Finding Speed Needed to Keep Particle on Rippled Surface

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In summary, the conversation discusses the problem of a particle moving on a rippled surface under the influence of gravity. The elevation of the surface is given by a cosine wave, and the particle starts with an initial velocity in the x direction. The question is to determine the values of the initial velocity that will keep the particle on the surface at all times. The conversation includes a hint and a step-by-step approach to solving the problem, with the final answer being that the particle will stay on the surface if the initial velocity is between 0 and \frac{1}{k}\sqrt{\frac{g}{d}}. The conversation also includes a discussion on the reasoning behind this answer and how it relates to the second derivatives of the particle's trajectory
  • #1
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Im having a little trouble starting this question

Consider a particle moving without friction on a rippled surface. Gravity acts in the negative [tex]h[/tex] direction. The elevation [tex]h(x)[/tex] of the surface is given by [tex]h(x) = d\cos(kx)[/tex]. If the particle starts at [tex]x = 0[/tex] with a speed [tex]v[/tex] in the [tex]x[/tex] direction, for what values of [tex]v[/tex] will the particle stay on the surface at all times?

There is a picture which shows the curve as a regular cosine wave with period [tex]\displaystyle{T = \frac{2\pi}{k}}[/tex].

Im a little bit unsure on how to go about answering the question. Should i be looking at the forces when the acceleration is a maximum (at the crest of the curve)? And then trying to set up an equation which involves the weight, normal force (which vanishes) and the 'centripetal' force (im not sure of the correct name for this type of force on a cosine wave)

Any help or a hint on how to go about it would be much appreciated.

Thankyou
 
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  • #2
I'm pretty sure I found a way to do this! :smile: I'm going to give you a hint first; see if you arrive at something. If not I'll show you what I got.

HINT: The particle stays on the "ramp" iif the rate of change of height of its natural trajectory (i.e. its trajectory due to gravity) with respect to position (i.e. dh/dx) is lesser (because of the minus sign) than the rate of change of height of the ramp itself with respect to position.

This is the reasoning with which I started the problem. I got the answer: "It will stay on the ramp, provided the initial speed is somewhere between 0 and [itex]\sqrt{g}/k[/itex].
 
  • #3
Hello!

Thankyou very much

Im a little confused with your idea. Are you saying that the velocity of the particle (which is tangent to the curve), must be less than [tex]\frac{dh}{dx}[/tex]?

Thankyou again.
 
  • #4
Not really the 'velocity' of the particle... velocity is the rate of change of position with respect to TIME. I'm talking about the rate of change of vertical position with respect to the horizontal position.

It makes sense: gravity must make the ball drop faster than the ramp does, otherwise it will be in midair right after it is lauched.

Start with the equations of kinematics at constant acceleration for the particle, and parametrize its y position with its x position.
 
  • #5
OK, I think i understand you. Would you be able to show me how you reached the answer you got in your first post please?
 
  • #6
I almost told you all there is to tell.

Do the first step as I posted it in my last post.

Of course dy/dx and dh/dx are both zero at the starting position x = 0... so you'll be interested to know for what value of v will dy/dx become non-zero faster than dh/dx... i.e. for what value of v is the SECOND derivative of y greater (in absolute value) than the second derivative of h.

The interval of v follows directly from this little analysis.

Post your work if you get stuck.
 
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  • #7
Ok,

Ive done this...

[tex]\frac{dy}{dt} = -gt[/tex]

and [tex]\frac{dx}{dt} = v[/tex] (is this right?)

so

[tex]\frac{dy}{dx} = -\frac{gt}{v}[/tex]

and

[tex]\frac{d^2y}{dx^2} = -\frac{g}{v} \frac{dt}{dx}[/tex]

[tex]\frac{d^2y}{dx^2} = -\frac{g}{v^2}[/tex]

And

[tex]\frac{d^2h}{dx^2} = -dk^2\cos(kx)[/tex]

This is where i get really unsure about where I am going.

