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Curved slit - angular motion

  1. Dec 3, 2008 #1
    1. The problem statement, all variables and given/known data
    The shape of the arch is defined by a sine function with a = 0.2 and the width of the arch is 1m. The velocity v is constant at 2m/s. What is the velocity at point P when x=0.25m?


    3. The attempt at a solution
    So since v is just in the x-direction, I figured that's the vx speed.
    Then I came to having to find an equation to vy and this is where I get stuck. I remember from high school physics that equation of velocity can be essentially found by taking a derivative of the position equation.

    In this case I get:

    From which I can find that at x=0.25m, vy = 0.444288 m/s
    But that combined with the horizontal speed of 2m/s gives me a resultant velocity of 2.05ish, instead of 2.19 that I'm supposed to get. Help please?
    Last edited: Dec 3, 2008
  2. jcsd
  3. Dec 3, 2008 #2


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    It is asking for acceleration, not velocity. You are adding apples to oranges.

    BTW, is that the exact wording of the problem? - it is appauling.
  4. Dec 3, 2008 #3
    Yeah, that's the exact wording.

    Sorry, the question is a two parter. I'll fix that. I just decided to not even tackle acceleration if I can't figure out velocity.
  5. Dec 3, 2008 #4


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    Where did I see acceleration?! I must be going mad. I am so sorry. Anyway, your velocity calculation is almost correct. Your problem is that vy is the derivative of y wrt t, not x. Think chain rule.
  6. Dec 3, 2008 #5
    Oh thank you, thank you! :) That allows me to get acceleration as well. Yay!
  7. Dec 3, 2008 #6
    Except... Now I'm doing a question that's basically the same, except

    y=0.23sin(pi*x) and Vx=2.4m/s and we have to find velocity at x=0.32

    I did the same thing as before and got:
    Vy = 0.552*pi*cos(2.4*pi*t)

    But the thing in the brackets ends up being >1 and that can't be. What did I do? :S
  8. Dec 3, 2008 #7


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    What thing in brackets? Do you mean 2.4πt? What's wrong with that being >1?
  9. Dec 3, 2008 #8
    Yes that, but cos can't be >1...
  10. Dec 3, 2008 #9


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    The cosine is not >1; its argument is >1. The domain of cos(x) is x any real number, (and actually this can be generalized even further). I think you're worried about the range. Don't worry about that; your calculator will take care of it.
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