Frictional forces, Maxiumum Velocity

[reb_01]

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

 

[fzero]

What is the magnitude of force required to keep the car on the track?

 

[reb_01]

Would that be the force of friction I got in part a? so 14626.71N?

 

[fzero]

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.

 

[reb_01]

Thank-you so much, I got the right answer :)