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Homework Help: Ground force applied to wheel in pure rolling motion at contact point

  1. Jan 5, 2014 #1
    1. The problem statement, all variables and given/known data
    Suppose a wheel with radius [itex]R[/itex] is resting on a non-inclined surface. A torque [itex]\tau[/itex] is applied to the wheel center. In an attempt to prevent wheel from spinning, the ground applies a static friction force to the wheel at the contact point (parallel to the surface), then the wheel starts rolling without spinning. The same friction force also acts as a torque on the wheel around its axis.
    This scenario is depicted below:


    I'm trying to find the magnitude of the force the ground applies to the wheel - that is, the force which drives the wheel forward (which should be the same in magnitude as the force the wheel applies to the ground at the contact point). And also the magnitude of the counter torque the ground is applying to the wheel (in counter clockwise direction). And also the final angular acceleration of the wheel

    2. Relevant equations
    The relation between linear acceleration [itex]a[/itex] and angular acceleration [itex]\alpha[/itex] for a pure rolling movement is given by
    a = \alpha \centerdot R

    The relation between torque [itex]\tau[/itex] and angular acceleration [itex]\alpha[/itex] is

    [itex]\tau = I \centerdot \alpha[/itex]

    where [itex]I[/itex] is the moment of inertia of the wheel around its axis.

    The relation between torque [itex]\tau[/itex], force [itex]f[/itex] and lever arm [itex]R[/itex] is:

    [itex]\tau = f \centerdot R[/itex]

    Being the engine torque [itex]\tau_e[/itex], the friction force [itex]f[/itex], the counter torque due to friction force [itex]\tau_f[/itex] and the moment of inertia of the wheel [itex]I[/itex] around its axis given by [itex]\frac{1}{2}mR^2[/itex]:

    The linear acceleration of the wheel is due to the friction force only:

    [itex]f = ma[/itex]
    [itex]a = \frac{f}{m}[/itex]

    3. The attempt at a solution
    The following equation is the counter torque the ground applies on the wheel's edge (negative, because it pointing in the opposite direction of [itex]\tau_e[/itex]):

    [itex]\tau_f = -f \centerdot R[/itex]

    The net torque causes angular acceleration on the wheel:

    [itex]\tau = \tau_e + \tau_f[/itex]
    [itex]\tau = \tau_e - f \centerdot R[/itex]
    [itex]\tau = I \centerdot \alpha[/itex]
    [itex]\alpha = \frac{\tau}{I}[/itex]
    [itex]\alpha = \frac{\tau_e - f \centerdot R}{I}[/itex]
    [itex]\alpha = \frac{\tau_e - f \centerdot R}{\frac{1}{2}mR^2}[/itex]

    Substituting [itex]\alpha[/itex] and [itex]a[/itex] in [itex](1)[/itex] gives:

    [itex]\frac{f}{m} = \frac{\tau_e - f \centerdot R}{\frac{1}{2}mR^2}R[/itex]


    [itex]f = \frac{2}{3}\frac{\tau_e}{R}[/itex]

    And that is the force [itex]f[/itex] from static friction which pulls the wheel forwards without making it spin or slip, and consequently, the force the wheel applies to the road surface at the contact point.

    But I found this link:
    http://www.asawicki.info/Mirror/Car Physics for Games/Car Physics for Games.html

    And it says:
    "The torque on the rear axle can be converted to a force of the wheel on the road surface by dividing by the wheel radius. (Force is torque divided by distance)."

    That statement doesn't match the approach I used above. If the force of the wheel on the ground was simply engine torque divided by radius (negative, since it pointing in the opposing direction):

    [itex]f = -\frac{\tau_e}{R}[/itex]

    then the counter torque applied to the wheel would be

    [itex]\tau_f = -f \centerdot R[/itex]

    that implies that [itex]\tau_f = -\tau_e[/itex].

    Wouldn't the counter-torque and the engine torque cancel each other?
    That means the net torque would be zero, and the wheel would just slip without rotating at all (zero angular acceleration, which makes no sense).

    In the same article there is a text below the heading "Torque on the drive wheels". It says the total torque on the rear axle is drive_torque + traction_torque + brake_torque. Let's disregard brakes. It also says the traction_torque equals to traction_force * radius. Then it uses this total torque to derivate the angular acceleration of the wheel (equals to total_torque / wheel inertia). The only way for the total torque to be non zero is if traction_force < drive_torque / radius.

    But the article doesn't explicitly mention how to calculate the traction force (friction force), but it states that drive force is the force the wheel exerts on the road (equals to drive_torque / radius). And from the Newton's third law, I suppose that the friction force should be equal to drive force. Then, from all above: traction_torque = drive_torque + traction_torque = drive_torque - (traction_force * radius) = drive_torque - (drive_torque / radius * radius) = zero. Thus, angular acceleration of the wheel = total_torque / wheel inertia = zero.

    Another argument: the ground force is what moves the wheel's center linearly forwards. If this force was just torque/radius, two wheels of same radius and mass but different moments of inertia (say, a disc and a ring) would accelerate forwards equally (since their moments of inertia aren't being taken into account when calculating the ground force), which is not true. The wheel with greater moment of inertia will roll slower than the other. This is why I think the ground force calculation should involve the moment of inertia of the wheel.

    I already know and understand well that the velocity of the point of contact is always zero for the non-slip condition.
    Am I doing something wrong on my calculations? What would be the magnitudes of force and counter-torque, the net torque and the final angular acceleration of the wheel?

    Last edited: Jan 5, 2014
  2. jcsd
  3. Jan 5, 2014 #2


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    Science Advisor
    Homework Helper

    hi ri_r! welcome to pf! :smile:
    your approach is correct

    the quote is wrong, for the reason you give

    as to the other parts of that link, the writer admits …
    "I've found somewhere that the inertia of a solid cylinder around its axis can be calculated as follows:

    inertia of a cylinder = Mass * radius2 / 2"​
    … so he can't remember where he found it, and he doesn't know how to work it out! :rolleyes:

    … and so i suggest you ignore the link completely

    btw, you can also do τ = Iα about the moving centre of rotation (because it – the point of contact – is moving in a straight line parallel to the centre of mass), which immediately gives you α = 2τe/3mr2 :wink:
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