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TomLikesPhysics Group Title

I have a parabola slide and I am looking for a function of the downhill force which only depends on the x and/or y coordinate. I came up with a function but I am not sure if this is correct. You can see everything on the attached picture.

  • 2 years ago
  • 2 years ago

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  1. TomLikesPhysics Group Title
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    Here is what I did so far.

    • 2 years ago
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  2. henpen Group Title
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    I assume that it is safe to assume that the parabola is y=x^2, with the tip at (0,0) dy/dx=2x |dw:1351340058220:dw|\[\tan^{-1}(\frac{dy}{dx})=\theta\] \[\theta=\tan^{-1}(2x)\] \[F=-mgsin(\theta)\]

    • 2 years ago
  3. henpen Group Title
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    So you seem in the right area

    • 2 years ago
  4. TomLikesPhysics Group Title
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    Hmmm... actually I do not know that. It is not specified so it might be better if I write: y=ax^2.

    • 2 years ago
  5. henpen Group Title
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    Then \[\theta=\arctan(a2x)\]

    • 2 years ago
  6. TomLikesPhysics Group Title
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    Yes, it just looks kind of ugly if we plug that into the equation for the force. F=-mgsin(arctan(2ax)

    • 2 years ago
  7. TomLikesPhysics Group Title
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    Is this really the finaly answer? Can´t we do something to make it simpler or look better? Is there perhaps a different way to determine the force in terms of x and/or y?

    • 2 years ago
  8. henpen Group Title
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    |dw:1351340440158:dw|No, its not

    • 2 years ago
  9. henpen Group Title
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    |dw:1351340462854:dw|

    • 2 years ago
  10. henpen Group Title
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    F=-mgsqrt(1+(2ax)^2)

    • 2 years ago
  11. TomLikesPhysics Group Title
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    Ah... Pythagoras... I see...Now it looks better. So this is the best we can do?

    • 2 years ago
  12. henpen Group Title
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    Simplify sqrt(1+(2ax)^2) maybe You could try to x/y decompose the force, but that's probably uglier

    • 2 years ago
  13. henpen Group Title
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    Although I think it's unsimplifiable

    • 2 years ago
  14. TomLikesPhysics Group Title
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    Ok. A big thank you to you, henpen. I really appreciate the help. Thx a lot.

    • 2 years ago
  15. henpen Group Title
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    http://fooplot.com/#W3sidHlwZSI6MCwiZXEiOiItc3FydCgxKyh4KV4yKSIsImNvbG9yIjoiIzAwMDAwMCJ9LHsidHlwZSI6MTAwMCwid2luZG93IjpbIi04IiwiNSIsIi02LjQyIiwiMS41OCJdfV0- It's probably correct- the force graph looks about right

    • 2 years ago
  16. TomLikesPhysics Group Title
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    The Force stands always normal on the slide, right? If so, it does not point at some special point all the time?

    • 2 years ago
  17. henpen Group Title
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    |dw:1351349861448:dw| The normal (restrictive) force is always at a normal to the curve

    • 2 years ago
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