Motion of a particle along a curve under gravity

[PhysiSmo]

[SOLVED] Motion of a particle along a curve under gravity

Homework Statement

Assume that a curve $$F(x,y)=0$$ is given. Find the condition that $$F$$ satisfies, under which a point particle of mass $$m$$ is moving along the curve (and never falls of it) under gravity.

Homework Equations

I suppose that Newton's second law plays a role somewhere in the problem.

$$m\ddot{r}=-mg\hat{y},$$

where $$\vec{r}$$ is the vector position of the particle.

The Attempt at a Solution

I have absolutely no clues on this one. The only idea I had was that since the particle is always on the curve, its velocity is tangential to the curve, thus perpendicular to a vector $$\vec{a}$$ that is perpendicular to the curve at a point (x,y).

Such a vector $$\vec{a}$$ is the gradient of the curve, that is $$\nabla F(x,y)$$. So that one condition could be

$$\vec{a}\cdot \dot{r}=0$$, or

$$\dot{x}\frac{\partial F}{\partial x}+\dot{y}\frac{\partial F}{\partial y}=0,$$

with $$\dot{x},\dot{y}$$ satisfying Newton's second law.

I don't really like my approach, any better ideas?

 

[tiny-tim]

Hi PhysiSmo!

It will come off if the reaction force is zero.

Does that help?

 

[PhysiSmo]

Thanks for answering tiny-tim!

I don't see how the reaction force enters the game, perhaps I didn't stated the problem clearly. Imagine a particle traveling on the surface of a sphere, or a cylinder, under gravity. Due to those objects's curvatures, the particle will definitely leave the surface at some point. And this will happen independently of the reaction force.

or not? :-)

 

[tiny-tim]

And this will happen independently of the reaction force.

No … there are two forces on the particle - its weight, and the reaction force - and they produce an acceleration.

That acceleration has to be enough to keep it on the curve.

It will leave the curve when the reaction force is zero.

 

[D H]

perhaps I didn't stated the problem clearly.

I think that perhaps you didn't state the problem clearly. What exactly do you mean by "moving along the curve (and never falls of it) under gravity"? Is this a real physical curve or surface that exerts a normal force (apparently tiny tim's interpretation) or is it the locus of points a particle will follow while subject to gravity and nothing else?

 

[PhysiSmo]

D H, it's a real physical curve, just like tim's interpretation. Taking the cylinder as an example, this curve is just a part of circle. The particle starts moving at $$\theta=\frac{\pi}{2}$$, given some initial energy $$E_0$$, and under the influence of gravity. At some angle, the particle leaves the curve.

What I'm looking for, is a condition satisfied by a random curve $$F(x,y)=0$$, so that the particle never leaves it.

 

[PhysiSmo]

No … there are two forces on the particle - its weight, and the reaction force - and they produce an acceleration.

That acceleration has to be enough to keep it on the curve.

It will leave the curve when the reaction force is zero.

So I have to find a relation for the reaction force produced by a random curve, right? A perpendicular surface has zero reaction force, while for a plane surface takes its maximum value. So I guess it has to do with the curve's curvature, right?!

I'll think it for a while...

Thanx again!

 

[tiny-tim]

So I guess it has to do with the curve's curvature, right?!

Hi PhysiSmo!

Yes, indirectly - but this is real physics, so it mostly has to do with good ol' Newton's second law, and the acceleration!

 

[PhysiSmo]

Hi PhysiSmo!

Yes, indirectly - but this is real physics, so it mostly has to do with good ol' Newton's second law, and the acceleration!

OK, now I'm confused! On the one hand, one could say that if acceleration is equal to zero, the particle stays on the curve, while on the other hand, one could say that acceleration does not need to be zero, but always tangential to the curve, the particle still stays on the curve!

Am I missing something here?

 

[tiny-tim]

… one could say that acceleration does not need to be zero, but always tangential to the curve, the particle still stays on the curve!

Am I missing something here?

Yes … you're thinking of acceleration relative to the curve.

Newton's law will only give us (directly) the actual acceleration.

