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also, if that works, than you dont need a physical rod, you can go with something else, electron, particle etc.

anyone know?

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- #1

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also, if that works, than you dont need a physical rod, you can go with something else, electron, particle etc.

anyone know?

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Another way of saying the same thing - the rod is not perfectly rigid,when you push on one end, the other end does not move right away. You can model the propagation of the displacement of the rod mathematically as a distributed spring-mass system. (The solution to this distributed spring-mass problem is the wave equation, with the characteristic speed of the wave being the speed of sound in the material.

- #3

Pengwuino

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At least i think thats how it goes.

@pervect

I think he means the physical displacement of the rod, not sending a sound wave through the rod.

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What pervect said is exactly right.

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wat if the rod was perfectly rigid?the rod is not perfectly rigid,

if thats possible... which i am doubting somewhat

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Pengwuino

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- #7

Integral

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When you strike (or attempt to move) a physical object shock waves are created which move at the speed of sound in that material. If you strike it with a force large enough so it is displaced locally with a speed greater then the speed of sound in the material the material will be destroyed (i.e. you will break it).

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Pengwuino

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- #9

jtbell

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I think that what Integral meant was that when you push a rod you push only a relativly small group particles (molecules, crystals) which make up the rod. The force acting between the particles that were pushed and their neighbouring particles then "pulls" the neighbouring particles, these particles then "pull" their neighbouring particles, etc.So integral, say you have a meter long piece of metal. If its levitating in mid air lol... and you grab the right end and move it to the left 10cm, the left side of the rod will move that 10cm not instantly, but only after a shockwave at v=speed of sound travels that 1 meter to the left side

This mechanism accurs almost instaneously in ordinary life. But for a rod a light-year long it will be far from that.

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Let voltages be velocities, and forces be currents.

Then I = c dv/dt is equvalent to F = m dv/dt = ma, so a mass is a capacitor.

Similarly, springs are inductors.

A bunch of masses connected by springs becomes a bunch of grounded capacitors connected by inductors.

This is, of course, the model of a transmission line.

Now, if we apply a constant force to one end of the mechanical system, this is equivalent ot forcing a constant current into one end of a transmission line.

THe boundary conditions at the "free" end are no force, so it's an open circuited transmission line.

We can then sketch out the general behavior of the solution (if we are familiar with how transmission lines act).

Initlally, there will be some voltage wave, of magnetiude v = z*i, propagating to the right, with Z being the impedance of the transmission line.

The mechanical analog is that there will be some initally velocity v which will propagate to the right at the speed of sound in the material. This velocity will be given by the force times the "impedance" of the bar, which will depend on the material it's made out of (it's bulk modulus and its density).

When the velocity wave reaches the right side of the transmission line, it reflects. Because the line is open circuted, the refleciton coefficient is unity.

This means that the velocity then doubles, and a new velocity wave heads back towards the left side, moving at the speed of sound in the material.

The result is a sort of "staircase" motion, where the velocity of individual points on the rod changes in discreete steps, rather than a smooth linear curve, a "staircase" approximation to v=a*t.

This analysis has ignored any damping effects, which would be modelled electrically by resistors. If you look at the length of the rod, you'd see it changing (oscillating) slightly with the above solution. The simple trnasmission line solution gives oscillations that never die out - a more realisitc solution would include some damping (due to unmodelled frictional forces and losses), so that eventually the oscillations in the length of the rod decay.

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Integral

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That is exactly what I meant. If you attempt to move the rod at a velocity which is close to the speed of sound in the material it will bend, if you attempt to move it faster then the the speed of sound in the material it will break.Pengwuino said:

Notice that I repeat the phrase "in the material" The speed of sound in solids is much higher then the speed of sound in a gas.

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Pengwuino

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If the following is a long rod, and you push on the left end at x=0

(x=0)|||||||||||||||||||||||||(x=d)

The right end of the rod at x=d will first move at a time t = d/cs, where cs is the speed of sound in the rod.

The velocity of any point on the rod will be a function of the position on the rod, and the time, i.e.

velocity = v(t,x)

For 0<t<T, with T = d/c, (the length of the rod over the speed of sound in the rod)

v(x,t) will be given by

0 if t < x/c

v if t >= x/c

v is a constant depending on the exact characteristics of the rod and the applied force - it will be proportional to the applied force.

The plot of the above function would require three dimensions, x,t, and the dependent variable v(x,t). It represents, however, a "velocity wave" propagating to the right along the rod at the speed of sound.

For T < t < 2*T, v(x,t) will be

1*v if (t-T) < (d-x)/c

2*v if (t-T) >= (d-x)/c

This represents a wave moving to the left at the speed of sound. Like the previous wave, this wave is a plot of velocity vs time.

Note that the rod will compress during the time interval 0<t<T, and the rod will expand during the interval T<t<2T. The total change in the length of the rod will be

delta-L = v*d/c, where 'v' is the constant velocity previously mentioned, d is the length of the rod, and c is the speed of sound in the rod.

If delta-L/d is too large a quantity, the material will break, but usally delta-L will be very small.

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Pengwuino

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- #16

russ_watters

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No. Atoms themselves are not rigid objects.Pengwuino said:

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Pengwuino

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Ok gotcha.

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i get the feeling you just cant do it...

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