Banked curve min/max velocities

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SUMMARY

The discussion focuses on calculating the range of speeds for a 930 kg car navigating a banked curve with a radius of 15 m and a coefficient of static friction of 0.300. The car's speed is initially set at 48.0 km/h, but the user encounters difficulties in determining the correct speed range due to the complexities introduced by friction. The equations of motion involving normal force and friction are analyzed, leading to incorrect speed calculations of 37.50 km/h and 56.58 km/h. The correct approach requires a comprehensive understanding of forces acting on the vehicle, including gravitational and centrifugal forces.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Knowledge of centripetal force and banking angles
  • Familiarity with static friction coefficients
  • Ability to manipulate trigonometric functions in physics equations
NEXT STEPS
  • Study the derivation of banking angle formulas in circular motion
  • Learn about the effects of friction on banked curves
  • Explore the application of Newton's second law in circular motion scenarios
  • Investigate real-world examples of vehicle dynamics on banked roads
USEFUL FOR

Physics students, automotive engineers, and anyone interested in vehicle dynamics and safety on banked curves.

ndoc
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Homework Statement



A curve of radius 15 m is banked so that a 930 kg car traveling at 48.0 km/h can round it even if the road is so icy that the coefficient of static friction is approximately zero. You are commissioned to tell the local police the range of speeds at which a car can travel around this curve without skidding. Neglect the effects of air drag and rolling friction. If the coefficient of static friction between the road and the tires is 0.300, what is the range of speeds you tell them?

Homework Equations



Nsinø = (mv^2)/r
N cos ø = mg

The Attempt at a Solution



Combining these and simplifying produces rg tan ø = v^2
Solving for theta I get 50.38

I then return to the equations, but instead add friction force

mgtanø = (mv^2)/r + μ(mg/cosø) and mgtanø = (mv^2)/r - μ(mg/cosø)
solving these for velocity, but this produces the wrong answer (I get 37.50 km/h and 56.58 km/h)

Thanks in advance for the help!
 
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Welcome to PF, ndoc.
It seems to me the situation is more complicated for the case with friction. I'm thinking of mg down and the CENTRIFUGAL force horizontal to my right. So the normal force is mg*cos(A) + mv^2/r*sin(A). Certainly it makes sense that as the car goes faster, it gets pressed harder against the banked road, increasing friction.
 

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