# Homework Help: Centripetal Force and Banked Curves

1. Dec 11, 2008

### KOKA

I've attempted to solve this problem but I'm not really sure if I'm doing it right. I've looked up other threads containing the same question, but they just don't have the answers I'm looking for. Thank you in advance to anyone who helps.

1. The problem statement, all variables and given/known data
A race-car driver is driving her car at a record-breaking speed of 225 km/h (62.5 m/s). The first turn on the course is banked at 15 degrees, and the car's mass is 1450 kg.

a) Calculate the radius of curvature for this turn.
b) Calculate the centripetal acceleration of the car.
c) If the car maintains a circular track around the curve (does not move up or down the bank), what is the magnitude of the force of static friction?
d) What is the coefficient of static friction necessary to ensure the safety of this turn?

2. Relevant equations
tan 15 degrees = v^2/rg
ac = v^2/r
mv^2/r = Fs (unsure)
Ff = uFN
FN = mgcos15 (unsure)

3. The attempt at a solution
a) r = v^2/gtantheta
= (62.5^2)/(9.8tan15)
= 1487.584027
= 1.49 x 10^3 m

b) ac = v^2/r
= (62.5^2)/(1.49 x 10^3)
= 2.625902086
= 2.63 m/s^2

c) mv^2/r = Fs
(1450)(62.5^2)/(1.48 x 10^3) = Fs
3807.558025 = Fs
Fs = 3.81 x 10^3 N

d) Ff/FN = u
3.81 x 10^3/ mgcos15 = u
3.81 x 10^3/ (1450)(9.8cos15) = u
0.277 = u

I'm pretty sure parts a and b are correct, but part c is where it gets confusing. Also, I'm getting a little confused in what is really the normal force for a banked curve. Is it mgcostheta as with an inclined plane, or mg/costheta?

Thanks a lot!

2. Dec 11, 2008

### Staff: Mentor

I don't see how you have enough information to solve for the radius.
Here you assumed that the track was banked for that particular speed so that no friction is needed. That would make parts c and d rather pointless.

What textbook is this problem from?

3. Dec 11, 2008

### KOKA

I'm learning independently right now so all I've got is a booklet, and it's made a few mistakes so far. This is one of the questions in there. I'm guessing I have to assume that it's banked for that particular speed or I wouldn't be able to solve it, or maybe I'm just going about it the wrong way? But then when you get to part c, there would be no static friction, right? I'd really like to understand this because it's part of my review for the final test, and if the answer to part c is so simple, I don't understand why you would even add part d. Thank you for replying again, you're so helpful.

4. Dec 11, 2008

### Staff: Mentor

If you assume, as you did, that the curve is banked for that particular speed, then your answer to a is fine. And that makes the friction in part c zero. (Which is why I said that it makes c & d rather pointless.)

I think that the problem is flawed--data is missing. (Who is the publish/author of the booklet? Maybe there's an errata listing somewhere.)

5. Dec 12, 2008

### KOKA

I checked, there's nothing. The booklet was just recently photocopied, so I don't think they'll have any listed for a while. I'll try looking into it a bit more, and then post the closest answer I can get in here so maybe other people who need it can look it up. Thanks anyway Doc Al.

6. Dec 12, 2008

### turin

I concur. Actually, this problem was posted not too long ago by another poster. I think it was this exact problem. I told the poster to ask their instructor for more info, but I don't know what happened with that. It's like you said: either you make an assumption that makes part c pointless, or you need more information; something is wrong with this problem.

7. Dec 16, 2008

### KOKA

Okay, I think I got it.
In part c, because the car does not move up or down the bank, the magnitude of the centripetal force is equal to the magnitude of the force of static friction. I think I got the answer to that the first time I attempted it. Then, in part d, I accidentally used normal force for an inclined plane, whereas it should be Fscostheta/mg = u, and the answer would be 0.259. The value seems a little high, assuming it's rubber on concrete, but it might just be due to the bank of the curve which adds to it.

8. Dec 16, 2008

### Staff: Mentor

The correct logic to use would be that since the acceleration in the vertical direction is zero, the net force in the vertical direction must be zero. The vertical components of the normal force and static friction must balance the weight. (The net force in the horizontal direction is what provides the centripetal force.)

Of course, as explained previously, this particular problem is flawed. Using your answer for part a should give you zero static friction, since you assumed that the road was perfectly banked for the given speed.

The problem would have made sense if part a read something like this:
"a) Calculate the radius of curvature for this turn, assuming its banking angle is designed for a speed of 200 km/hr."​