The Physics Teacher, Vol. 41, No. 6, pp. 355–361, September 2003
©2003 American Association of Physics Teachers. All rights reserved.
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Analysis

For the quantitative analysis, the data are pasted into a spreadsheet. We first determine the air drag and rolling friction coefficients from the coasting run. We fit velocity-versus-time graphs with a calculated curve as shown in Fig. 8.

Figure 8.

We assume the standard forms for the air drag force and rolling frictional force:

<i>F</i><sub>d</sub>  =  (1/2) <i>C</i><sub>d</sub>  <i>rho</i> <i>A</i>(<i>v</i>  +  <i>v</i><sub>w</sub>)<sup>2</sup> <i>F</i><sub>r</sub>  =  <i>µ</i><sub>r</sub><i>m</i><i>g</i>,

where Cd is the drag coefficient, rho the air density, A is the cross-sectional area, v is the velocity, vw is the wind velocity, µr is the coefficient of rolling force, and mg is the weight of the car.

The sum of these forces divided by the mass gives the acceleration of the car. Because the forces are a nonlinear function of the speed, there is no simple expression for the velocity versus time. So the resulting motion can't be analytically solved but it is easily solved numerically. We use the straightforward Euler method of dividing time into short intervals and assuming constant acceleration within each interval. The velocity at the end of the interval is used to determine the drag forces and acceleration in the next interval. The free parameters are: the starting speed, the coefficient of rolling friction, and the air drag coefficient. If you look at the details, you will see various bumps and wiggles in the data. These are due to humps and dips in the road, and variations in wind speed.

The torque and power of the engine are both determined from the acceleration runs. Using a spreadsheet, we take differences in velocities and divide by the time interval to get the acceleration. So the acceleration is given by

<i>a</i><sub>n</sub>  =  ((<i>v</i><sub>n + 1</sub>  –  <i>v</i><sub>n</sub>)/((1/2)(<i>Delta</i> <i>t</i><sub>n + 1</sub>  +   <i>Delta</i> <i>t</i><sub>n</sub>))).

We assume the only forces acting on the car are the forward driving force by the road on the wheels, an air-drag force, and a rolling friction force. Solving Newton's second law for the force on the wheel by the road, we have (omitting the index n):

<i>F</i><sub>w</sub>  =  <i>m</i><i>a</i>  +  <i>F</i><sub>d</sub>  +  <i>F</i><sub>r</sub>,

where the drag and rolling forces are calculated from the velocities using the parameters determined above. Ignoring the mass of the wheel, this force at the perimeter of the wheel is created (and opposed) by the torque on the axle. So the torque on the wheel is just the force on the wheel times the wheel radius:

<i>tau</i><sub>w</sub>  =  <i>F</i><sub>w</sub><i>R</i>.

The engine spins faster and has less output torque than the wheels, both of these changed by a factor of the overall gear ratio G. So (neglecting transmission losses) the engine output torque is:

<i>tau</i><sub>EO</sub>  =  ((<i>tau</i><sub>w</sub>)/<i>G</i>).

Since we are interested in how torque and power vary with the speed of the engine, the engine speed is found from:

<i>omega</i><sub>E</sub>  =  <i>G</i> <i>omega</i><sub>w</sub>  =  <i>G</i>(<i>v</i>/<i>R</i>).

Once the torque is found, the engine power can be directly calculated using the angular equivalent of force times velocity:

<i>P</i>  =   <i>tau</i><sub>E</sub><i>omega</i><sub>E</sub>


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