Oscillation of a particle on a parabolic surface [equation of motion]

[happyparticle]

Homework Statement

equation of motion

Relevant Equations

F = ma
$\ddot{\theta} + \frac{k}{m} x = 0$

Hi,

I have a particle on a parabolic surface $$y = Ax^2$$ and I have to show that the frequency is $$\omega = \sqrt{2Ag}$$

I don't know how to deal with a parabola. I don't think I can use the polar coordinates like a circle.

I don't see how to start this problem and in which coordinates system.

For a circle I can use the polar coordinates and then use $F = ma => -mg sin \theta = mR\ddot{\omega}$

I have to get the equation of motion $$\ddot{\theta} + 2Ag \theta = 0$$

thus, $2Ag = \omega^2$

 

[haruspex]

which coordinates system.

It’s straightforward enough in Cartesian.
Draw a diagram of the particle at some (x,y) and analyse the forces.

 

[happyparticle]

I think using energy $u(y) = mgy$ can get the job done. Someone pointed out to me. I didn't notice.

However, is it the only way?

 

[haruspex]

I think using energy $u(y) = mgy$ can get the job done.

What about KE?

 

[happyparticle]

What about KE?

$y_{max}, E = mgy_{max}$

$K = E - U, \frac{1}{2}m\dot{x}^2 = mgy_{max} - mgy$

$\dot{x} = \sqrt{2g(y_{max} - Ax^2)}$

I'm stuck here, Am I in the right path?

I'm wondering if $y_{max} = Ax_{max}^2$

 

[haruspex]

$\frac{1}{2}m\dot{x}^2 = mgy_{max} - mgy$

What about $\dot y$?

 

[happyparticle]

What about $\dot y$?

$\dot{y} = 2A\dot{x}$

$\dot{x} = \frac{\sqrt{2g}}{2A}$

That doesn't work and I can find why.

 

[haruspex]

$\dot{y} = 2A\dot{x}$

$\dot{x} = \frac{\sqrt{2g}}{2A}$

That doesn't work and I can find why.

No, I mean as part of the KE.

 

[happyparticle]

No, I mean as part of the KE.

That's what I did.

$2A\dot{x} =\sqrt{2g} (y_{max} - y)$

$\dot{x} = \frac{\sqrt{2g}}{2A}(y_{max} - y)$

 

[haruspex]

$\dot{y} = 2A\dot{x}$

That equation is wrong. You have not differentiated x² correctly. Use the chain rule.

$\frac{1}{2}m\dot{x}^2 = mgy_{max} - mgy$

That equation is wrong because it omits the KE embodied in the vertical component of motion.

 

[happyparticle]

If $\frac{1}{2}m\dot{y}^2 = mgy_{max} - mgy$

I get $x = \frac{\sqrt{2g}}{2A}(y_{max} - y)$

or $\dot{y} = \sqrt{2g}(y_{max} - y)$

 

[haruspex]

If $\frac{1}{2}m\dot{y}^2 = mgy_{max} - mgy$

Now you are only counting the vertical component of KE. Both contribute.

 

[happyparticle]

Now you are only counting the vertical component of KE. Both contribute.

Ah, I see, but how I do that.
$K = \frac{1}{2}m\dot{y}^2 + \frac{1}{2}m\dot{x}^2$

$U = mgy$

$E = mgy_{max} + mgx_{max}$Which gives me

$\dot{y}^2 + \dot{x}^2 = 2g(y_{max} + x_{max} - y)$

I'm not sure what to do with the left side.

 

[haruspex]

$E = mgy_{max} + myx_{max}$

I assume you meant mgx_max, but I don't understand the reason for that term. What aspect of GPE is not covered by mgy?

 

[happyparticle]

E = total energy

When the particle is at $y_{max}$ then all the energy = U, thus $E = mgy_{max}$

 

[haruspex]

E = total energy

When the particle is at $y_{max}$ then all the energy = U, thus $E = mgy_{max}$

Sorry, missed out a crucial word ("not") in post #14.
So, yes, $E = mgy_{max}$.

What dynamic energy equation does that give you (instead of the one at the end of post #13), and what equation connects $\dot y, \dot x$ ( it is not $\dot y=2A\dot x$)?

 

[happyparticle]

What dynamic energy equation does that give you (instead of the one at the end of post #13), and what equation connects $\dot y, \dot x$ ( it is not $\dot y=2A\dot x$)?

$\dot{y}^2 + \dot{x}^2 = 2g(y_{max} - y)$

$\dot{y} = 2Ax$

$\dot{x} = \frac{1}{2\sqrt{A}\sqrt{y}}$

sorry, I can't find a connection between $\dot{x},\dot{y}$

 

[haruspex]

$\dot{y} = 2Ax$
sorry, I can't find a connection between $\dot{x},\dot{y}$

That equation is wrong. You have differentiated the LHS wrt t but the RHS wrt x. How do you differentiate Ax² wrt t?
The relationship I am asking for is the correct form of that.

