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Angular velocity and moment of inertia

  1. Apr 7, 2009 #1
    Could you let me know if my answers are correct?

    1. The problem statement, all variables and given/known data
    A light rope is wrapped several times around a large wheel with a radius of 0.400m. The wheel rotates in frictionless bearings about a stationary horizontal axis. The free end of the rope is tied to a suitcase with a mass of 0.15kg. The suitcase is released from rest at a height of 4.00m above the ground. The suitcase has a speed of 3.50 m/s when it reaches the ground.
    Calculate:
    A) the angular velocity of the wheel when it reaches the ground and
    B) the moment of inertia of the wheel


    2. Relevant equations
    A) angular velocity (w) = speed/radius
    B) mgh = 1/2 mv2 + 1/2Iw2


    3. The attempt at a solution

    After using these equations I found angular velocity to be 8.75 rad/s and the moment of inertia to be 12.9 kg*m2.

    Correct or incorrect? Thanks for you help.
     
  2. jcsd
  3. Apr 7, 2009 #2

    LowlyPion

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    With ω = v/r then 8.75 looks ok.

    But your I looks a little large.

    m*g*h is total energy to start with dividing by ω2 where ω2 looks a lot bigger than m*g*h doesn't look right.
     
  4. Apr 7, 2009 #3
    Thanks for the reply. I also though I was rather large, but with my calculations that's what I ended up getting... strange.
     
  5. Apr 7, 2009 #4

    LowlyPion

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    I think you've done something wrong.

    Maybe show your numbers and how you got there.
     
  6. Apr 7, 2009 #5
    A) w = v/r = (3.50)/(0.400) = 8.75 rad/s

    B) Ki + Ui = Kf + Uf
    Ki and Uf = 0

    Ui = Kf
    mgh = 1/2 mv2 + 1/2Iw2
    (15)(9.8)(4) = 1/2 (15)(3.5)2 + 1/2 (I) (8.75)2
    588 = 91.9 + 38.3(I)
    496.1 = 38.3(I)
    I = 12.9 kgm2
     
  7. Apr 7, 2009 #6

    LowlyPion

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    Whoa. SI units please.

    m = .15kg
     
  8. Apr 7, 2009 #7
    Oh my gosh I can't believe I did that *facepalms*

    Thanks for your help, I really appreciate it
     
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