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"(adsbygoogle = window.adsbygoogle || []).push({});

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/s^{2}

If the car on the road is accelerating at 1 m/s^{2}then, on the ideal roller, it should be accelerating the roller circumference at the same rate (1 m/s^{2}).

Converting this acceleration to radians/s^{2}is 1 m/s^{2}/0.025 m giving 40 radians/s^{2}.

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/s^{2}) therefore the moment of inertia (I) = t/a = 0.025 N.m / 40 rad/s^{2}= 0.000625 N.m/s^{2}(units equivalent to SI units for I which are kg.m^{2})

Moment of inertia (I) for a solid cylinder = (mr^{2})/2 = 0.000625 kg.m^{2}

Therefore. 2.I/r^{2}= m, which is 2 x 0.000625 kg.m^{2}/ 0.000625 m^{2}= 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.

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# Trying to link linear and rotational dynamics

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