Uniform Circular Motion and Centripetal Force

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SUMMARY

The discussion focuses on calculating the maximum speed a car can safely navigate a banked curve with a radius of 11.0 meters and an angle of 8.00°. The coefficient of static friction between the tires and the road is 0.870. The user initially calculated a maximum speed of 10.4 m/s but was informed that the correct answer is 11.1 m/s. Key factors in the calculation include the forces acting on the car, such as gravity, friction, and the normal force, which must be accurately represented in a free-body diagram.

PREREQUISITES
  • Understanding of centripetal force and its equation Fc=mv²/r
  • Knowledge of static friction and its coefficient, specifically μ=0.870
  • Ability to draw and interpret free-body diagrams
  • Familiarity with trigonometric functions, particularly sine and cosine
NEXT STEPS
  • Study the principles of centripetal acceleration in banked curves
  • Learn how to accurately draw and analyze free-body diagrams in physics
  • Research the role of static friction in circular motion
  • Explore the effects of varying angles and radii on maximum safe speeds in banked curves
USEFUL FOR

Traffic safety engineers, physics students, and anyone involved in vehicle dynamics and road safety design will benefit from this discussion.

AJayS
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Hey guys, first post here! Hoping to get a little help.

Homework Statement



You are a traffic safety engineer in charge of determining safe speeds for roads. A particular banked curve has a radius of 11.0 meters and is banked at an angle of 8.00°. The coefficient of static friction between common tires and this road is 0.870. What is the maximum speed that a car can drive this curve? Use both the bank of the curve and the friction on the tires in determining your answer.

Homework Equations


f=μn
Fc=mv^2/r

The Attempt at a Solution



So what I did was split the force of gravity, mgcos8 in the direction perpendicular to the ramp and mgsin8 parallel. Also, the force of friction towards the center of rotation and set all forces towards the center of rotation to mv^2/r.

Essentially, I had mgsin8 + (μ X mg X cos8)=mv^2/r.

Common factor of m cancels and I solve for V as everything else is given. I receive an answer of 10.4 m/s. The correct answer is apparently 11.1 m/s, so close, but not close enough for a rounding issue I believe. My only other guess is that somehow I've split the vector wrong. Any help's much appreciated, thanks very much in advance.
 
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Welcome to PF;
Notice that gravity acts straight down - so there is no component of the gravity force acting directly towards the center of the turn.

Did you draw a free-body diagram?
You need the vector sum of the forces to point horizontally towards the center - the forces are gravity, friction and the normal force from the road and they all act in different directions.
 
Hi Simon, thanks for the response.

Now I see it a little clearer. I did draw a free body diagram but it was incorrect. I was treating the parallel surface of the bank as straight horizontal. I'm still getting an incorrect answer however.

As I have it, I have the sum of the horizontal forces as the normal force in the x direction (Fnsin8). I am confused as to where friction fits into all of this. If I'm correct, doesn't friction always oppose the direction of motion? In other words, would the ∑Fx= Fnx - friction in the x direction?
 
If I'm correct, doesn't friction always oppose the direction of motion?
Nope - a car accelerating forward in a straight line has a net friction force acting on it pointing in the same direction as the acceleration.

In this case, this is static friction. Static friction acts one stationary objects too - more accurately: between surfaces that are instantaneously stationary wrt each other.

It opposes the direction the surfaces would move if there were no friction.

Consider:
http://t0.gstatic.com/images?q=tbn:ANd9GcSfjwcftYv3-4OgKxziSLn1cHg850F5nLClXo9lngoNk7IXNoqNvQ
 
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