 

[happyparticle]

I'm sorry...
$\dot{y} = A\dot{x}^2$

$\dot{y}^2 + \frac{\dot{y}}{A} = 2g(y_{max} - y)$

I can't have $\sqrt{2Ag}$

 

[haruspex]

I'm sorry...
$\dot{y} = A\dot{x}^2$

Still wrong.
Use the chain rule.

 

[happyparticle]

A is a constant
$\frac{dy}{dt} = A \frac{d(x^2)}{dt}$

 

[haruspex]

A is a constant
$\frac{dy}{dt} = A \frac{d(x^2)}{dt}$

Yes, but what do you get when you apply the chain rule to the RHS?
You've so far tried $2Ax$ and $A\dot x^2$.
(You are familiar with the chain rule?)

 

[happyparticle]

(You are familiar with the chain rule?)

I think so, but I don't see how to use the chain rule here.
I mean, I can use the chain rule if I have something like $\sqrt{(x^2y-xy)}$

 

[haruspex]

I think so, but I don't see how to use the chain rule here.
I mean, I can use the chain rule if I have something like $\sqrt{(x^2y-xy)}$

$\frac d{dt}f(x(t))=\frac d{dx}f.\frac d{dt}x$.

 

[happyparticle]

x is a function?

x = x(t) ?

 

[haruspex]

x is a function?

x = x(t) ?

Yes. In your problem x and y both depend on t, and there is a fixed relationship between x and y, so you can think of it as y depends on x depends on t.

 

[happyparticle]

Alright,

$f(x) = Ax^2$

$\frac{d}{dx}f = 2Ax$
$\frac{d}{dt}x = \dot{x}^2$

$2Ax \cdot \dot{x}^2$

Is that right?

 

[haruspex]

$\frac{d}{dt}x = \dot{x}^2$

Inventive.

 

[happyparticle]

I though if $f(x) = Ax^2, x(t) = x^2$

why not $x^2$ ?

 

[haruspex]

I though if $f(x) = Ax^2, x(t) = x^2$

What if I were to define another function g(x)=x³? Would that suddenly make x(t) equal x³?
No, x(t) just means that x is some function of t. It doesn't tell you what the function is.
$\frac{d}{dt}x=\dot x$.

 

[happyparticle]

Alright, so I got

$2Ax \cdot \dot{x} = 2g(y_{max}- y)$

I still don't see how to get
$\omega = \sqrt{2gA}$

$t = \int\frac{x}{\frac{g(y_{max}-y)}{Ax}} dx$

$\omega = \frac{2\pi}{t}$

$\omega = 2\pi \cdot \frac{g}{Ax^2}$

 

[haruspex]

Alright, so I got

$2Ax \cdot \dot{x} = 2g(y_{max}- y)$

Have you studied dimensional analysis? It's a powerful way to check your equations.
In the present case we can say x and y have length dimension (L). We write that as [x]=[y]=L.
To be consistent, A has dimension 1/L, so that $y=Ax^2$ works out correctly. In the notation: [A]=L^-1,
Derivatives are like division: $[\dot x]=LT^{-1}$, etc.
In your equation above, on the left you have
$L^{-1}L.LT^{-1}=LT^{-1}$, i.e. velocity.
On the right: $LT^{-2}.L=L^2T^{-2}$, i.e. a velocity squared.
So the equation has to be wrong. This technique would have picked up some of your previous errors.

Anyway, you now have correctly that $\dot y=2Ax.\dot x$.
Plug that into

$\dot{y}^2 + \dot{x}^2 = 2g(y_{max} - y)$

 

[happyparticle]

Have you studied dimensional analysis? It's a powerful way to check your equations.

No, I don't. Maybe that's why I don't fully understand. However, it should be a way to get the answer by using the small oscillations and the simple harmonic motion.

I was trying to get something in this form $$\dot{\theta} = \sqrt{2Ag}[elliptic integral]$$ or $$\ddot{\theta} + \frac{g}{l} sin \theta = 0$$

Using F = Ma or E = K+U

However, as I said I don't know how to deal with a parabola.

 

[haruspex]

No, I don't. Maybe that's why I don't fully understand. However, it should be a way to get the answer by using the small oscillations and the simple harmonic motion.

I was trying to get something in this form $$\dot{\theta} = \sqrt{2Ag}[elliptic integral]$$ or $$\ddot{\theta} + \frac{g}{l} sin \theta = 0$$

Using F = Ma or E = K+U

However, as I said I don't know how to deal with a parabola.

Well, you seem to be prone to algebraic errors, so I strongly recommend you study dimensional analysis. It's not hard.
Meanwhile, please proceed as I instructed at the end of post #32. You need to get to an equation in which the only variables are x and its time derivatives.

 

[happyparticle]

Alright, thanks for your help and your patience.