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Effect of friction on a wheel

  1. Jul 18, 2011 #1
    Could someone verify that my thinking is correct, or point out the mistakes if it is incorrect?

    I'm trying to figure out where energy goes when rolling a cylinder down an incline. It is initially stationary.

    1) First, assume there is no air resistance. If the surface of the cylinder and the surface of the incline do not deform (assume they are rigid), there can be no rolling resistance. There can still be static friction though, which will prevent the wheel from slipping (sliding). And since no work is performed, the speed of the wheel at the bottom of the incline can be calculated using this energy balance:

    Initial potential energy = final translational kinetic energy + final rotational kinetic energy

    (translational velocity, v, is set to equal [itex]\omega r[/itex] in order to calculate individual energies)

    2) Now assume that the wheel does in fact deform slightly (let's say it's made out of rubber). In this case, energy will be converted to heat. To find out how much energy,

    [itex]F_{friction} = C_{rr}mg\cos{\theta}[/itex]

    ,where [itex]C_{rr}[/itex] is rolling resistance. Calculating the work,

    [itex]\int F_{friction} dx = (C_{rr}mg\cos{\theta})(x)[/itex]

    where x is the length of the incline. This is the energy lost in the form of heat, so:

    Initial potential energy = heat from friction + final translational kinetic energy + final rotational kinetic energy

    3) Now assume air resistance is added. I don't know what to do here -- is energy lost because of air resistance?

    Sorry for such a weird question, but I'm on summer break and doing some bicycle-related physics for fun and I'm not really in "school mode".
     
  2. jcsd
  3. Jul 19, 2011 #2
    yes, clearly, you will need to spent some energy in overcoming drag.

    Airplanes, cars, they all are put into wind tunnels for studies and figure out their drag coefficients...

    typically, the front of moving objects is not as important as the back end...at the front they can turn pretty thick pretty quickly, but at the back a longer, pointier profile is prefered...like the solar/photovoltaic race cars that then to have a water drop shape (a drop on its side, actually).

    where exactly does the energy go? well, I guess some work is done in moving the air out of the way, warming it up...
     
  4. Jul 19, 2011 #3

    sophiecentaur

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    If there is no actual slippage or the cylinder, overcoming the static friction between cylinder and slope, there will be no energy loss. There is no Force times Distance to constitute Work done against the friction. It's just as if the cylinder were the wheel of a vehicle, held on the slope by a brake of some kind. That's assuming no slippage and no distortion of surfaces etc . - the ideal case. Where there's distortion, the hysteresis of the material becomes relevant.
     
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