Think of something sliding down a curved surface … it will have a non-zero tangential acceleration, obviously, because it's picking up speed.

But it will also have a non-zero normal (radial) acceleration, won't it?

And yet it still stays on the surface!

The question is, at any particular speed, what does its normal acceleration need to be more (or less) than, to keep it on?

 

[PhysiSmo]

The question is, at any particular speed, what does its normal acceleration need to be more (or less) than, to keep it on?

What could one say about the normal acceleration's magnitude? Doesn't it only need to be pointing "downward", so that the particle won't leave the curve?

 

[tiny-tim]

No, because there could be a "downward" acceleration that is too small compared with the rate at which the curve is "falling away".

The downward acceleration must be enough to "keep up with" the curve.

Hint: if the radius of curvature of the curve at a particular point is r, what is the normal acceleration?

 

[PhysiSmo]

So the normal acceleration must be more than the rate at which is the curve is "falling away". The normal acceleration is

$$a_N=\frac{u^2}{r}=ku^2,$$

$$k$$ being the curvature, $$r$$ the radius of the curve at a point x.

Hmmm...I still can't see how to incorporate my curve F(x,y)=0 though...ps. thank you very much for your answers..

 

[tiny-tim]

The normal acceleration is

$$a_N=\frac{u^2}{r}=ku^2,$$

That's right!

And what is u, in terms of y … ?

 

[PhysiSmo]

For a given position vector $$\vec{r}=x\hat{i}+y\hat{j}$$, the velocity is

$$u=\sqrt{\dot{x}^2+\dot{y}^2}...$$

 

[tiny-tim]

Yes, but what is its value?

This is a physics problem!

Hint: conservation of … ?

 

[PhysiSmo]

Let's conserve the energy then!

$$E=T+V=\frac{1}{2}mu^2+mgh=\frac{1}{2}mu^2+mgy.$$

Thus we obtain a relation $$u=u(y),$$ namely

$$u^2=\frac{2E}{m}-2gy,$$

so that

$$a_N=2\frac{\frac{E}{m}-gy}{r}.$$

As we said before, the normal acceleration must be more than the rate at which is the curve is "falling away", which depends on the curvature $$k$$. Is this correct?

 

[tiny-tim]

$$u^2=\frac{2E}{m}-gy,$$

so that

$$a_N=\frac{\frac{2E}{m}-gy}{r}.$$

As we said before, the normal acceleration must be more than the rate at which is the curve is "falling away", which depends on the curvature $$k$$. Is this correct?

Hi PhysiSmo!

Yes, that's right … but it's neater if we say let h be the height at which it was at rest.

Then: $$a_N=\frac{g(h\,-\,y)}{r}\,.$$

It will lose contact when the reaction force is zero … in other words, when this acceleration is exactly enough to balance the normal component of gravity.

Which is … ?

 

[PhysiSmo]

Hi Tim!

Applying all of the above for the case of a particle moving on a circle, one finds that the particle leaves the curve when

$$a_N=G_N,$$

that is when the acceleration is equal to the normal component of gravity. One finds then the angle at which the particle falls off. In our case, though, the normal component of gravity can be written as

$$G_N=mgcos\theta.$$

So the particle never leaves the curve if $$a_N\geq mgcos\theta.$$

But how can we express $$\theta$$ in terms of known variables? The direction of $$G_N$$ is the same as the direction of the curvature radius $$r$$. That should be a clue, but I couldn't find something useful to proceed..

 

[tiny-tim]

So the particle never leaves the curve if $$a_N\geq mgcos\theta.$$

But how can we express $$\theta$$ in terms of known variables?

Hi PhysiSmo!

Well, the tangent of the slope of the normal is dx/dy, isn't it?

