Physics Q&A Game: Exploring Faraday's Law with Marcus' Polar Route Question

In summary, a physics Q&A game was suggested in a forum, similar to the Astronomy Q&A game. One question was posed about the voltage difference between the wingtips of a plane flying a polar route. The conversation discussed Faraday's law and the Hall effect, and eventually led to a true story about choosing between an exact answer and an approximation due to the cost of CPU time. The physical motivation for choosing the approximation was not revealed.
  • #36
pressure of the tire to the surface is force exerted per unit area. the normal force is not a point source force but one distributed evenly over the area of interaction. greater area means more friction // less slipping.
 
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  • #37
Originally posted by On Radioactive Waves
pressure of the tire to the surface is force exerted per unit area. the normal force is not a point source force but one distributed evenly over the area of interaction. greater area means more friction // less slipping.

The frictional force f is calculated as μN. Where N is the normal force [the weight of the car], and μ is the coefficient of friction. Sorry.:smile:
 
  • #38
are you talking about the static or kinetic coefficient of friction?

ffk = [mu]kN

ffs is less than or equal to [mu]sN

sorry i don't know how to do subscripts
 
  • #39
Originally posted by On Radioactive Waves
are you talking about the static or kinetic coefficient of friction?

ffk = [mu]kN

ffs is less than or equal to [mu]sN

sorry i don't know how to do subscripts

Both apply but it doesn't matter. In either case here, we only need to total weight of the car, and the applicable coefficient of friction in order to calculate the frictional forces. :smile:
 
  • #40
so the answer is not greater traction? sorry i didnt explain my answer well. peace
 
  • #41
Originally posted by On Radioactive Waves
so the answer is not greater traction? sorry i didnt explain my answer well. peace

How does the performance of the vehicle benefit from wide tires? Clearly we do get some traction advantage but why; by what physical mechanism or process? A simple application of the fritional force theory seems to yield no advantage. Peace on you too.
 
  • #42
hmm is it something to do with torque?
 
  • #43
Originally posted by Ivan Seeking
How does the performance of the vehicle benefit from wide tires? Clearly we do get some traction advantage but why; by what physical mechanism or process? A simple application of the fritional force theory seems to yield no advantage. Peace on you too.

wider tires allow one to inflate the tires to a lower pressure
and still support the car


the weight of the car is equal to the footprint area of the tires
multiplied by the air overpressure

with wider tires, it is more practical to have a large footprint area and thus a lower pressure (for the same weight of car)

intuitively lower pressure will contribute to better adherence to the road (small irregularities won't cause vibration that bounces the tire out of contact)
 
  • #44
Originally posted by marcus
intuitively lower pressure will contribute to better adherence to the road (small irregularities won't cause vibration that bounces the tire out of contact)

Can you elaborate on this direction of thought?
 
  • #45
Originally posted by marcus
(small irregularities won't cause vibration that bounces the tire out of contact)

What does this matter if the frictional force calculated is independent of surface area?
 
  • #46
Originally posted by Ivan Seeking
Can you elaborate on this direction of thought?

hi Ivan, I don't really believe the normal force is constant

I think irregularities in the road cause the wheel to bounce up and down and at times nearly leave the road

If I were designing a racing vehicle I would want low pressure in the tires so I could have a tight suspension (with little or no play). And so you could pick up the tire a ways before it would actually lose contact with the road.

i would probably want a large footprint to even out little irregularities in the road and to retain contact even with a lot of up/down motion

this is guessing, you asked me to elaborate.
 
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  • #47
Originally posted by marcus
lower pressure will contribute to better adherence to the road (small irregularities won't cause vibration that bounces the tire out of contact)

Well, I think I'll call this good enough. I was really looking for any of the following the references:

1).The coefficient of friction is higher for a soft tire:
Buried in this argument are any number of ideas that ultimately lead to the final value of the coefficient of friction. For example, the tire deforms and can then push against vertical surfaces. But this is all that happens at the microscopic level. So really we end up with a number of different levels of "contact" characteristics between the surfaces. After rationalizing this a bit we realize that all arguments boils down to the measured coefficient of friction for a particular tire configuration; e.g. pressure, temperature, stickiness, surface relaxation times, etc. etc. etc. The more one thinks about this, the more variables seem to crop up.


