Normal Forces on Tyres while cornering


by HairyButtock
Tags: cornering, forces, normal, tyres
HairyButtock
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#1
Mar24-12, 05:04 PM
P: 5
I'm trying to derive an expression for the normal forces on a cars tyres travelling around a circular banked track. Initially I was going to assume the centre of gravity was located longitudinally and laterally centre between all four tyres (so front and rear had the same forces on them). Also the car will be travelling at constant angular velocity.

However even with an unbanked track, I seem to be missing something, I know that the normal force must be velocity dependent (since it is going to decrease on the inside tyres and increase on the outside tyres as velocity increases).

In the image uploaded the axis of rotation is to the left.

Taking moments about the centre of gravity you get...

[tex] \sqrt{\frac{x^{2}}{4}+h^{2}}(N_2sinθ-N_1sinθ-F_1cosθ-F_2sinθ)=0 [/tex]
and I know F1+F2=mrω2
as well as the coefficient of friction being the same for both tyres.

I'm struggling to combine these two to give a velocity dependent expression for N1 & N2.

Confusingly when I take moments about the base of each tyre I get the same value for N1 & N2 after rearranging...
[tex] \sqrt{\frac{x^{2}}{4}+h^{2}}\frac{mgsin\phi }{x} [/tex]
which is not velocity dependent or correct.

Cheers for any help.

EDIT-Sorry, I can't figure out how to move this post over to the homework/coursework-type questions forum or how to delete it.
Attached Thumbnails
flat circular track.png  
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McLaren Rulez
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#2
Mar24-12, 10:51 PM
P: 261
I'm not sure of this but I'll give it a go.

When you go around a turn, choose the frame of the car. The CG is pushed out due to a centrifugal force and that's [itex]mv^{2}/r[/itex]. Now, we balance the torques using say, the outer wheels as our axis. The torque from the centrifugal force would be [itex]mv^{2}h/r[/itex]. The torque equation reads [itex] mgx/2-N_{1}-mv^{2}h/r=0[/itex].

Do the same analysis with the inner wheels as your axis to get [itex] mgx/2-N_{2}+mv^{2}h/r=0[/itex]

I think the force of friction doesn't matter since those go through our axis of rotation anyway. At first glance it looks correct because it is velocity dependent and the normal force for the outer wheels (wheel 2) is indeed larger.
HairyButtock
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#3
Mar25-12, 06:43 AM
P: 5
I was trying to keep it in an inertial reference frame ideally (sorry, should have said). Although I agree with your working.

Surely it's possible to do this in an inertial reference frame; since all those equations are literally just Normal the torque created by centripetal force.
In this scenario is centripetal force not just the friction forces? and isn't the torque created by friction not acting through the centre of gravity?

Sorry, realised I was being an idiot above.
It should read:
if you take moments about the centre of gravity you get:
[tex] \frac{x}{2}(N_2-N_1) - h(F_1+F_2)=0 [/tex]
and if you take moments about either tyre you get:
[tex] \frac{x}{2}mg-N_2=0 \,\,\,\,\,\, or \,\,\,\,\,\, \frac{x}{2}mg-N_1=0 [/tex]
although this is the same problem, so I'm still obviously missing something.

McLaren Rulez
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#4
Mar25-12, 05:46 PM
P: 261

Normal Forces on Tyres while cornering


Okay your first equation is identical to what I posted earlier because [itex]F_{1}+F_{2}=mv^{2}/r[/itex]. By the way, small mistake in my earlier post, it should be [itex] mgx/2-N_{1}x-mv^{2}h/r=0[/itex] and [itex] mgx/2-N_{2}x+mv^{2}h/r=0[/itex]. So that's settled.

For your second set of equations, I can't quite explain why the [itex]mv^{2}/r[/itex] term needs to be there but it definitely does. Hopefully someone else can explain it.
jim hardy
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#5
Mar25-12, 10:26 PM
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I see your first equation


but on moments around individual tyres, pardon me if i'm dumb. it's been a long time.

you wrote : ##\frac{x}{2}mg-N_2=0 \,\,\,\,\,\, or \,\,\,\,\,\, \frac{x}{2}mg-N_1=0##


If i sum moments about bottom of right hand tire taking CCW positive

i see: ##mg\frac{x}{2}-N_1x - \frac{mv^2}{r}h=0 ##

exactly what McLaren said.
McLaren Rulez
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#6
Mar25-12, 11:38 PM
P: 261
Jim, I think the OP wants to do it in an inertial frame of reference. Even I'm not sure where the [itex]mv^{2}h/r[/itex] term comes from when this is analyzed in an inertial frame. I have a feeling that I am missing something obvious because it is a pretty simple problem but it still evades me.

Could you explain again why you have that term? When we see it going around the turn from an inertial frame, the centripetal force is provided by the friction so this has no torque right?
jim hardy
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#7
Mar26-12, 09:24 AM
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I don't quite understand inertial frame - so my bad .
Could you explain again why you have that term?
Anyhow, i just summed moments about the bottom of right hand tire, as in the dynamics course i took ~1965..

the centripetal force indeed is developed by the tires and goes through my summing point, ergo zero torque .
But the centrifugal force acts on CG at distance h above my summing point and is coupled through the automobile suspension so produces torque, ergo nonzero moment.

I should know what you mean by inertial frame but i don't. I seem to remember something called "Center of Mass Coordinates" ...

My bad, i'll look it up.

