What factors affect a car's maximum speed on a circular race track?

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Homework Help Overview

The discussion revolves around understanding the factors that affect a car's maximum speed on a circular race track, specifically focusing on the role of friction and centripetal force in this context.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the forces acting on a car in circular motion, including friction, gravitational force, and centripetal force. There are attempts to identify the necessary equations and the relationships between these forces. Questions are raised about the role of mass and the direction of forces involved.

Discussion Status

The conversation is ongoing, with participants exploring various interpretations of the forces at play. Some guidance has been offered regarding the identification of forces and the importance of free-body diagrams, but there is no explicit consensus on the correct approach or solution yet.

Contextual Notes

Participants mention the need for specific values, such as the mass of the car and the coefficient of friction, while also discussing the implications of different surface conditions on the car's performance.

alessandro
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Friction provides the force needed for a car to travel around a flat, circular race track. What is the maximum speed at which a car can safely travel if the radius of the track is 80 m and the coefficient of friction is 0.40 ?

I'm totally lost...hope u guys can help me
 
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It is traveling around a circle, so centripetal force comes into play. What provides the centripetal force?
 
ok. to start off this question.
you need to think of the equation and plot in the knowns.
what equation do you use in this case ? ( hint: Force equation)
now this is in a circular race track, so it must be a centripeteal force.

the ultimate question in this question is what really supplies the neccesary force for the car to stay on the track ?

let me know your progress.
 
This is a supply demand problem:
Supply force: .4 friction
demand = v^2
Fnet = MV/r

solving algebraically:
V=\sqrt{}Force of friction (radius)

note you may want to get the equation yourself, because I'm doing this by memory and it may be wrong.
 
rock.freak667 said:
It is traveling around a circle, so centripetal force comes into play. What provides the centripetal force?

the agent provides the centripetal force...the agent proivdes also the acceleration right?
 
lovemake1 said:
ok. to start off this question.
you need to think of the equation and plot in the knowns.
what equation do you use in this case ? ( hint: Force equation)
now this is in a circular race track, so it must be a centripeteal force.

the ultimate question in this question is what really supplies the neccesary force for the car to stay on the track ?

let me know your progress.

the friction provides the force necessary right? But usually to find this force i need also the mass...
 
alessandro said:
the friction provides the force necessary right? But usually to find this force i need also the mass...

No you do not need the mass, what is the normal reaction in this case equal to?
 
rock.freak667 said:
No you do not need the mass, what is the normal reaction in this case equal to?

i don't really know..i'm lost..
 
First, draw a free-body diagram of the car, and identify the forces on the car. Tell us what forces you think are on the car and which way they point.
 
  • #10
vela said:
First, draw a free-body diagram of the car, and identify the forces on the car. Tell us what forces you think are on the car and which way they point.

the forces are the gravitazional force which points downward, friction force which acts in opposite direction of the car...
 
  • #11
alessandro said:
the forces are the gravitational force which points downward, friction force which acts in opposite direction of the car.
You're still missing a force. If gravity pulls the car down, and there's no counteracting force, the car will accelerate downward (through the road), which doesn't happen, so there must be another force on the car that balances the car's weight. What is that force and which way does it point?

Also, for simplicity, you can assume that the car isn't speeding up or slowing down, so there's no frictional force parallel to the velocity of the car, but there is still another frictional force, the one that keeps the car from sliding off the curve. Which direction does that frictional force point?
 
  • #12
vela said:
You're still missing a force. If gravity pulls the car down, and there's no counteracting force, the car will accelerate downward (through the road), which doesn't happen, so there must be another force on the car that balances the car's weight. What is that force and which way does it point?

Also, for simplicity, you can assume that the car isn't speeding up or slowing down, so there's no frictional force parallel to the velocity of the car, but there is still another frictional force, the one that keeps the car from sliding off the curve. Which direction does that frictional force point?
that frictional force point downward ?
 
  • #13
alessandro said:
that frictional force point downward ?

What's your reasoning for saying it points downward? It would help if you explained your answers, otherwise we can't tell if you're just making random guesses.
 
  • #14
because the gravitazional force points upward and the frictional force which is against the gravitazional force...it would be pointing downward
 
  • #15
1 force of gravity > down
2 Centrifugal force > outwards
3 Lift (aerodynamic force)(like a wing of a airplane) > up
4 Drag (air resistance) > backwards
5 Trust > forwards

so if trust is to much your car will lift more, therefore centrifugal fore will prevail and your car will want to go straight forwards.

so you would need the value for how much lift the car has, the weight, the and also take in consideration the friction (on soap the car will slide faster than on concrete.) of the road.

it all depend of how accurate you want the result to be.
 

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