2).There is more rubber to remove:
To insure maximum friction, we typically [historically] want to remove as much rubber from the tire as possible. Obviously, as "On Radioactive Waves" almost started to touch on, if we add rubber by making a thicker tire, we create greater angular inertia than with a wider tire. This then affects the acceleration performance of the car, in addition perhaps to making for an impractical tire design.

I am sure other variables exist that I have never even considered. In any event, good enough, here's to you marcus.
 
  • #48
Originally posted by Ivan Seeking
good enough, here's to you marcus.

In or around 1990 the BIPM, under the international authority of the General Conference on Weights and Measures (CGPM), established new Electrical Standards known as the 1990 volt and the 1990 ampere.

These are defined independently of the metal kilogram prototype in Paris and are in fact based solely on the atomic clock together with two quantum effects.

What are the two (low temperature) quantum mechanical effects used to define V90 and A90 ?

Or, equivalently if you prefer to look at it that way, the effects used to define V90 and Ω90 .

The definitions are based on ADOPTING exact values for two fundamental physical constants. What are they?

PS: it is sort of analogous to defining the meter by adopting an exact value for the speed of light---the official way since 1983.
You can't measure the speed of light in vacuo any more because it has a decreed exact value---in the 1990 electrical context there are two more constants which you can't measure.
'These are the electrical standards used in practice, though not yet finally made official.

The adopted values are among those listed at the NIST fundamental constants website in the "adopted values" section.

To sum up----the 1990 electrical standards have no logical connection to the metal kilogram and the Newton force based on it---or on the old force between infinite parallel wires definition of the ampere. How are they defined?
 
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  • #49
The quantum magentic flux unit is used, and I believe that is defined with h-bar and e.


JMD
 
  • #50
Originally posted by nbo10
The quantum magentic flux unit is used, and I believe that is defined with h-bar and e.


JMD

Yes! but that is only one fundamental constant and only enough to define one unit-----the 1990 ampere.

be more complete and specific.

How, for example, is the 1990 volt defined? Using the atomic clock plus what quantum device?

good, but not yet complete



BTW if you do dimensional analysis of hbar/e2
you will find that (if you carefully do it in SI units only and
follow SI conventions) you will get a ratio of voltage to current.
But that sort of ratio is called a resistance.
check out the adopted constants at the NIST website
nist.gov, if I remember right---the world's premier "fundamental
constants" website
 
  • #51
They mostly likely use a squid or josephson junction. With the use of the quantum hall effect both a volt and amp and be "defined".

JMD
 
  • #52
Originally posted by nbo10
They mostly likely use a squid or josephson junction. With the use of the quantum hall effect both a volt and amp and be "defined".

JMD

http://physics.nist.gov/cuu/Constants/

these adopted values define the current (1990) electrical standards. I'm quoting from this section of the site

http://physics.nist.gov/cgi-bin/cuu/Category?view=html&Adopted+values.x=102&Adopted+values.y=11

conventional value of Josephson constant
Value 483 597.9 GHz V-1
Standard uncertainty (exact)


conventional value of von Klitzing constant
Value 25 812.807 Ω
Standard uncertainty (exact)


Specifying exact values of these two is equivalent to specifying exact values of hbar and e.

Your turn JMD---you pose a question that you know the answer to
and whoever gets it right has the next turn.
 
  • #53
Lets try an easy one,

What is the diference in first and second order phase tranistions?


JMD
 
  • #54
Originally posted by Ivan Seeking

Why do racing cars use wide tires?

Simply because wide tires wear less than narrow ones - thus they last longer.
 
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  • #55
Originally posted by nbo10
Lets try an easy one,

What is the diference in first and second order phase tranistions?


JMD

For starters here are two examples-----the melting of ice is a first order transition.
The change in magnetic state at a critical temperature is a second order transition.

there are different definitions around----one is that first order involves a substantial supply of heat or change in free energy content----and second order just happens without a big change in heat content either way.

In a second order transition the specific heat may change and various other properties (like magnetisation). The change may be less obvious.



