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What is centre of gravity and stability?


scilearner

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Friend asked me if there are 2 racing cars, one racing car has a centre of gravity of 2 feet and other 1 feet, which is more stable? I have learned physics long time ago, but i can not recall learning about this. Can any of you explain the answer to me simply and tell me which area of physics I have to read. Thanks :)

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The physicist in me wants to tell you about torque. The pedant wants to rip your head off for statements like "has a center of gravity of two feet" (which presumably means "has a center of mass two feet above ground") and assuming that "stability of a car" was a well-defined term (while in reality the location of a car's center of mass is rather irrelevant when I drop a large rock on it).

Edited by timo
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Friend asked me if there are 2 racing cars, one racing car has a centre of gravity of 2 feet and other 1 feet, which is more stable? I have learned physics long time ago, but i can not recall learning about this. Can any of you explain the answer to me simply and tell me which area of physics I have to read. Thanks :)

depends.

 

if the car with the center of gravity two feet above the ground is 20 feet wide and the car with a CoG 1ft above the ground is 3ft across then the one with a CoG 2 feet above the ground is more stable.

 

stability is a parameter with multiple factors. you can't just take the height of the CoG and magically work everything else out.

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I don't even understand the question here. What kind of stability are they referring to? I mean generally why is a car with lower centre of gravity more stable (what do you mean by car been stable here, isn't any car resting on ground stable.

Edited by scilearner
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I don't even understand the question here. What kind of stability are they referring to? I mean generally why is a car with lower centre of gravity more stable (what do you mean by car been stable here, isn't any car resting on ground stable.

Presumably since we are talking about racing cars we have to also consider stability when the car is in motion. Some vehicles with a high cener of gravity have been known to roll during emergency maneuvers.

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Thanks for replies. Now what has torques got to do with. In torque why do we take the centre of gravity as the pivot point. In this situation, why would a torque cause more change in a car with higher centre of gravity.

 

EDIT: Oh wait I read more about torques, and it seems that at centre of gravity there is no net torque on the object. Ok then I can see why it taken as the pivot point. So simply in a racing car, if the centre of mass is close to the wheels, does that mean the force applied by wheels is less likely to destablise the car?

Edited by scilearner
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In addition to what IA said, there is another effect that makes the lower center of mass more stable with regards to tipping: For the same given centrifugal force (assuming tipping while driving through a curve) the torque is lower.

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well if you have two cars with exactly the same proportions but with a weight distribution that causes a difference in the height of the CoG then the one with a higher CoG will require less torque to tip over as you have to do less work against gravity.

 

Why does it have to do less work against gravity. Can anyof you point to me to a diagram or describe in what direction is the torque we are referring to. Is in the direction of wheels or where? Thanks

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a) The proper term is "center of mass", not "center of gravity".

b) When a car drives through a curve then there are two relevant forces acting on it, that can be considered as acting on the center of mass: gravity and the centrifugal force (the latter is not a gravitational force which is why the term "center of gravity" that you guys all seem to like so much is inappropriate - as I already mentioned in my first post).

c) The gravitational force points down, the centrifugal force horizontally towards the left or the right of the car (depending on the curve direction).

d) The point around which the car could tip is the contact point of the appropriate wheel with the road.

e) Gravitational force tries to tip the car in the one direction, centrifugal force into the other. Stronger torque wins.

 

I think that should be enough information for you to draw the diagram yourself.

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right,

 

imagine we have a square sitting on a surface (taking this as a 2 dimensional problem for simplicity). the center of gravity is in the center.

 

in order to tip the square over on its 'side' we need to rotate it, this will meanyou need to raise the center of gravity as the squate pivots on its corner. this means doing work against gravity and increasing the amount of force you need to apply to generate sufficient torque.

 

if you compare to a tall slender rectangle the center of gravity doesn't need to be moved very far before it is beyond the pivot and gravity will do the rest for you.

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Guest Rollin78

Centre of gravity is said to be concentrated on that point of any object,where actually the whole weight is concentrated. If an object rotates whenthrown, the center of gravity is also the center of rotation. When an object issuspended so that it can move freely, its center of gravity is always directlybelow the point of suspension. An object can be balanced on a sharp pointplaced directly beneath its center of gravity. It is important for automobilesand trucks to have their centers of gravity located close to the road, becausea low center of gravity gives them stability.

 

 

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  • 3 weeks later...
Center of gravity is said to be concentrated on that point of any object,where actually the whole weight is concentrated. If an object rotates when thrown, the center of gravity is also the center of rotation. When an object is suspended so that it can move freely, its center of gravity is always directly below the point of suspension. An object can be balanced on a sharp point placed directly beneath its center of gravity. It is important for automobiles and trucks to have their centers of gravity located close to the road, because a low center of gravity gives them stability.

 

your answer I believe is the best.

 

I have car. If I may in laymen terms. You first must weigh the car at each of the four points each wheel contacts the ground.

This will determine the actual center of mass, and in the perfect scenario all 4 contact points should be equal, placing the heaviest component in the center of the car, lowest to the ground.

 

You must take care to equal all 4 contact points, and a scale is needed. Moving components of the car to locations that result in more balance. Weight distribution. The drivers weight should also be included.

 

Its not enough to have a low center of mass if there are points that are very disproportionate to the other points of contact.

 

Another factor when relating to roll, or G force around turns, Is to heavily modify the chassis, so it becomes as rigid as possible.

 

Such as connecting the struts by use of strut tower brace. The rear suspension by traction bars, and also the length of the chassis with sub frame connection bars.

 

That should get you a more stable platform. then you must determine what springs to use for launch, so you don't have excessive wheel spin under acceleration.

 

I hope that helps you build a killer rod.

 

Cheers super-ball

Edited by superball
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