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The rider will be treated as a .

The rider is subject to external influence from the earth (gravity) and from the track (normal force). We assume friction is not present, since we are told to determine the speed at which friction is unnecessary to complete the turn.

and .

Notice that we have used a true vertical y-axis and true horizontal x-axis. The reason will become clear in a moment.

\begin{large}\[\ F_{res} = \sum F_{x} = N \sin\theta\]\end{large}

\begin{large}\[N \sin\theta = \frac{mv^{2}}{2} \]\end{large}

The situation here is very different than that of an object sliding down an inclined plane. In the case of an object moving along the plane, the acceleration will have both x and y components if our untilted coordinates are used. For an object moving along a banked curve, the object will not be moving up or down and so a_y must be zero. The x-component of the acceleration will of course not be zero because the object is following the curve. This difference between motion along a banked curve and motion down an inclined plane is the reason we use tilted coordinates for the incline and traditional coordinates for the curve.

It is not appropriate to assume N = mg cosθ. Note that we have not tilted our coordinates to align the x-axis with the ramp. Thus, the y-direction is not perpendicular to the ramp.

\begin{large} \[ N = \frac{mg}{\cos\theta} \]\end{large}

Note both the similarity to the standard inclined plane formula and the important difference.

\begin{large} \[ v = \sqrt{gr\tan\theta} = 15 \: {\rm m/s} = 33 \:{\rm mph} \]\end{large

}

Note that our result is independent of the mass of the rider. This is important, since otherwise it would be impractical to construct banked curves. Different people would require different bankings!

The car plus contents will be treated as a .

The car will be subject to external influences from the earth (gravity) and from the track (normal force and friction).

and .

In this problem, it is not clear a priori which way friction should point. Under certain conditons friction will point up the incline while in others it will point down (can you describe the conditions for each case?). This ambiguity should not prevent you from solving. Simply guess a direction. If you guess wrong, your answer will come out negative which will indicate the correct direction.

\begin

\begin
{large} \[F_{f} = \mu_{s} N \] \end{large

}

Whenever static friction applies, it is important to justify using the equation F_f = μN, since it is also possible that F_f < μN.

\begin
{large} \[ \sum F_{x} = \mu_{s}N \cos\theta + N \sin\theta = \frac{mv^{2}}{r} \]\[ \sum F_{y} = N \cos\theta - \mu_{s} N \sin\theta - mg = 0\] \end{large

}

\begin
{large} \[ N = \frac{mg}{\cos\theta - \mu_{s} \sin\theta} \]\end{large}

\begin
{large}\[\mu_{s} \ge \frac{v^{2}\cos\theta - gr \sin\theta}{v^{2} \sin\theta + gr \cos\theta}\]\end{large

}

\begin
{large}\[\mu_{s} \ge 1.3\]\end{large

}

Our answer is slightly in error. The normal force exerted on the car is actually increased beyond our assumed value by the presence of airfoils at the front and back of the car that generate negative lift (a downward force, often simply called "downforce"). This force is specifically intended to increase the maximum friction force available from the tires. As a result, the friction coefficient can likely be lower than our answer and still keep the car on the track. How large (as a fraction of the car's weight) would the downforce have to be keep the minimum acceptable coefficient of friction below 1.0? Assume the downforce presses straight into the surface of the track. (IndyCars can generate more than three times their weight in downforce when traveling near top speed.)