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Homework Help: Maximum Acceleration along Parabolic Curve

  1. Jan 10, 2012 #1
    1. The problem statement, all variables and given/known data


    So i was just doing a few past papers for an upcoming exam and i got stuck on this one.

    i have got to the stage where i have 1/p all in term of x and y, but i can't see how you would go about getting the maximum acceleration just from x and y values.

    2. Relevant equations


    3. The attempt at a solution

    so here is how i went about it:

    y^2 = 4ax

    therefore by implicit differentiation

    2ydy/dx = 4a

    thus dy/dx = 2a/y

    for the equation for 1/p we need (dy/dx)^2

    which is,

    4a^2/y^2 →4a^2/4ax since y^2 = 4ax

    thus we get (dy/dx)^2 = a/x

    now we need d^2y/dx^2

    so by implicit differentiation again 2a*y^-1 → -2a*y^-2*dy/dx

    (-2a/y^2)(2a/y) → (-2a/4ax)(2a/y) → -a/xy

    therefore 1/p = ((1+a/x)^(3/2))/(-a/xy)

    This is where I found it difficult to continue. I tried by saying a is a maximum when p is a minimum (a=v^2/p) so da/dp = -v^2/p^2, but i couldn't really get any further any help would be greatly appreciated because it is just annoying me now haha.
  2. jcsd
  3. Jan 10, 2012 #2
    Try this:

    for a in a=v^2/ρ to be min or max,
    da/dρ = 0
  4. Jan 10, 2012 #3

    Simon Bridge

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    Isn't the radius of curvature a minimum at the apex of the parabola?
  5. Jan 10, 2012 #4
    yeah i did that and got it down to (1+a/x)^(9/4) = 0 which i assume is wrong
  6. Jan 10, 2012 #5
    yep but i dont know how this would help me to find the max acceleration or how to go about it
  7. Jan 10, 2012 #6
    ok i went about it a completely different way. this time rearranging it to x= y^2/4a now i will differentiate with respect to y instead of x.

    so dx/dy = 2y/4a → y/2a

    d^2x/dy^2 = 1/2a

    therefore curvature p = ((1+(y/2a)^2)3/2)/(1/2a)

    at the vertex which is (0,0)the curvature is 1*1/2a therefore curvature = 2a

    and max acceleration is v^2(1/2a)

    does this seem legitimate?
    Last edited: Jan 10, 2012
  8. Jan 10, 2012 #7

    Simon Bridge

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    You have derived an equation for the acceleration as a function of the (x,y) position.
    The max acceleration is where the radius of curvature is smallest.
    The radius of curvature is smallest at the apex of the parabola.
    You have the equation of the shape of the parabola.
    Therefore, you know the position of the apex of the parabola...
    What happens when you plug that into your equation for acceleration?

    Oh I see you've done something like that.
  9. Jan 10, 2012 #8
    the problem was i was dividing by 0 when i differentiated with respect to x
  10. Jan 10, 2012 #9

    Simon Bridge

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    That doesn't make sense - lets see how I do:

    [tex]\frac{d}{dx}\big [y^2=4ax\big ] \Rightarrow 2y\frac{dy}{dx} = 4a[/tex]which tells us that [tex]\bigg ( \frac{dy}{dx} \bigg )^2 = \frac{4a^2}{y^2}[/tex]
    [tex]\frac{d}{dx}\big [y\frac{dy}{dx} = 2a \big ] \Rightarrow \bigg ( \frac{dy}{dx} \bigg )^2 + y\frac{d^2y}{dx^2} = 0[/tex]which tells us that[tex]\frac{d^2y}{dx^2}=-\frac{4a^2}{y^3}[/tex]
    [tex]\frac{1}{\rho} = -\frac{4a^2}{y^3(1+\frac{4a^2}{y^2})^{3/2}} = -\frac{4a^2}{(y^2+4a^2)^{3/2}}[/tex]

    We can put y=0 safely here. Agrees with your [itex]v^2/2a[/itex].
  11. Jan 11, 2012 #10
    very nice! good answer
  12. Jan 11, 2012 #11

    Simon Bridge

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    You did pretty well yourself.
    I should have spotted the problem with the final stage in your calc in post #1 earlier - you just needed to substitute for x at the end (or not substitute for y earlier). That's the only difference between mine and yours.

    Since you'd already worked it out I figured it'd do no harm to model the direct approach.

    Notice how LaTeX makes the equations really clear?
    It is very much worth learning.
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