Finding Speed on Slippery Curves: R, Theta, Mu

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Homework Help Overview

The problem involves a car navigating a slippery curve, with parameters including the radius of curvature (R), banking angle (theta), and coefficient of friction (mu). The original poster seeks to determine the car's speed to avoid frictional force between the car and the road.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the role of friction and banking in providing centripetal force, with some attempting to relate the forces involved through equations. Questions arise regarding the nature of centripetal acceleration and its components on an incline.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the forces at play. Some guidance has been offered regarding the resolution of forces, but no consensus has been reached on the approach to the problem.

Contextual Notes

Participants express uncertainty about the relationship between banking angle and centripetal acceleration, indicating a need for clarification on these concepts. There is also mention of the challenge in recalling relevant equations and principles.

vivekfan
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Homework Statement



A car rounds a slippery curve. The radius of curvature of the road is R, the banking angle is theta and coefficient of friction is mu. What should be the cars speed in order that there is no frictional force between the car and the road?

Homework Equations


F=mv^2/r


The Attempt at a Solution



In general, I have no idea how the components or forces work for banking curves, so an explanation and help would be greatly appreciated.
 
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im not sure about what to do with the bank, but since the force of friction (along with the banking) is providing the centripetal force, set mu*m*g= mv^2/r I'm sorry I can't quite remember what to do with the bank angle.
 
Centripetal acceleration is a horizontal force. Resolve it into the components on the incline.

Gravity is vertical. Resolve its force components.

If you are gong to ignore friction then the component of gravity down the incline must be balanced by centripetal force up the incline.
 
is centripetal acceleration always a horizontal force?
 
looking back from what I recall you should use:


tan(theta)= v^2/rg

I can't really explain it, I am going to review the concept now
 
vivekfan said:
is centripetal acceleration always a horizontal force?

Yes, for this type problem you should take it as horizontal.

On a roller coaster though it will be radial but in the vertical plane.
 

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