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Curvilinear Motion

  1. Jan 26, 2008 #1
    1. The problem statement, all variables and given/known data
    A particle moves along the curve [itex]y(x)=x^2-4[/itex] with constant speed of 5 m/s. Determine the point on the curve where the max magnitude of acceleration occurs and compute its value.



    2. Relevant equations radius of curvature
    [tex]|a|=\sqrt{a_n^2+a_t^2}[/tex]



    3. The attempt at a solution
    I have assumed that 'constant speed' means [tex]v_t=5 \Rightarrow a_t=0[/tex]

    So now I have to find where a_n is maximum. Any hints to get me going here?

    Thanks,
    Casey
     
  2. jcsd
  3. Jan 26, 2008 #2
    So I got the right answer. But I don't know how to spell out a proper procedure for future problems.

    I found y'(x)=0; at x=0.

    So x=0 is a critical point. Then I just plugged that into [tex]a_n=\frac{[1+\frac{dy}{dx}^2]^{3/2}}{\frac{d^2y}{dx^2}}[/tex] and that worked.

    But do I know that x=0 is a MAXIMUM, if I do the second derivative test, I get positive 2, so I would have thought it was a minimum.
     
  4. Jan 26, 2008 #3
    Can anyone point out how I would know it was a maximum? Shouldn't the 2nd derivative test work here?
     
  5. Jan 26, 2008 #4
    But really. Do it. Do it. (Dodgeball reference)
     
  6. Jan 26, 2008 #5

    Hootenanny

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    You should note here that x = x(t), that is, position is a function of time. And that the maximum acceleration would occur when x'''(t)=0 and not when y''(x)=0.
     
    Last edited: Jan 26, 2008
  7. Jan 26, 2008 #6
    I see. So how would I determine, if and where a maximum would be if I am given y(x)?
     
  8. Jan 26, 2008 #7
    Any takers? 'cause I don't know...
     
  9. Jan 26, 2008 #8
    Have you studied parametric equations? I'm guessing that's exactly what you're doing right now, from the looks of this problem.

    If you can restate the problem by writing independent eqations for y(t) and x(t) (i.e. parametrically), then you'll be set. The key is that you know how x and y are related, and you know that the magnitude of the velocity (i.e. the speed) is constant. Do you know how to express the velocity vector for parametric equations?
     
  10. Jan 26, 2008 #9
    I have actually never studied them. This is for a dynamics course. I have probably studied them, but never heard them worded as "parametric". Give me an example, if you would.
     
  11. Jan 26, 2008 #10
    Oh, then maybe that's not the way to go. It's where you have separate equations for x(t) and y(t) and then you use the chain rule a lot to express dy/dx in terms of dx/dt and dy/dt.

    If you haven't seen it, then you probably have to stick with expressing the normal acceleration as a function of the curvature. Do you know how to do that? I'm not sure I remember, myself ... :uhh:
     
  12. Jan 26, 2008 #11
    No...you're right, it does have to do with the chain rule. I am just confused by it....that, is I do not have a nice procedure as to where to start or how to set it up.

    I have y as a function of x. Now what?
     
  13. Jan 26, 2008 #12
    Can I jump in? :smile:
    First of all constant velocity means [itex]u^2=u_x^2+u_y^2=C[/itex]. Thus first take the derivative of [itex]y(t)=x(t)^2-4[/itex] to find the relation bewteen [itex]u_x,\,u_y[/itex] and plug it in [itex]u^2[/itex]
     
  14. Jan 26, 2008 #13
    You lost me. I am given y as a function of x... not time. i'll look it over some more, though.
     
  15. Jan 26, 2008 #14
    The [itex]y,\,x[/itex] are actually [tex]y(t),\,x(t)[/itex]. Can you take the derivative of [itex]y(t)=x(t)^2-4[/itex] with respect to [itex]t[/itex]?

    Hint. [itex]u_x=\dot{x}(t),u_y=\dot{y}(t)[/itex]
     
  16. Jan 26, 2008 #15
    I am sorry, I cannot follow any of your notations. I have y(x) not y(t). . . where are you getting t? What is u? Where am I?
     
  17. Jan 26, 2008 #16
    Ok! :smile:

    The partile is moving in the plane x-y. Thus x,y varies with time, i.e. [itex]x(t),y(t)[/itex] and the velocity [itex]\vec{u}[/itex] has two compontents, i.e. [itex]u_x=\frac{d\,x(t)}{d\,t},\,u_y=\frac{d\,y(t)}{d\,t}[/itex] and the magnitude of the velocity is

    [tex]u^2=u_x^2+u_y^2[/tex]

    Do you follow up to now?
     
  18. Jan 26, 2008 #17
    Not this part. I get all the maths after that, but why do i automatically assume that it varies with time? It says nothing about time.
     
  19. Jan 26, 2008 #18
    The partile moves on the parabola [itex]y=x^2-4[/itex] doesn't it?
    That means that it's coordinates [itex](x,y)[/itex] are not constant thus they vary with time. Correct?
     
  20. Jan 26, 2008 #19
    Although I cannot think of a situation in which position is a function independent of time, it is still not clear to me why it is so obvious that x and y DO vary with time.

    Are you trying to say that since [tex]\frac{d}{dx}(y)=v(x)[/tex] and

    [tex]v=\frac{dx}{dt}[/tex]

    I can do this: [tex]\frac{dy}{dt}=\frac{dx}{dt}*\frac{dy}{dx}[/tex]?

    Or am getting close?
     
  21. Jan 26, 2008 #20
    First of all

    if you throw a stone from height [itex]H[/itex] with horizontial velocity [itex]u_o[/itex] then it will moves both horizontal and vertical, i.e. [itex]x(t),y(t)[/itex] and the orbit of the stone is
    [tex]y=H-\frac{g}{2\,u_o^2}\,x^2[/tex]
    Your "stone" is moving on the orbit
    [tex]y=x^2-4[/tex]

    Is it clear now?
     
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