Equations of Motion for an Object Falling in a Parabolic Bowl

[Binaryburst]

I have a bowl in the form of a parabola (say y=x^2, motion is in 2d! On the graph of the parabola) and i place an object on the edges of the bowl and let if fall in the bowl. I have gravity, and no friction. My question is plain simple: what are it's equations of motion? I know it has to oscillate. I know calculus quite well, but that doesn't seem to help. Is there something I'm missing out? I would simply love to see the correct equations :D. Thank you very, very much in advance.

 

[tiny-tim]

Welcome to PF!

Hi Binaryburst! Welcome to PF!

Start by using conservation of energy, and that will give you v as a function of y.

Carry on from there. ​

 

[Binaryburst]

Thanks for the tip :D but I've already gotten the speed with respect to x or y. I was wondering how I could get it as a function of time :)

 

[tiny-tim]

ok, now use cos or sin to get the x or y component of the speed …

you now have x' as a function of x (or y' as a function of y) ​

 

[Binaryburst]

I'm on it.

 

[Binaryburst]

I get elliptic integral !?

 

[tiny-tim]

very likely!

well, if you will start with a parabola! ​

 

[Binaryburst]

Wow... Thanks a lot! :D this is great!

 

[Binaryburst]

If I take the whole speed, not just it's components I get a sin(t) :D

 

[tiny-tim]

sorry, not following you …

you'll need to show the equations ​

 

[rollingstein]

I get elliptic integral !?

I was trying too. Do you get:

dx/dt=sqrt(2g) sqrt((1-4*x^2)*(x1^2-x^2))

 

[Binaryburst]

Nope. I'll post the equations right away.

 

[Binaryburst]

I get this: (2g(x1^2-x^2))^0.5. Then if i integrate it with respect to t i get x=x0*sin(t(2g)^0.5).

 

[rollingstein]

I get this: (2g(x1^2-x^2))^0.5. Then if i integrate it with respect to t i get x=x0*sin(t(2g)^0.5).

Somehow my expression for v_x has an additional (1-4x)^0.5 factor. Not sure what I'm doing wrong.

 

[rollingstein]

I get this: (2g(x1^2-x^2))^0.5. Then if i integrate it with respect to t i get x=x0*sin(t(2g)^0.5).

This term (2g(x1^2-x^2))^0.5

Is that your v or v_x?

 

[Binaryburst]

My equation is for v alone.

 

[Binaryburst]

I'm thinking how could i get it without using the conservation of energy and using forces.

 

[rollingstein]

My equation is for v alone.

What did you do next?

 

[Binaryburst]

I had v=f(x). Rewritten it as follows: v/f(x)=1 ; 1/f(x)*dx/dt=1. ; Integrate with respect to t
Int( 1/f(x)* dx/dt * dt ) = int( 1 dt )

 

[rollingstein]

I had v=f(x). Rewritten it as follows: v/f(x)=1 ; 1/f(x)*dx/dt=1. ; Integrate with respect to t
Int( 1/f(x)* dx/dt * dt ) = int( 1 dt )

That's where I think you are wrong.

You've put v=dx/dt.

Only v_x = dx/dt

But I could be talking nonsense! Be warned. But I'd love to see your opinion.

Your solution does look super tempting! :)

 

[Binaryburst]

Hmmm.. That's interesting I actually got the vx. Sorry for the blunder. I was too excited :) correcting the mistake.

 

[rollingstein]

Hmmm.. That's interesting I actually got the vx. Sorry for the blunder. I was too excited :)

Funny point is your solution seems to satisfy all the boundary conditions etc. I'm puzzled.

My solution integrates to something super messy. :(

 

[Binaryburst]

I am super puzzelled as well. I can't figure out what I did.

 

[rollingstein]

Another reason why your solution seems wrong to me.

take dx/dt and dy/dt

$v_x^2 + v_y^2 = v^2$

But yours don't seem to sum up to v. Try.

 

[Binaryburst]

Actually i got the total speed dependent only on the x-axis so saying that v.total is dx/dt is correct because it's no longer the slope of the parabola but the slope of f(x).

 

[rollingstein]

Actually i got the total speed dependent only on the x-axis so saying that v.total is dx/dt is correct because it's no longer the slope of the parabola but the slope of f(x).

I don't agree.

The only way v = dx/dt is if $\vec{v}$ was directed along the x axis.

i.e. v_y was zero.

But here the velocity does have a y component so I don't see how you are allowed to do what you did.

OTOH, your elegant solution form tempts me but still...

PS. Maybe I am wrong. I'm no expert.

 

[Binaryburst]

Omg! What a humongous mistake in my formulation of the conservation of energy!

 

[rollingstein]

Omg! What a humongous mistake in my formulation of the conservation of energy!

Conservation is fine. Your v is ok.

It's where you put v=dx/dt that the problems start.

 

[rollingstein]

Here's what I get :

$\frac{dx}{dt} = \sqrt{2g} \sqrt{\frac{x_1^2-x^2}{1+4x^2}}$

Sad part: If this is right the integral is ugly.

 

[Binaryburst]

It is right. That's what I got too... That's just v*cos(theta). It results that v is equal to f(x).

 

[rollingstein]

It is right. That's what I got too... That's just v*cos(theta). It results that v is equal to f(x).

Now try writing x(t). :)

 

[Binaryburst]

In the consevation of Energy formula you got the speed with respect to the height only which is x^2 so we actually got vx :~> when you replace y with x^2 and vy when you leave y unchanged. PS: I'm not an expert either. :D

 

[rollingstein]

In the consevation of Energy formula you got the speed with respect to the height only which is x^2 so we actually got vy :~>

Nope. I don't think so.

 

[Binaryburst]

I did a simulation with vx and it matched the sinusoidal, it seems. Any help from the experts?

 

[mfb]

Where do you consider the motion along the bowl? It will lead to accelerations both in vx and vy, to follow the shape.I'll try to start from scratch - with energy conservation, but it is possible to avoid this word if necessary.

If the object is at rest at position $x_0>0$, it has a total energy of $gx_0^2$, setting its mass and a meter to 1.
At position x, $0<x<x_0$, energy conservation leads to $v^2=2g(x_0^2-x^2)$.
The derivative of the parabola there is 2x, therefore $v_x=\frac{1}{2x}v_y$. It follows that $v^2 = v_x^2 + v_y^2 = v_x^2 (1+4x^2)$. Combining both,
$$\frac{dx}{dt}= v_x = \sqrt{\frac{v^2}{1+4x^2}} = \sqrt{\frac{2g(x_0^2-x^2)}{1+4x^2}}$$
The sign is arbitrary and depends on the current part of the oscillation.
This leads to ugly elliptic integrals. It can be written as
$$\frac{dt}{dx}=\sqrt{\frac{1+4x^2}{2g(x_0^2-x^2)}}$$
While this gives elliptic integrals as well, it allows to determine t(x) via numerical integration.

g=10 and x₀=1 leads to a period of T=2.36s.
For large x₀ (>>1), most of the parabola is like a free fall, and the period should approach 4 times this free-fall time. As an example, x=sqrt(500) leads to T=40.06, whereas a free fall would give T=40s.
For small x₀ (<<1), the deflection of the bowl is negligible, and we get a harmonic oscillator. The numerator for dt/dx can be approximated as 1, and the period approaches $T \approx \frac{2\pi}{\sqrt{2g}}$.