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?

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.

You'll painfully come to realize guys here aren't really interested in helping with real physics. They're more into thrashing small time 'cranks' such as yours trully.

You are only a crank because you don't do physics, and yet voice opinions on physics.

On the track:
car mass = m = 1 kg
car propulsion = F = 1 N
car track acceleration = a = F/m

On the dynometer:
cylinder mass = M
cylinder radius = R
(solid) cylinder moment of inertia = I = (1/2) M R^2
cylinder torque = t = F R (only if an arm holds the car in fixed position!!!)
cylinder angular acceleration = A = t/I = 2 F/(MR)
cylinder circumferential acceleration = A R = 2 F/M

So if you want A R = a so you can accurately correlate v and a on a simulated straight run, then
2 F/M = F/m
or
M = 2 m -- regardless of the radius. (But you have to hold the car in fixed position.)

Another way to do this:

For all speeds v << c :

Kinetic energy of a car at speed v : (1/2) m v^2
Kinetic energy of a rotating solid cylinder at circumferential speed v : (1/2) I ω^2 = (1/2) (1/2) M R^2 (v/R)^2 = (1/4) M v^2

Since these two are the same when M = 2 m, the same amount of work is done one the cylinder by the fixed car as is done by the car on itself on the track to build up the speed from 0 to v. So the cylinder is a type of model for the road.

See? It worked. They can scent us from miles away. If you're lost and need a real Physicist to find you, just plant 'us'. We have our usefulness and this is how: As they come to trap this skunk they notice your plight at the side. How nice. Kudos to all 'cranks'. Who said " I love 'Cranks' "?


PS: The real giants in this forum are Rpenner, Alphnumeric,FBM..... Others you'll find out by yourself. These you can take seriously. How sure am I? Just know even the village tramp knows who the king is.

@rpenner - Thanks for the confimation and the alternative approach.

DamoPhys