What is the Physics behind a Car Flipping on a Roundabout?

[kudoushinichi88]

I am asking this question on behalf of a forummer from Gendou.com's Physics Board... Sorry if a similar question has been posted before...

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Here is a mechanics question which totally stumped me:

A car together with its passengers has a combined mass of 2000kg. Its width, wheel-to-wheel is 2m and its center of gravity is 1m from the ground and midway between the wheels. It enters a roundabout which has a diameter of 10m.
Calculate the speed below which it must travel to ensure that it will not flip over about the outer wheels. Is the car were to travel at half this max speed, what is the combined frictional forces on the wheels?

Here are a couple of things that I'm quite confused about if you don't want to bother with the calculations:
a) what causes the car to flip in the first place?
b) in this case, if centripetal force is the resultant force of circular motion, what causes it in the first place?
c) does the difference of normal force on both sides of the car cause the car to overturn?
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Now, I understand that when the car accelerates around the roundabout, the center of gravity of the car moves away from the center of the roundabout due to it's tendency to move in a straight line. And when the center of gravity goes over the outer wheels, a torque is produced causing the car to flip.

But I don't know how to describe this mathematically. What puzzles me the most is, how do you explain mathematically what exactly causes the center of mass to move?

 

[cryptoguy]

a) On the roundabout the centripetal force is provided by the friction force alone. At the threshold speed, (mv^2)/r = $$\mu$$Fn. As soon as the centripetal force is greater then the max Friction (provided by the tires), the car will skid, or, flip.

Now I'm not completely sure about this but here's my try... The tires on the outside of the bend have a larger Fc (because v^2 is larger) so they do not immediately lose traction. The tires on the inside of the curve, however do lose traction which effetively means they are partially in air. This shifts the center of gravity toward the outer part of the car.

Not sure if this is a correct/adequate explanation, maybe someone will correct me.

 

[kudoushinichi88]

Yes, I initially thought that the centripetal force is provided purely by the frictional force. Does that mean in this question, we are missing the coefficient of friction between the tires and the road?

 

[rl.bhat]

When a vehicle is moving along unbanked road with forces of friction f1( inner tire) and f2( outer tire) and normal reaction R1 and R2, then taking moments about the center of gravity, we get (f1+f2)*h = (R1-R2)*b/2, where h is the height of cg from ground and b is the distance between the wheels. f1 +f2 = mv^2/r and R1 + R2 = mg. Solving the two equations we get R1 = 1/2*[ mg - 2mv^2h/rb] As velocity increases R1 goes on decreasesing and a stage is reached when R1 = 0 and inner wheel leaves the ground.

 

[Shooting Star]

A small correction: in the above, (f1+f2)*h = (R2-R1)*b/2. The final result is correct.

 

[Shooting Star]

Yes, I initially thought that the centripetal force is provided purely by the frictional force. Does that mean in this question, we are missing the coefficient of friction between the tires and the road?

You thought correctly. It's only the frictional force, preventing the car from slipping sideways, which provide the centripetal force.

Conceptually, perhaps going to the rotating frame may make things easier to understand. In that frame, there is a centrifugal force mv^2/r acting outward at the CM, weight acting downward at the CM, forces of friction acting inward at the points of contact, and normal reactions at the points of contact of the tires. The car does not topple over because the torque about the point of contact of the outer tire due to the three forces, viz., the weight, the normal reaction on the inner tire, and the centrifugal force, all balance out, as long as the speed is less than a certain value.

 

[rl.bhat]

A small correction: in the above, (f1+f2)*h = (R2-R1)*b/2. The final result is correct.
Moment due to R1 is clockwise and moment due to R2 is counterclockwise.

 

[Shooting Star]

Exactly the reason I have mentioned the correction. f1 and f2 both have CW moments in your picture, You are equating the magnitude of the CW moment (LHS) with the magnitude of the CCW moment (RHS), as a condition for the car not to rotate, right?

 

[rl.bhat]

Yes.

