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Motion down a curve

  1. May 10, 2009 #1
    I have a project about analyzing motion down a curved ramp. I am quite confused about how to approach it. Can someone please point me in the right direction? (Is this the right place?)
     
  2. jcsd
  3. May 11, 2009 #2

    tiny-tim

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    Hi oxnume! :smile:
    Conservation of energy will usually do it. :smile:

    Show us how far you get, and where you're stuck, and then we'll know how to help! :wink:
     
  4. May 21, 2009 #3
    This is what my partner has came up with so far. But I can't seem to understand it and there are some parts that just appear from nowhere. I would've approached it a bit differently but I'm not sure if I'm right. So can you please look at this and tell me if it makes any sense?

    The attachment is .zip because the original .doc file is too big (358kb), no virus, promise!

    Also, in the file we used an example with a block on a ramp, would the effect be the same if we used a ball instead of a block?
     

    Attached Files:

    Last edited: May 21, 2009
  5. May 21, 2009 #4
    Since this involves kinematics and calculus, would the physics section be better or is this fine?
     
  6. May 23, 2009 #5
    Can someone please take a look at this?
     
  7. May 24, 2009 #6

    tiny-tim

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    sorry, oxnume, i don't like .doc files

    can't you type it out for us?
     
  8. May 24, 2009 #7
    There's a lot of equation editor stuff that I can't reproduce here. I printed them into images. That's fine right?
     

    Attached Files:

  9. May 24, 2009 #8

    tiny-tim

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    Yes, that's fine :smile:but it'll take ages for the attachments to be approved.

    What equation editor stuff do you need?

    With a few copy-and-paste symbols like θ, and the X2 tag just above the Reply box, you should be able to type any equations necessary. :wink:
     
  10. May 24, 2009 #9
  11. May 25, 2009 #10
    Ok the attachments are approved.
     
  12. May 25, 2009 #11

    tiny-tim

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    ok, I can se them now (sideways :frown:) …

    but what are you asking us about? :confused:
     
  13. May 25, 2009 #12
    I'm not completely sure of the correctness of this. I lose understanding right where it starts talking about the second portion of the normal force. From what I understand, the whole curve can be taken as a combination of tiny pieces of straight edges that have a certain slope (the derivative of the function of the curve). So the overall force at any one point can be calculated by adding up the different forces in the freebody(-ish) diagram.

    Q1. The second normal force (FN,2) is what throws me off (and the rest of the sin and cos stuff with derivatives). The equation of that normal force is given as FN,2 = mv2/R
    That looks awfully like the centripetal force equation for a circle with radius R. But in this case where is the circle? I originally thought that other force would consist of the inertia force from the previous "piece" of the curve, but I have no idea how to express that in terms of math. Is this correct at all?

    Q2. The "trig identities" on page 3, I have never seen those identities in my life and my partner refuses to explain where she got them from. Are they correct at all? I just get completely lost from that point on...

    Q3. If we were to do an experiment of this. Would having a ball roll down a curve be the same as having a piece of block slide down the curve (like the diagram)?
     
  14. May 26, 2009 #13

    tiny-tim

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    Hi oxnume! :smile:
    Yes, FN,2 is the "centripetal force", mv2/R.

    Here, R is the radius of curvature of the curve at that point … the radius of the circle that most closely fits the curve.

    However, I think most members of PF would strongly disagree with calling it a force (your book calls it the "second normal force") … it's really the mass times the centripetal acceleration, and comes on the RHS of F = ma, not the LHS …

    btw, this is a matter of geometry, not physics … if an object goes at speed v along a curve with radius of curvature R, then its acceleration is automatically v2/R
    Yes, as I said, it isn't part of the normal force (except possibly for a frame of reference moving with the object, for which fictitious forces such as this have to be invented so that Newton's first law still works)
    The tangent of the slope of the curve, tanθ, is dy/dx, ie f'(x).

    The line-element is ds = dx√(1 + (dy/dx)2) = dx√(1 + (f'(x))2) (so distance along the curve is ∫ds = ∫dx√(1 + (f'(x))2))

    (there's some "squareds" missing in your formulas :frown:)

    (and so sinθ = dx/ds and cosθ = dy/ds)
    No, because
    i] there's no friction force impeding a rolling object (the point of contact is stationary, so there is no https://www.physicsforums.com/library.php?do=view_item&itemid=75" by the reaction force)
    ii] you'd have to add angular momentum and energy to the equations.
     
    Last edited by a moderator: Apr 24, 2017
  15. May 27, 2009 #14
    Thanks so much for your help.

    Is there another more appropriate name for it?
     
  16. May 27, 2009 #15

    tiny-tim

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    Well, it's not a force, so just the mass times the centripetal acceleration.

    But you'd better call it whatever your professor does, if you want to pass the exams! :smile:
     
  17. May 27, 2009 #16
    Heh, we weren't actually taught this. This is just an "explore by yourself" project that we chose.

    Is there no friction force or is µ just dramatically reduced? Because there would still be a nonconservative force acting on the ball or else it would just go up and down forever right?
     
    Last edited by a moderator: Apr 24, 2017
  18. May 28, 2009 #17

    tiny-tim

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    There is no friction force impeding the motion!

    However, there is air resistance, and also something called "rolling resistance", which essentially is loss of energy through deformation of the ball … see wikipedia. :smile:
     
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