Grand Prix Racing - | The Science of Fast Pinewood Cars |

Splat! That's the sound of a Grand Prix car falling off the track. Well, that's NOT the way to race your car. It was the way the first race for truth about the force of gravity was conducted! That race corrected centuries of errors in thinking about gravity. This model of that race, is one of the simplest presented in this manual.

In order to avoid the complications of friction for now, assume there is no air to cushion the fall (that is, this model requires a vacuum). Also assume there is no energy lost to heat or spinning, etc.. This is the simplest model of a Grand Prix race with some physical reality.

It was at Pisa in Italy that Galileo Galilei proved to the world that objects are accelerated at the same rate no matter what their masses are. Mathematically, the impact greatly simplified the possible models for the action of gravity on an object. From it's simple form F = mg, the general force relation, Newton's 2nd law, F = ma would be inspired.

A model for free fall in a vaccuum is derived below beginning with the energy approach. Then the Newtonian equations of motion approach is used.

E | Total conserved energy |

U | Potential energy |

T | Kinetic energy |

F | Force |

m | Mass |

g | Acceleration by gravity |

v | Velocity (strickly speaking, speed) |

H | Height of release |

h | Height of object as it falls |

t | Time from start of fall |

Important expressions derived for this model include:

Energy Relation | E = mgH = mgh + mv2/2 |

Equation of Motion | F = m a = -mg |

Time of fall | t(h) = \[2(H-h)/g] |

Speed of fall | v(t) = -gt |

E = U(h) + T(h) for E the conserved energy, U the gravitational potential energy and T the kinetic energy of the object.

A car is at height H off of the floor.

The gravitational potential energy is the force of weight mg (mass of the car, m, times gravitational acceleration at sea level, g = 386.088 in/s2) times the distance to fall.

So U = mgh at some height, h, above the floor.

Then E = mgH + 0 at the point of release where H is the initial height above the floor.

At the floor, the potential energy has been converted into kinetic energy. In a fall with no spinning, all the kinetic energy is linear motion. This is written as,

T = mv2/2

Then E = 0 + mv2/2 at the floor where the "0" is a reminder that there is no potential left (since we measured H from the floor).

Since E is constant, (top) mgH = mv2/2 (floor)

Rearranging algebraically, we get the speed with which the car hits the floor,

v = \[2gH]

Notice the dependance on mass has cancelled out. Since g is essentially constant, there is no dependance on weight (W = mg, remember weight is a force!).

Likewise, at any height, h from the release point until it hits the floor, we also have,

E = mgh + mv2/2

So E = mgH = mgh + mvv/2 which leads to,

v(h) = \[2g(H-h)], the speed of the car at a given height between the release point and the floor.

Note that this expression for the speed is in terms of the position h, not time, t.

How long did it take to fall? We can use Newton's second law to find out.

F = ma = -mg

First, divide through by m to get

a = -g

But acceleration is the change in the speed over time, which we write as

a = dv/dt = -g

Rearranging we get dv = -gdt

Now we integrate to get the speed:

v(t) = -gt + v0

Velocity, v(t), is the change is position, h, over time. v0 is the initial velocity, which is zero in this model.

So v(t) = -gt = dh/dt

Again we integrate to get

h(t) = -gtt/2 + v0t + h0

The initial position, h0, is H so

h(t) = -gtt/2 + H

Solving for t we get,

t = \[2(H-h)/g]

One problem with using the equations of motion to solve for model behavior is that it won't always be possible to integrate the forces twice to solve for the time it took the car to "fall". But it may still be possibile to solve for time by integrating the v(h) we determined from considering the conservation of energy relations.

From above, dh/dt = v(h) = \[2g(H-h)] so / t / h \[2g] | dt = | dh/\[H-h]. / 0 / H (starts falling at h=H, not 0)

Let's evaluate each integral separately.

/ t \[2g] | dt = \[2g](t-0) = \[2g]t and / 0 / h | dh/\[H-h] = -2\[H-h] - -2\[H-H] = -2\[H-h]. / H

So together again we get,

\[2g]t = -2\[H-h] and

t = -\[2(H-h)/g]

Don't let the sign bother you, we can pick the negative branch of the square root and make t positive. This is the same expression we obtained from considering the equation of motion.

Rearranging the equation we can get h(t):

t2 = 2(H-h)/g -> gt2 = 2H - 2h -> 2h + gt2 = 2H -> 2h = 2H - gt2 and

h = H - gt2/2

Now we differentiate to get v(t) and again to get a(t).

v(t) = -gt and a(t) = -g as above!

Once again, this is the way we will solve for v(t), t and h(t) in most of the following models. We'll call our trajectory variable y instead of h. h will be reserved as a name for the variable of the vertical axis.

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Grand Prix Racing - | The Science of Fast Pinewood Cars |

Copyright © 1997, 2004 by Michael Lastufka, All rights reserved worldwide. |