1. Nov 18, 2007

### jigs90

1. The problem statement, all variables and given/known data
I don't even know where to start on either of these two problems

A car of mass 474 kg travels around a flat,
circular race track of radius 197 m. The co-
efficient of static friction between the wheels
and the track is 0.123.
The acceleration of gravity is 9:8 m=s2 :
What is the maximum speed v that the car
can go without flying off the track?

It has a follow up too which I would really appreciate some help on!

The same car now travels on a straight track
and goes over a hill with radius 101 m at the
top.
What is the maximum speed that the car
can go over the hill without leaving the road?

A step by step outline would make my day!
Thanks

2. Relevant equations

A= V^2/ r
F= MV^2/ r

3. The attempt at a solution

I'm not sure how to go about tackling that problem, I tried just substituting 101 m in for the radius but that doesn't work.

2. Nov 18, 2007

### jigs90

I tried doing the f= mv^2/r and when I entered in my answer for the first problem, it was wrong, I don't know how to take into consideration the static friction because I then used the equation mus (static friction) times N and subtracted that from the weight and then reworked the problem but my answer was still incorrect

3. Nov 18, 2007

### dynamicsolo

Both problems will involve the centripetal force equation, but centripetal force is not a physical force, so in each situation, you need to think about what force provides the centripetal force.

For this first one, the car is on a horizontal curve. What force is acting on the tire surfaces to hold it on the turn? How do you calculate that force? That is the force which provides the centripetal force, so you can set it equal to F_centripetal.

In this problem, the centripetal force is toward the center of the hill, which we are to take as having a semicircular cross section. At the top of the hill, what forces are acting on the car? The centripetal force will be equal to the net force on the car. What is true about these forces when the car is on the verge of losing contact with the road surface when it is moving fast enough?

4. Nov 18, 2007

### jigs90

on the first one I set the equation of
mus(static)N = mv^2/ r but that still isn't right....Isn't the static friction providing the centripital force?

5. Nov 18, 2007

### dynamicsolo

It does sound weird to use static friction when talking about a moving car, but consider that, in order to hold the turn, we do not want the tires to slide along the road surface. (Likewise, when we deal with rolling wheels, we will also use static friction rather than kinetic friction.)

So, yes, the static friction is providing the centripetal force for a level road curve. How do you find N?

6. Nov 18, 2007

### jigs90

isn't N equal to mg?

7. Nov 18, 2007

### dynamicsolo

Yes, for an object on a horizontal surface. What do you get when you put these pieces together?

8. Nov 18, 2007

### jigs90

isn't that what I had been using before though?
mus(static) mg= mv^2/s and the m's cancel and you solve for v^2 but I still keep getting it incorrect

9. Nov 18, 2007

### dynamicsolo

What are the values you're putting in to calculate v^2?

10. Nov 18, 2007

### jigs90

.123(474)9.8

11. Nov 18, 2007

### dynamicsolo

What is the radius of the curve?

12. Nov 19, 2007

### jigs90

Nevermind...I figured it out. I was entering in the wrong decimal place. Thanks for all your help though. :)