# Center of Gravity and Centrifugal Force

1. Mar 23, 2013

### jpman90

Hi, I have this problem that I'm trying to solve and it has been a few years since I've taken Physics. I want to know how you would approach and solve this problem, and what equations you would use. Hopefully somebody here can help me out.

The Problem:
A vehicle is on a road that is V inches wide and the vehicle has a center of gravity of X inches. It then starts to go around a bend of Y degrees at an angle of inclination of Z degrees at a velocity of W. The angle of inclination is present so that the centrifugal force does not push the vehicle off of the road. The gravitational force will balance the centrifugal force assuming the velocity is appropriate.

My Questions:
-I know there is the concept of centripetal force as well, but how does it come into play in this case?
-As the vehicle starts to go around the bend, how will the center of gravity start to shift? I want to know what will cause the vehicle to flip over? There should be two factors. They are the angle of inclination and the velocity. The vehicle should flip if the it moves too fast (topples outwards) or too slow (topples inwards).

Lots of variables, I know, but unfortunately I don't have any numbers at this point. I can come up with some estimates if you require any to put some perspective to this problem.
Your help is greatly appreciated. Thanks.

2. Mar 23, 2013

### Emilyjoint

You need to know the radius of the bend.
This is more important than the width of the road (V inches)
You also need to know the width of the vehicle to make use of the centre of gravity (X inches)
Have you looked at basic texts about forces in circular motion?

3. Mar 23, 2013

### jpman90

Hi Emily,

Ok, the radius of the bend is something that I can obtain. I'll call it variable R.
Assume that the width of the vehicle is equal to the width of the road (also V inches). How would I proceed in this case?

Unfortunately I've sold all of my textbooks so I've had to resort to reading papers on the internet. I know the different formulas that you can use in cases like this but I don't understand how to apply them in my specific scenario?

Thanks,

J

4. Mar 23, 2013

### Emilyjoint

Does W2/R have any meaning for you... That would be my starting point.
Can you draw a diagram showing the forces acting on the car.... The force through the centre of gravity and the force at the wheels.
Don't forget to take the angles into account.
You have not mentioned friction so I am assuming you are going to ignore friction.

5. Mar 23, 2013

### Staff: Mentor

Oh, two threads. Anyway, this is what I wrote in the other thread:

That does not make sense. X inches relative to what, in which direction?
What do you want to calculate? The angle? Assuming R does not change to quick (otherwise it will be hard to define "balance"), draw a force diagram and you get a relation between Z, W g and R. Y and V do not matter.

Centrifugal <-> centripetal force are just a question of taste. In a non-rotating coordinate system, you need the former, in a rotating system, you get the latter.
Does not matter, but it is given by the geometry.
The incline prevents this in your setup. The net force, attached to the center of gravity, has to be supported by the wheels (i. e. has to "hit" the road somewhere between the wheels).

6. Mar 23, 2013

### jpman90

I know V^2/R, but not that one?

I have attached a diagram. I have it from two different viewpoints, one from behind and one from above.

You are correct, ignore friction forces

#### Attached Files:

• ###### Problem Diagram.bmp
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7. Mar 23, 2013

### jpman90

Hi mfb,

Apologies for the double post. I saw a similar post to this one in the general forum and thought I posted this in the wrong area.

1) Center of gravity relative to its rest position on flat ground. Assume that most of the weight is concentrated near the top of the vehicle, so the center of gravity will not be near the midpoint.

2) In this case I want a definite curve. Please check the diagram I posted recently..if Y is small for a large R then the turn would be very sharp. The chance of flipping over would be higher. However, if Y is close to 180 then there is next to no curve at all and it would be very tough to flip over.
I would like to calculate the CG required to ensure that I do not flip over. Guess I should have specified that in the beginning, sorry lol.

3) Got it, thank you

4) So you are saying that if the angle of inclination changes drastically the center of gravity remains in a fixed position? Is that really how it works??

4) At a very high or very low velocity and a high center of gravity, wouldn't it be easier to flip? And how do I calculate the net force that is specifically attached to the Center of Gravity as opposed to just any point on the vehicle?

