Trying to link linear and rotational dynamics

In summary: I realize this is more complicated than just finding the mass of the roller, but that is the goal.Thanks for the input - I will definitely consider the effects of drag when I am further along with this project.In summary, according to the author, the ideal roller mass for a radio controlled car is 2 kg.
  • #1
DamoPhys
3
0
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.
 
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  • #2
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.
 
  • #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).
 

1. How are linear and rotational dynamics related?

The relationship between linear and rotational dynamics is defined by Newton's second law, which states that the sum of all forces acting on an object equals the mass of the object multiplied by its acceleration. This applies to both linear and rotational motion, with linear forces causing linear acceleration and rotational forces causing rotational acceleration.

2. What is the difference between linear and rotational motion?

Linear motion refers to an object moving in a straight line, while rotational motion refers to an object rotating around a fixed axis. Linear motion can be described using variables such as velocity and displacement, while rotational motion can be described using variables such as angular velocity and angular displacement.

3. How can linear and rotational dynamics be linked mathematically?

The mathematical link between linear and rotational dynamics is through the concept of torque. Torque is the rotational equivalent of force and is equal to the product of force and the distance from the axis of rotation. This means that a force applied at a distance from the axis of rotation will cause rotational motion, just like how a force applied to an object will cause linear motion.

4. What is the role of inertia in linking linear and rotational dynamics?

Inertia is the tendency of an object to resist changes in its motion. In the context of linking linear and rotational dynamics, inertia plays a crucial role as it determines how an object will respond to forces and torques. The higher an object's inertia, the more difficult it is to change its linear or rotational motion.

5. How does the concept of center of mass relate to linear and rotational dynamics?

The center of mass is the point at which an object's mass is evenly distributed, and it is the point about which an object's mass is balanced. In terms of dynamics, the center of mass is important as it is the point at which all linear and rotational forces can be applied to an object to cause motion. This allows us to simplify the analysis of complex systems by treating them as a single point rather than considering each individual particle.

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