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The Proof is 'not-so' Trivial - Physics Edition

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Reply 80
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Phichi
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Original post by lerjj
()


Have you gone far enough in maths to understand the integrating factor :smile:?

Original post by langlitz
Problem 15 **

A particle of mass m is projected from an initial height h0h_0 in the vertical direction, with an initial velocity v0v_0. It is subject to a gravitational force mgmg and a linear resistance proportional to its velocity −mkv-mkv.
Show v(t)=−u+(v0+u)e−ktv(t)=-u+(v_0+u)e^{-kt} where u=g/ku=g/k is the terminal velocity of free fall.
Also find the equation of motion h(t)h(t).


Just to check, are you saying that the resistance is in the same direction as the motion? I'm asking because your terminal velocity (should) have a negative sign no? You're currently saying the free fall is upwards.

x¨=−kx˙−g  ⇒x¨=0   ⇒  x˙T=u=−gk\ddot{x} = -k\dot{x} -g \, \, \Rightarrow \ddot{x} = 0 \, \, \, \Rightarrow \, \, \dot{x}_T = u = \dfrac{-g}{k}
(edited 9 years ago)
Reply 81
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lerjj
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Original post by Phichi
Have you gone far enough in maths to understand the integrating factor :smile:?



Just to check, are you saying that the resistance is in the same direction as the motion? I'm asking because your terminal velocity (should) have a negative sign no? You're currently saying the free fall is upwards.

x¨=−kx˙−g  ⇒x¨=0   ⇒  x˙T=u=−gk\ddot{x} = -k\dot{x} -g \, \, \Rightarrow \ddot{x} = 0 \, \, \, \Rightarrow \, \, \dot{x}_T = u = \dfrac{-g}{k}


I don't know how to solve any but the most trivial differential equations (i.e. the ones which you can jut integrate to solve). I will google around though, this doesn't look too difficult...

Regards to signs, I believe he's got down as positive, hence a positive weight and negative resistance term.
Reply 82
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Phichi
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Original post by lerjj
I don't know how to solve any but the most trivial differential equations (i.e. the ones which you can jut integrate to solve). I will google around though, this doesn't look too difficult...

Regards to signs, I believe he's got down as positive, hence a positive weight and negative resistance term.


The mass is moving vertically upwards, the weight and resistance will both be directed downwards, their signs will be the same.
(edited 9 years ago)
Reply 83
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lerjj
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Original post by Phichi
The mass is moving vertically upwards, the weight and resistance will both be directed downwards, their signs will be the same.


You're right, I misread it as free-fall from some height.
Reply 84
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Phichi
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Original post by langlitz

Spoiler



Solution 15


Let f(t)=v(t)f(t) = v(t)

Equation of motion:

x¨+kx˙=−g   ⇒   f′(t)+kf(t)=−g\ddot{x} + k\dot{x} = -g \, \, \, \Rightarrow \, \, \, f'(t) + kf(t) = -g

Laplace transform of both sides:

L{f′(t)+kf(t)}=L{−g}\mathcal{L}\{{f'(t)+ kf(t)}\} = \mathcal{L}\{{-g}\}

L{f′(t)}+kL{f(t)}=−gL{1}\mathcal{L}\{{f'(t)}\} + k\mathcal{L}\{{f(t)}\} = -g\mathcal{L}\{{1}\}

Leading to:

[sF(s)−f(0)]+kF(s)=−gs   ⇒   (s+k)F(s)=f(0)−gs\left[ sF(s) - f(0) \right] + kF(s) = \dfrac{-g}{s} \, \, \, \Rightarrow \, \, \, (s+k)F(s) = f(0) - \dfrac{g}{s}

As f(t)=v(t)f(t)=v(t) we know that f(0)=v0f(0) = v_0

F(s)=v0s+k−gs(s+k)F(s) = \dfrac{v_0}{s+k} - \dfrac{g}{s(s+k)}

F(s)=v0s+k−gks+gk(s+k)F(s) = \dfrac{v_0}{s+k} - \dfrac{g}{ks} + \dfrac{g}{k(s+k)}

