changing constants... physics

[SaberKitty]

a friend of mine had an assignment in her ap physics class- she had to change a constant and state how the change would effect everyday life...
she said "lets say we can change the constant of the elastic potential energy, and its a decrease from the normal one, therefore the elastic potential energy is less, a clock (that has springs) would not work or slow down?"
after a bit of research we dicided that the springs in clocks only are part of the winding mechanism ao it probably wouldn't make any difference in the other functions of the clock- just the amount of times it needed to be rewound.

but- how would this change in elastic potenetial energy impact-pun inteneded because that's just the way i am- car crashes? would it make the dammage more or less? why?

-i have no idea as to the answer, but since all you guys are spartipants- i figured you might know it or at least how to find out.

thanks so much
SaberKitty

- oh and she's not at my HS- so next time i see her on aim i'll ask what her teacher said.

[ZutAlors!]

Possibly an interesting question here. Interesting to me at least because this is very mechanical-engineering oriented.

First, some clarification: I assume that when you say "constant of the elastic potential energy" what you really mean is "modulus of elasticity" E (also known as Young's modulus). The modulus of elasticity is the constant that relates stress to strain, or (more simply), force to distance. In other words, if I have a block of rubber of height L and area A, and squish it with a force F, it will move a distance D = FL/(EA), where E = modulus of elasticity.

Now, E relates to potential energy because, in the same rubber block, PE = (1/2)FD = (F^2)L/(2EA) = (D^2)EA/L. So if you *decrease* the modulus of elasticity, squishing the rubber with the same *force* will give you *more* potential energy, and squishing the rubber to the same *distance* will give you *less* potential energy. This is exactly what you said would happen in the clock: *decrease* the modulus, and you get less potential energy storage if you wind the spring the same amount.

However, there may be two flies in the ointment, depending on what you want to do with the assignment. 1) the modulus of elasticity for different materials is different (rubber is more squishy than steel, right?). So do you want to change E for only *one* material, or *one* part, or *every single material in the world*?

Second, the modulus of elasticity only applies in the elastic region, not the plastic region. What does this mean? Think of a sheet of rubber and a sheet of chewing gum. If you stretch the rubber, it bounces back when you stop stretching; this is *elastic*. If you stretch the gum, however, it will stay stretched; this is *plastic*. Most materials, including steel, will stretch elastically (like a spring) for a while, but if you stretch them too much, they will start deforming plastically. You can see this with a paper clip: bend it just a little and it bounces back, bend it a lot and it stays bent.

And now, to try to answer your question: To determine the effect of a different E in a car crash, you also have to make one more assumption: at what point does the deformation of the bumper change from elastic to plastic? At the same *force* as before, or the same *deflection* as before? If you choose the same *force*, then (looking above at the equations) the bumper can absorb more potential energy in a crash before it plastically deforms (plastically deforms = takes damage). If you choose the same *deflection*, then the bumper absorbs *less* potential energy in a crash before it plastically deforms. I am not really sure which of these assumptions is the most reasonable, but I would guess that choosing the seme deflection is the most reasonable.

As a side note, *plastic* deformation is extremely complex, different for different materials, and not really related to elastic deformation. So changing the modulus of elasticity probably would not change the plastic damage to the bumper, although, like I said, it's pretty complex.

Any more questions? I'm not quite sure if this explanation is too technical or not technical enough.

-ZutAlors!, PhD in ME

[fetus]

here's a reply from someone who isn't so far removed from high school and doesn't have a phd (yet :]). while i am a mechanical polymer engineer (well, studying to be one, technically, but its a minor formality) and know what the previous poster is talking about, i also remember my high school physics class. we didn't talk about modulous of elasticity or anything like that. what it seems to me you're refering to (and i'll explain in some detail for those who don't know what i'm talking about) is the amount of "bounce" or the amount of "stick" two things have when they collide. i forget the name (its most likely the coefficient you're talking about) but its a number used in calculating how much energy is lost (not transferred from one to the other) in a collision between 2 objects. if memory serves me right, it is a value from 0 to 1, 0 meaning it is inelastic and all energy is lost/absorbed in the collision (ie, they stick together), 1 being they are completely elastic and no energy is lost (they bounce apart without losing energy). i think thats right but i could have my 1 and 0 backwards. (the rest of this assumes i'm right about 0 being inelastic and 1 being elastic. if thats incorrect, just switch my 0's and 1's...) anyway, 2 good examples: 1) dropping a piece of wet clay on the floor compared to a rubber bounce ball. the clay has a coefficient of close to 0 and just hits the ground and stays. the bounce ball has a higher coefficient, probably around .9 or so. its high, so the ball still bounces pretty close to the original height, but the fact it isn't actually 1 makes it lose energy, so it doesn't actually bounce forever. the other is an experiment in which you load little cars with weights and experiment with having velcro on them compared to a spring and stuff like that (this is pretty closely tied to inertia/momentum and stuff, btw).
now on to your actual question: if you lowered the coefficient, there would be no "theoretically perfect" collisions, for starters, and in addition, things would simply just "bounce" less. the rubber ball wouldn't bounce as high each time, etc. a rubber ball bouncing isn't exactly integral to you life, but you get the idea.

[This message has been edited by fetus (edited 04-11-2002 09:42 AM).]

[ChienFou]

But what we want to know is: Would we be more likely to be injured?

CF is CEng BSc(Eng)ACGI (Mechanical Engineering 1969 Imp. Coll London)

I think it makes no difference as you'd design to the new Young's modulus

[fetus]

or maybe i was in the "dumb" physics class and saber wanted to know about young's mod... :]

[ZutAlors!]

fetus, what you're referring to is the coefficient of restitution, and you do have the one and zero correct. Using this coefficient is a simplified, but very useful, approximation in collisions. I'm thinking this wasn't what SaberKitty was originally referring to because it doesn't apply to her original example of clock spring (no collisions, here). Seems to me that this might be a difficult coefficient upon which to perform the type of thought experiment SaberKitty proposes, because the value of the coefficient depends on each of the materials, and the structure of the colliding objects, to begin with...although perhaps considering the case where all coefficients of restitution were amgically 1, or magically 0, would be interesting.

ChienFou, I think that the answer to your question might be a little more involved: assuming, say, a much lower Young's modulus but the same yield stress would probably result in not much design difference, other than the happy circumstance of more damage-resistant bumpers. Assuming a much lower Young's modulus but the same yield strain would imply that the bumper structure would not absorb near as much energy in the plastic region, sot he structure would have to be beefed up accordingly.

[Rollercoaster]

It is quite possible, as well, that the original question was referring to the 'coefficient of prostitution', a variable ranging from 5 to as high as 500. It incorporates the characteristics of colliding bodies, in addition to elasticity and compressibility properties.

[One Skunk Todd]

Wouldn't that also involve harmonic motion, relative humidity, friction and lubricity?

[impossibleroot]

And, of course, currency.