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Confirming my answer

  1. Dec 9, 2013 #1
    A 0.70-kg disk with a rotational inertia given by MR2/2 (M) is free to rotate on a fixed horizontal axis suspended from the ceiling. A string is wrapped around the disk and a 2.0-kg mass (m) hangs from the free end. If the string does not slip, then as the mass falls and the cylinder rotates, the suspension holding the cylinder pulls up on the cylinder with a force of:

    A. 6.9N
    B. 9.8N
    C. 16N
    D. 26N
    E. 29N
    ans: B

    My solution:

    ma = mg - F
    Fr = Ia/r
    mg - ma = Ia/r2

    mg - ma = Ma/2

    a = (mg)/(m + 0.5M)

    ƩFy = FN - Mg - Ma

    FN = M (g + a) = ~9.77N

    Please point out any errors since I really want to ensure I understood every process throughly and correctly. If you have anything to add, in terms of helpful steps or things to consider in general, please reply! :)
     
  2. jcsd
  3. Dec 9, 2013 #2

    haruspex

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    You were doing well until here:
    Look at the free body diagram for the disk (cylinder?).
    Where is there an Ma in there?
     
  4. Dec 9, 2013 #3
    [ignore/]
     
    Last edited: Dec 9, 2013
  5. Dec 9, 2013 #4
    Sorry, I meant to add ƩFy = 0, isn't that a correct assumption?
     
  6. Dec 9, 2013 #5

    haruspex

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    Yes, but you don't have the right contributors to ƩFy.
     
  7. Dec 9, 2013 #6
    Ohh... yeah, that's a totally correct assumption. I misread the scenario!

    I got the same answer, 9.77... but not sure where you have the Ma term as haruspex mentioned. If you consider the mass is falling with F = ma, how much tension remains on the line? This is a downwards force on the cylinder.

    ƩFy = FN - Mg - tension
     
  8. Dec 10, 2013 #7
    Okay, thank your for clearing that up. Yes, I used the wrong variable name but calculated it as ma was tension.
     
  9. Dec 10, 2013 #8

    haruspex

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    OK, so you meant ƩFy = FN - Mg - m(g-a), right?
    But in the next line of the OP you wrote M (g + a), which is not so easily explained?
     
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