Question 12.CS.1: In 2001, drivers practicing for the Firestone Firehawk 600 C......

In 2001, drivers practicing for the Firestone Firehawk 600 CART series race (see racecar in CS Photo 12.1) complained of being dizzy and disoriented. Due to concerns over driver safety, the race was canceled (resulting in a huge loss of revenue and some ensuing lawsuits).
The turns at the Texas Motor Speedway have a radius of 750 feet, as shown in CS Fig. 12.1, and are banked at 24 degrees. This bank angle is much higher than that found at Indianapolis (9 degrees) or at Michigan (18 degrees), and allowed the cars to reach speeds of 230 mph (337.3 ft/s). To investigate why the drivers were feeling disoriented, determine (a) the normal acceleration of the driver, (b) the normal force from the seat, assuming that the driver weighs 150 lbs and that the seat harness applies a sideways force on the driver.

Additionally, knowing that the weight of the car and driver is 1750 lbs, determine the minimum coefficient of sideways friction necessary to keep the car from skidding up the track. We can estimate that the downforce caused by the spoilers of the car is 2.5 times the car’s weight at these high speeds.

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STRATEGY: For the first part of the problem, we are analyzing the driver. Using normal and tangential coordinates, we can determine the normal acceleration of the driver. Then, we will analyze the forces on the driver by using Newton’s second law. For the second portion of the problem, we will analyze the forces on the car by again using Newton’s second law, but now will isolate the racecar for our system.

MODELING (THE DRIVER): Draw the free-body and kinetic diagrams of the driver; this includes the normal force N from the seat, the weight mg of the driver, and the side force F_{s} due to the harness belt. Note that we are only looking at the forces in the vertical and normal direction and ignoring the forces in the tangential direction.

ANALYSIS: You can determine the normal acceleration using

a_{n}=\frac{v^{2}}{R}=\frac{(337.33\;\mathrm{ft}/s)^{2}}{750\,\mathrm{ft}}=151.7\;\mathrm{ft}/s^{2}

This is 4.71 times the gravitational constant, 32.2 ft/s². Now, summing forces in the n and y directions:

\begin{array}{c l}\underrightarrow{+}{{\Sigma F_{n}=m a_{n}=m\displaystyle\frac{v^{2}}{R}\qquad}}&{{F_{s}\cos\theta+N\sin\theta=m a_{n}=m(151.7\ f t/s^{2})}}\\ {{\qquad}}&{{}}\end{array}            (1)+↑\Sigma F_{y}=0\qquad\qquad\qquad N\cos\,\theta-F_{s}\sin\,\theta-m g=0           (2)

Where m = W/g = 150 lb/32.2 ft/s²= 4.658 lbs²/ft. Solving (1) and (2), you find N = 424.5 lbs and F_{s} = 584.7 lbs. This normal force is 2.83 times the person’s weight, which means that the driver will be pulling almost 3 g’s for prolonged periods during the race. This can result in blood pooling in the lower extremities, causing the driver the discomfort reported.

MODELING (THE CAR): Now you can analyze the car. Drawing the freebody and kinetic diagrams yields:

ANALYSIS: You can use the diagrams to set up your equations of motion.

\underrightarrow{+}\Sigma F_{n}=m a_{n}=m\frac{v^{2}}{R}\qquad F\cos\theta+(N_{C}-D)\sin\theta=m a_{n}=m(151.7\ \mathrm{ft}/s^{2})               (4)

+↑\Sigma F_{y}=0\qquad\qquad (N_{C}-D)\cos\theta-F\sin\theta-m g=0               (5)

Solving (4) and (5) for F and N_{C} gives F = 7146 lb and N_{C} = 8597 lb. The minimum friction coefficient to allow this to occur is μ = F/N_{C} = 0.831.

REFLECT and THINK: You could also solve for the rated speed of the curve, exactly like what was done in Sample Prob. 12.7 (you get an answer of 70.7 mph when you have no friction and no downforce). The Texas Motor Speedway was originally designed for NASCAR racing, and these cars typically reach speeds of 190 mph during qualifying. The turbocharged engines and large downforce of the CART racers allowed much greater speeds, which put the racers at risk. If you were designing a new track for the CART racecars, what is the minimum radius of curvature that you should use to keep the normal force on the drivers under two times their weight?

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