What is the Coefficient of Static Friction for a Car on a Circular Track?

[Bearbull24.5]

Homework Statement

To test the performance of its tires, a car travels along a perfectly flat (no banking) circular track of radius 412 m. The car increases its speed at uniform rate of, at= 4.14m/s^2, until the tires start to skid. If the tires start to skid when the car reaches a speed of 37.7 m/s, what is the coefficient of static friction between the tires and the road? The acceleration of gravity is 9.8 m/s2.

Homework Equations

u(s)=(V^2)/(r*g)

The Attempt at a Solution

I believe the above equation is the one I am supposed to be using. I simply rearranged the equation to solve for Vmax to solve for the static coefficient. I have no idea why I need the tangential acceleration and I believe that 37.7 is the maximum velocity.

 

[kuruman]

What must be true for the car to start skidding? Where does the formula that you plan to use come from?

 

[Bearbull24.5]

For the car to start skidding doesn't that mean it overcomes the force of static friction? I got the formula from my notes and it was originally Vmax=sqrt((radius)(u(s))(g))

 

[kuruman]

For the car to start skidding doesn't that mean it overcomes the force of static friction?

That is correct, but in what direction is the force of static friction that needs to be overcome in the present case?

I got the formula from my notes and it was originally Vmax=sqrt((radius)(u(s))(g))

Careful here! The formula from your notes was derived for a car that goes around a circle at constant speed, i.e. there is no tangential acceleration. When there is tangential acceleration, the net force of static friction that needs to be overcome is "mass times resultant magnitude of acceleration."

 

[Bearbull24.5]

But there is no mass given

 

[kuruman]

If no mass is given, call it m and proceed with the algebra. If you do things correctly, maybe you will not need to know the mass in the end.

 

[Bearbull24.5]

Okay, I think I may be onto something here. f(s)=u(s)N. If I substitute ma(t) in for f(s) and solve for u(s) I end up getting that u(s) is equal to tangential acceleration times gravity since the mass in the forces cancel out.

 

[Bearbull24.5]

Edit** It would be tangential acceleration divided by gravity

 

[Bearbull24.5]

And that was not correct. Extremely lost, confused and only 2 more attempts left before I get no points at all for this problem

 

[Bearbull24.5]

https://www.physicsforums.com/showthread.php?p=2595490

I finally figured it out. Had to find the centripetal acceleration and use that and the tangential acceleration to find the total magnitude of acceleration and then plug it into the formula ma=u(s)N. While solving for u(s) the m cancels out and I get a right answer for once. Yay