Frictionless banked turn? What is wrong with my thinking?

  • Thread starter hey123a
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  • #1
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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
 

Answers and Replies

  • #2
ehild
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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

Omit the minus signs, as you use magnitudes.

Check the figure. Is N=mgcosθ?



ehild
 
  • #3
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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
ehild
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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
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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
ehild
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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
 
Last edited:
  • #7
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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
 

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