 

[kudoushinichi88]

(f1+f2)*h = (R1-R2)*b/2, where h is the height of cg from ground and b is the distance between the wheels. f1 +f2 = mv^2/r and R1 + R2 = mg. Solving the two equations we get R1 = 1/2*[ mg - 2mv^2h/rb] As velocity increases R1 goes on decreasesing and a stage is reached when R1 = 0

So, let me do the calculations step by step, so I can familiarise myself with LaTeX...

$$(F_1 + F_2)h = (R_2 - R_1) \cdot \frac {b}{2}$$

$$F_1 + F_2 = \frac {mv^2}{r}$$

$$R_2 - R_1 = mg$$

Solving for R1,

$$R_1 = \frac {1}{2} \left (mg - \frac {2mv^2h}{rb}\right)$$

Plugging in the values,

m = 2000kg,
r = 5 m,
b = 2 m,
and R1 = 0 N,

speed limit for car not to flip $$v = \sqrt {5g} \ \mbox{ms^{-1}$$

Combined frictional force on tires,

$$F = \frac {mv^2}{r} \\ = \frac {2000 \times \left (\frac {\sqrt {5g}}{2} \right)^2 }{5} \\ = 500g \ \mbox {N}$$

Is this correct?

 

[Shooting Star]

Correct, if the value of g is in MKS.

 

[kudoushinichi88]

Wow, thanks a lot!

 

[Volcano]

I am very happy to find this topic. I was asking an almost the same question soon. But worrying about explaining the problem because of my percect English :) By the way, big thanks for helpers too.

May i add one more question? As Shooting star explain,

1. In rotating frame: We use the torque about the "point of contact of the outer tire"

2. In Earth frame: We use the torque about the point of "center of mass"

I can not understand the reason of choice of this two different points. As i know, if a body is balanced, the point of moment which we may prefer is arbitrary. Even the point is anywhere in space but surely we pick an easy point for math operations.

 

[Shooting Star]

You can use either point – both will give the same result. I had intuitively mentioned moment about the point of contact of the outer tire because it would be the point which will remain fixed (in the rotating frame) once the car does topple. That’s all.

 

[Volcano]

Sorry, i am a bit late. Shooting star, as you said, once the car does topple, the forces rest on car is; mg, f2, R2. By the way, certainly if i choose Earth reference frame, mg and R2 are balanced and f2 is the unbalanced force.

If i choose the torque about the point of contact of the outer tire, the torks of f2 and R2 are zero. So "mg" is the single force. If i choose this point how can i explain the moment(car) equilibrium as the choose of CW? Of course, we can not say, the car is in moment or force equalibrium, there is a net force(centripetal force) on car but for CW the tork is zero like the moment for rotating frame. This is really confusing me.

I am sorry, maybe i couldn't explain myself. If so please ask me.

 

[Shooting Star]

If i choose the torque about the point of contact of the outer tire, the torks of f2 and R2 are zero. So "mg" is the single force.

Where is mv^2/r? Or maybe I didn't understand the question?

 

[Volcano]

Thank you. Do you mean centripetal or centrifugal force? My problem is the point of moment for inertia reference system. I mean centripetal force. So CP(centripetal) force must be the f2 force because this is the net force as i know,

F=ma -> f2 = ma -> f2 = m(v^2/r)

thus, f2 is the centripetal force(on the ground). If i was prefer rotating frame, there must be another force(inertia force=centrifugal force) on car which effect about center of weight. At this point, both centripetal and centrifugal forces completely different forces and also effects from different points(ground and center of mass). What am i missing?

 

[Volcano]

Sorry for reply myself but is this threat forgotton? I can not solve this myself, any idea?

 

[Shooting Star]

Yes, it was temporarily forgotten. Sorry. But I'd have eventually answered, I have a huge list of "to be answered", which, unfortunately, grows everyday.

I'll answer soon, maybe tomorrow. I have to re-read the whole thread.

 

[Doc Al]

May i add one more question? As Shooting star explain,

1. In rotating frame: We use the torque about the "point of contact of the outer tire"

2. In Earth frame: We use the torque about the point of "center of mass"

I can not understand the reason of choice of this two different points. As i know, if a body is balanced, the point of moment which we may prefer is arbitrary. Even the point is anywhere in space but surely we pick an easy point for math operations.

