Frictional forces, Maxiumum Velocity

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The discussion revolves around calculating the maximum frictional force and maximum velocity of a car rounding a circular turn. The maximum frictional force was correctly calculated as 14626.71 N using the coefficient of static friction and the car's weight. To find the maximum velocity, the centripetal force equation is applied, which relates the frictional force to the required centripetal force for circular motion. The user initially confused the frictional force with the centripetal force but clarified that the frictional force is indeed what keeps the car on track. The conversation highlights the importance of understanding the relationship between friction and centripetal force in circular motion problems.
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Homework Statement



A 2100-kg car rounds a circular turn of radius 19 m. The road is flat and the coefficient of static friction between tires and road is 0.71.

a)What is the maximum frictional force the tires can share with the pavement before the car begins to skid?

b)What is the maximum velocity the car can move without skidding?

c) If the car had half the mass, what would the maximum velocity be?


The Attempt at a Solution



For part a I got the right answer : Ff= .71 * 2100 * 9.81 = 14626.71 N
but I don't understand how to calculate the maximum velocity for parts b and c. I know we are looking for V2 and that V1 = 0
 
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What is the magnitude of force required to keep the car on the track?
 
Would that be the force of friction I got in part a? so 14626.71N?
 
reb_01 said:
Would that be the force of friction I got in part a? so 14626.71N?

Not exactly. It's true that the static friction keeps the car on the track, but the magnitude of the force needed to keep the car in circular motion is given by the centripetal force.
 
Thank-you so much, I got the right answer :)
 
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