Now:

f(t)=L−1{F(s)}=L−1{v0s+k}−L−1{gks}+L−1{gk(s+k)}f(t) = \mathcal{L}^{-1}\{{F(s)}\} = \mathcal{L}^{-1}\{\dfrac{v_0}{s+k}\} - \mathcal{L}^{-1}\{\dfrac{g}{ks}\} + \mathcal{L}^{-1}\{\dfrac{g}{k(s+k)}\}

f(t)=v0e−kt−gk+gke−ktf(t) = v_0e^{-kt} - \dfrac{g}{k} + \dfrac{g}{k}e^{-kt}

f(t)=−gk+(v0+gk)e−ktf(t) = \dfrac{-g}{k} + \left(v_0 + \dfrac{g}{k}\right)e^{-kt}

Using the initial equation of motion, u=−gku = \dfrac{-g}{k}

f(t)=u+(v0−u)e−ktf(t) = u + (v_0 - u)e^{-kt}

The question does state u=gku = \dfrac{g}{k}, leading to f(t)=−u+(v0+u)e−ktf(t) = -u + (v_0 + u)e^{-kt} as required :s-smilie:

Hopefully someone can attempt it some other way (lerjj :wink:), I'll leave the second part for now :smile:
(edited 9 years ago)
Original post by Phichi
x

You can actually do this without laplace transforms or integrating factors. It's just a first order separable ODE
Reply 86
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Phichi
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Original post by langlitz
You can actually do this without laplace transforms or integrating factors. It's just a first order separable ODE


I can think of over 5 ways to do it. I.F would be the quickest, hence why I suggested it, separating the variables wouldn't make it slightly exciting now would it :wink: Purposely did it using Laplace so lerjj could have a go, we don't get many questions here which are A Level friendly.
A thread for physics questions only which have nothing to do with homeworks or school anyway? no matter what kinds of physics? as long as its possible to answer? So I guess this is the right thread to ask my question:

I have busied myself with Heisenberg's uncertainty principle and have found different equations for that:

http://www.physicoro.de/fml/14101.png
http://www.steffen-grimm.de/zufallundseele/unschaerfe.gif
http://upload.wikimedia.org/math/9/2/6/926af8ecb0b2e3374460f0e26864f47b.png
http://www.cobocards.com/pool/data/tex/eabaaa51f7bdeff3a5165a2e469db4ec.gif

My question is simple: Why are there so many different equations for Heisenberg's uncertainty principle?
Reply 88
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Phichi
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Original post by Kallisto
A thread for physics questions only which have nothing to do with homeworks or school anyway? no matter what kinds of physics? as long as its possible to answer? So I guess this is the right thread to ask my question:

I have busied myself with Heisenberg's uncertainty principle and have found different equations for that:

http://www.physicoro.de/fml/14101.png
http://www.steffen-grimm.de/zufallundseele/unschaerfe.gif
http://upload.wikimedia.org/math/9/2/6/926af8ecb0b2e3374460f0e26864f47b.png
http://www.cobocards.com/pool/data/tex/eabaaa51f7bdeff3a5165a2e469db4ec.gif

My question is simple: Why are there so many different equations for Heisenberg's uncertainty principle?


This thread is better suited in the main forum, it's not really a 'proof' question. But, here you go:

Heisenberg initially stated that:

ΔxΔp≳h\Delta x \Delta p \gtrsim h

Which is shown near enough by these two:

http://www.physicoro.de/fml/14101.png
http://upload.wikimedia.org/math/9/2/6/926af8ecb0b2e3374460f0e26864f47b.png

However, since, it's been proven to actually be:

ΔxΔp≥ℏ2\Delta x \Delta p \geq \dfrac{\hbar}{2}

Which is the same as both of these:

http://www.steffen-grimm.de/zufallundseele/unschaerfe.gif
http://www.cobocards.com/pool/data/tex/eabaaa51f7bdeff3a5165a2e469db4ec.gif

Noting that ℏ=h2π\hbar = \dfrac{h}{2\pi}
(edited 9 years ago)
Original post by Phichi
This thread is better suited in the main forum (...)


