# Homework Help: Uniform Circular Motion of a car

1. Sep 18, 2010

### uchicago2012

1. The problem statement, all variables and given/known data
If a curve with a radius of 60 meters is properly banked for a car traveling 60 km/hr, what must the coefficient of friction be for a car on the same curve traveling 90 km/hr?

2. Relevant equations
Fmax = u * Fn
where Fmax = force causing the friction, u = coefficient of friction, and Fn = normal force

3. The attempt at a solution
I'm beginning to think I don't really understand uniform circular motion. I read through my book's examples but they're not very helpful. What does a car traveling around a curve 60 km/hr tell us about a car traveling around the curve at 90 km/hr? Is the problem saying that the coefficient of friction was zero in the first case, or that it just increased? I don't understand where to begin with these problems.

2. Sep 18, 2010

### zorro

In the question, 'properly banked' curve indicates that there is no frictional force coming in to play to provide centripetal acceleration. Only the component of normal force (horizontal) on the car provides the necessary centripetal force for its motion in curved path.

So don't involve any equation which includes friction force.

'What does a car traveling around a curve 60 km/hr tell us about a car traveling around the curve at 90 km/hr?'

You first derive the equation. Then you can easily figure out the answer to your question.

3. Sep 18, 2010

### Staff: Mentor

Hint: Use this data to figure out the angle that the road is banked. See Abdul Quadeer's tips.

4. Sep 19, 2010

### uchicago2012

So I drew a free body diagram that had the normal force, Fn, coming away at an angle from the vertical. I decided the angle between Fn and the vertical was the angle that the curve was banked, since if it wasn't banked Fn would presumably be completely vertical. I found the angle theta between the vertical and Fn to be about 25 degrees but I'm still confused as to how to solve the rest of the problem. I have some vague idea that if I can find what Fn would be for the car traveling 90 km/hr then I might be able to find the change in Fn and sub that into F = u * Fn, where the change in Fn would be Fn and F would be the net forces, or Fn + Fg, and u would be the coefficient of friction. Only I don't see how to find what Fn would be for the car traveling at 90 km/hr and I'm not particularly sure if that idea is even remotely correct.

5. Sep 19, 2010

### Staff: Mentor

How did you solve for the banking angle? You'd solve the rest of the problem in a similar manner.

Hint: Draw your free body diagram and consider vertical and horizontal forces separately, applying Newton's 2nd law to each direction. Then combine the two equations.

6. Sep 19, 2010

### uchicago2012

I found the angle theta by:

FNx = FN sin theta
FNy = FN cos theta
Fg = mg

Fnet,y = FN cos theta - mg
Fnet,x = FN sin theta = m (v2/ R) <- this is from Fnet = ma, and a for centripetal motion equals (v2/R)

So I rewrote them to be::
FN cos theta = mg
FN sin theta = m (v2/ R)

by trig, I saw tan theta = FNx/ FNy
so
tan theta = m (v2/R) / mg
and then I solved for theta.

Using this method, I can find the angle of the road would need to be banked with no coefficient of friction for a car traveling 90 km/hr, but I don't see how theta has anything to do with the coefficient of friction.

7. Sep 19, 2010

### Staff: Mentor

Use the same basic idea, only now you have an additional force--the friction.

Find the net force in the x-direction; apply F = ma.

Find the net force in the y-direction; apply F = ma. (Of course, a = 0 for the y-direction.)

Set up those equations and solve for the coefficient of friction.

8. Sep 19, 2010

### uchicago2012

But if I do that I get:

Fnet,x = Fk + Fn sin theta = ma
= m(v2/R)​
Fnet, y = FN cos theta - mg = ma
= FN cos theta - mg = m(0)​
= FN cos theta = mg​

FN sin theta = m (v2/R) - Fk
FN cos theta = mg

then
tan theta = (m(v2/R) - Fk) / mg
which doesn't help because the masses don't cancel out, since I added in an extra force. Unless there's some algebraic trick or substitution I'm missing

9. Sep 19, 2010

### cepheid

Staff Emeritus
I'm confused. Don't you KNOW everything in this expression except for Fk?

EDIT: Nevermind, that's not true. What I should have said was: I'm confused: doesn't the mass still cancel out because of the nature of Fk?

EDIT 2: I didn't actually check any of your work, I was just addressing that specific point under the assumption everything else was right (which it might not be).

Last edited: Sep 19, 2010
10. Sep 19, 2010

### uchicago2012

Er. What exactly is the nature of Fk? It's the friction force, it opposes the direction of movement of some force, in this case, the anti-centripetal force.

I'm not sure what you mean.

11. Sep 19, 2010

### cepheid

Staff Emeritus
Yeah, but how does it depend upon MASS, and on the coeff. of friction, which I believe is what you are trying to solve for?

12. Sep 19, 2010

### uchicago2012

I don't really know. I don't understand it well, I believe.

Fk = u * Fn
and it seems probable that Fk generally increases as mass increases, if u remains the same.

I'm just confused. Even if I solved the equation for Fk I would still need to find Fn in order to be able to apply the two quantities to the only equation I know relating Fn and Fk.

13. Sep 20, 2010

### Staff: Mentor

Careful! What direction does friction point? (And we're talking about maximum static friction, not kinetic.) Consider the x and y components of the friction force when applying Newton's 2nd law.

Also, express friction in terms of the normal force. Once you get your equations correct, the mass will drop out and your only unknown will be the coefficient of friction.

14. Sep 20, 2010

### uchicago2012

I redrew my free body diagram so that Fr (the friction force, I renamed it) is in the third quadrant pointing down. I put it there because Fr is supposed to be parallel to the surface, which would make it perpendicular to Fn. Theta2 would then just be negative theta, measured from the x axis.

Fnet,x = Fn sin theta + Fr cos theta2 = ma
Fn sin theta = ma - Fr cos theta2

Fnet,y = Fn cos theta - Fr sin theta2 - mg = ma
Fn cos theta = Fr sin theta2 + mg​

I feel as if this has something to do with Fr. Am I putting it in the wrong place? I don't know why it would go elsewhere- I thought I finally figured out that Fr was always supposed to be parallel to the surface.

15. Sep 20, 2010

### Staff: Mentor

Sounds good to me.

Good! What is the acceleration?
Skip this step.

Good! What is the acceleration?
Skip this step.
You're doing fine. Express the friction in terms of the normal force. Then see if you can combine those two equations in a manner that eliminates the unknowns you don't care about. (Hint: Move all mass-containing terms to one side.)