Centripetal Acceleration of curved exit ramp

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SUMMARY

The discussion focuses on the design of a curved exit ramp for a toll road, emphasizing the need for banking the road to ensure that cars do not rely on friction to navigate the curve. The key equation derived is tan(Θ) = v²/(g * r), where v is the speed of the vehicle, g is the acceleration due to gravity, and r is the radius of the curve. The centripetal acceleration (a_c) is defined as a_c = v²/r, and the normal force components are analyzed to derive the necessary banking angle. The final conclusion confirms that the centripetal force must be correctly identified to ensure the design meets safety standards.

PREREQUISITES
  • Understanding of centripetal acceleration and its formula a_c = v²/r
  • Knowledge of forces acting on an object in circular motion
  • Familiarity with trigonometric functions, particularly tangent
  • Basic principles of road design and vehicle dynamics
NEXT STEPS
  • Study the effects of different banking angles on vehicle stability
  • Learn about frictionless motion in circular paths
  • Explore the implications of varying speeds on ramp design
  • Investigate real-world applications of centripetal force in civil engineering
USEFUL FOR

Civil engineers, transportation planners, and students studying physics or engineering who are interested in roadway design and vehicle dynamics.

rash219
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Centripetal Acceleration !

Homework Statement



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 force causing the centripetal acceleration will be supplied by the component of the normal force toward the center of the path
a. Show that for a given speed v and radius r the curve must be banked at an angle [tex]\Theta[/tex] such that tan[tex]\Theta[/tex] = v^2/r * g


Homework Equations



a_c (centripetal acceleration) = V^2 / r
[tex]\Sigma[/tex]F_y = m * a = 0

The Attempt at a Solution



th_00959_untitled_122_662lo.JPG


i hope this diag. makes sense to you...

According to the question a_c = n * Sin[tex]\Theta[/tex] ---- (1)

Then

[tex]\Sigma[/tex]F_y = m * a = 0
(n * Cos [tex]\Theta[/tex]) - (m * g) = 0
n = (m * g) / (Cos [tex]\Theta[/tex]) -------- (2)

substitute 2 in 1 for n

a_c = (m * g) / (Cos [tex]\Theta[/tex]) * Sin[tex]\Theta[/tex]
= (m * g) Tan [tex]\Theta[/tex]

now a_c (centripetal acceleration) = V^2 / r

therefore (V^2 / r) = (m * g) Tan [tex]\Theta[/tex]

and Tan [tex]\Theta[/tex] = (V^2) /(m * g * r)

what am i doing wrong ?.?
 
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Check your equation (1) again, you have identified the centripetal force, not the centripetal acceleration.
 
Thanks! worked out right...
 

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