Really? Oh, I am sorry. Next time I will make sure to post my question in the right thread carefully.

Original post by Phichi
(...)
Heisenberg initially stated that:

ΔxΔp≳h\Delta x \Delta p \gtrsim h

Which is shown near enough by these two:

http://www.physicoro.de/fml/14101.png
http://upload.wikimedia.org/math/9/2/6/926af8ecb0b2e3374460f0e26864f47b.png

However, since, it's been proven to actually be:

ΔxΔp≥ℏ2π\Delta x \Delta p \geq \dfrac{\hbar}{2\pi}

Which is the same as both of these:

http://www.steffen-grimm.de/zufallundseele/unschaerfe.gif
http://www.cobocards.com/pool/data/tex/eabaaa51f7bdeff3a5165a2e469db4ec.gif

Noting that ℏ=h2π\hbar = \dfrac{h}{2\pi}


I see, I guess so. The following euqations were a better approximation for the uncertainties, so a single true equation came into being, namely:

ΔxΔp≥ℏ2π\Delta x \Delta p \geq \dfrac{\hbar}{2\pi}
You have a factor of π \pi wrong in your mathematical statement of Heisenberg's Uncertainty principle.

Original post by Phichi
This thread is better suited in the main forum, it's not really a 'proof' question. But, here you go:

Heisenberg initially stated that:

ΔxΔp≳h\Delta x \Delta p \gtrsim h

Which is shown near enough by these two:

http://www.physicoro.de/fml/14101.png
http://upload.wikimedia.org/math/9/2/6/926af8ecb0b2e3374460f0e26864f47b.png

However, since, it's been proven to actually be:

ΔxΔp≥ℏ2π\Delta x \Delta p \geq \dfrac{\hbar}{2\pi}

Which is the same as both of these:

http://www.steffen-grimm.de/zufallundseele/unschaerfe.gif
http://www.cobocards.com/pool/data/tex/eabaaa51f7bdeff3a5165a2e469db4ec.gif

Noting that ℏ=h2π\hbar = \dfrac{h}{2\pi}
Reply 91
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Phichi
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Original post by Kallisto
Really? Oh, I am sorry. Next time I will make sure to post my question in the right thread carefully.



I see, I guess so. The following euqations were a better approximation for the uncertainties, so a single true equation came into being, namely:

ΔxΔp≥ℏ2π\Delta x \Delta p \geq \dfrac{\hbar}{2\pi}


ΔxΔp≥ℏ2\Delta x \Delta p \geq \dfrac{\hbar}{2}

Original post by WishingChaff
You have a factor of π \pi wrong in your mathematical statement of Heisenberg's Uncertainty principle.



Thanks for spotting, fixed, cheers.
(edited 9 years ago)
Original post by Phichi
x


Oh, thanks for correction, Phichi. But I have thought about the equation again. Here are my considerations:

If ℏ=h2π\hbar = \dfrac{h}{2\pi}, so the equation must be ΔxΔp≥ℏ2≥h2π2≥h4π\Delta x \Delta p \geq \dfrac{\hbar}{2} \ge \dfrac{h}{\dfrac{2\pi}{2}} \ge \dfrac{h}{4\pi}, is that right? I think so.

But what means the π\pi in the equation exactly? is it a better approximation?
(edited 9 years ago)
Ok, stability of water bottles based on total angle to tip it is worst either when empty or full of water due to the high centre of mass.

The answer I want is the maths to estimate the height the water needs to be for lowest centre of mass and the reasoning behind it.

