Maximum velocity on a banking curve formula

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Discussion Overview

The discussion revolves around finding the maximum velocity for a car turning on a banked curve, focusing on the equations governing centripetal force and the normal force acting on the vehicle. Participants explore the derivation of these equations and the implications of different assumptions in their calculations.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant describes their initial approach using a free-body diagram and a formula for centripetal force, but encounters discrepancies with the results found in HyperPhysics.
  • Another participant explains that the normal force equals mg cosθ only when there is no perpendicular acceleration component, suggesting that this condition does not apply in the current scenario.
  • A participant seeks clarification on whether the normal force can be expressed as N - mg cosθ due to the net force applied towards the center of the curve.
  • Another participant confirms that the perpendicular component of the net force is indeed related to the normal force, reinforcing the earlier explanation.

Areas of Agreement / Disagreement

Participants generally agree on the need to consider the net force and its components when calculating the normal force, but there is no consensus on the specific values or equations to use in the context of the problem.

Contextual Notes

There are unresolved assumptions regarding the conditions under which the normal force is calculated, particularly in relation to the presence of acceleration components and their effects on the forces acting on the vehicle.

oranboron
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http://hyperphysics.phy-astr.gsu.edu/hbase/mechanics/imgmech/carbank.gif

I had a problem in my textbook asking me to find the maximum velocity for a car turning on a banked curve.

After i drew my freebody diagram i had the same formula for centripetal force as hyperphysics above (Fx).

However when i tried getting my result my answer was wrong... I didn't know what i was doing wrong so i searched hyperphyics and found their vmax formula turned incredibly different from mine. Then i saw that they had a second equation. I was able to finally figure out that they isolated for the normal force, n = mg/(costheta - usintheta) in Fy and substituted it into the Fx formula.

BUT the value for the normal force is different from what i had before. I had n = mgcostheta

Why couldn't i use that value for N in my version of Fx = Fcentripetal ?

Thank you for your help.
 
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The normal force would only equal mg cosθ if there were no component of acceleration perpendicular to the incline, which is not the case here.
 
Ok, so if i understand it correctly. Since there is now a net force being applied to the centre and it has a component perpendicular to the incline. This perpendicular force itself is the value of N - mgcosθ ?

Thank you for your patience and help.
 
oranboron said:
Since there is now a net force being applied to the centre and it has a component perpendicular to the incline. This perpendicular force itself is the value of N - mgcosθ ?
Right, that's the component of the net force perpendicular to the incline.
 

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