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Kinetic energy of wheel/tyre when braking

  1. Jun 1, 2013 #1
    I have been doing some working on braking energies, and have hit a bit of confusion with regards to the kinetic energy of the wheels and tyres.

    Obviously, the kinetic energy of the moving car itself follows the old rule of 1/2mv2.

    For a 1000kg car at 30ms that's then 450kJ.

    I have then calculated the kinetic energy of a 12kg wheel/tyre with a rolling diameter of .575m, assuming the 'centre of mass' to be at 75% of the rolling radius as 57kJ. (30ms = 105 rad/s)

    This seems substantial addition to the total kinetic energy over four wheels. Now, assuming I have not made a big mistake (I may well have done), should that 228kJ be added to the 450kJ of the moving vehicle, giving me a total of 678kJ of energy that must be removed from the car to bring it to a standstill? Or is it somehow absorbed into the 450kJ?
  2. jcsd
  3. Jun 1, 2013 #2


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    Hi Kozy! :smile:

    I haven't checked your calculations,

    but they must be wrong, since the "rolling mass" of the wheels should be half the actual mass of the wheels.
  4. Jun 1, 2013 #3

    jack action

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    This might help: http://hpwizard.com/rotational-inertia.html

    Rule of thumb, the wheels, tires, brakes and half-shaft rotational effects should add the equivalent of 5% of the total mass of the car.
  5. Jun 7, 2013 #4
    Could you expand on this? I am not familiar with that...
  6. Jun 7, 2013 #5


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    If you do all the calculations, you find that a rolling object of mass m, moment of inertia I, and rolling radius r, behaves like an object of total mass m + mr,

    (eg a force F will give it a linear acceleration of F/(m + mr))

    where mr is the rolling mass, or equivalent mass, equal to I/r2 :smile:

    (and so mr for a uniform disc = m/2)
  7. Jan 13, 2015 #6
    a hollow disc , yes, but a solid disc, no. I =1/2Mr^2 vs I =MR^2
    hollow disc acts like 2x the mass Is sitting in the car,
    solid might be as low as 1.3x the mass as if it was sitting in the car, depending on the mass distribution of the disc.
  8. Jan 27, 2015 #7
    The mass of the wheels is included in the mass of the car, so calculate the linear KE @ ½ * 1000 * 30 ²
    Now calculate the rotational KE of 1 wheel at 30 m/s:
    Start with the rolling radius (r) : = 0.2875 m
    The radius of gyration (k) = 0.75 * 0.2875 = 0.2156 m
    The moment of inertia (i) = m * k ² = 0.5579 kg-m ²
    The rortation rate (w) at 30 m/s = v/r = 30 / 0.2875 = 104.35 rad/sec
    The rotational KE at 30 m/s = ½ * i * w ² = 3,037.5 J
    Multiply by 4:
    12,150 J
    Add to linear KE
  9. Jan 27, 2015 #8
    You have calculated all that weight to be at the tire, not the wheel. if the wheel and it is distributed like a disc, then "I" is 1/2mr^2
    if you did it correct, the KE of the weight at the outer diameter vs weight in the car would be equal and additive.
  10. Jan 28, 2015 #9
    It cant be considered a disc. The mass distribution isnt that simple.
    But if you have the radius of gyration and mass you can calculate the moment of inertia, with that and the rolling radius you can calculate the rotational KE.
    Each wheel has both linear and rotating KE.
  11. Jan 28, 2015 #10
    I know ,that's why I said it the inertial of the weight if on the tire, is like a hollow disc, and I would be I=MR^2. that weight would be like if the mass was attached on a mass less radius cord, and its center was close to the OD of the tire. (so its going to be a little off due the width (thickness of the tire... or in other words, all the mass is not at the outer diameter, its down toward the center slightly, thus it is not a 'slippless" calculation vs the linear KE equiv. (so slightly less than the linear KE)
    However, the wheel could be considered a Disc, and use I=1/2MR^2, if the mass is evenly distributed . best case even if it was, the rotational KE would be less than 1/2 that of the linear KE. (depending on the wheel diameter vs the overall wheel and tire diameter)
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