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Author Topic: Ornery U: Calculus and Mechanics (from the Physics thread)
OrneryMod
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This creates a new topic from an offshoot on another thread.

Originally stated by the Drake:

There's little point to describing mechanics without the use of Calculus.

Until you realize that

v=at

and

d = v0 + 1/2 at^2

for a good reason, basic dynamic motion will elude you.

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canadian
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As does Calculus! Is this 101?
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The Drake
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I'm happy to respond in this thread to any questions that people might have about physics or calculus in general, now that it exists.

I originally took physics as a high school junior, and I had not yet taken calculus. It was memorization of a confusing jumble of equations that I found highly unsatisfying. Understanding the calculus makes it far easier, because you have fewer equations to memorize.

Also, if you look at the equations and understand them you can map what appear to be very different problems into the same equations. Like the fact that springs and shocks are very much like capacitors and resistors, and the equations are built the same way. One set of equations describes mechanical energy, the other electrical energy.


Even the basic algebra is quite fun.


Using the equations

Velocity = Original Velocity + Acceleration * Time

or

V = V0 + AT

If you walk off the top of a building, you'll accelerate at 9.8 m/s^2 (meters per second per second)

But how long will it take you to *SPLAT*?

D = D0 + T*V0 + 1/2AT^2

D0 is the height that you start at.
V0 is the velocity that you start at.

D0 would be the height of the building, while V0 would be *zero* because even though you are walking off the building, we are only worried about your downward velocity.

So now, you solve for T, assuming a 100m building.

100 = 1/2*A*T*T

Gravity is 9.8 m/s/s (ignoring drag)

100 / 4.9 = T^2

T = 4.5 seconds


Not bad, and it can be a lot of fun to just plug in different numbers and see how it turns out.

What if you jump UP off the building?

V0 = -0.5 m/s

What if you have a piece of cardboard?

A = 9.8 m/s/s - 2.1 m/s/s (depending on how much cardboard - thanks Mythbusters!)

The reason that I make the claim that physics is pointless without mathematics, is that the results are often quite counter-intuitive.

For instance, how long does it take to fall from a building twice as high as our first one if you don't hit terminal velocity?

original building
100 / 4.9 = T^2 ++> 4.5 seconds

taller building
200 / 4.9 = T^2 ++> 6.4 seconds

So, it takes you 4.5 seconds to fall the first 100m, but only 1.9 seconds to fall the second!

The equation involves a square root, which aren't intuitive for anyone I've met. You must master the mathematics to find out everything that is cool about our universe.

Then you start layering in things like wind resistance and terminal velocity. It just gets better and better.

It is tremendously satisfying to run the equations to predict physical behaviour, then drop something from your apartment window and see it match.

That's why I love science and math. You can't have one without the other.

[ February 21, 2008, 06:00 PM: Message edited by: The Drake ]

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canadian
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erg...this is already too advanced for my math neglected mind.
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Everard
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*nod*
I agree with drake that the math can make the science more elegant, and the connections more obvious.

Where I disagree is that understanding the world requires math. The concepts of physics can be understood, and applied, by everyone, opening up a better understanding of the world.

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martel
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I sort of agree with Ev, and also think that mathematical physics with algebra can be valuable, although with calculus it is obviously more accurate.

I took physics without calculus last year as a high school junior, and this year I'm taking AP physics, which is basically as much calculus as you can do with only one variable (we're in our first year of calc.) What it looks like to me is that you can construct a facsimilie, if you will, of the natural world and its rules using no math, a more accurate one using algebra, a still more accurate one using single-variable calculus, and probably more accurate with more advanced calculus.
Take projectile motion--in my first year, we learned the basic equations involving acceleration, time, and position. This year, we can describe an object with variable acceleration, which you can't do without calculus.
That's just a basic example--doing electricity and magnetism, I see it even more--for example, electric fields are almost useless unless you know calculus.

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DonaldD
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Not only that, but you don't even need math to understand the basic concepts of calculus...

On a side note, shouldn't this be "Calculus and Dynamics"?

oops, I see you've graduated to dynamics on the other thread and are moving on from there. I'll just shut up now

[ February 21, 2008, 07:54 PM: Message edited by: DonaldD ]

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The Drake
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It depends on the completeness of the understanding, naturally. We all know gravity pulls you down at an early age.

I'm obviously far behind real physicists, who might tell me that I can't really understand the world without knowing quantum equations.

I guess my point is that I think there would be very little value to gain from a non-numerical statics and dynamics.

Sure, you could learn to understand why hanging a heavy object at the end of a horizontal pole deforms the pole. But you couldn't know how heavy an object you could hang, nor the deflection to expect.

