Road Bank Top Speed Calculation | G & \mu

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Homework Help Overview

The problem involves calculating the top speed of a vehicle on a banked road with given parameters, including a specified speed, radius, and coefficient of friction. The context is centered around concepts of circular motion and forces acting on the vehicle.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculation of the angle of the bank and the forces acting on the vehicle, including normal force and friction. There are attempts to express these forces in terms of gravitational force and centripetal force. Questions arise regarding the orientation of the centripetal force relative to the bank and ground.

Discussion Status

The discussion is ongoing, with participants providing insights and questioning the setup of the problem. Some guidance has been offered regarding the components of forces, but there is no explicit consensus on the approach or solution yet.

Contextual Notes

Participants are working under the assumption of no friction initially and are exploring how the introduction of friction affects the calculations. There is also a focus on understanding the relationship between the forces involved in the scenario.

dlingo
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Advanced Road Bank question

Homework Statement


There is a road bank designed for 60km/hr (16.7 m/s) with a radius of 70m and no friction. What is the top speed with a mew of 0.8[tex]\mu[/tex] ?
G=gravitational constant
V=16.7m/s

Homework Equations



Tan[tex]\theta[/tex]=V^2/(R*G)
Fc=(M*V^2)/R

The Attempt at a Solution


First I calculated the angle at which the car won't move with no friction and I got 22.124
Then I started filling in what I knew:
Fnormal:=(G*sin([tex]\theta[/tex]))+(Fc*Sin(90-[tex]\theta[/tex])) The component of gravity added to a component of Centripetal force.
then:
Ffriction:=Mew*Fnormal
Fc:=Ffriction+the component of fnormal pushing back against the Fc.
(M*V^2)/R=Mew*((G*sin([tex]\theta[/tex]))+(Fc*Sin(90-[tex]\theta[/tex])))the component of fnormal pushing back against the Fc.
But I don't know if I'm on the right track or completely wrong.
 
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dlingo said:

The Attempt at a Solution


First I calculated the angle at which the car won't move with no friction and I got 22.124
Then I started filling in what I knew:
Fnormal:=(G*sin([tex]\theta[/tex]))+(Fc*Sin(90-[tex]\theta[/tex])) The component of gravity added to a component of Centripetal force.

You might want to recheck where the angle is. I suspect you want [tex]M g cos(\theta) + F_c sin(\theta)[/tex]

then:
Ffriction:=Mew*Fnormal
Fc:=Ffriction+the component of fnormal pushing back against the Fc.
(M*V^2)/R=Mew*((G*sin([tex]\theta[/tex]))+(Fc*Sin(90-[tex]\theta[/tex])))the component of fnormal pushing back against the Fc.
But I don't know if I'm on the right track or completely wrong.

See if you can find the components of -centripetal force (well, I suppose we could call it centrifugal force if we whisper softly) and gravitational force that lie along the road surface.
 
There is one thing I don't understand, Is the Fc parallel to the ground or the bank?
 
dlingo said:
There is one thing I don't understand, Is the Fc parallel to the ground or the bank?

Fc is in the same direction as the radius vector for the moving object describing the circle. In this case, it's parallel to the ground.
 

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