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Max turning velocity for a car as a function of centre of mass and axle width

I'm trying to solve this problem. Show that a car is more likely to overturn going around a sharp corner if it has a high centre of mass.

[Hint: consider the maximum velocity for the turn as a function of centre of mass and axle width].

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Attempts and Ideas

I attempted to equate the turning moments of a lever joining the left wheel to the centre of mass and solved for $\color{red}{v_{\text{max}}}$.

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$$\tau = \text{Force} \times \text{Distance}$$ $$\tau_1 = \tau_2 \text{ at Equilibrium}$$ $$m a h = \color{green}{\frac{1}{2} w}\color{red}{m g}$$ $$a h = \color{green}{\frac{1}{2} w}\color{red}{ g}$$ but $\color{blue}{a} = \color{blue}{\frac{v^{2}}{r}}$ $$\color{blue}{\frac{v^{2}}{r}}h = \color{green}{\frac{1}{2} w}\color{red}{ g}$$ $$\color{blue}{v^2} =\color{blue}{r} \color{green}{ \frac{1}{2} \frac{w}{\color{black}{h}}}\color{red}{ g}$$ $$\color{blue}{v_{\text{max}}} = \sqrt{\color{blue}{r} \color{green}{\frac{1}{2}}\color{red}{g}} \sqrt{\frac{\color{green}{w}}{\color{black}{h}}}$$


But then I realised that the left wheel can come off the ground but the car may not necessarily tip over.

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  • $\begingroup$ What you calculated is the velocity that gives the centrifugal force to take the wheel out of the ground. When the car start to overturn, the lever arm will change. You may also want to note that the centre of mass will go higher. It means that the car will have more potential energy. The forces have to do enough work to provide this energy. $\endgroup$ – toliveira Jan 4 '18 at 18:52
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There are some inconsistencies in your solution attempt.

Assume the car is turning to the right. The force diagram is:

force diagram

$K_A$ and $K_B$ are wheel friction forces creating the centripetal force which accelerate the car to the right. a designates the radial acceleration but is not a force. Solution in the inertial frame is tough. You can adopt the accelerating reference frame by adding a D'Alembert force (centrifugal force) as follows:

enter image description here

Point B is now at "at rest" and will rotate if the torque arising from $F_D$ is higher than the stabilizing torque arising from mg. Friction forces and $N_B$ are not relevant for this axis of rotation and $N_A$ is zero if the rotation initiates.

Now you can apply your own calculations which correctly relate velocity to w and h.

Wheel A will indeed come off the ground, as expected for a sharp right turn. This will not change the calculation. What's more important, is that once the turn-over initiates, things just become worse and worse. The torque due to $F_D$ increases and the stabilizing torque decreases. You are bound to overturn.

Why do cars still survive a slight lift-off of wheel A? An experienced driver will quickly turn the drivers wheel to the left, thereby increasing the radius and decreasing D'Alembert force along with its related torque.

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Your calculation is correct. Your 'doubt' is incorrect. It does not apply because (as npojo points out) the driver either steers out of the turn by increasing $r$, which increases $v_{max}>v$, or the driver brakes to make $v<v_{max}$.

If the driver keeps $r$ and $v>v_{max}$ constant then the vehicle topples. As it starts to topple the lever arm for $ma$ increases while the lever arm for $mg$ decreases. So the torque causing the vehicle to topple increases. It does not decrease. The fact that work has to be done to raise the CM is irrelevant. This does not prevent toppling.

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This is the opposite case of a car that's driving in a circle while leaning "ïnwards". So the car "stands" not on the outside wheels but on but the inside wheels (w.r.t. the circle's center). In this case, there are two opposite forces: the horizontal component of the gravitational "force", which tends to pull the car towards the center of the circle, and the centrifugal "force" which tends to pull the car outwards. In this case, the car can drive in a circle on two wheels, when the two forces are equal.

When the car drives in a circle on the outside wheels, the two forces are directed in the same direction (that is, the CF "force" and the outward horizontal component of the "force" of gravity), so these two forces cooperate and the car will always topple over and fall to the ground (if it stays on the same distance to the center of the circle).

I think it's obvious that how lower and wider the car, the more force it take to lift the outer wheels. The ideal situation is of course that the height is zero and the width infinite, but I think you understand this is not feasible and certainly not practical! That's why, for example, Formula 1 cars are as low as possible in practice and the width as wide as possible; there is a certain ratio between the both for which the car's stability is "as good" as possible.

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Modern cars and truck have both active and passive roll-over control systems. Such as the linked twisted crank bars connecting left and right suspension, or computer controlled hydraulics that adjust the distribution of suspension loads.

Many new cars have suspensions that under impact or over-turning moments deform to a geometry more suitable to cope with imbalanced loads and moments and then gradually go back to their optimal default configuration. These designs help to increase the balance of the car and its handling of sharp turns and deep potholes.

There are after market racing suspension packages that will improve the corner taking and steering of the cars.

My Honda Accord has built in angles and expansion arcs of the entire suspension that will increase the wheel-bas and even the distance between the front and rear axels if called for, in coping with a sharp turn or small jump over a bump in the road, till the car regains its balance. These are proprietary frame and suspension configurations that give each car model its specific feel of handling!

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