- #1

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l^2*((dθ/dt)/dt)-mg*d(cosθ)/(dθ/dt)=dL/dθ

The part I want to focus on is d(cosθ)/(dθ/dt). I know d(cosθ) is not a constant, because it depends on dθ/dt, since it's its integral with respect to dt, but how do I find what it is?

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- Thread starter Andreas C
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- #1

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l^2*((dθ/dt)/dt)-mg*d(cosθ)/(dθ/dt)=dL/dθ

The part I want to focus on is d(cosθ)/(dθ/dt). I know d(cosθ) is not a constant, because it depends on dθ/dt, since it's its integral with respect to dt, but how do I find what it is?

- #2

BvU

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Why don't you show us the full problem statement and the way you found this 'mess'(which I don't know if it's right):

- #3

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BvU

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Furthermore -- and this is rather important -- you don't really mean d(cosθ)/(dθ/dt) ##\qquad## (or rather $$ {d\cos\theta \over {d{d\theta\over dt} }} $$ but instead you mean $$

{\partial \cos\theta \over {\partial {d\theta\over dt} } }$$ The ##\partial## derivative is different from the ##d## derivative (I think it's called the total derivative). Look it up - it appears a lot more in physics. The Euler Lagrange equation is a

Meaning you consider ##\mathcal{L}## as a function of

{ \partial \mathcal{L} \over \partial q} - {d\over dt}\left (\partial L\over\partial \dot q \right ) = 0

$$And what do you get when you do a partial differentiation wrt ##\dot \theta## of your $$\mathcal L = {m\over 2} \left (\ell\dot\theta\right)^2 - mg\ell\left (1-\cos\theta\right ) \ \ \rm ? $$

- #5

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Furthermore -- and this is rather important -- you don't really mean d(cosθ)/(dθ/dt) ##\qquad## (or rather $$ {d\cos\theta \over {d{d\theta\over dt} }} $$ but instead you mean $$

{\partial \cos\theta \over {\partial {d\theta\over dt} } }$$ The ##\partial## derivative is different from the ##d## derivative (I think it's called the total derivative). Look it up - it appears a lot more in physics. The Euler Lagrange equation is apartialdifferential equation.

Meaning you consider ##\mathcal{L}## as a function ofindependentvariables ##q## and ##\dot q## and take partial derivatives (differentiate to the one while keeping the other constant):$$

{ \partial \mathcal{L} \over \partial q} - {d\over dt}\left (\partial L\over\partial \dot q \right ) = 0

$$And what do you get when you do a partial differentiation wrt ##\dot \theta## of your $$\mathcal L = {m\over 2} \left (\ell\dot\theta\right)^2 - mg\ell\left (1-\cos\theta\right ) \ \ \rm ? $$

That is actually exactly what I meant. That symbol is called the partial derivative, but I didn't know how to insert it, and I thought it would be the same, thanks :) I'm gonna work it out.

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Hint: shouldn't take long !

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no, you still want to differentiate it !I got (m*l^2*(dθ/dt))/2, is that right?

The hint was a nudge to make you see (which you did) that ##\cos\theta## is not a function of ##\dot\theta## so the partial derivative of ##V## wrt ##\dot\theta## is zero.

Yes you should. It's the difference beteween a stable equilibrium at ##\theta = 0## and an unstable one.for ∂L/∂θ I got -mgl*sinθ, should I worry about the minus sign?

Kudos! Not an easy subject all on your own. Good thing there is PF !Don't worry, it's not for homework, I'm teaching myself classical mechanics.

- #10

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no, you still want to differentiate it !

The hint was a nudge to make you see (which you did) that ##\cos\theta## is not a function of ##\dot\theta## so the partial derivative of ##V## wrt ##\dot\theta## is zero.

Yes you should. It's the difference beteween a stable equilibrium at ##\theta = 0## and an unstable one.

Kudos! Not an easy subject all on your own. Good thing there is PF !

Ηmmm... I'm a bit confused now... Are you saying I have to differentiate (m*l^2*(dθ/dt))/2 again?

Yes you should.

So I should have a plus sign instead of a minus one?

- #11

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- #12

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Not 'again'. m*l^2*(dθ/dt))/2 is the kinetic energy T in ##\mathcal L = T - V\ ##. You haven't differentiated it yet.Are you saying I have to differentiate (m*l^2*(dθ/dt))/2 again?

[edit] My mistake, I misread the /2 for ^2. Anyway, you found it and fixed it.

Certainly not. You got a minus sign for ##{d\mathcal L\over d\theta}## and you want to keep it. The Euler Lagrange equation is $${ \partial \mathcal{L} \over \partial q} - {d\over dt}\left (\partial L\over\partial \dot q \right ) = 0$$ and if you do it right you automatically end up with ##\ddot \theta + \omega^2 \sin\theta = 0##.So I should have a plus sign instead of a minus one?

