Banked Circular Motion without friction

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SUMMARY

The discussion focuses on the physics of banked circular motion, specifically for a roadway designed for vehicles traveling at 28 m/s on a circular arc with a radius of 190 m. The key equation used is the centripetal force equation, F_C = mv²/r, which relates the mass, velocity, and radius of the curve. The participants clarify the relationship between normal force and gravitational force, ultimately determining that using mg tan(β) = mv²/r yields more accurate results than assuming N = mg cos(β). Careful attention to coordinate systems is essential for accurate calculations.

PREREQUISITES
  • Understanding of banked curves in physics
  • Familiarity with centripetal force equations
  • Knowledge of trigonometric identities, specifically sin(2β)
  • Basic principles of forces acting on objects in circular motion
NEXT STEPS
  • Study the derivation of the centripetal force equation F_C = mv²/r
  • Learn about the effects of banking angles on vehicle dynamics
  • Explore the application of trigonometric identities in physics problems
  • Investigate the role of friction in non-banked versus banked curves
USEFUL FOR

Physics students, educators, and engineers involved in roadway design and vehicle dynamics will benefit from this discussion, particularly those focusing on the principles of circular motion and force analysis.

hqjb
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Homework Statement



A roadway is designed for traffic moving at a speed of 28 m s . A curved section of the
roadway is a circular arc of 190 m radius. The roadway is banked so that a vehicle can go
around the curve with the lateral friction forces equal to zero

Homework Equations



[itex]F_C = \frac{mv^2}{r}[/itex]

The Attempt at a Solution



FBD.jpg


[itex] N\sin\beta = \frac{mv^2}{r}[/itex]
[itex] mg\cos\beta\sin\beta = \frac{mv^2}{r}[/itex]
[itex] 2\sin\beta\cos\beta = \frac{2v^2}{rg}[/itex]
[itex] \sin(2\beta) = \frac{2(28)(28)}{(190)(9.8)} [/itex]

I got the right answer if I didnt assume [itex]N = mg\cos\beta[/itex]

Edit : Nevermind, careless mistake I was using different coordinate systems.
 
Last edited:
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hqjb said:

Homework Statement



A roadway is designed for traffic moving at a speed of 28 m s . A curved section of the
roadway is a circular arc of 190 m radius. The roadway is banked so that a vehicle can go
around the curve with the lateral friction forces equal to zero

Homework Equations



[itex]F_C = \frac{mv^2}{r}[/itex]

The Attempt at a Solution



FBD.jpg


[itex] N\sin\beta = \frac{mv^2}{r}[/itex]
[itex] mg\cos\beta\sin\beta = \frac{mv^2}{r}[/itex]
[itex] 2\sin\beta\cos\beta = \frac{2v^2}{rg}[/itex]
[itex] \sin(2\beta) = \frac{2(28)(28)}{(190)(9.8)} [/itex]

I got the right answer if I didnt assume [itex]N = mg\cos\beta[/itex]

Edit : Nevermind, careless mistake I was using different coordinate systems.

I assume you found that using [itex] mg\tan\beta = \frac{mv^2}{r}[/itex] was more fruitfull?
 

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