Wild assumptions about the bottle are fine. Call it 30cm high, 50g weight, water, 7cm diameter. Or whatever.
Reply 94
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Phichi
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Original post by Kallisto
Oh, thanks for correction, Phichi. But I have thought about the equation again. Here are my considerations:

If ℏ=h2π\hbar = \dfrac{h}{2\pi}, so the equation must be ΔxΔp≥ℏ2≥h2π2≥h4π\Delta x \Delta p \geq \dfrac{\hbar}{2} \ge \dfrac{h}{\dfrac{2\pi}{2}} \ge \dfrac{h}{4\pi}, is that right? I think so.

But what means the π\pi in the equation exactly? is it a better approximation?


ΔxΔp≥ℏ2≥h2π(2)≥h4π\Delta x \Delta p \geq \dfrac{\hbar}{2} \ge \dfrac{h}{2\pi(2)} \ge \dfrac{h}{4\pi}

Slight error in your arithmetic.

The π\pi is in the proven expression for the principle. It happens to be when going through the proof, the 4π4\pi arises in the expression, however without going into some complexity, it's hard to show you why. Google some proofs for H.U.P, you'll see :smile:
Reply 95
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Phichi
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Original post by Motorbiker
Ok, stability of water bottles based on total angle to tip it is worst either when empty or full of water due to the high centre of mass.

The answer I want is the maths to estimate the height the water needs to be for lowest centre of mass and the reasoning behind it.

Wild assumptions about the bottle are fine. Call it 30cm high, 50g weight, water, 7cm diameter. Or whatever.


Need to clear some things up before even attempting this. Are we assuming the bottle has a certain shape? The structure of the bottle would be crucial, we know, for a uniform bottle, the centre of mass will lie on the axis going through the centre of the bottle, parallel to it, however ridges or such on the side of the bottle will change the effectively y co-ordinate for the COM. Are we allowed to assume it's a uniform hollow cylinder? Perhaps with negligible mass or such, needs some specifics.

Ultimately, the COM depends of the ratios of mass due to the air, water, and if decided bottle too. We know for a uniform bottle, the COM of bottle full of air would be in the same position as the bottle filled with water, assuming both are uniform.
(edited 9 years ago)
Original post by Phichi
Need to clear some things up before even attempting this. Are we assuming the bottle has a certain shape? The structure of the bottle would be crucial, we know, for a uniform bottle, the centre of mass will lie on the axis going through the centre of the bottle, parallel to it, however ridges or such on the side of the bottle will change the effectively y co-ordinate for the COM. Are we allowed to assume it's a uniform hollow cylinder? Perhaps with negligible mass or such, needs some specifics.

Ultimately, the COM depends of the ratios of mass due to the air, water, and if decided bottle too. We know for a uniform bottle, the COM of bottle full of air would be in the same position as the bottle filled with water, assuming both are uniform.


Assume a hollow uniform cylinder of a set mass.

And Yea, com is highest in either full or empty bottle. I want the equation for combined com, then the minimum point of this equation.

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Reply 97
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Phichi
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Original post by Motorbiker
Assume a hollow uniform cylinder of a set mass.

And Yea, com is highest in either full or empty bottle. I want the equation for combined com, then the minimum point of this equation.

Posted from TSR Mobile


Specific densities of air and water, or just leaving ρw\rho _w & ρa\rho _a in the equation?
Original post by Phichi
Specific densities of air and water, or just leaving ρw\rho _w & ρa\rho _a in the equation?


I'd assume nothing for air, would be negligible in this volume and 1 gram per cm^3 for water. So just working out the volume of water would give the effective mass of the liquid.

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Reply 99
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Phichi
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Original post by Motorbiker
I'd assume nothing for air, would be negligible in this volume and 1 gram per cm^3 for water. So just working out the volume of water would give the effective mass of the liquid.

Posted from TSR Mobile


So you are wanting it purely in terms of the cylinder mass and water mass? I've just had a go, including the mass of air.

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