(or, of course ##- \omega^2 \sin\theta - \ddot \theta = 0## ) .

- #13

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Not 'again'. m*l^2*(dθ/dt))/2 is the kinetic energy T in ##\mathcal L = T - V\ ##. You haven't differentiated it yet.

Certainly not. You got a minus sign for ##{d\mathcal L\over d\theta}## and you want to keep it. The Euler Lagrange equation is $${ \partial \mathcal{L} \over \partial q} - {d\over dt}\left (\partial L\over\partial \dot q \right ) = 0$$ and if you do it right you automatically end up with ##\ddot \theta + \omega^2 \sin\theta = 0##.

(or, of course ##- \omega^2 \sin\theta - \ddot \theta = 0## ) .

Ok, I think I've got it: ##\ddot \theta##=-mgl*sinθ. Is that what you mean?

- #14

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Yes! Well done.

(you lost a factor ml^{2} on the left, however...)

(you lost a factor ml

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- #17

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If my calculations are correct, then ##ml^2 \ddot θ =-mgl*sinθ##, which means that ##\ddot θ = -(g*sinθ)/l##, which doesn't sound right at all...

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- #19

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The period is ##T = {2\pi\over \omega} = 2\pi\sqrt{\ell/g}##.

- #20

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The period is ##T = {2\pi\over \omega} = 2\pi\sqrt{\ell/g}##.

Ah, ok, so it's a special case. However, the equation of motion I found here is for any amplitude, right?

Also, say I now want to find the x and y components of acceleration and velocity. How do I do that? I had a couple of ideas but I'm not sure if they are going to be fruitful...

- #21

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the equation of motion is a second order differential equation. A solution needs two boundary conditions, e.g. an amplitude and a phase (or the a and b in the expression above).However, the equation of motion I found here is for any amplitude, right

For that you need to express x and y in ##\theta## ; rather straightforward !Also, say I now want to find the x and y components of acceleration and velocity

- #22

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the equation of motion is a second order differential equation. A solution needs two boundary conditions, e.g. an amplitude and a phase (or the a and b in the expression above).

For that you need to express x and y in ##\theta## ; rather straightforward !

So I have to know the initial θ in the case of a pendulum starting from a stationary position, because that would be the amplitude. What about the phase?

Also, what do you mean by "For that you need to express x and y in ##\theta## ; rather straightforward !"?

Does that mean that ##\ddot x = l*sin\ddotθ##?

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- #23

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Never done a mass and spring oscillator before ?So I have to know the initial θ in the case of a pendulum starting from a stationary position, because that would be the amplitude. What about the phase?

##\ \ddot\theta = -\omega^2\theta\ ## with ##\ \theta(0) = \theta_0\ ## and ##\ \dot \theta(0) = 0\ ## gives ##\ \theta(t) = \theta_0\cos(\omega t) = \theta_0\sin(\omega t + {\pi\over 2})##

As long as ##\sin\theta\approx\theta## you have ##x=\ell\theta## so ##\dot x = \ell\dot\theta## and ##\ddot x = \ell\ddot\theta## etc. With ##x=\ell\sin\theta## things get more complicated (24A here).Does that mean that ##\ddot x = l*\sin\ddot\theta \rm ?##

- #24

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Never done a mass and spring oscillator before ?

##\ \ddot\theta = -\omega^2\theta\ ## with ##\ \theta(0) = \theta_0\ ## and ##\ \dot \theta(0) = 0\ ## gives ##\ \theta(t) = \theta_0\cos(\omega t) = \theta_0\sin(\omega t + {\pi\over 2})##

As long as ##\sin\theta\approx\theta## you have ##x=\ell\theta## so ##\dot x = \ell\dot\theta## and ##\ddot x = \ell\ddot\theta## etc. With ##x=\ell\sin\theta## things get more complicated (24A here).

Yeah, but that would be the phase for sinθ approaching θ, and I'm looking to find the equations for any amplitude, not just very small ones... It also bugs me how for some reason, the angular acceleration always turns out to be negative, shouldn't it be positive when the angular speed is increasing, like after it reaches the maximum height?

- #25

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Well, what of the link? There are more of those (google mathematical pendulum, physical pendulum or just pendulum). If you just want the equations: yoeru can differentiate ##\ x =\ell\sin\theta\ ## yourself, I expect ? Is there a specific question ?I'm looking to find the equations for any amplitude, not just very small ones...

Where did you get that idea ? The minus sign only indicates the restoring force (and hence the acceleration) is opposite to the deviation from equilibrium ! If the deviation is negative, the acceleration is positive.It also bugs me how for some reason, the angular acceleration always turns out to be negative, shouldn't it be positive when the angular speed is increasing, like after it reaches the maximum height?

A thorough understanding of harmonic oscillators is very important for many areas of physics. If you have energy left over, studying things like damping is more useful than studying the implications of ##\ \sin\theta\ ##not ## \ \approx\theta\ ## (my opinion).

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