2
$\begingroup$

I am not sure if this is the right place to ask it, but this is a question that I thought today, and it gave me some curiosity to understand. Imagine that a car will curve, we can say the turn is a bit tight , what are the factors that can help it to flip? I was wondering about some aspects:

if the car has mass, it has inertia, so while it is curving it tends to keep the motion in the same direction that it was instants before the turn. right? So, if the car has more mass, it has more inertia, and since there is friction, one heavier car would flip easier then one lighter, considering that all other possible variables were equal.

Center of gravity, a car with an higher center of gravity would flip easier. The whole inertia of the car distributed to higher heights would be further of the tires(where friction acts), creating angular momentum.

The car being thin because it has less surface contact with the ground;

The car being lighter. This opposes what I've said in "a)", but a heavier car is more difficult to get off the road. A lighter car has more instability.

Am I wrong? In what I'm wrong? What do you think?

Thanks for helping. :)

$\endgroup$
1
$\begingroup$

A car will flip as soon as the acceleration vector from its center of mass points outside the "footprint" defined by connecting the tire contact patches with lines.

Whether or not this occurs has a lot to do with how much lateral force the tires can generate through their friction with the ground — or if they catch on something like a curb.

Generally speaking, the mass has little to do with it, because it affects the "grip" of the tires in the same proportion that it contributes to the inertia of the vehicle.

The geometry of the vehicle — the length and width of the wheelbase vs. the height of the center of gravity — is the primary factor determining stability. A car with a wide/long stance and a low CG will generally slide rather than flip, unless the wheels catch on something.

| improve this answer | |
$\endgroup$
  • $\begingroup$ I agree with you. For a specific amountof mass, the car gains inertia proportional to stability due weight. Thanks for help. :) $\endgroup$ – Vitor Aguiar Sep 8 '16 at 12:40
0
$\begingroup$

A heavier car will not flip easier than a lighter one, while going through a curve (unless you are considering strong side winds).

To understand why this is the case we have to look at what makes a car flip. A car flips because the sum of moments is not zero. Normally when you push against the side of a car, its weight will get distributed differently between the tires, which creates a moment which opposes the moment created by you pushing against the car. When the entire weight/normal force of the car has to be placed at the tires on one side of the car, then the car will be at the tipping point before it starts to flip. When a car goes through a curve at high speed, then the "push" from the side can be replaced by a centrifugal force (which is a pseudo force, only present in the reference frame of the car). However both the centrifugal force and the maximum opposing moment, due to gravity, are both proportional to the mass of the car. So changing the mass of the car, assuming identical mass distributions, does not affect how easier it is for the car to flip in a curve.

The properties that do affect the ease with which a car can flip going through a curve are the height of the center of mass of the car and the width between the tires of the car. Namely the height of the center of mass of the car acts as the lever arm of the centrifugal force and (half) the width between the tires acts as the lever arm of gravity, the bigger the arm the bigger the moment the force generates.

| improve this answer | |
$\endgroup$
  • $\begingroup$ Yea, before I read your comment I've thought the same thing.. You're right, the bigger the mass, the bigger the inertia, so the bigger will be centrifugal force, but the bigger the mass, more weight the car has, the more stable it is, so these both conditions balance each other. However, a car with more mass, despite is it more stable, the bigger inertia at high speeds can help it rollover $\endgroup$ – Vitor Aguiar Sep 8 '16 at 12:19
0
$\begingroup$

When a car makes a turn with a radius of R it is subject to an acceleration force, vector, called centripetal force $$ F=m V^2/R $$
this force is applied to the center of gravity of the car and is directed horizontally opposite of center of turn.
This Vector adds to the vector of weight of the car which is directing vertically down.

The sum of these two vectors is a vector drifting out and down diagonally, call it: R fore roll .
If R falls outside of the rectangular area between four wheels the car will start to roll. But it takes a few seconds for it to start from just lifting the inside tires off the road to pass the critical angle of no return. Experienced drivers could maneuver the car back to balance during those early moments.

A car with lower CG and wider wheel-base needs much larger centripetal force to go off balance. A car with stronger roll control rods will be more stable but will have a hard ride. Fancy cars have computer controlled roll prevention!
having worn out tires on the rear promotes skidding and roll. Road condition on a tight turn such as bumps or puddles promotes roll. Finally sudden jerky braking to avoid a pot-hole is dangerous too. Some race cars have suspension designed to lower the CG and open the wheel-base in a tight roll!

| improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.