Simple Physics

Arcade Fire · 18 Apr 2006 at 15:04

I think I'll let you off this time - the fees would be over a quarter of what you'd be giving me and I don't like the thought of giving evil corporations money for nothing

Tommy B · 18 Apr 2006 at 15:07

This mechanics paper is driving me insane. I found the quantum one a joke and apparently that's a lot more difficult than this. I've now been asked to explain the horizontal motion of a ball shows that air resitance is negligible. It then gives me a stupid graph with vertical and horizontal distance. How in the love of God can you state air resistance is negligible. I take it I just put that mavity pushes the ball down at exactly 9.81 ms-2

Tommy B · 18 Apr 2006 at 15:08

Well thanks for your help Arcade. Really appreciated, and thanks Sab too.

daz · 18 Apr 2006 at 15:21

Tommy B said:
This mechanics paper is driving me insane. I found the quantum one a joke and apparently that's a lot more difficult than this. I've now been asked to explain the horizontal motion of a ball shows that air resitance is negligible. It then gives me a stupid graph with vertical and horizontal distance. How in the love of God can you state air resistance is negligible. I take it I just put that mavity pushes the ball down at exactly 9.81 ms-2

Because for a lot of objects, air resistance is neglible. Furthermore, the calculations required for it probably triple the amount of calculations, perhaps only increase the the accuracy by less than 10% for a lot of objects.

I suppose the trajectory is a parabola, and you have to calculate how far the ball has travelled etc?

Visage · 18 Apr 2006 at 15:23

daz said:
Because for a lot of objects, air resistance is neglible. Furthermore, the calculations required for it probably triple the amount of calculations, perhaps only increase the the accuracy by less than 10% for a lot of objects.

Its not so much that, more that as soon as you introduce air resistance the equations become non-linear, so they're pretty much unsolvable analytically except for the simplest cases.

daz · 18 Apr 2006 at 15:27

Visage said:
Its not so much that, more that as soon as you introduce air resistance the equations become non-linear, so they're pretty much unsolvable analytically except for the simplest cases.

Like you say, for most simple objects, like a ball, you would be fine.

Visage · 18 Apr 2006 at 15:31

daz said:
Like you say, for most simple objects, like a ball, you would be fine.

No, by simple, I mean cases where the air resistance is a constant term. For cases where resistance is a function of velocity you end up with equations like:

mdv/dt = F - f(v)

so you have to solve t = Int(m/(F-F(v)))

Where Int = Integral.

For any cases where f is not linear in v there is no general solution, regardless of the composition of the object.

//Mike · 18 Apr 2006 at 17:02

Visage is correct

.

Only restive (or damping) forces proportional to velocity can be solved analytically. Any hint of a non linear model can only be solved using numerical methods (and hence no general solution can be formed).

Visage · 18 Apr 2006 at 17:10

//Mike said:
Visage is always correct .

Thats better.....

//Mike · 18 Apr 2006 at 17:16

Modest!

ArmyofHarmony · 18 Apr 2006 at 17:43

You doing physics a level i guess?

I got C's in that module A (mechanics) paper twice... then i finally got an A third time round

I hate physics, and im doing a physics degree now... don't fall for it

Arcade Fire · 18 Apr 2006 at 18:06

Hey, I have a (possibly stupid) physics question. Say you have a particle falling with air resistance proportional to its velocity, governed by the equation

m dv/dt = -mg - kv

where k is some constant that makes the dimensions match. You rearrange and integrate, to get

-m Int [ 1/(mg+kv) ] dv = Int dt

The left hand sides integrates to a logarithm, and the right integrates to give t plus an integration constant which is -m log(mg), so you have

-m log(mg+kv) = t - m log(mg)

but isn't there something wrong here? For a start we can't take the logarithm of a quantity with dimensions, and secondly we seem to have a dimension mismatch on the right hand side of the equation. It seems that we can partially resolve the problem by rearranging the logarithms to give

m log[mg/(mg+kv)] = t

so that the quantity we're taking the logarithm of is dimensionless, but then it seems we can rearrange to give

mg/(mg+kv) = exp(t/m)

so we're just taking the exponential of something with dimensions instead. Rearranging further gives us a sensible equation for v in terms of t, namely

v = mg/k * [exp(-t/m) - 1]

so our particle starts at v=0 at t=0, and then falls downwards until it reaches its terminal velocity at v=-mg/k. But we still have the annoying niggle of that exponential of a dimensionful quantity. I'm sure this is the sort of thing I used to do all the time without thinking about it - what am I missing here!

daz · 18 Apr 2006 at 18:12

Our mechanics lecturer went over this, you have to introduce a constant of the same dimension as the thing you want to take the logarithm of. And then it all works out in the end... i can't remember properly.

He said if he saw any of us take a logarithm of something with dimensionless in the exam, he would give us half marks... even if the rest of the answer was right.

Arcade Fire · 18 Apr 2006 at 18:15

Hmm, I suppose that we could equally well integrate 1/v up to give log(Av) instead of log(v), and just say that A has the right dimensions to make Av dimensionless. I'm not sure that helps here though...