Part A
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Photo by Keith Finlay, courtesy of Wikimedia Commons |
Velodromes are indoor facilities for bicycle racing. Olympic velodromes are usually ovals 250 m in circumference with turns of radius 25 m. The peak banking in the turns is about 42°. Assuming a racer goes through the turn in such a velodrome at the optimal speed so that no friction is required to complete the turn, how fast is the racer moving?
System:
The rider will be treated as a point particle.
Interactions:
Model:
Approach:
Diagrammatic Representation
Since we are assuming no friction is needed, we have the free body diagram and coordinate system shown above.
Mathematical Representation
The corresponding equations of Newton's Second Law are:
Notice that we have used a true vertical y-axis and true horizontal x-axis. The reason will become clear in a moment.
\begin
[\sum F_
= N \sin\theta][\sum F_
= N \cos\theta - mg = 0 ] \end
Where r is the radius of the turn, v is the speed of the racer and m is the racer's mass (including bike and gear).We have set the force in the x-direction equal to the centripetal force, because this is what the force must be in order for the object (the racer and his bicycle) to execute circular motion.
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 ay 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.
From the y-component equation, we find:
\begin
[ N = \frac
]\end
Note both the similarity to the standard inclined plane formula and the important difference.
Substituting into the x-component equation then gives:
\begin
[ v = \sqrt
= 15 \:
= 33 \:
]\end
Is this speed reasonable for a bike race?
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!
Part B
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Photo courtesy Wikimedia Commons |
The Indianapolis Motor Speedway (site of the Indianapolis 500 race) has four turns that are each 0.25 miles long and banked at 9.2°. Supposing that each turn is 1/4 of a perfect circle, what is the minimum coefficient of friction necessary to keep an IndyCar traveling through the turn at 150 mph from skidding out of the turn?
System:
The car plus contents will be treated as a point particle.
Interactions:
Model:
Approach:
Diagrammatic Representation
The free body diagram from Part A is modified to include friction, as shown above.
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.
Mathematical Representation
The resulting form of Newton's Second Law is:
\begin
[ \sum F_
= F_
\cos\theta + N \sin\theta = \frac{mv^{2}}
][ \sum F_
= N \cos\theta - F_
\sin\theta - mg = 0] \end
We have three unknowns (N, Ff, and m) but only two equations. To solve, we must develop another constraint. To do so, we must notice a key phrase in the problem statement. We are asked to find the minimum coefficient of friction. The minimum coefficient will be the value such that the static friction force is maximized, satisfying:
\begin
[F_
= \mu_
N ] \end
Remember that the point of contact of tires with the road surface is static (unless the car is in a skid) so static friction applies here. That is the reason that there is a minimum coefficient.
Whenever static friction applies, it is important to justify using the equation Ff = μN, since it is also possible that Ff < μN.
With this substitution, we have:
\begin
[ \sum F_
= \mu_
N \cos\theta + N \sin\theta = \frac{mv^{2}}
][ \sum F_
= N \cos\theta - \mu_
N \sin\theta - mg = 0] \end
Proceeding as in Part A:
\begin
[ N = \frac
{\cos\theta - \mu_
\sin\theta} ]\end
which is then substituted into the y-component equation to give:
\begin
[\mu_
\ge \frac{v^
\cos\theta - gr \sin\theta}{v^
\sin\theta + gr \cos\theta}]\end
The substitutions here require some thought. The turns are quarter-circle segments with a length of a quarter mile each. Thus, they have the same radius of curvature as a full circle with circumference 1 mile. The corresponding radius is 256 meters. Then, converting the speed to m/s and using the formula derived above gives:
\begin
[\mu_
\ge 1.3]\end
Given our answer to Part A, we expect that μs can be as small as zero if v2 = gr tanθ. Is that fact reflected in our formula?
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.)
[Examples from Dynamics]