Circular motion problem no idea what's going wrong ><

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SUMMARY

The discussion focuses on calculating the correct banking angle for a highway curve with a radius of 560 meters, designed for vehicles traveling at a speed of 73.0 km/hr (approximately 20.28 m/s). The initial attempt at solving the problem incorrectly concluded a banking angle of 90 degrees. The correct approach involves analyzing the forces acting on the vehicle, specifically using the equations of motion, where the net force in the radial direction is equal to the centripetal force required for circular motion. The correct formula to determine the banking angle is derived from the relationship tan(θ) = v²/(r*g), leading to a more accurate calculation of the angle.

PREREQUISITES
  • Understanding of circular motion dynamics
  • Familiarity with Newton's laws of motion
  • Knowledge of trigonometric functions
  • Ability to manipulate equations involving forces
NEXT STEPS
  • Study the derivation of the banking angle formula in circular motion
  • Learn about centripetal acceleration and its applications
  • Explore the effects of friction on banking angles
  • Investigate real-world applications of banking angles in road design
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and circular motion, as well as civil engineers involved in road design and safety assessments.

irNewton
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Homework Statement



A highway curve of radius 560m is designed for traffic moving at a speed of 73.0km/hr (1216m/s). What is the correct banking angle of the road?


Homework Equations



a=v^2/r
F=ma


The Attempt at a Solution



Forces in t direction:
Fnet=Fncos[tex]\vartheta[/tex]-Fg=0
Fn=mg/cos[tex]\vartheta[/tex]=0


Forces in r direciton:

Fnetr=Fnsin[tex]\vartheta[/tex]=mv^2/r
mgsin[tex]\vartheta[/tex]/cos[tex]\vartheta[/tex]=mv^2/r
tan[tex]\vartheta[/tex]=v^2/r*g
[tex]\vartheta[/tex]=tan-1(1216^2/(9.81*560))
[tex]\vartheta[/tex]= 90 degrees?

which is wrong... = (
 
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double check the way you're assigning your forces, is the Fny component which is cancelling vertical gravity the hypotenuse of the triangle? I find it helpful in these cases to rotate the coordinate system so that the x-axis is actually the sloped road. That way your Fn has only 1 component, and you break down gravity into y & x where x points towards the center of the turn.
 

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