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Homework Help: Moments of Inertia, Kinetic Energy and rotations

  1. Nov 24, 2008 #1
    1. The problem statement, all variables and given/known data
    A bicycle racer is going downhill at 11.0 m/s when, to his horror, one of his 2.25 kg wheels comes off when he is 75.0 m above the foot of the hill. We can model the wheel as a thin-walled cylinder 85.0 cm in diameter and neglect the small mass of the spokes.

    How fast is the wheel moving when it reaches the foot of the hill if it rolled without slipping all the way down? How much total kinetic energy does the wheel have when it reaches the bottom of the hill?

    2. Relevant equations
    I think...
    mgy = 1/2*m*v^2 + 1/2*I*w^2
    v=rw ("w" is supposed to be angular velocity)

    3. The attempt at a solution

    I'm totally stumped by part A. I just don't see how i can solve for its final speed without knowing the time or the degree. I'm also assuming the question is asking for its linear speed since it wants the answer in m/s.

    I thought about using mgy = 1/2*m*v^2 + 1/2*I*w^2
    = squareroot{4gy} = v

    but this didn't work and I was thinking that this didn't work because the wheel isn't falling straight down from a height of 75 m, but rather rolling down from what would be the hypotenuse of a triangle with that height...but i understand a triangle isn't needed in this problem. (but i'm not sure)

    I then tried finding the angular velocity by using the given speed of the wheel, 11 m/s (which i assumed to be equal to the linear velocity of the wheel) by using the equation v=rw ([11m/2]/.425 = w) which gave me a value of 25.88 rad/s to plug into the equation: 1/2*2.25kg*11^2 + 1/2(2.25*.425^2)25.88^2 = 136.30 J for total kinetic energy... but this was incorrect. I'm also not sure why the total kinetic energy isn't just mgy which would equal 1653.75 J; this number is a lot bigger than 136.30 and i'm just really confused :mad:
  2. jcsd
  3. Nov 24, 2008 #2


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    This doesn't look right. Conservation of energy means that

    Initial energy = final energy

    So for starters, what is the initial energy? Remember that the bike was going 11.0 m/s initially.

    Yes, that is helpful to relate v and w.
  4. Nov 24, 2008 #3
    I was thinking that its total initial energy would be the sum of its translational and rotational kinetic energy so that is why i thought i could use that equation.

    I don't know what to use if thats not it.
  5. Nov 24, 2008 #4


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    In addition to that, potential energy should be part of the sum also.
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