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Rotational Kinetic Energy of disc brakes

  1. Nov 5, 2012 #1
    The disc brakes of a high performance car are often made of carbon fiber instead of iron, thereby reducing the mass. If both types of discs are of the same size and shape, and each iron disc has a mass of 4 kg and each carbon disc has a mass of 1 kg, what is the reduction in rotational kinetic energy at 72 km/h if all the four iron discs in the car are replaced with carbon discs?

    M1 = 4 kg
    M2 = 1 kg
    v = 72 km / h = 20 m / s
    I = (1/2)MR2
    K = (1/2)Iω2 + (1/2)Mv2
    ω = v / R
    ΔK = KFe - Kcf

    My equation:
    4[.5(.5M1R2)(v2 / R2) + .5M1v2] - 4[.5(.5M2R2)(v2 / R2) + .5M2v2]

    The R's should cancel and and from that point on it's just plug and chug. Hoever I get 3600 J as the difference in KE. What's wrong with it?

    [Edit] The answer is 1200 J by the way. That's why I don't like the 3600 J. [Edit]
     
  2. jcsd
  3. Nov 5, 2012 #2

    Delphi51

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    The question asks for the reduction in rotational energy.
    You found the reduction in (rotational + translational) energy (correctly).
     
  4. Nov 5, 2012 #3
    So then if my equation is really:

    4[.5(.5M1R2)(v2 / R2)] - 4[.5(.5M2R2)(v2 / R2)]

    I get 600 J as the difference, which is another 600 J short of the answer.
     
  5. Nov 5, 2012 #4

    Delphi51

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    The expression looks good! I put in 4 for M1, 1 for M2and it worked out to 3v².
    v = 72000/3600 = 20 and it evaluated to 1200 J.
    Did you lose the minus sign on the 2nd term?
     
  6. Nov 5, 2012 #5
    What I put in my TI-84:

    4(.5(.5*4*200) - .5(.5*1*200))
     
  7. Nov 5, 2012 #6
    Never mind. I see what I'm doing. I'm putting the v2 in parentheses with the .5M1R2. Thanks. I appreciate the help.
     
  8. Nov 5, 2012 #7

    Delphi51

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    Easier to do that calc in your head than with the calculator! And no problems with brackets. Most welcome.
     
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