Rotational Kinetic Energy of bicycle wheels

  1. Mar 6, 2008 #1
    A bicycle has wheels of radius 0.33 m. Each wheel has a rotational inertia of 0.082 kg* m2 about its axle. The total mass of the bicycle including the wheels and the rider is 74 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?

    I thought this: Rotational KE = (1/2)Iw^2
    =(1/2)(second bold number)w^2

    Linear KE= (1/2)mv^2
    = (1/2)(third bold number)(radius*w)^2 (i.e. plug in r*w for v)
    Total KE is equal to Rotational KE + Linear KE
    add the two eqns
    (1/2)Iw^2/ (some # * w^2)
  2. jcsd
  3. Mar 7, 2008 #2
    i think you are correct
    eventually one takes away the w^2 in both numerator and denominator, then gets a result independent of w
  4. Mar 7, 2008 #3
    i keep getting .041/4.07 and that is not right
  5. Mar 7, 2008 #4
    How many wheels does a bicycle have?

    Please post in the HW forums.
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