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Angular speed of rotating hoop

  1. Mar 14, 2015 #1
    1. The problem statement, all variables and given/known data
    http://imgur.com/jcZRQdu [Broken]

    2. Relevant equations


    3. The attempt at a solution
    So I know how to solve this with conservation of energy, but I can't seem to get the correct answer using kinematic equations.

    τnet = Iα = (MR2*(a/R))
    -Tension = MRa

    Fnet = ma
    w-Tension = ma
    -Tension = ma - mg

    Plugging this into the first equation:
    ma-mg = MRa
    a-g = Ra
    a = g/(1-R)
    a = 9.8/(1-0.0800) = 10.65 m/s2

    vinitial = 0
    a = 10.65 m/s2
    Δy = 0.45 m
    vfinal = ?

    vfinal2 = vinitial2 + 2aΔy
    vfinal = sqrt(2aΔy)
    vfinal = 3.096 m/s
    v = rω
    ω = v/r = 3.096/0.0800 = 38.7 rad/s

    Correct answer: 26.3 rad/s

    Again, I know this can be solved with conservation of energy, but I'm trying to figure it out with kinematic equations.
     
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. Mar 14, 2015 #2

    ehild

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    What is the question?

    What are the relevant equations?
     
    Last edited by a moderator: May 7, 2017
  4. Mar 14, 2015 #3

    TSny

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    Dimensional analysis of the second equation above shows that it can't be correct.
     
  5. Mar 14, 2015 #4
    I forgot to include the questions in the picture:
    After the hoop has descended 0.45 m, calculate the angular speed of the hoop.

    Relevant equations:
    τ = Iα
    F = ma
    Kinematic equations
     
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