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Trying to link linear and rotational dynamics

  1. Jun 9, 2010 #1
    I have been working on a dynamometer project for radio controlled cars. You can see my early (and somewhat crude) attempts here: http://www.youtube.com/watch?v=qEhnFd8bwqM"

    A question has come up in some of the discussions I have had with other enthusiasts. Given the diameter of the rollers and the mass of the car, what is the ideal roller mass needed to simulate "real world" conditions. (When I say "real world" it is assumed that we are not accounting for aerodynamic drag or the difference between track and roller friction or contact patch size differences etc).

    I got a headache trying to directly relate the concepts of linear acceleration, rotational acceleration, moments of inertia etc, so I tried the following numerical approach. Given the caveats outlined above can anyone advise as to whether the following approach is valid for approximating the recommended roller mass?

    Given
    Mass of car = 1 kg
    Assume car produces mechanical output force of 1 N for a time of 1 second (s)
    Roller is a solid cylinder of radius 0.025 meters

    Linear acceleration:
    F = m.a => a = F/m = 1 m/s2

    If the car on the road is accelerating at 1 m/s2 then, on the ideal roller, it should be accelerating the roller circumference at the same rate (1 m/s2).
    Converting this acceleration to radians/s2 is 1 m/s2/0.025 m giving 40 radians/s2.

    Torque(t) = F.r
    The Force applied to the roller by the car is the same as the mechanical output of the car, i.e. 1 N.
    Therefore the Torque (t) applied to the roller is 1 N x 0.025 meters = 0.025 N.m

    Torque(t) is also moment of inertia x angular acceleration (t = I.a). We already know the Torque (0.025 N.m) and we know the angular acceleration (40 rad/s2) therefore the moment of inertia (I) = t/a = 0.025 N.m / 40 rad/s2 = 0.000625 N.m/s2 (units equivalent to SI units for I which are kg.m2)

    Moment of inertia (I) for a solid cylinder = (mr2)/2 = 0.000625 kg.m2
    Therefore. 2.I/r2 = m, which is 2 x 0.000625 kg.m2 / 0.000625 m2 = 2 kg

    Is this approach valid? Is 2 kg the "ideal" mass for the roller to simulate "real world" conditions in this example?

    Any thoughts or comments would be appreciated.
     
    Last edited by a moderator: Apr 25, 2017
  2. jcsd
  3. Jun 9, 2010 #2

    rcgldr

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    Homework Helper

    Dyno's don't worry about matching angular inertia to a cars mass. Instead they just make sure there's enough angular intertia and friction (plus cars are strapped down) to deal with high powered cars. The higher the angular inertia in the roller, the less the angular inertia in all the cars rotating parts will have in the measurement.
     
  4. Jun 9, 2010 #3
    Thanks rcgldr but that is not quite the question I was asking. I am not worried about the angular rotation of the car parts - I am trying to calculate what roller dimensions will offer the car the most realistic situation on the dyno. For example, with light rollers, the car will spool up to speed quickly without really challenging the batteries, speed control circuitry, motor etc. With rollers that are too heavy, the cars electronic components will be put under stresses that they otherwise would not encounter. There is an "ideal" roller setup that would put the car under the same stresses that it would experience on the road - that's what I am after (even though at higher speeds drag forces stat to become significant which I am ignoring for now).
     
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