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Motion on Function

  1. Apr 21, 2009 #1
    I was wondering about the problem of having a "ramp" defined as y=f(x), and a ball (or other object) beginning at "X_0" with initial velocity "V_0" and being allowed, under acceleration due to gravity, to move frictionlessly on the ramp's surface.
    I thought the easiest way to approach this is to take the component of its acceleration parallel to the x-axis (as it is the point of the x-axis which will through differentiation enable us to calculate the gradient of the slope at that point).
    This is what i ended up with:

    a = gf`(x)/(1+f`(x)^2)

    Where g= -9.81 and f`(x) is the gradient of the slope at that point.

    This gives me the acceleration as a function of the position on the slope...and i have no idea how to solve this. What I would ideally like to end up with is something in the form:

    x=g(t)

    where "g" is any function of t. But an implicit form wouldn't be too un-elegant...
     
  2. jcsd
  3. Apr 21, 2009 #2

    tiny-tim

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    Welcome to PF!

    Hi coddy! Welcome to PF! :wink:
    Why not just use conservation of energy? :smile:
     
  4. Apr 21, 2009 #3
    I say conservation of energy. This would also give you the advantage of being able to account for rotational ennergy of the ball should you choose to do so. However, this gives you V(y) where y is your height and V your velocity tangential to the ramp assuming you stay on the surface.
     
  5. Apr 21, 2009 #4
    Is your expression for acceleration as a function of f'(x) missing a square root?
     
  6. Apr 22, 2009 #5
    Now i'm more confused, I tried deriving the formula again and can't remember how i came to the original one. This is what i did this time:

    On a triangle where "theta" is equal to tan^-1 (f`(X)), the component of "g" down the slope =g/sin(theta). Then draw a second similar triangle where the hypotenuse = g/sin(theta). The component of this parallel to the x-axis (horizontal) is gcos(theta)/sin(theta) = g/tan(theta) = g/f`(x), which gives me:

    d^2 x / d t^2 = g/f`(x)

    Which is is correct? The original, the original with a square root over the "1+f`(x)^2" or the one above??
     
  7. Apr 22, 2009 #6
    Draw a right triangle with y vertical and x horizontal. Using f'(x) as the vertical y, and the value "1" as the horizontal x. Calculate Pythagorean theorem x2 + y2 = z2, or hypotaneus z = sqrt(x2 + y2). Now calculate sine(theta) = opposite over hypotaneus.
     
  8. Apr 23, 2009 #7
    That gives me:

    a=g.f`(x)/sqrt(1+f`(x)^2)

    Which is the acceleration of the "ball". However, this is the acceleration of the ball parallel to the slope... I am only interested in acceleration which will affect the ball's x-coordinate, as this determines the balls change in x-position. Basically it's modelled as a one-dimensional line which the ball travels on...once i have a "x=p(t)" where "p" is a function, i can use the fact that "y=f(x)" and determine the balls actual postion (x,y)coordinates using parametric-style stuff...
    Obviously i havn't done this, but i'm sure it would go a bit mental for 1-to-many functions...
     
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