Car on circular turn with friction; finding max. velocity

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SUMMARY

A car with a mass of 1800 kg can round a circular turn with a radius of 10 m and a coefficient of static friction of 0.40. The maximum speed without skidding is determined by the balance of centripetal force and frictional force. The correct approach involves using the equation F_f = μF_n, where F_f is the frictional force and μ is the coefficient of friction. The maximum speed is calculated as v = √(μgR), resulting in a maximum velocity of approximately 9.8 m/s.

PREREQUISITES
  • Understanding of centripetal force and its relation to circular motion
  • Knowledge of frictional forces and coefficients of friction
  • Familiarity with Newton's laws of motion
  • Basic algebra for solving equations involving square roots
NEXT STEPS
  • Study the derivation of centripetal force equations in circular motion
  • Learn about the role of friction in vehicle dynamics
  • Explore the effects of varying coefficients of friction on vehicle performance
  • Investigate real-world applications of these principles in automotive engineering
USEFUL FOR

Students studying physics, automotive engineers, and anyone interested in understanding vehicle dynamics and frictional forces in circular motion.

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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 \mu F_n, \mu being friction coefficient. In our case we want our friction force to be as big as it can, hence F_f = \mu F_n.
 
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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