It seems ultimately frustrating, like watching a foreign film without subtitles.

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Rallan
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Graphs! We need graphs! It's all well and good to let the good people of Ornery know that displacement, velocity, and acceleration can all be derived from one another with calculus, but they're gonna be struggling without some sort of visual aid. Dunno about everyone else, but "area under the curve" was the magical phrase that unlocked the basics of calculus for me, complete with wildly out of scale graphs on the blackboard.

EDIT: Oh and to answer canadian's question, yes, this will be Calculus 101. Not because we have to dumb it down, but because the relationship between acceleration, velocity, and displacement is based on the absolute basic fundamentals of calculus. There are no simpler integrals and differentials you can do in the entire subject than the stuff you'll be shown here.

(unless of course someone decides to explain the mathematical side of what happens when things go round and round in circles, in which case you may want to run screaming for the hills)

[ February 22, 2008, 09:26 AM: Message edited by: Rallan ]

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The Drake
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Yes, calculus is rather difficult in ascii, you can't even write a decent derivative. I'm wondering whether people would find more value from an Algebra 101.

Or, we can post links to well-written calculus primers on the net, and discuss them here.

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Rallan
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And one of us might need to throw in a Geometry 101 sooner or later, unless we want all the physics examples to be bodies moving backwards and forwards along a straight line (and we don't want to keep things that simple, because then we'll never be able to knock things over [Smile] ).
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RickyB
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I tend to agree with Drake. You need, at the very least, to be willing to deal with equations in order to be able to deal with physics. You can understnad principles without it, but not particulars. I'm saying this as a total liberal arts kinda guy. [Smile]
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Psudo
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Here's the answer to your web-mathematical prayers: http://www1.chapman.edu/~jipsen/mathml/asciimathdemo.html

That little gadget lets you type equations in ASCII and see them as they should be seen. Just start and end the equation with backticks (the key left of number 1). It works automagically in Firefox/Mozilla type browsers. There's a plugin you need for at least some versions of Internet Explorer, which is linked from the page.

Some common equation for use as examples:
`E=mc^2`
`e^(pii)-1=0`
`(-b+-root(2)(b^2-4ac))/(2a)`
`sin(x)/cos(x)=tan(x)`
`sum_(i=1)^n i^3=((n(n+1))/2)^2`

Hope that helps ya out.

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scifibum
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One thing I never quite grokked was that kinetic energy and momentum seem to be on different scales.

Kinetic energy is proportional to the square of velocity:
E = .5(m(v*v))

Momentum is linear with velocity:

P = mv

So, doubling velocity doubles momentum, but quadruples kinetic energy. I don't think I understand how/why that's the case.

Teach me!

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RickyB
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"Where I disagree is that understanding the world requires math. The concepts of physics can be understood, and applied, by everyone, opening up a better understanding of the world."

Perhaps. I don't presume to state for fact how the world makes sense to others, especially since I have overwhelming evidence that be it as it may, it is decidedly different than the way I do it.

That said, a lot of the limited extent I DO understand the physical world, would be impossible to me without understanding basic equations. Even more... spiritual concepts I arrived at flowed more easily - or at all - into place because they made equation sense.

That's just me.

scifi - compound interest? [Smile]

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scifibum
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Vague financial analogy doesn't quite close the gap for me Ricky, but thanks. [Wink]
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IrishTD
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Scifi - I'm awful at mechanics, but if you look at your equation for KE, you'll see a momentum term buried in it. Without doing some reading, I'm not going to be able to help much beyond that [Smile]

E = .5(m(v*v)) = .5(P*v)

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Everard
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*nod*
Energy is a different beast then momentum. Momentum is inertia in motion. When we're talking about momentum we're talking about "what force do I need to apply in order to bring this beast to a stop, applied over how long a period of time?" i.E. Ft=delta mv.<---Impulse=change in momentum, impulse momentum theorem. Another way of looking at this is through newtons second law- Force = dp/dt is another way of saying F=ma.

Kinetic Energy, on the other hand, is "How much work could we do on something with this?" The work-energy theorem says "The change in kinetic energy is equal to the work done," in other words, if I take an object, smash it into another object, it will do some work. Work is the dot product of force and displacement. Force shows up again, but this time, instead of multiplying by time, we're multiplying by distance.

If you take objects and smash them together, the change in the kinetic energy of one object will eventually get you the distance that the object moved while in contact with the other. The change in momentum will get you the time they were in contact.

F=ma. F=mdv/dt. or, Fdt=mdv. Thats impulse momentum.

Define work as Fx.
Fx
F=mdv/dt.
x=vdt
mdv/dt times vdt=mvdv=del 1/2mv^2
Define that as kinetic energy, and you have
Work=Change in kinetic energy.