An online dictionary I gave a slightly different take on it: first order is abrupt and involves a gross change in properties like from solid to liquid

Second order (this dictionary indicated) was continuous as the critical temperature was approached----rather than abrupt---and likely to involve more subtle properties of the material.

the dictionary practically equated first order with discontinuous transition and second order with continuous.

I think there must be several alternative ways of classifying phase transitions by first or second order, and perhaps higher orders than two.

******for reference here are the dictionary entries*****


Phase transition
A change of state such as occurs in the boiling or freezing of a liquid, or in the change between ferromagnetic and paramagnetic states of a magnetic solid. An abrupt change, characterised by a jump in an order parameter is known as first order; a change in which the order parameter evolves smoothly to or from zero is called continuous or second order.

Order parameter
A variable such as ... the magnetisation in an Ising model... used to describe the degree of order in a phase above (below) its critical point. In a continuous phase transition (second order phase transition), the order parameter goes continuously to zero as the critical point is approached from above (below).
 
  • #56
Almost what I'm getting at.

What happenes to various properties, during a phase change. ie volume, entropy, compressibility, heat capacity. All of which can be expressed from the expression of the free energy. There is a distinct behavior for first and second order tranistions.

JMD
 
  • #57
Originally posted by nbo10
Almost what I'm getting at.

What happenes to various properties, during a phase change. ie volume, entropy, compressibility, heat capacity. All of which can be expressed from the expression of the free energy. There is a distinct behavior for first and second order tranistions.

JMD

In Type I there is an abrupt, discontinuous change in these properties (vol, entropy, heat capacity)

In Type II the change is continuous.

www.mit.edu/~levitov/8.334/lec2.pdf[/URL]

Some MIT lecture notes on statistical mechanics, by Levitov.
Discussing the Landau model of phase transition and
Type I/Type II also called first order/second order transition.

Has a concise summary of the difference at the end of lecture 2, which I will try to copy for extra clarity. Nice question "nbo10".

BTW symmetry is an important feature of Type II. In type I (most common type) it can happen that there is no significant change in symmetry. But in type II (in magnetization for example) there can be an abrupt change in the amount of organization or symmetry.
The notes talk about this.
 
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  • #58
want to copy in the end-of-lecture summary from those notes

www.mit.edu/~levitov/8.334/lec3.pdf[/URL]

Type I phase transitons are discontinuous, accompanied by jumps in physical quantities, such as density or entropy, latent heat, and volume change. The symmetry is typically not changed in such transitions.

Type II phase transitions are very different. In such transitions, physical quantities vary continuously, while symmetry changes abruptly. The phase transformation in a type II transition is described by spontaneous symmetry breaking.
 
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  • #59
In first order phase tranistions, there is a discontiniuty in the first derivative in the free energy.

second order phase tranistions have a disconitiuty in the second derivative in the free energy.

In ice the volume changes as the sturcture changes, First order PT

The tranistion from normal metal to a superconducting state there is a jump in the specific heat, a second order PT

There is no gradual change in symtmetry, there is an abrupt change in going from one state to another.

JMD
 
  • #60
Originally posted by nbo10
In first order phase tranistions, there is a discontiniuty in the first derivative in the free energy.

second order phase tranistions have a disconitiuty in the second derivative in the free energy.

In ice the volume changes as the sturcture changes, First order PT

The tranistion from normal metal to a superconducting state there is a jump in the specific heat, a second order PT

There is no gradual change in symtmetry, there is an abrupt change in going from one state to another.

JMD

Well I can't say scored a complete hit---in terms of what you expected. Would you like to have a second turn?
(You have to receive and acknowledge a right answer for the turn to move on to another person.)
 
  • #61
It was close.
I have one more that is semi-interesting, a engineering proferssor asked this to my class. By the way his solution wasn't even close to being correct.

When you clap your hands, what keeps your hands from passing through each other? Or when anything touches anything, what keeps it from mergeing together?

Bonus if you can guess how the professor answered this question.

JMD
 
  • #62
If you don't mind a broad brush answer

I would say the Exclusion Principle. It prevents matter from being any more "crowded together" than it is.

The professor's answere might be electrical repulsion.
 
