Frictionless banked turn? What is wrong with my thinking?

In summary, the Daytona 500 is a major NASCAR event held in Daytona, Florida at the Daytona International Speedway. The turns on the oval track have a maximum radius of 316 meters and are steeply banked at an angle of 31 degrees. If these turns were frictionless, the cars would need to travel at a speed of approximately 43 meters per second to maintain the circular motion. The normal force in this case would be equal to the horizontal component of the weight, opposing the centripetal force needed for the circular motion.
  • #1
hey123a
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Homework Statement


The Daytona 500 is the major event of the NASCAR season. It is held at the Daytona International Speedway in Daytona, Florida. The turns in this oval track have a maximum radius (at the top) of r = 316, and are banked steeply, with θ = 31°. Suppose these maximum-radius turns were frictionless. At what speed would the cars have to travel around them?


Homework Equations


Fc = mv^2/r



The Attempt at a Solution


εfx = Fac = Nx
(mv^2)/r = sinθN
(mv^2)/r = sinθ(-wcosθ)
(mv^2)/r = sinθ(-mgcosθ)
v = √[sinθ(-gcosθ)r]
v = √[sin31°(-(-9.8))(cos31°)(316)]
v = 36.9m/s

answer is actually 43m/s
 
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  • #2
hey123a said:

Homework Statement


The Daytona 500 is the major event of the NASCAR season. It is held at the Daytona International Speedway in Daytona, Florida. The turns in this oval track have a maximum radius (at the top) of r = 316, and are banked steeply, with θ = 31°. Suppose these maximum-radius turns were frictionless. At what speed would the cars have to travel around them?

Homework Equations


Fc = mv^2/r

The Attempt at a Solution


εfx = Fac = Nx
(mv^2)/r = sinθN
(mv^2)/r = sinθ(-wcosθ)
(mv^2)/r = sinθ(-mgcosθ)
v = √[sinθ(-gcosθ)r]
v = √[sin31°(-(-9.8))(cos31°)(316)]
v = 36.9m/s

answer is actually 43m/s

Omit the minus signs, as you use magnitudes.

Check the figure. Is N=mgcosθ?



ehild
 
  • #3
ehild said:
Omit the minus signs, as you use magnitudes.

Check the figure. Is N=mgcosθ?



ehild

yes, on an incline the normal force is equal to the magnitude of the y component of the weight. the y component of the weight is equal to mgcosθ
 
  • #4
hey123a said:
yes, on an incline the normal force is equal to the magnitude of the y component of the weight. the y component of the weight is equal to mgcosθ

It is true if an object slides on a slope. Now, the car does not slide but stays at a constant hight. The vertical component of the normal force opposes the weight. The horizontal component of the normal force provides the centripetal force.
ehild
 
  • #5
ehild said:
It is true if an object slides on a slope. Now, the car does not slide but stays at a constant hight. The vertical component of the normal force opposes the weight. The horizontal component of the normal force provides the centripetal force.
ehild

why is it only true if the object slides on a slope?
 
  • #6
The normal force is a force of constraint. It ensures that the body performs the constrained motion. Its magnitude depends on the motion.

In case of sliding along a slope, the body can not rise from the surface, neither can it penetrate into the surface: there is a force balancing the normal component of gravity: the normal force is equal to -mgcosθ.

Here the car has to travel along a horizontal circle with speed V. If the normal force was equal to -mgcosθ, the resultant force would be parallel to the slope, and downward, with magnitude mgsinθ. But the circular motion to be maintained, the resultant force has to be horizontal, and equal to mV2/R.

ehild
 
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  • #7
ehild said:
The normal force is a force of constraint. It ensures that the body performs the constrained motion. Its magnitude depends on the motion.

In case of sliding along a slope, the body can not rise from the surface, neither can it penetrate into the surface: there is a force balancing the normal component of gravity: the normal force is equal to -mgcosθ.

Here the car has to travel along a horizontal circle with speed V. If the normal force was equal to -mgcosθ, the resultant force would be parallel to the slope, and downward, with magnitude mgsinθ. But the circular motion to be maintained, the resultant force has to be horizontal, and equal to mV2/R.

ehild
ah okay thanks
 

1. What is a frictionless banked turn?

A frictionless banked turn is a type of motion in which a vehicle, such as a car or a bicycle, is able to make a turn without the use of friction. In this scenario, the only forces acting on the vehicle are its weight and the normal force from the banked surface. This allows the vehicle to maintain a constant speed and follow a curved path without slipping or skidding.

2. How does a frictionless banked turn work?

In a frictionless banked turn, the banked surface is angled in such a way that the normal force from the surface provides the centripetal force required to keep the vehicle moving in a curved path. This means that the vehicle does not need to rely on friction between its tires and the road to make the turn. Instead, it can maintain a constant velocity and follow the banked curve smoothly.

3. What factors affect a frictionless banked turn?

The main factors that affect a frictionless banked turn are the speed of the vehicle, the angle of the banked surface, and the radius of the turn. As the speed increases, the angle of the banked surface must also increase to provide enough centripetal force. Similarly, a tighter turn will require a steeper banked surface to keep the vehicle on its path.

4. What are the advantages of a frictionless banked turn?

A frictionless banked turn has several advantages over a standard turn with friction. First, it allows for a smoother and more stable turn, as there is no risk of the vehicle slipping or skidding. Additionally, it can be made at higher speeds, as the centripetal force is provided by the banked surface rather than the friction between the tires and the road. This can be beneficial in certain racing or sports scenarios where high speeds are desired.

5. What are the limitations of a frictionless banked turn?

One limitation of a frictionless banked turn is that it requires a specific angle and radius of the banked surface in order to work effectively. If these factors are not properly calculated or if the surface is not smooth, the turn may not be successful. Additionally, a frictionless banked turn can only be used in certain situations, such as on a banked race track or a curved road with a banked surface. It may not be applicable in everyday driving scenarios on flat roads or surfaces.

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