia.org/wiki/Wien%27s_displacement_law) which says λ_max=b/T.

lambda_max = b/T

where b = 2.9 *10^6 nm-K.

Wien's displacement law gives rise to the concept of color temperature. As a body gets hotter, the wavelength of the emmitted light gets shorter (it shifts from red at 750nm to blue at 390nm; UV is shorter wavelengths). An incandescent light has a temperature of 3900°K and is a red-brown color with λ_max=743nm; the sun has a temperature of 5,500°K and is yellow with λ_max=527nm;; the blue sky is blue light scattered from sunlight and if it were to come from a black body, would have a temperature of 9000°K with λ_max=322nm. You can tell the temperature (and hence radiance) of an object from its color (e.g. the temperature of red hot metal, the sun, stars).

The eye/brain has a mechanism for color balancing, so that a white surface looks white whether it's illuminated by incandescent light, the sun, fluorescent light or blue sky. Cameras (film, CCD, MOS) have an equivalent mechanism; you tell the camera the color temperature of the light, or later, after taking the photo, you can correct the image. A photo taken in a room with incandescent light looks red when taken by a camera/film which is expecting daylight illumination.

A black body at 26,000°K would have its maximum radiation at

lambda_max = b/T

b = 2.9 *10^6 nm-K
T = 26000K

lambda_max = 2.9*10^6nm-K/26*10^3K
           = 111nm

The meteorite would be emitting mostly UV light.

Here's more useful information.

With the surface of the meteorite at 26,000°, we could expect that Fe would boil off the surface of the meteorite.

	Note
	Meteorites pass too quickly through the atmosphere to heat up the insides. Only the outer 1cm shows any sign of heating.

Assume that instead of heating the atmosphere to 26,000° the energy was used to boil off Fe (the meteorite then would be at 3000° or so and you would see a dull red ball streaking through the sky).

Here's some more info.

How much Fe was boiled off ^[189] ? The energy available could have boiled off 21kTonnes of Fe. Here's the estimated loss of mass (I don't know how they know this).

The meteorite must have been loosing mass from other mechanisms as well. It's likely that the meteorite was not a single fused piece of iron. It was more likely a loose aggregate of lumps of iron.

At what rate (mass/sec) was Fe/Ni ablating from the meteorite ^[190] ?

This material wasn't just falling off the side of the meteor after being blown off by the passing breeze. The meteorite was travelling at 40times the speed of sound and was preceded by a shock wave at about 10⁵° ( The Magnetic Effect and Shock Wave of a Meteor http://adsabs.harvard.edu/full/1965SvA.....8..890I). This would have been a very bright 0.65secs.

Calculate the K.E. of the meteorite at impact ^[191] . Compare this to the energy that the meteorite lost in passing through the atmosphere - 130TJ (0.13 * 10¹⁵J). This is only about 1% of the meteorite's K.E. For impacts and explosions, energy released is usually converted into tons of TNT. The energy released by detonating TNT is 2.72MJ/kg (see strength of explosives http://en.wikipedia.org/wiki/Strength_(explosive)). How many tonnes of TNT is the Barringer Meteor Crater worth ^[192] ? Websites estimate the energy at 2-20 MTonnes. One that does a comparable calculation is Tswaing Crater S.A. (http://www.hartrao.ac.za/other/tswaing/tswaing.html)

Note

The chemical reaction of explosives is not particularly energetic compared to combustion. The useful property of explosives is that they denotate. A shock wave moving through the explosive sets it off. The result is that all the material explodes at once, rather than the material being blown apart by the explosion at the initiating site, before it can explode. Here's some heats of combustion (http://en.wikipedia.org/wiki/Heat_of_combustion) fuel heat of combustion, MJ/kg hydrogen 142 methane 56 gasoline 47 TNT 2.27 (energy on detonation)

Here's some more info

Calculate the temperature of the meteorite immediately after impact, assuming no energy is transferred to the ground (all stays in the meteorite). ^[193] . So we now know why Barringer never found the meteorite - it would have vaporised on impact.

Assuming the K.E. of impact, instead of heating up the meteorite, was transferred to excavated rock as K.E. Assume that the rock was all ejected at 45° (i.e. travelling the furthest distance for its initial velocity) (in the actual impact, the rock would get a distribution of energy, some travelling further and some travelling less distances). How far would the ejecta travel? Here's some input data.

First calculate the velocity of the ejected material ^[194] .

Note

Escape velocity (http://en.wikipedia.org/wiki/Escape_velocity) from Earth is 11.2km/sec. The velocity of the ejecta would have a wide distribution. It's reasonable to expect that some of it would be blasted into space, never to return. Escape velocity from the Moon (2.4km/sec) and Mars (5.0km/sec) is lower than Earth's and for a similar impact, more ejecta would escape from these planet's gravitational field. In fact ejecta from other planets e.g. Lunar meteorites (http://en.wikipedia.org/wiki/Lunar_meteorite) and Martian meteorites (http://en.wikipedia.org/wiki/Mars_meteorite) have been found on the Earth's surface. finding a meteorite is hard enough. Realising that it's from the Moon or Mars is beyond my understanding.

The next part is how far away did the material, ejected at 196m/sec, fall ^[195] ?