I have been working on a dynamometer project for radio controlled cars.
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?
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
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.