[ March 17, 2009, 04:18 PM: Message edited by: Everard ]

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scifibum
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Thanks for that explanation, Everard. I'll chew on it and see if I can internalize it.
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RickyB
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I think it's the same problem with grasping E=MC^2. The potential energy of objects is out of limited-rational whack with their mass. Therefore, the kinetic energy is greater than the mere momentum.
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Everard
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I'm not... entirely sure what you just said ricky, but I think you're getting stuff crossed up.

Kinetic energy can be less than the momentum, equal to the momentum, greater than the momentum. It depends on the situation. But the relationship doesn't have anything to do with the potential energy of the object.

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RickyB
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OK, sorry. [Smile]

So what does it depend on?

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Everard
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Well, as noted, momentum is mv, and kinetic energy is 1/2mv^2. If you set those equal to each other, mv=1/2mv^2. do some canceling, and you get 1=1/2v. In other words, when the velocity of a given object is 2 m/s, its KE and momentum will be equal to each other. (In MKS units). If v is >2, kinetic energy is larger. if v<2, momentum is larger.
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RickyB
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Hmm. Thanks.
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vulture
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(Just noticed the thread)

But whether the values are numerically greater than each other or not depends on the units used only - there is nothing physically significant at all about whether KE > momentum or vice versa. (And you can always find a reference frame to make the ratio between the tow whatever you like; physical behaviour is however independent of reference frames).

Back on scifi's original question, there isn't really an answer per se.

You could say (correctly) that the change in momentum is force times time while the change in energy is force times distance, and that that brings in an extra factor of velocity to the energy. But that doesn't really address the question of why delta_p = F * t and delta_E = F * x.

The only answer I can give, which is still a non-answer, is that the quantities of momentum and energy, thus defined, happen to be conserved quantities. Thanks to Noether's theorem, you can in fact derive the fact that they are conserved from symmetries of physics.

Spatial invariance (physics doesn't depend on your absolute position - hence there is no absolute position) leads to conservation of momentum (force x time).

Time invariance (physics doesn't depend on the absolute time, which also doesn't exists) leads to conservation of energy (force x distance).

I admit that I have a tendency to think of momentum as being 'real' (a real mass m moving at a real speed v - it is a physical phenomenum), and energy as a 'made up' book keeping device that is useful for making calculations. But I couldn't really defend that point of view [Smile]

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scifibum
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"You could say (correctly) that the change in momentum is force times time while the change in energy is force times distance, and that that brings in an extra factor of velocity to the energy."

I like this. This is something I can just about wrap my head around. I still can't grok the math of it - not like I grok 2 + 2 - but "force times distance brings in an extra factor of velocity" serves nicely to reassure me. [Smile]

Also, thank you for validating my discomfort with the relationship between momentum and kinetic energy. [Big Grin]

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The Drake
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Um, how are you setting values equal to each other when they have different units, ev?

2 joules is not equal to 2 kg m/s.

I stumbled a few times in working through this, but then found a nice reference so I didn't have to.

quote:
name David B.
status educator
age 50s

Question - In the kinetic energy formula, why do we divide by 2? My
6th graders want to know and I'm not really certain.s
----------------
David B.,
When we push on something in the direction of motion, we make it speed up.
This adds energy, kinetic energy. The force we push with, times the
distance we push, is called the work we do. Kinetic energy is set up so
that this work equals the change in kinetic energy. It turns out that the
one-half in the kinetic energy formula is necessary to make this happen.

When doing motion, there are four major constant-acceleration formulas. One
that we see is v^2-v0^2=2ax. Remember, x is displacement. Multiply
everything by mass: mv^2-mv0^2=2max. The net force on an object equals
mass times acceleration: mv^2-mv0^2=2Fx. For constant acceleration work is
W=Fx. Also divide by two: (1/2)mv^2-(1/2)mv0^2=W.

Ken Mellendorf
Math, Science, Engineering
Illinois Central College
=====================================================
Because otherwise energy would not be conserved!

To do the derivation quickly and roughly, starting with a mass m at rest
with a constant force F acting on it. Then F = ma, or, if we multiply by x,
Fx = max. Fx is just what is defined as the work done on the mass m by the
force F acting over a distance x. The velocity of the mass after this
acceleration is given by v^2 = 2ax. Remember the kinematics?

Then work = W = Fx = m v^2/2. And now it is clear that that pesky factor of
2 comes from the kinematics. You can get v^2 = 2ax from x = at^2/2 and
v=at, which would be a good exercise to derive.

Best, Dick Plano, Professor of Physics emeritus, Rutgers U
=====================================================


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