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Fictitious forces, and motion

  1. Jan 21, 2007 #1
    1. The problem statement, all variables and given/known data

    The first question asks:
    Which (fictitious) force will cause the sledge to accelerate along the rails? Give the modulus and direction of the sledge at distance r<R from the centre.

    The next question asks:
    Obtain an expression for the displacement r(t) for r<R, starting from the acceleration, and the initial condition r(0) = [tex]r_0[/tex].

    2. Relevant equations

    3. The attempt at a solution
    a) Centrifugal force is the fictious force that causes the sled to accelerate. The acceleration is given by [tex]\omega^2 * r[/tex].

    b) Now this is where I get most stuck.

    So far I have said that [tex]a = \frac{d^2r}{dt^2}[/tex], so I integrated twice and ended up with the basic kinematic equation [tex]r = ut + \frac{1}{2}at^2[/tex] (which I would have guessed, but integrated as I assumed thats what the question was hinting at).
    From here I said that u = 0, and was left with [tex]r = \omega^2rt^2 + r_0[/tex], but I'm a little confused about this. Surely r cannot be on both sides of the equation at the same time. As I've been writing this I've also thought about using [tex]\frac{v^2}{r}[/tex] instead of [tex]\omega^2r[/tex], but on further thought I end up with the same problem of having r on both sides of the equation (it would now be multiplied by the constant term), as well as having the v term that is never mentioned in the question itself (I know I probably could use it, but further questions relate to omega so I have this gut feeling that I should stick with omega in this question).

    Any thoughts about this?
  2. jcsd
  3. Jan 21, 2007 #2


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    The accleration is not constant. You integrated [tex] \frac{d^2r}{dt^2}[/tex] assuming it was constant.

    You gave the correct acceleration in part (a):

    [tex] \frac{d^2r}{dt^2} = \omega^2 r [/tex]
  4. Jan 21, 2007 #3
    I just had the thought, that going from [tex]\frac{d^2r}{dt^2} = \omega^2r[/tex] that you could say that [tex]r=e^{at}[/tex] and put it back into the equation. From this and the initial conditions it emerges that [tex]r=r_{0}e^{\omega t}[/tex].

    I think that makes a little more sense to me.
    Last edited: Jan 21, 2007
  5. Jan 21, 2007 #4


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    Yes, that's how to solve linear second order differential equations :approve:
  6. Jan 21, 2007 #5
    Never dawned on me that thats what the problem was reffering to. I naturally assumed that the acceleration was constant. Makes lots of sense now.
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