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Rotational kinetic energy of a bike wheel

  1. Oct 29, 2007 #1
    1. The problem statement, all variables and given/known data

    A bicycle has wheels of radius 0.29 m. Each wheel has a rotational inertia of 0.091 kg·m^2 about its axle. The total mass of the bicycle including the wheels and the rider is 79 kg. When coasting at constant speed, what fraction of the total kinetic energy of the bicycle (including rider) is the rotational kinetic energy of the wheels?


    2. Relevant equations

    translational KE = 1/2 mv^2

    rotational KE = 1/2 I (omega)^2 = 1/2 I (v/r)^2

    rotational KE / total KE

    = (I/r^2) / (I/r^2 + m)

    3. The attempt at a solution

    Moment of inertia (I) is given: 0.091 kg·m^2

    Radius of the wheel is .29 meters

    Mass is 79kg

    so:

    rotational KE / total KE

    = (I/r^2) / (I/r^2 + m)

    =(0.091kg·m^2/0.29^2m) / (0.091kg·m^2/0.29^2m + 79kg)
    =.0135 which is not the right answer.

    Anyone know where I messed up?
     
  2. jcsd
  3. Oct 29, 2007 #2

    rl.bhat

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    Rotational motion

    Your formula is true when the axis of rotation passes through CM. In this case it is not true.
    Using moment of inertia and radius of the wheel and total mass, find the mass of the wheel(Mw) and the rest of the mass (Mr)
    Now the total energy = 1/2*(Mr)*v^2 + 2x[1/2{(Mw)*(r^2)/2}(v/r)^2 + 1/2(Mw)*v^2]
     
    Last edited: Oct 30, 2007
  4. Oct 29, 2007 #3
    Thanks for your reply. I tried to get the velocities (v) to cancel algebraically but I could not. I think it is beyond my ability to derive a speed. Any hints on how to handle to velocity?
     
  5. Oct 29, 2007 #4

    Dick

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    The answer is a lot simpler than that. There are two wheels.
     
  6. Oct 29, 2007 #5
    I am intrigued. Are you implying that the answer could be something such simple as .5 or 1.2?
     
  7. Oct 29, 2007 #6

    Dick

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    Not with an odd number for the rotational inertia. But it's still pretty simple. In your original calculation you only put in the kinetic energy for one wheel. Double it.
     
  8. Oct 30, 2007 #7
    Thanks for your help.

    I am a bit confused. Would I double the answer or would I double just rotational or total kinetic energy?

    = 2((I/r^2) / (I/r^2 + m))
    or
    = (I/r^2) / 2(I/r^2 + m)

    I am not sure what to do.
     
  9. Oct 30, 2007 #8

    Dick

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    Think. You did it correctly for a monocycle. Linear KE doesn't change. Just rotational.
     
  10. Oct 30, 2007 #9
    Thanks for your help. That gave me just enough information to finally figure that one out.
     
  11. Oct 30, 2007 #10

    Dick

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    I strive to give "just enough", which means you figured it out on your own. You're welcome.
     
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