centripetal force and static friction

[levodopa]

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A 1000 kg car wants to round a curve on a flat road of radius 100 m at a speed of 72 km/hr (~20 m/s) without slowing down. What will happen to the car if the pavement is dry and the coefficient of static friction is 0.60?

A.
The car will make the turn only if it speeds up.
B.
The car will make the turn if it stays the same speed.
C.
The car will make the turn only if it slows down.
Correct Answer
D.
There is not enough information to determine what will happen to the car.

This is confusing to me conceptually. I know that you're supposed to find the force due to static friction (I got that, and it's 6000N) and also the centripetal force (which is 4000N).

If force of friction > centripetal force, why would the car make the turn only if it slows down? If it slows down, then the centripetal force goes down even more, but it's still less than the force of static friction. I just don't get it! My answer was B.

 

[MedPR]

Are you sure you copied down the numbers correctly? Seems like it would be B.

 

[levodopa]

Seems like I'm not the only one to find an issue with this problem. I think they meant B. Stupid TPR.

 

[chiddler]

How did you guys solve this question?

I solved for friction coefficient with given conditions and it was lower than 0.6. So I decided that it can make the turn because more friction would just give more leeway for speed.

Does this work? have a more efficient way?

 

[MedPR]

How did you guys solve this question?

I solved for friction coefficient with given conditions and it was lower than 0.6. So I decided that it can make the turn because more friction would just give more leeway for speed.

Does this work? have a more efficient way?

I set frictional force = mv^2/r

 

[chiddler]

I set frictional force = mv^2/r

and you found v, yes?

 

[folktale]

I'm also getting a lower centripetal force than frictional force, with a ratio of Fc to Fs of 4:6

I would have gone with B here as well.

 

[levodopa]

and you found v, yes?

They give v in the passage, you solve for the centripetal force.

So F_cent=mv2/r
=(1000*400)/100
=4000 N

Force due to static friction:
F=us*normal force
= 0.6 * 1000*10
= 6,000 N

 

[MedPR]

and you found v, yes?

The problem gives you v.

Unless we are considering 20m/s to be the angular velocity? So then velocity is actually r*w = 20*100?

Then mv^2/r=4*10^5 and static friction is only 6*10^3.. So in that case C would be the answer. But if they give velocity in km/hr or m/s, that must be linear velocity, not angular velocity right?

 

[chiddler]

No i agree. I considered it linear velocity not angular. Angular would be rad/s.

What I did is find what u (friction coeff) should be with all those given variables. I found that it was very low compared to 0.6 (it's 0.4 iirc) and therefore the car can make it at this speed because that is the minimum friction that allows uniform circular motion.

What I was thinking, in asking "did you solve for v", is that you can do the same thing by solving for velocity. I now realize that you cannot because solving for v would assume that the car is using all possible friction force which is not true.

Eh. I don't think what I wrote is very clear. Difficult to explain.

So the conclusion that you guys made is that centripetal force is less than friction force therefore it can go at the same speed?

 

[MedPR]

No i agree. I considered it linear velocity not angular. Angular would be rad/s.

What I did is find what u (friction coeff) should be with all those given variables. I found that it was very low compared to 0.6 (it's 0.4 iirc) and therefore the car can make it at this speed because that is the minimum friction that allows uniform circular motion.

What I was thinking, in asking "did you solve for v", is that you can do the same thing by solving for velocity. I now realize that you cannot because solving for v would assume that the car is using all possible friction force which is not true.

Eh. I don't think what I wrote is very clear. Difficult to explain.

So the conclusion that you guys made is that centripetal force is less than friction force therefore it can go at the same speed?

Yup, I calculated friction > centripetal force and since the force exerted by the road on the car is equal to the centripetal force (newton's 3rd law), the frictional force will prevent the road from "pushing" the car so much that it slips.

 

[Ineedhopenow]

Can someone explain the concept behind this problem?

(i.e. if frictional force is greater than centripetal force, it'll turn)

I'm confused on why this is.

 

[MedPR]

Can someone explain the concept behind this problem?

(i.e. if frictional force is greater than centripetal force, it'll turn)

I'm confused on why this is.

Friction prevents slipping. Imagine trying to make a left turn on ice, a frictionless surface. Your wheels would slip out to the right because there is no friction to allow them to grip the road. Friction is pushing (or pulling, however you want to think about it) you into the turn. If the force is greater than friction, you will slide because there is no other force preventing you from sliding.

 

[Ineedhopenow]

Friction prevents slipping. Imagine trying to make a left turn on ice, a frictionless surface. Your wheels would slip out to the right because there is no friction to allow them to grip the road. Friction is pushing (or pulling, however you want to think about it) you into the turn. If the force is greater than friction, you will slide because there is no other force preventing you from sliding.

So if the centripetal force equals the frictional force, you'd turn left. Then when the frictional force is greater than the centripetal force, what happens? You slide inward while making the left turn?

 

[MedPR]

So if the centripetal force equals the frictional force, you'd turn left. Then when the frictional force is greater than the centripetal force, what happens? You slide inward while making the left turn?

No. Friction only acts to oppose motion, never to induce it. If friction is greater, you don't slide. It's like making left/right turns everyday when you drive somewhere. The friction in the road is great enough to prevent you from sliding, but it has never pushed you into the curb or wall or whatever. It's like normal force. It is lazy and only exerts enough force to prevent something.