Thanks !! old jim
HairyButtock
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#8
Mar26-12, 10:18 AM
P: 5
By "inertial reference frame" I meant an observer watching the car with constant velocity. From the reference frame of the car (i.e. considering the car as stationary) you have to introduce pseudo-forces (such as centrifugal forces) to explain why the car isn't moving in that reference frame.

However, after ALOT of searching I stumbled across this site on car rollover. It says that an acceptable thing to do, even in an inertial reference frame, is to consider the friction forces acting radially outwards from the centre of gravity since it creates the same torque that the friction creates from the base of the wheels. So according to that website both your calculations are correct (in terms of inertial and non-inertial reference frames, I guess ).

The paragraph in question:
In basic Physics, forces can be analyzed for the separate effects of translation (straight-line motion) and rotation. This can be done equally correctly in different frames of reference. So far, we have been considering the force acting ON the vehicle as seen from the ground, and so we would draw our horizontal arrow TO THE LEFT in this drawing at a height of the bottom of the tires. This would represent the ACTUAL centripetal force acting on the vehicle by the road, and would be analyzable as to both translation and rotation effects. I think it may be easier to understand if we look at this in a different, but exactly equivalent way. Instead of saying that the centripetal force is acting to the left (in our drawing), at the height of the tires, we can say that there is an identical force to the right, acting ON the center-of-gravity. The effect is exactly the same, regarding the action to rotate/twist the vehicle, but I am hoping that the arrows and discussion might better show why this acts to roll the vehicle over sideways.
jim hardy
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#9
Mar26-12, 10:56 AM
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By "inertial reference frame" I meant an observer watching the car with constant velocity.
okay

from Wikipedia
In physics, an inertial frame of reference (also inertial reference frame or inertial frame or Galilean reference frame) is a frame of reference that describes time homogeneously and space homogeneously, isotropically, and in a time-independent manner.[1]
so if in my alleged mind i "freeze frame" the car and replace centrifugal force with a rope pulling horizontally on CG

which of course makes tires pull other direction aka centripetal force

have i achieved thought equivalent of inertial reference?

They've assigned names to a lot of things we 'just did' when i was in school and sliderules were king.
HairyButtock
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#10
Mar26-12, 11:17 AM
P: 5
The thing is that centrifugal force doesn't actually exist, it's a 'pseudo-force' that emerges when you build Newton's laws of motion in an accelerating reference frame. You shouldn't need it to explain the turning forces experienced by some objects in circular motion when being watched by an observer whose acceleration is zero (i.e. in an inertial reference frame).

see: http://en.wikipedia.org/wiki/Fictiti...ircular_motion

I'm not referring to definition of centrifugal force as the reaction force to a centripetal force by the way (which in this instance would be the friction expierienced by the track as supposed to that expierienced by the tyres)
jim hardy
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#11
Mar26-12, 11:47 AM
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Thanks for that link !

I think i've got it. It's one of those 'fine points'.

Road pushes car sideways
##\sum{F} = ma##
and that's a dynamics problem

and in my school days only difference between statics and dynamics was, in statics ##\sum{F} = 0##

So there'd be a statics problem also to test whether car successfully turns or rolls over
and that's the one where we'd include ##\frac{mv^2}{r} ##

Thanks for your patience .

Circular motion

See also: Centrifugal force, Reactive centrifugal force, and Coriolis force

A similar effect occurs in circular motion, circular from the standpoint of an inertial frame of reference attached to the road. When seen from a non-inertial frame of reference attached to the car, the fictitious force called the centrifugal force appears. If the car is moving at constant speed around a circular section of road, the occupants will feel pushed outside by this centrifugal force, away from the center of the turn. Again the situation can be viewed from inertial or non-inertial frames (for free body diagrams, see the turning car):
1.From the viewpoint of an inertial reference frame stationary with respect to the road, the car is accelerating toward the center of the circle. This acceleration is necessary because the direction of the velocity is changing, despite a constant speed. This inward acceleration is called centripetal acceleration and requires a centripetal force to maintain the circular motion. This force is exerted by the ground upon the wheels, in this case thanks to the friction between the wheels and the road.[13] The car is accelerating, due to the unbalanced force, which causes it to move in a circle. (See also banked turn.)
2.From the viewpoint of a rotating frame, moving with the car, there is a fictitious centrifugal force that tends to push the car toward the outside of the road (and to push the occupants toward the outside of the car). The centrifugal force balances the friction between wheels and road, making the car stationary in this non-inertial frame.
HairyButtock
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#12
Mar26-12, 01:10 PM
P: 5
Cheers guys, think I've cracked it now.
Now onto the joys of sprung, unsprung weight and lateral load transfer.
McLaren Rulez
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#13
Mar26-12, 07:05 PM
P: 261
HB, I'm glad you cracked it. This thread might be helpful because it's a simplified version of your problem with an excellent analogy from A.T. in post 11.

The key point that I missed earlier is that in an inertial frame, the net change in angular momentum about either tyre is not zero. The car is accelerating towards the center of its circular path and this is a change in angular momentum whose magnitude is exactly the missing [itex]mv^{2}h/r[/itex]

In uniform circular motion, there is an acceleration towards the center but there is never any velocity along the radial direction. Similarly, in our problem, there is a torque (and a change in angular momentum) about the axis at the tyre but the axis moves such that there is never any angular momentum with respect to it at any time.
PAllen
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#14
Mar26-12, 07:19 PM
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