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Homework Help: Rolling downhill

  1. Nov 11, 2007 #1
    A bicycle racer is going downhill at 11.0m/s when, to his horror, one of his 2.25kg wheels comes off when he is 75.0m above the foot of the hill. We can model the wheel as a thin-walled cylinder 85.0cm in diameter and neglect the small mass of the spokes.

    I need to get the velocity at the bottom of the hill and KEtotal at the bottom

    I know Kf=1/2Mv^2 + 1/2(I)w^2

    Vo=11m/s
    M=2.25kg
    r=.425m

    I know I=MR^2
    I=.4064kg*m^2

    would you use Ki+Ui=Kf+Uf to get w? or is there another way?
     
  2. jcsd
  3. Nov 11, 2007 #2
    No, that looks correct.
     
  4. Nov 11, 2007 #3
    here was my attempt
    Ki=136.125J
    Ui=1653.75J
    Kf=1/2mv^2+ 1/2Iw^2
    Uf=0

    for im stuck on Kf
    Its a thin cylinder so I=MR^2 for w did i calculate it correct? v=wr so 11/.425= 25.8rad/s
    Ki +Ui=Kf + Uf
    so 136.125J +1653.75J= 1/2mv^2 + .4064*(25.8)^2 +0

    from here i solved for the v i plugged in 2.25kg into m and i got 38.8m/s as Vf
     
  5. Nov 11, 2007 #4
    am i messing up on calculating the ang velocity (w)?
     
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