Designing a Curved Exit Ramp: Angle Calculation for Speed & Radius

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SUMMARY

The discussion focuses on the design of a curved exit ramp for a toll road, emphasizing the importance of banking the road to prevent skidding without relying on friction. The necessary banking angle (theta) is derived from the equation tan(theta) = v²/(rg), where v represents the speed of the vehicle, r is the radius of the curve, and g is the acceleration due to gravity. The calculations confirm that the centripetal force required for a vehicle to navigate the curve is provided by the normal force's component directed towards the center of the circular path.

PREREQUISITES
  • Understanding of centripetal acceleration
  • Knowledge of trigonometric functions, specifically tangent
  • Familiarity with the concepts of normal force and gravitational force
  • Basic physics principles related to motion on curved paths
NEXT STEPS
  • Study the derivation of centripetal force in circular motion
  • Explore the effects of different banking angles on vehicle dynamics
  • Learn about frictionless motion and its applications in civil engineering
  • Investigate real-world examples of banked curves in road design
USEFUL FOR

Civil engineers, transportation planners, and anyone involved in road design and safety optimization will benefit from this discussion.

ballahboy
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An engineer wishes to design a curved exit ramp for a toll road in such a way that a car will not have to rely on friction to round the curve without skidding. He does so by banking the road in such a way that the necessary force causing the centripetal acceleration will be supplied by the component of the normal force toward the center of the circular path. Show that for a given speed v and a radius of r , the curve must be banked at the angle (theta) such that tan(theta)=v^2/r*g
 
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What have you tried? think about what tan(theta) is in terms of sine and cosine and what sine and cosine would represent in this case.
 
this is what i was thinking...

Fn sin(theta) = mv^2/r
(mg/cos(theta))sin(theta) = mg tan(theta) = mv^2/r
tan theta = v^2/(rg)

does that look right?
 

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