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Marble Speed on a Track

  1. Dec 13, 2011 #1
    Hi all--Thanks for any help you might be able to provide, I've been lurking in the forum for a while and find everyone's comments to be extraordinarily helpful.

    1. The problem statement, all variables and given/known data
    The spring with a constant 103N/m launches a marble (m=5 g, r=.5cm) on a horizontal track with a loop (like a roller coaster) followed by a brief horizontal area and then a 35° ramp. The height of the ramp is 15 cm which is also the radius of the loop. The marble rolls without slipping, with I=5 x 10-8kgm2.

    a)What should the spring compression be if the marble just makes it through the loop?
    b)What is the speed of the marble at the top of the ramp?

    2. Relevant equations
    Probably more?

    3. The attempt at a solution

    I found the minimum speed at the top of the track in order for the marble to not fall:
    mv2/r=mg → v=√rg
    so v=√(.15m)(9.8m/s2)=1.21 m/s minimum

    Then I set Us equal to Kr
    (1/2)kl2=(1/2)I(v2/r2) → l=sqrt[(I*v2/r2)/k)
    and solved with v=1.21 m/s
    so l=2.93x10-6m

    I may have missed something in part a, but I have no idea where to go for part b. Should I use the energy difference from the top of the loop to the top of the ramp? Or should I calculate the velocity when the marble has gone through the loop and then use that to calculate the velocity at the top of the ramp? This seems like a huge mess of kinematics, and like there would be an easier solution using energy, but I'm not sure.

    Thank you so much!
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Dec 13, 2011 #2


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    Wow, complicated question!

    You say spring energy = rotational kinetic energy at the top of the loop.
    What about the energy used to get up to the top of the loop?
    And the translational kinetic energy?

    (b) is easier. Ignore the loop - it does not absorb any of the marble's energy.
  4. Dec 13, 2011 #3
    yes - if it's both moving and rotating and up high it has all three types of energy

    You know also that the ramp is only half as high as the loop so it's going to be going faster than it did through the loop. There's no energy loss so just try conservation of energy to solve it.
  5. Dec 13, 2011 #4
    What I'm wondering is why they gave the spring constant???
  6. Dec 14, 2011 #5
    Alright, so I tried it again:
    Us = Kr + Kt + Ug (I think there is also gravitational potential energy once the marble is at the top of the loop)
    Giving me
    (1/2)kΔl2 = (1/2)I(v2/r2) + (1/2)mv2 + mgh
    And solved for l (h=30cm for the top of the loop).
    Which gave me 6.3 x 10-3m or .63 cm.

    I used the same equation, but instead I solved for V (with h=15cm for the top of the ramp and the Δl from part a), which gave me 1.89 m/s.

    I don't have the solutions until Friday, but if anyone does see a glaring error I'd love to hear about it. Thanks for the help, though!
  7. Dec 14, 2011 #6
    duhh because they asked for the spring compression is why they gave K

    Ok so your elastic potential energy on the left is equal to rotational potential plus kinetic plus gravitational for a and b so OK there. The two heights are different. You got that. Just plug your answers back into the equation to make sure you didn't make computational error.

    Good Job!
  8. Dec 14, 2011 #7


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    I agree with your answer for (a).
    For (b), I get a slightly larger answer. (.0199 for the total E, .00735 for mgh and .003v²)
    It's a grand tour of all the types of energy taken so far! Great question.
  9. Dec 14, 2011 #8
    I'm just glad they didn't throw in an explosion or two
  10. Dec 14, 2011 #9
    Thanks for the help.

    I checked my calculations, and I got 1.89m/s again. My work is:

    v = sqrt[(r2(kΔl-2mgh))/(r2m + I)]
    v = sqrt[(.005)2(103*.00632 - 2*.005*9.8*.15))/(.0052*.005 + 5x10-8)]
    I left out units so as to simplify things, but they work out to be m/s.

    Should I be using a different r?
  11. Dec 14, 2011 #10


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    I don't understand the numerator. Why -2mgh? We had the total energy in part (a) as .0199 J, so I just used E =mgh + ½mv² + ½Iw².
  12. Dec 14, 2011 #11
    For part b, The only r you should use is the radius of the marble to get the moment of inertia and from that, the rotational energy. You used a different r to calculate the velocity at the top of the loop because your mv^2/r centripetal force had to be equal to your force normal at that point. That is not the same r you use for part b where the only r is the radius of the marble.

    Delphi didn't the 2 come from multiplying through to eliminate the 1/2 that was part of some of the energy expressions? Part a didn't ask for the total energy did it? So you can compute it but you don't have to.
  13. Dec 14, 2011 #12
    So I got -mgh from solving the equation for v:
    (1/2)kΔl2 = (1/2)I(v2/r2) + (1/2)mv2 + mgh
    then pull all the terms with v to one side and get rid of the 1/2:
    I(v2/r2) + mv2 = kΔl2 - 2mgh
    get rid of the r2 fraction (not necessary, really, but I did it):
    v2(I + mr2) = r2(kΔl2 - 2mgh)

    Then divide by (I + mr2) and do a square root:
    v = sqrt[(r2(kΔl2 - 2mgh))/(I + mr2)]

    I did do the rs correctly, it seems, though.
  14. Dec 14, 2011 #13
    I haven't checked your calculations but Good job!
  15. Dec 14, 2011 #14


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    My mistake, the 1.89 looks good!
  16. Dec 14, 2011 #15
    The difficulty I have with helping on homework is remembering which of the many similar problems I'm currently working on. I live in mortal fear of doing something so dumb it will further confuse an already confused student, which I have already done. It's a lot harder to correct problems one hasn't selected nor solved than it is to grade papers on a test when you've already made out the answer key, :-)
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