We are tasked with finding the maximum velocity of a 4-wheeled vehicle undergoing circular motion without tipping (assuming that the force of gravity will act at the car's center of mass and the the force of friction between the tires and the road will prevent slipping).

We know that the center of mass is at height h, there is distance w for the length of the axles (the track), and that the car is making a turn with radius r.

I am using the general approach of setting the torque's of the gravitational force = to the torque caused by the apparent "centrifugal force" (m*(v^2/r)). However, I am having difficulty with the pivot point selection, the distance from the pivot point to the application of the force, and the angles. So far, I have the apparent force also applied to the center of mass, and this makes the distance to the pivot point of the "outside" wheels of the turn (h^2 +.5w^2)^(1/2). I also have the angles being 45 degrees and 135 degrees since my diagram has the vehicle as a square. Since sine of both of these angles is root 2/2, everything cancels, leading me to v = (rg)^1/2 - which I suspect is incorrect. Is there a better pivot point, or are my distances wrong?

  • $\begingroup$ The pivot point is the outer wheel of the turn $\endgroup$
    – Bob D
    Commented Dec 1, 2023 at 2:44
  • $\begingroup$ There's a difference between tipping and rollover. Tipping means the outer wheel just lifts off the ground. You seem to be looking at rollover. $\endgroup$
    – Bob D
    Commented Dec 1, 2023 at 3:07
  • $\begingroup$ @BobD Thank you for that clarification. The problem does say rollover. I am still concerned about identifying the radius vector. The problem hints at using the pythagorus theorem to find distances from pivot point, but does that apply to both the torque caused by gravity and the centripetal acceleration? $\endgroup$
    – naebkcaj
    Commented Dec 1, 2023 at 4:48
  • $\begingroup$ Once tipping begins the vehicle becomes unstable and cannot maintain a single maximum velocity without eventually rolling over. This is because both the CG will continue to rise and the moment due to the centrifugal force continue to increase eventually leading to rollover. I will post an answer in an attempt to explain. $\endgroup$
    – Bob D
    Commented Dec 1, 2023 at 12:45

1 Answer 1


However, I am having difficulty with the pivot point selection, the distance from the pivot point to the application of the force, and the angles.

First of all, I am not an expert on vehicle dynamics. So at the risk of oversimplifying the problem, see the Figures below.

The "pivot point" where tipping/rollover occurs is point A in the figures, i.e., the outer wheels of the turn. See FIG 1. The centripetal force, which is the static friction forces of the wheels is not shown. This analysis in the rotating frame assumes the maximum possible static friction force is not exceeded so that sliding does not occur, which is the assumption you stated. (Whether or not this is a reasonable assumption is another matter.)

The moments about the pivot point A are (1) the clockwise moment due to the centrifugal force, and (2) the counter-clockwise moment due to gravity. Impending tipping occurs when these two moments are equal. When the moment due to the centrifugal force exceeds that due to gravity, tipping occurs. The equation for the maximum velocity where this occurs is shown in the figure and is.

$$V_{max}=\sqrt {\frac{Rgw}{2h}}\tag{1}$$

The maximum centripetal acceleration is given by


Substituting eq(2) into eq(1)


This equation can be found in the link below where it makes the following statement (where $T$ equals $w$):

"It follows directly from Relationship 1 and Equation 2 that if, for a sustained period of time,


the vehicle model will predict rollover"

In the literature the ratio


is referred to as the Static Stability Factor (SSF)

The meaning of the last part of the statement can be understood if one considers the following (refer to FIG 2)

Once the speed (or centripetal acceleration) exceeds the maximum for impending tipping two things occur each of which contributes to continued instability unless the speed is sufficiently reduced to re-establish equilibrium:

  1. The height of the CG increases which increases the clockwise moment contribution of the centrifugal force
  2. The distance between the CG and the pivot point A decreases resulting in a decrease in the counter-clockwise moment contribution of the weight of the vehicle. When the CG is overhead the pivot point A, the distance is zero.

In FIG 2 where you reach the point where the CG is directly over the pivot point A, $d=0$ and rollover is imminent at zero velocity.


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  • $\begingroup$ Thank you - this is very helpful. My last questions are: 1.) In the torque cross products, what happens to the sine theta component? How do you define the lever arm to remove that component? 2.) Why is (w/2) associated with the weight if Fg is in the -y direction (and likewise for the apparent force that is in the x direction being associated with height)? Thanks again. $\endgroup$
    – naebkcaj
    Commented Dec 1, 2023 at 20:50
  • $\begingroup$ @naebkcaj (1) The "lever arm" for each force is the perpendicular distance from the line of action of each force to point A about which the moments are calculated. In FIG 1 it is $h$ for the centrifugal force $F_c$ and It is $d$ for the weight $mg$. Since it's the perpendicular distance the angle is 90 deg and therefore the sine is 1. $\endgroup$
    – Bob D
    Commented Dec 1, 2023 at 22:03
  • $\begingroup$ (2) In FIG 1, when the vehicle is level, $d=w/2$. That's because the line of action of $F_g$ bisects the distance $w$ between the wheels. For the apparent force it is $h$. Note that when the vehicle rotates clockwise about A, $h$ increases while $d$ decreases. Ultimately, as shown in FIG 2, the lever arm $d$ for the weight vanishes and there is no longer a counterclockwise moment to oppose the moment due to the apparent force, and the vehicle is about to roll over. $\endgroup$
    – Bob D
    Commented Dec 1, 2023 at 22:04
  • $\begingroup$ That is a helpful explanation. Thank you again! $\endgroup$
    – naebkcaj
    Commented Dec 2, 2023 at 0:40

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