So the only remaining thing to convert to x and y is the value of the radius of curvature, for which there's an extremely well-known formula …

which I've forgotten! ​

 

[PhysiSmo]

Well, if y=f(x), the radius of the curvature is simply

$$r=\frac{\left(1+\left(\frac{dy}{dx}\right)^2\right)^{3/2}}{\left|\frac{d^2y}{dx^2}\right|}.$$

For the case of $$F(x,y)=0,$$ though, how can one generalize this relation, without introducing a time parametrization, namely without writing

$$r=\frac{(x'^2+y'^2)^{3/2}}{|x'y''-y'x''|}.$$

Is it correct to write

$$r=\frac{(1+(\nabla F)^2)^{3/2}}{|\nabla^2F|},$$

or not...?

ps. thank you very much for your replies!

 

[tiny-tim]

For the case of $$F(x,y)=0,$$ though, how can one generalize this relation, without introducing a time parametrization, namely without writing

$$r=\frac{(x'^2+y'^2)^{3/2}}{|x'y''-y'x''|}.$$

hmm … I hadn't thought of that.

(I'd forgotten what the original question was! )

Let's see … write it F(x,f(x)) = 0.

Then, differentiating wrt x:

∂F/∂x + f´(x)∂F/∂y = 0.

So y´ = f´(x) = -(∂F/∂x)/(∂F/∂y).

And y´´= … ?

 

[PhysiSmo]

Well, the tangent of the slope of the normal is dx/dy, isn't it?

Hi Tim!

What exactly do you mean by that? I'm not sure I understand the terminology, do you mean that $$tan\theta=dx/dy$$?

Let's see … write it F(x,f(x)) = 0.

Then, differentiating wrt x:

∂F/∂x + f´(x)∂F/∂y = 0.

So y´ = f´(x) = -(∂F/∂x)/(∂F/∂y).

And y´´= … ?

Differentiating once more wrt x, we get

$$\frac{\partial}{\partial x}\frac{\partial F}{\partial x}+\frac{\partial}{\partial x}\left(f'(x)\frac{\partial F}{\partial y}\right)=0$$

$$\frac{\partial^2F}{\partial x^2}+f'(x)\frac{\partial^2F}{\partial x\partial y}+\frac{\partial F}{\partial y}f''(x)=0.$$

So that y''=f''(x) is also known in terms of partial derivatives of F...

 

[tiny-tim]

Hi PhysiSmo!

Yes, that's it!

(and yes, $tan\theta=dx/dy$ , where theta is the angle from the horizontal.)

 

[PhysiSmo]

Hmm...it seems quite bizarre that one has to combine the equations

$$a_N\geq mgcos\theta.$$

and

$$tan\theta=dx/dy$$

so that

$$a_N\geq mg cos(arctan(dx/dy)).$$

Isn't it odd?

 

[tiny-tim]

:rofl: The way you put it is odd! :rofl:​

The way I put it is:

$$a_N^2\,\geq\,m^2g^2 cos^2\theta\,=\,\frac{m^2g^2}{(1\,-\,tan^2\theta)}$$
and $$tan\theta=dx/dy$$ ! ​

(erm … are we using the same theta? … I've lost track … )

 

[PhysiSmo]

:rofl: The way you put it is odd! :rofl:​

The way I put it is:

$$a_N^2\,\geq\,m^2g^2 cos^2\theta\,=\,\frac{m^2g^2}{(1\,-\,tan^2\theta)}$$
and $$tan\theta=dx/dy$$ ! ​

(erm … are we using the same theta? … I've lost track … )

:rofl: I'd say that this relation is somehow more elegant than mine :rofl:

My theta appeared in the normal component of gravity, g_N, which is in the same direction along with the radius of curvature!

Edit: Why did the squares got into the game? (a_N^2...)

 

[tiny-tim]

… elegance … !

Why did the squares got into the game? (a_N^2...)

Elegance, my dear chap! ​

I just don't like equations with square roots in them!

 

[PhysiSmo]

Elegance, my dear chap! ​

I just don't like equations with square roots in them!

Fair enough! What about theta then, are we using the same theta?

 

[tiny-tim]

My theta was from the horizontal, and I think your theta is from the vertical.

 

[PhysiSmo]

Hi Tim!

Everything is clarified! Thank you very very much for your time!