  • #63
The professors responce was Pauli's Exclusion Principle. but that is not correct. Pauli's Exclusion Principle imposses symmetry on the wavefunctions.

"electrical repulsion" is also incorrect. You might be thinking along the correct path.

Another bonus, restate the question in a scientific way.

JMD
 
  • #64
Originally posted by nbo10
It was close.
...
When you clap your hands, what keeps your hands from passing through each other? Or when anything touches anything, what keeps it from merging together?

JMD

Fluids do merge together so the idea is what makes a solid solid
and related issues

So most solids have some lattice of bonds
or an effective system of intermolecular forces
that gives them integrity

cells in your hands are bags of fluid enclosed in membrane

if you ask why don't my hands merge when they press together
it must have to do with the strength of membranes

that is like crystal lattice bond-work in two dimensions.

these are amusing questions. i wish you were prepared to
come up with some more
 
  • #65
Where is nbo? I want to know if I got the answer right or not.

He asks why don't things merge on contact and the answer is that some things (fluids) do tend to mix on contact so it is a question about solids---how they maintain integrity.

Having one solid pass thru another would involve breaking a lot of bonds. So the bond energy is the answer.

but also, given that the lattice is going to hold, solids are still mostly empty space so there is a further question of why one atom cannot just pass thru another atom.

Even when gasses mix, atom does not pass thru atom and that (which is not what the question is about) may have to do with the exclusion principle. So I stick with my original answer:

Originally posted by marcus
Fluids do merge together so the idea is what makes a solid solid
and related issues

So most solids have some lattice of bonds
or an effective system of intermolecular forces
that gives them integrity

cells in your hands are bags of fluid enclosed in membrane

if you ask why don't my hands merge when they press together
it must have to do with the strength of membranes

that is like crystal lattice bond-work in two dimensions.

these are amusing questions. i wish you were prepared to
come up with some more
 
  • #66
I have to sleep sometime

Thats good enough marcus. Your question

To restate the question

Why do solids have a bulk modulus?

Electrostatic potentials forming a lattice.

JMD
 
  • #67
Originally posted by nbo10

Thats good enough marcus. Your question

It is more interesting for me when you ask the questions.
hope you answer this and return with a question.

One seldom sees the Einstein equation written in form
compatible with SI metric units, but suppose one did.
then the equation would be written this way:

Gmu,nu = 8pi (G/c4) Tmu,nu

In what SI units are the quantities on the left expressed?

In what SI unit would (c4/G) be expressed? I am intentionally looking at the reciprocal of what appears in the equation. Of course G is Newton's constant and c is the speed of light.

In what SI units are the Tmu,nu terms on the right expressed?

dont be offended if this is too trivial a question. the fact is that people usually see the equations written in units with c = G =1 so in my experience they don't think concretely about how things would look in familiar SI terms
 
  • #68
Momentum-energy stress T is in N(ewtons), curvature metrics g is unitless.

G/c4 unit is N-1.

In some papers curvature g is expressed in m-2 and momentum-energy tensor T in N/m2=J/m3
 
  • #69
Originally posted by nbo10
The professors responce was Pauli's Exclusion Principle. but that is not correct. Pauli's Exclusion Principle imposses symmetry on the wavefunctions.


JMD

Actually vise versa, the (anti)symmetry of wave function results in Pauli exclusion principle (repulsion of fermions) as a mathematical by-product.
 
  • #70
Originally posted by Alexander
...
G/c4 unit is N-1.

I was asking about (c4/G) which is a force and the answer I was looking for is that in SI it is expressed in Newtons, the SI force unit.

However you said something just as good! You said the reciprocal is expressed in reciprocal Newtons, or as you say N-1.

Originally posted by Alexander
...

In some papers curvature [Gmu,nu, not g] is expressed in m-2 and momentum-energy tensor T in N/m2=J/m3

The curvature Gmu,nu is expressed in 1/area units----in other words---and the Tmu,nu is expressed in energy density units----joules per cubic meter.

As you point out, joules per cubic meter is also equivalent to the unit of pressure, Newton per sq. meter.

And as you say this happens in some papers, while in other papers the curvature is made dimensionless, so it is not expressed in SI units or in any kind of units.

OK Alexander, your turn. Have a good one!
 

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