Uniform Circular Motion of a car

[uchicago2012]

Homework Statement

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?

Homework Equations

F_max = u * F_n
where F_max = force causing the friction, u = coefficient of friction, and F_n = normal force

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.

 

[zorro]

In the question, 'properly banked' curve indicates that there is no frictional force coming into 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.

 

[Doc Al]

If a curve with a radius of 60 meters is properly banked for a car traveling 60 km/hr,

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

 

[uchicago2012]

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

 

[Doc Al]

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.

 

[uchicago2012]

I found the angle theta by:

F_Nx = F_N sin theta
F_Ny = F_N cos theta
F_g = mg

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

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

by trig, I saw tan theta = F_Nx/ F_Ny
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.

 

[Doc Al]

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.

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.

 

[uchicago2012]

But if I do that I get:

F_net,x = F_k + F_n sin theta = ma

= m(v²/R)​

F_{net, y} = F_N cos theta - mg = ma

= F_N cos theta - mg = m(0)​

= F_N cos theta = mg​

F_N sin theta = m (v²/R) - F_k
F_N cos theta = mg

then
tan theta = (m(v²/R) - F_k) / 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

 

[cepheid]

then
tan theta = (m(v²/R) - F_k) / mg
which doesn't help because the masses don't cancel out

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

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 F_k?

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).

 

[uchicago2012]

Er. What exactly is the nature of F_k? 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.

 

[cepheid]

Er. What exactly is the nature of F_k? 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.

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?

 

[uchicago2012]

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

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

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

 

[Doc Al]

But if I do that I get:

F_net,x = F_k + F_n sin theta = ma

= m(v²/R)​

F_{net, y} = F_N cos theta - mg = ma

= F_N cos theta - mg = m(0)​

= F_N cos theta = mg​

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.

 

[uchicago2012]

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

F_net,x = F_n sin theta + F_r cos theta₂ = ma

F_n sin theta = ma - F_r cos theta₂​

F_net,y = F_n cos theta - F_r sin theta₂ - mg = ma

F_n cos theta = F_r sin theta₂ + mg​

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

 

[Doc Al]

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

Sounds good to me.

F_net,x = F_n sin theta + F_r cos theta₂ = ma

Good! What is the acceleration?

F_n sin theta = ma - F_r cos theta₂​

Skip this step.

F_net,y = F_n cos theta - F_r sin theta₂ - mg = ma

Good! What is the acceleration?

F_n cos theta = F_r sin theta₂ + mg​

Skip this step.

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

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.)