Thanks,

J

8. Mar 23, 2013

### Staff: Mentor

So X is the height of the CoG above the floor?

Y is just the length of the curve, it does not influence the balance at any point of the curve. For every inclination and radius, there is some velocity range where the car does not fall over. The same is true if you fix two other parameters (like radius and velocity) and look for a range of the third (here: inclination).
A high center of gravity and a small width reduces the size of those ranges.
I don't understand your question, and the relation to my post.
You can restrict the analysis to the center of gravity.

9. Mar 23, 2013

### jpman90

I realized something that I might have missed in the question. X is the center of gravity above the floor when the vehicle is on flat ground. On the curved inclination I'm not sure how X would change. Is there an equation or way to model that?

I agree that there are always fixed parameters on the curve that would prevent the vehicle from flipping over. I want to be able to model those parameters with equations.

For the other question, lets take it from the top:
I asked - As the vehicle starts to go around the bend, how will the center of gravity start to shift?
You said - Does not matter, but it is given by the geometry.
I took that to mean that the center of gravity was not dependent on the curvature or angle of inclination, but I guess I misunderstood you. What did you mean when you said "does not matter but it is given by the geometry"?

Lastly, you say that I can restrict the calculation to the center of gravity. How do I do that? In a calculation of forces, how do I incorporate the center of gravity height?

10. Mar 23, 2013

### Staff: Mentor

It is just geometry: It is X away from the corresponding point on the floor (between the wheels), and the line to this point is orthogonal to the surface.

If inclination and curve speed exactly cancel in their forces, the position of the center of gravity does not matter, the car is always stable if it is stable on a flat track without a curve.*

You do it automatically, don't worry.
The height is not relevant to determine the forces.*

*depending on your definition of "radius", there can be a small influence: Do you mean the radius the center of gravity has, or the radius the central point between the wheel has? For realistic curves, those two are nearly the same, so I would not worry about that.

11. Mar 23, 2013

### jpman90

What I'm trying to determine is the center of gravity (X) that will cause the vehicle to flip at a given Y, Z, W, V, and R. Would I really need to differentiate between the radius from the center of gravity and the radius from between the wheels in order to figure that out?

I'm struggling because it seems like this is a combination of forces (mg, v^2/R) and physical distances (X, R, V).

12. Mar 23, 2013

### jpman90

Haha, remembered moments of forces. Fxd. Thanks

13. Mar 24, 2013

### BobG

Your center of gravity won't change, at least in relation to itself. I think you mean the location of the center of gravity relative to some point that you haven't specified very well. In fact, you might mean the location of the center of mass (they aren't quite the same).

Flat curve (no banked curves) - centrifugal force horizontal to the surface at some magnitude (v^2/r). Gravitational force perpendicular to the surface. To flip or not flip, you want to look at the torques those forces create on the outside wheels.

In other words, I think you want the location of the center of mass relative to the outside wheels, which are your pivot point.

You want to calculate the torque created by gravity, which will depend on the angle of the gravitational force to the radius (outside wheels to center of mass). You need the component of the gravitational force that's perpendicular to your radius. Force times radius gets you torque.

Once the torque from the centrifugal force is greater than the torque from the gravitational force, you flip.

Toss in a banked curve and you can see how it obviously becomes harder to flip, since there will come a point where your centrifugal force could actually make it harder to flip (which isn't to say that your centrifugal force couldn't still make it easier to slide before you get the idea that it's always better to go as fast as possible through a curve - in fact, usually a vehicle will slide before it flips unless you're running around the curve on a rail or something that makes it impossible to slide).

14. Mar 27, 2013

### jpman90

I solved it, so please let me know if my solution is correct.

((V/(2*cosz) + (x-(v*tanz)/2)*sinz)*m*g = ((x-(v*tanz)/2)*cosz)*(m*w^2)/R

This would give me the moment of force from gravity (left hand side) on the outer wheel and the moment of force from centrifugal (right hand side of equation) on the outer wheel. I would then solve for w to find the velocity at which the car would start to tip.