Centripetal acceleration speeding car problem

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SUMMARY

The discussion focuses on calculating the maximum and minimum speeds of a car navigating a banked curve with a 30-degree angle and a radius of 80 meters, where the coefficient of friction is 0.3. The centripetal acceleration is derived from both the normal force and frictional force acting on the car. Key equations include the balance of forces in both vertical and horizontal directions, specifically using the relationships μR - mgcosθ = mv²/r and mgsinθ - μR = mv²/r. The importance of correctly defining the normal force (N) versus reaction force (R) is emphasized, as confusion arises from the terminology used in various textbooks.

PREREQUISITES
  • Understanding of centripetal acceleration and its formula: Cp = v²/r
  • Knowledge of forces acting on an object in circular motion, including friction and normal force
  • Familiarity with trigonometric functions and their application in physics problems
  • Ability to set up and solve equations based on Newton's laws of motion
NEXT STEPS
  • Study the derivation of centripetal force equations in banked curves
  • Learn about the role of friction in circular motion and how it affects speed limits
  • Explore the differences between normal force and reaction force in physics contexts
  • Practice solving problems involving banked curves with varying angles and coefficients of friction
USEFUL FOR

Physics students, educators, and anyone interested in understanding the dynamics of vehicles on banked curves, particularly in the context of centripetal acceleration and frictional forces.

Dumbledore211
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Homework Statement


A car is traveling round a bend which is banked at an angle of 30 to the horizontal. The bend is assumed to be in the shape of an arc of a circle of radius 80m. the surface of the road is rough and the coefficient of friction between the tyres of the and the surface of the road is 0.3. Find the greatest speed and the least speed without slipping occurring

Homework Equations



Cp= v^2/r

The Attempt at a Solution


So, here the bend is assumed to be an arc of a circle which is why the angle θ=30 is assumed to be formed at the center of the circle. The centripetal acceleration is also caused by the maximum frictional force which has a coefficient of 0.3. The equation that best describes the motion of the car is as follows frictional force - mgcosθ= mv^2/r
or, fs-mgcos=mv^2/r
or, μR- mgcosθ=mv^2/r
What befuddles me about this problem is the value of R which is supposed to be the normal reaction force. How do I go about solving for R?? Can any of you guys drop a hint as to where I am going wrong.??
 
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All your equations look strange. Did you mix horizontal and vertical forces there?

How did you define R?
 
Dumbledore211 said:
So, here the bend is assumed to be an arc of a circle which is why the angle θ=30 is assumed to be formed at the center of the circle.
No, the arc is of unknown length. The arc is in the horizontal plane. The 30 degrees is the angle that the road is banked laterally, at all points along the arc.
The centripetal acceleration is also caused by the maximum frictional force which has a coefficient of 0.3.
That would be true on an unbanked road, but here the normal force also supplies some centripetal force. Correspondingly, only the horizontal component of the frictional force contributes to the centripetal force.
Write out the ∑F=ma equation for both vertical and horizontal. Don't forget that the frictional force can act up the slope, not just down the slope.
 
The 30 degrees is actually the angle created by the arc with the horizontal plane. So, the vertical force in this case is the normal force i.e R=mgcos30 and the horizontal component of the forces in this case are as follows mgsin30-fs=mv^2/r
or, mgsin30-μR=mv^2/r
or, mgsin30-0.3mgcos30=mv^2/r
or, gsin30-0.3gcos30=v^2/r
Are my set of equations correct?
 
Dumbledore211 said:
The 30 degrees is actually the angle created by the arc with the horizontal plane.
No, I believe you are still misreading it. The arc is horizontal, but the road surface slopes from one side to the other at an angle of 30 degrees. That's what 'banked' means.
 
I noticed recently that people seem to be using R for the normal force instead of the usual N. I don't know where this new convention started. First of all R is also the gas constant, so it can be confusing in some cases. Second, if R is intended to refer to the "Reaction" force to the body's weight and/or an outward force, it introduces a new level of confusion to Newton's third law. It is really just the perpendicular component of a stress in the surface that is equal and opposite to the stress in the body that is in contact with the surface.

AM
 
@Andrew mason It is just not me. Almost all my textbooks use R to denote the normal reaction force
 
Dumbledore211 said:
@Andrew mason It is just not me. Almost all my textbooks use R to denote the normal reaction force
By calling it a "reaction" force they are referring to another force (ie. its Third Law pair). It can be very complicated to determine what, exactly, it is a "reaction" to. It is not a third law pair to the weight of the body, for example.

All forces are third law "reaction" forces so singling out this as R "the Normal reaction force" is misleading and confusing at best.

AM
 
Dumbledore211 said:
The 30 degrees is actually the angle created by the arc with the horizontal plane. So, the vertical force in this case is the normal force i.e R=mgcos30 and the horizontal component of the forces in this case are as follows mgsin30-fs=mv^2/r
or, mgsin30-μR=mv^2/r
or, mgsin30-0.3mgcos30=mv^2/r
or, gsin30-0.3gcos30=v^2/r
Are my set of equations correct?
What does the sum of all forces (friction, normal force and gravity) have to equal?

AM
 

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