## Banked Curves Analysis

All roads have embankments, especially in turns, to assist drivers in comfortably maneuvering through the path. This embankment usually slopes towards the center of the turn to allow gravity to aid the car turning. The three forces acting upon the car in the turn are centripetal force, the normal force of the car, and friction between the car and the road. Centripetal force only acts upon the car while it is turning. “The centripetal force needed to turn the car depends on the speed of the car.” (Stranbrough) Greater speed requires greater centripetal force that the car must handle. When a car is on a flat surface, the normal force is a vertical force because it is perpendicular to the road’s surface. Since the normal force is always perpendicular to the surface of the road, the normal force on a car will have a horizontal component. This force is towards the center of the turn and gives the car the centripetal force needed to turn the car. The speed of the vehicle also dictates the force of friction exerted between the tires and the road. If the centripetal force is greater on a car, a greater friction force is exerted to keep the car in the turn. The combination of the horizontal normal force, centripetal force, and the friction make it possible for cars to drive through turns.

The formula for measuring acceleration is v2/r. Where “v” is the velocity of the car, in m/s, when it enters the turn. “R” is the radius of the turn in meters. The track we are racing on is the Talladega Superspeedway. This track is a tri-oval course with a 2.6-mile diameter and 33 degree banked turns. Each of the two identical turns has a radius of 1,100 feet, which is 335.3 meters.

67.1

335.28

33°

13.43

78.2

335.28

33°

18.24

89.4

335.28

33°

23.84

100.6

335.28

33°

30.18

208.8

89.4

38.28

304.8 (Alabama, 11/01/11)

89.4

26.22

335.28

89.4

23.84

400

89.4

19.98

As the velocity of the car increases, the centripetal acceleration increases. As shown in the data, when the car was traveling 67.1 m/s (150 mph) the car experienced a centripetal acceleration of 13.43 m/s. When the car’s speed was 100.6 m/s (225 mph), the centripetal force was 30.18 m/s. This is quite a significant amount of force that friction must overcome to allow the car to turn. The centripetal acceleration does change depending on the relationship of the slope and radius of the turn. The data shows that a car moving at a constant speed of 89.4 m/s in a turn with a radius of 208.8 meters experiences an acceleration of 38.28 m/s. This is high compared to the accelerations of other turns, but it makes since that a car going through a 200-meter turn at 89 m/s is going to need a lot of centripetal acceleration to make the turn. For a car to complete a 400-meter turn at 89 m/s, there is only 19.98 m/s of acceleration acting on the object.

Reference:

Stranbrough, J. (2006, February 6). A Banked Turn With Friction. In

"Alabama."

The formula for measuring acceleration is v2/r. Where “v” is the velocity of the car, in m/s, when it enters the turn. “R” is the radius of the turn in meters. The track we are racing on is the Talladega Superspeedway. This track is a tri-oval course with a 2.6-mile diameter and 33 degree banked turns. Each of the two identical turns has a radius of 1,100 feet, which is 335.3 meters.

**Table 1: Effect of Acceleration based on Speed****Trial****Velocity (m/s)****Radius (meters) (Alabama, 11/01/11)****Slope of Turn****Centripetal Acceleration (m/s)****1**67.1

335.28

33°

13.43

**2**78.2

335.28

33°

18.24

**3**89.4

335.28

33°

23.84

**4**100.6

335.28

33°

30.18

**Table 2: Effect of Radius of Turn on Centripetal Acceleration****Trial****Radius (meters)****Velocity (m/s)****Centripetal Acceleration (m/s)****1**208.8

89.4

38.28

**2**304.8 (Alabama, 11/01/11)

89.4

26.22

**3**335.28

89.4

23.84

**4**400

89.4

19.98

As the velocity of the car increases, the centripetal acceleration increases. As shown in the data, when the car was traveling 67.1 m/s (150 mph) the car experienced a centripetal acceleration of 13.43 m/s. When the car’s speed was 100.6 m/s (225 mph), the centripetal force was 30.18 m/s. This is quite a significant amount of force that friction must overcome to allow the car to turn. The centripetal acceleration does change depending on the relationship of the slope and radius of the turn. The data shows that a car moving at a constant speed of 89.4 m/s in a turn with a radius of 208.8 meters experiences an acceleration of 38.28 m/s. This is high compared to the accelerations of other turns, but it makes since that a car going through a 200-meter turn at 89 m/s is going to need a lot of centripetal acceleration to make the turn. For a car to complete a 400-meter turn at 89 m/s, there is only 19.98 m/s of acceleration acting on the object.

Reference:

Stranbrough, J. (2006, February 6). A Banked Turn With Friction. In

*Physics at BHS*. Retrieved November 28, 2012, from http://www.batesville.k12.in.us/physics/phynet/mechanics/circular%20motion/banked_with_friction.htm"Alabama."

*Motorsports*. GOBUILDALABAMA.COM, 01 Nov. 2011. Web. 18 Dec. 2012. <http://www.al.com/motorsports/index.ssf?resources/speedway_stats.html>.