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What is the total rotational kinetic energy of the wheels?

  1. Oct 20, 2006 #1
    A 50kg rider on a moped of mass 75kg is traveling with a speed of 20 m/s. Each of the two wheels of the moped has a radius of 0.2m and a moment of inertia of 0.2 kg*m^2. What is the total rotaional kinetic energy of teh wheels?

    KEr = 1/2Iw^2
    = 1/2(.02)(20)^2
    = 4
    Two wheels = 2 *4 = 8 Joules

    I think I am wrong as the angular velocity (w) has to be in rad/s. For some reason, I can not figure out how to get that. What is the formula for w?
  2. jcsd
  3. Oct 20, 2006 #2

    Doc Al

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    Staff: Mentor

    Yes, that formula requires angular speed, not linear speed. They are related by: v = wr.

    (A complete circle, one circumference, is equivalent to [itex]2 \pi[/itex] radians.)
  4. Oct 20, 2006 #3


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    Science Advisor
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    Yeah, you're using an incorrect value for [itex]\omega[/itex].

    If you're confused by the radians, it might be helpful to figure out how many degrees per second the wheels are turning, and then use the conversion:
    [tex]1 \rm{degree}=\frac{\pi}{180} \rm{radians}[/tex]
  5. Oct 20, 2006 #4
    thanks. I knew I was forgetting something. It just didn't look right.
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