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Moving particle

  1. Mar 6, 2005 #1
    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.

  2. jcsd
  3. Mar 6, 2005 #2


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    I'm pretty sure I found a way to do this! :smile: I'm gonna 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].
  4. Mar 7, 2005 #3

    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.
  5. Mar 7, 2005 #4


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    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.
  6. Mar 7, 2005 #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?
  7. Mar 7, 2005 #6


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    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.
    Last edited: Mar 7, 2005
  8. Mar 8, 2005 #7

    Ive done this...

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

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


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


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

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


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

    This is where i get really unsure about where im 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??
  9. Mar 8, 2005 #8


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    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]


    [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).
  10. Mar 8, 2005 #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.
  11. Jan 20, 2010 #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]
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