You'll never go wrong if you take torques about the center of mass. From the rotating frame, since the car is not accelerating in that frame, you can take torques about any point. (Of course you must then include the fictitious noninertial forces.)

 

[Volcano]

Shooting star,

Thank you for all help, i guessed. But worried about completely forgotton. The forum is growing incredible.

Doc Al,

Thank you too. I approve the explanations what you mentioned. And also understand the "center of mass" approach for inertia reference system instinctively and right for me too.

I will always prefer center of mass from now on if i use Newton's second law. OK. But, hmm, pfff! but why this is the only point? I guess,

1. All stable inertia axis always pass through center of mass. So most(all ?) of the given inertia moments for bodies are for the point about center of mass.(cylinder, wheel, disc, sphere). Bodies rotates itself an axis which pass through the center of mass and this point is the most(only ?) quiet point on body.

2. Moment equalibrium is only for bodies which is in force equalibrium too. So, we can not use moment equalibrium for a body which has a net force. We must admission this bodies like in effect of a force-pair.

I suppose improvising a bit. Is it? These are ideas and guess to understand.

 

[Shooting Star]

Hi Doc Al,

Let me answer this, even though it may take a bit of time. I don't want Volcano to lose faith in me and PF.

Shooting star,

Thank you for all help, i guessed. But worried about completely forgotton.

Not completely...

 

[Shooting Star]

As I understand it, you were at one point confused that why in the Earth frame we have taken the moments about the CM but in the rotating frame I had mentioned taking moments about the point of contact of the outer tire. (In this context, at least refer to post #14.)

(We’ll use the terminology of post #4, i.e., suffix 1 pertains to inner tire.)

Using the Earth frame, we had obtained as a condition for stability the equation:
R1 = 1/2*[mg - 2mv^2h/rb] (Post #4). This was done by taking the moments about the CM, and the fact that the frictional forces provide the centripetal force.

In the rotating frame, there is only centrifugal force acting outward at the CM, but no centripetal force. If we take moments about the point of contact of the outer tire, we will get:
mg*b/2 = R1*b + (mv^2/r)*h (since the moments of f1, f2 and r2 is zero about that point), which is the same equation as above derived in Post #4.

I hope this sets your mind at rest to some extent, that you can take in this case whichever point you like. Note that the 2nd method gave the equation a bit more directly, but that is beside the point. The moment about the point of contact of the outer tire could also have been taken in the Earth frame.

I am adding another post to your latest questions, in as much as I’ve understood them.

 

[Shooting Star]

Let me mention a “theorem”.

The total external torque about the CM equals the time rate of change in angular momentum about the CM. This means that N = dL/dt holds not only for inertial frames, but for any frame moving with the CM. That is the reason why you can always take the torque about the CM, and not go wrong, as Doc Al had mentioned.

The CM frame has many other special properties, as will become evident to you as you do more mechanics problems. The reasons for that will become clear too.

About your 1st point, I am not able to understand it very well. If a rod is hanging by a hinge at one end and oscillating, I don’t think you would chose the axis passing through the CM to deal with it. But for bodies on which the total external force is zero, the axis passing through the CM is a “good” axis.

2. Moment equalibrium is only for bodies which is in force equalibrium too. So, we can not use moment equalibrium for a body which has a net force. We must admission this bodies like in effect of a force-pair.

About your 2nd point, you are not right in saying that “Moment equilibrium is only for bodies which is in force equilibrium too.” If the total force is zero, there may still exist a couple acting on the body which increases the angular momentum. Also, we can use moment equilibrium for a body which has a net force acting on it, as was done to find the stability condition of the car in the Earth frame.

 

[Volcano]

Shooting star,

Thank you very much. Just read. I will read and work again but so far, everything is much more clear. I think i learned(will learn too) much thing from this treasure which you give me, i think so.

Thank you again, for giving your time for me. You are very kind and good physicist.

Best Regards

 

[Shooting Star]

Oh, shucks...

That was the bestest thing anybody has ever said about me. Good cheer to you too.