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Potential Energy and Loop-de-Loops

  1. Oct 27, 2015 #1
    1. The problem statement, all variables and given/known data
    A bead slides without friction around a loop–the–loop (see figure below). The bead is released from rest at a heighth = 3.60R.
    8-p-005.gif
    (a) What is its speed at point circleA.gif ? (Use the following as necessary: the acceleration due to gravity g, and R.)
    v = _________________


    (b) How large is the normal force on the bead at point circleA.gif if its mass is 4.70 grams?
    Find the magnitude and direction.

    2. Relevant equations
    u = mgh
    k = .5mv^2
    ui + ki = uf + kf
    g = 9.8 m/s^2
    h = 3.60R

    3. The attempt at a solution
    For a, I don't know how much farther I can go. the farthest I've gotten is this:

    where h = 3.60R

    ui + ki = uf + kf
    mgh + 0 = .5mv^2 + mgh
    gh = .5v^2 + gh

    If I subtract gh from the right though, I'll get 0... which is obviously not the answer here. So I need major help there.

    As for b, I know the direction is downward. It's upside down so if I flip the entire diagram, I see that normal force is really pointing down.

    As for the first part, I just converted grams to kg, then multiplied by 9.8 m/s^2. However, the website I'm entering the answer into keeps saying I'm wrong when I type in when I input my answer 0.04606 N, so I'm not sure what's up there... I also tried 0.0461 N, and still no dice...

    Thanks for any help!
     
    Last edited by a moderator: Apr 16, 2017
  2. jcsd
  3. Oct 28, 2015 #2

    haruspex

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    What does the variable uf represent? What is the 'final' state in the calculation you are trying to do? (I.e., where is the bead at the end?)
     
  4. Oct 28, 2015 #3
    uf is final potential energy, which I'm counting as the top of the loop at point A. So, although I don't know if the set up is correct, I'm making the final stop to be at the loop-de-loop point A.
     
  5. Oct 28, 2015 #4

    haruspex

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    Ok, so what should uf equal? Why have you substituted mgh?
     
  6. Oct 28, 2015 #5
    where (initial height) = 3.60R

    and I fixed the initial h and h at a, as I just realized how bad of a mistake that was

    ui + ki = kf + uf
    mg(initial height) + 0 = .5mv^2 + mg(height at a)
    3.60Rg = .5v^2 + g(height at a)

    I don't know how to get height at A based on the 3.60R stuff so I am totally lost there. My book keeps saying that height at a is 2R, but I don't understand that conceptually and I don't just want to put down something to answer the question without fully understanding it :(
     
  7. Oct 28, 2015 #6

    haruspex

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    For the purposes of gravitational PE, you have to choose a common baseline. The initial height h is measured from the bottom of the loop,so the bottom of the loop is the baseline. The loop has radius R, so what is the height of the top of the loop?
     
  8. Oct 28, 2015 #7
    OH MY G THAT MAKES SO MUCH MORE SENSE!

    Thank you!

    The height is 2R, as the diameter is the height of the loop.

    3.60Rg = .5v^2 + g2R
    3.60Rg - 2Rg = .5v^2
    2Rg(3.60-2) = v^2
    v = sqrt(Rg(3.20))

    So then, proceeding on

    Fn + Fg = sum of forces = ma

    So would ma be centripetal acceleration?
     
  9. Oct 28, 2015 #8

    haruspex

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    ma would be the centripetal force, a the centripetal acceleration, which I assume is what you meant.
     
  10. Oct 28, 2015 #9
    Yes, thank you, that's what i meant

    Fn + Fg = sum of forces = ma = mv^2/R

    Fn + Fg = m (3.2gR)/R
    Fn + Fg = 3.2gm
    Fn + mg = 3.2mg
    Fn = 3.2mg - 1mg
    Fn = mg(2.2)
    Fn = (4.70 x 10^-3)(9.8)(2.2)
    Fn = 0.101332 N

    Okay, I hope I got it now!
     
  11. Oct 28, 2015 #10

    haruspex

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    That looks right.
     
  12. Oct 28, 2015 #11
    Awesome, thanks so much!
     
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