If i set these equal to each other and assume that the particle will first leave the surface at the max value of cos, 1 i get

[tex]\frac{g}{v^2} = dk^2[/tex]

[tex]v = \frac{1}{k}\sqrt{\frac{g}{d}}[/tex]

Have i gone on the wrong track entirely or is any part of this right??
 
  • #8
You've got it all right (almost). It was I who forgot about 'd' in solving for v.

The only detail you're missing is that once you've found the second derivatives, you must not set them equal, you must set an inequality between them. It's kind of hard to explain why, but I feel intuitively that if the particle is ever going to leave the ramp, it will be at the very first instant after it is lauched. And so, if the ball drop faster than the ramp at x=0, it also will at any other point, and therefor will not leave the ramp. So any value of v such that the ball drop faster than the ramp at x=0 is good. In math, we want v such that

[tex]\left|y''_{xx}(0)\right| \geq \left|h''_{xx}(0)\right|[/tex]

You correctly found the derivatives:

[tex]\frac{d^2y}{dx^2} = -\frac{g}{v^2}[/tex]

and

[tex]\frac{d^2h}{dx^2} = -dk^2\cos(kx)[/tex]

All you have to do is evaluate them at x=0, plug them in the inequality, and retrieve the range of v (which is [itex][0, \frac{1}{k}\sqrt{g/d}][/itex] rather than the [itex][0,\sqrt{g}/k][/itex] that I wrote in first post).
 
  • #9
Ah rite thankyou! :smile:

I understand what you mean about the inequality bit.

Also, i feel that the insertion of x = 0 is a valid one as it can be shown that the maximum acceleration to keep the particle on the plane occurs at the crests of the curve. So if it stays on the plane there, it will stay on the plane everywhere else.

Thankyou again for your help, it is much appreciated.
 
  • #10
for just intuition purposes I offer this explanation. If at x=0 there was a cliff then the particle would be dropping on a parabolic path. However, now there is sinusoidal hill. If the curvature of the sinusoid is greater than the curvature of the parabola then the particle would fly off the hill. If it is the other way around then the particle would be forced to stay on the hill. Since second derivative describes the curvature or concavity, hence [tex]\left|y''_{xx}(0)\right| \geq \left|h''_{xx}(0)\right|[/tex]
 

1. What is the "Finding Speed Needed to Keep Particle on Rippled Surface" experiment?

The "Finding Speed Needed to Keep Particle on Rippled Surface" experiment is a scientific investigation that aims to determine the minimum speed required to keep a particle on a rippled surface without it falling off. This experiment can help us understand the dynamics of particles on uneven surfaces and can have applications in fields such as material science, physics, and engineering.

2. How is the experiment conducted?

The experiment involves setting up a rippled surface and releasing a particle at different speeds onto the surface. The speed is gradually increased until the particle stays on the surface without falling off. The speed at which this occurs is recorded and used as the minimum speed required to keep the particle on the rippled surface.

3. What factors can affect the results of the experiment?

There are various factors that can affect the results of the experiment, such as the shape and size of the particle, the shape of the ripples on the surface, the material of the surface, and the viscosity of the medium the particle is moving through. These factors should be controlled and kept consistent throughout the experiment to ensure accurate results.

4. What are the potential applications of this experiment?

This experiment can have various applications in different fields. In material science, it can help understand the behavior of particles on uneven surfaces and design materials with better grip and stability. In physics, it can provide insights into the dynamics of particles on rippled surfaces. In engineering, it can aid in the development of more efficient and stable structures, such as bridges, roads, and buildings.

5. What are the limitations of this experiment?

One of the limitations of this experiment is that it is based on ideal conditions and does not take into account external factors such as wind or other disturbances. Additionally, the results may vary depending on the specific parameters of the experiment, such as the size and shape of the ripples and the type of particle used. Further research and testing may be needed to validate the results and make more accurate conclusions.

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