Car on circular turn with friction; finding max. velocity

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Homework Help Overview

The problem involves a car of mass 1800 kg navigating a circular turn with a radius of 10 m, where the coefficient of static friction between the tires and the road is 0.40. The objective is to determine the maximum speed at which the car can travel without skidding.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • The original poster attempts to apply the net force equation for circular motion and considers the role of friction in maintaining the car's path. Participants discuss the relationship between frictional force and centripetal force, questioning the assumptions about normal force and its implications for maximum speed.

Discussion Status

Participants are actively engaging with the problem, exploring different interpretations of the forces involved. Some guidance has been offered regarding the relationship between friction and centripetal force, but there is no explicit consensus on the correct approach yet.

Contextual Notes

There is a discussion about the assumptions regarding the normal force and its role in the context of maximum speed without skidding. The participants are navigating the constraints of the problem while considering the effects of friction.

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Homework Statement



A car of mass 1800 kg rounds a circular turn of radius 10 m. If the road is flat and the coefficient of static friction between the tires and the road is 0.40, how fast can the car travel without skidding?

Homework Equations





The Attempt at a Solution


I thought to do Fnet(c) = (mv^2)/r
Fn-mg= (mv^2)/r
but i thought that for max speed without going off road is when Fn=0
so it would just be mg= (mv^2)/r
then v=radical gr
which gave me v=9.998 m/s

Was wrong...I don't know what else to do and how to include friction. Any good help will be greatly appreciated. --awe.g
 
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Imagine your car taking a turn. It can, as long as force of the friction serves as a centripetal force (this is indeed the force that makes your car turn ;)). Hence they both have to be equal. Now, the force of the friction is the force that the car applies to the surface of the road multiplied by the coefficient of the friction.
 
ok so then my frictional force is equal to Fn-mg?
but how does that help exactly?
 
Why would your frictional force be equal to Fn-mg? your friction force is lesser than or equal to [tex]\mu F_n[/tex], [tex]\mu[/tex] being friction coefficient. In our case we want our friction force to be as big as it can, hence [tex]F_f = \mu F_n[/tex].
 
oh right. but I thought I want Fn to be 0 because you want to go as fast as possible without going off the road?
 
Yep, but your car ain't going to fly :P. I mean, increasing velocity increases force neaded for a car to turn and it works only in XY plane. Fn is constant all the time.
 
lol too bad... I got the question now, thanks for all your help!
 

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