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jason.farnon Mar7-12 01:25 PM

outward radial motion in unbanked turn
 
Sorry for not following the template, this is a follow up to a few other old posts.

In solutions to problems where a car negotiates an unbanked turn, I usually see that the centripetal force is supplied by friction. My understanding is that friction applies in the direction opposite motion. So there is some outward radial motion the friction is counteracting? What force supplies this or is my analysis incorrect? Is it the wheels of the car?

Ericv_91 Mar7-12 01:53 PM

Re: outward radial motion in unbanked turn
 
These centripetal force questions are rather difficult to visualize, so here maybe one of the diagrams here will help you a bit: http://hyperphysics.phy-astr.gsu.edu/hbase/corf.html

PeterO Mar8-12 02:18 AM

Re: outward radial motion in unbanked turn
 
Quote:

Quote by jason.farnon (Post 3803434)
Sorry for not following the template, this is a follow up to a few other old posts.

In solutions to problems where a car negotiates an unbanked turn, I usually see that the centripetal force is supplied by friction. My understanding is that friction applies in the direction opposite motion. So there is some outward radial motion the friction is counteracting? What force supplies this or is my analysis incorrect? Is it the wheels of the car?

Your mind is [like most people] stuck in the inertial frame - one which is not accelerating.

When the car travels in a circle, it is accelerating towards the centre of the circle. Even if it travels an constant speed [note speed has no direction; it is velocity when we add direction]

In order to accelerate towards the centre, something has to push it that way. it is the friction force that provides that.

The force is achieved by the driver turning the steering wheel. This angles the front wheels, meaning they would tend to slide across the surface, rather than merely roll along. If friction is large enough, instead of sliding across the surface, that friction results in circular motion.

Note that if the friction is not big enough, the car instead goes straight ahead or, at best, travels in a circle of greater radius than is necessary to follow the desired path.
Such a motion is generally referred to as "a skid".

If you want to see the effect of such a skid, draw the following on some graph paper:

Using a compass, with the origin as centre, draw a circle of radius 5, and another of radius 6 units. That gives you the "road" you wish to negotiate.

Now using (-2,0) as the centre, draw a circle of radius 7, starting at (5,0) and going anticlockwise.
That path shows the car "skidding wide" on the corner due to in sufficient friction.

PeterO Mar8-12 02:34 AM

Re: outward radial motion in unbanked turn
 
Quote:

Quote by jason.farnon (Post 3803434)
Sorry for not following the template, this is a follow up to a few other old posts.

In solutions to problems where a car negotiates an unbanked turn, I usually see that the centripetal force is supplied by friction. My understanding is that friction applies in the direction opposite motion. So there is some outward radial motion the friction is counteracting? What force supplies this or is my analysis incorrect? Is it the wheels of the car?

Lets look at a straight line example of basically the same thing.

A brief-case is placed on the floor of a train, then the train moves off from the station.
The friction [between the case and the floor] supplies the force to accelerate the case so that it "keeps up with" with the accelerating train.

That frictional force clearly acts forward - but that seems to be in the direction of motion!

That friction force does act in the direction the case would move [relative to the train] if there was no friction - like for example a skate board placed on the floor of the train beside the brief-case.

NOTE: The train provides an accelerated frame of reference [it is, after all, accelerating].
In that accelerated frame of reference it is common to "find" a fictitious force the one accelerating the skateboard towards the rear of the train.
The friction "overcomes" that fictitious force.
Of course there is no fictitious force: that friction force is merely supplying the required force to accelerate the brief-case.

EXAMPLE 2: When you are a passenger in a very powerful car, and the driver guns the motor, you are "pushed back into the seat" as the car accelerates away.
Look carefully and you will find there is nothing there to actually push you into the seat.

what actually happens is that the seat - attached to the car - accelerates down the road. You inertia means that you briefly do not accelerate.
That mans you stay behind, and the seat back gets squashed between a stationary you, and a moving seat frame.
Once squashed, the springs/padding/frame of the seat NOW apply sufficient force to you, to make you ALSO accelerate with the car.
The only option was for you to break through the seat, through the back structure of the car and end up sitting on the road, as the (now damaged) car accelerates away from you.

jason.farnon Mar8-12 11:49 AM

Re: outward radial motion in unbanked turn
 
PeterO,

Thanks for the explanation and examples. I think you got me close but not quite there. I can see the operation of friction in your briefcase on a train example. In the absence of friction the briefcase moves to the back of the train relative to the train's motion, so the direction of friction must be in the same direction as the train. I can follow your example about the passenger in the car. Still, I don't see the relevance to the question of a car going in a circle. My understanding of inertial/non-inertial frames may be weak. But I can see that those examples involve objects in accelerating objects, so the coordinate system set in that object is accelerating. How does that apply to the radial motion of a car going in a circle? I'm not talking inside the car, but isn't the car going in a circle on a speedway, with the speedway assumed fixed in space, have the speedway as an inertial frame?

I think what you said about the wheels agrees with my sense of the answer. If I turn the wheels theta>0 degrees off the path I'm going in, which is tangential to the circular path, friction will act in that direction theta off the path, so it will have an inward radial component. Then the answer to my question, what is the outward radial motion friction is counteracting, would be, it is the relative motion of the turned tires to the ground. Those tires, relative to the ground, move outward. Does that sound right?

PeterO Mar9-12 04:38 AM

Re: outward radial motion in unbanked turn
 
Quote:

Quote by jason.farnon (Post 3804870)
PeterO,

Thanks for the explanation and examples. I think you got me close but not quite there. I can see the operation of friction in your briefcase on a train example. In the absence of friction the briefcase moves to the back of the train relative to the train's motion, so the direction of friction must be in the same direction as the train. I can follow your example about the passenger in the car. Still, I don't see the relevance to the question of a car going in a circle. My understanding of inertial/non-inertial frames may be weak. But I can see that those examples involve objects in accelerating objects, so the coordinate system set in that object is accelerating. How does that apply to the radial motion of a car going in a circle? I'm not talking inside the car, but isn't the car going in a circle on a speedway, with the speedway assumed fixed in space, have the speedway as an inertial frame?

I think what you said about the wheels agrees with my sense of the answer. If I turn the wheels theta>0 degrees off the path I'm going in, which is tangential to the circular path, friction will act in that direction theta off the path, so it will have an inward radial component. Then the answer to my question, what is the outward radial motion friction is counteracting, would be, it is the relative motion of the turned tires to the ground. Those tires, relative to the ground, move outward. Does that sound right?

The friction with the circular motion is a parallel to the brief case in the train.

In the train, without friction, the case would slide straight back along the floor. Friction makes it accelerate forward with the train.

On the flat track, the car would proceed straight ahead. Friction makes it accelerate towards the centre of the circle - and thus travel in circular motion.
Note:
If the acceleration was too small, the car would spiral out.
If the acceleration was too big, it would spiral in.

Like all good friction forces in action, the friction is just the right size to achieve the desired circular motion [unless to alter the steering wheel position, and thus the size of the friction force].

REMEMBER: CIRCULAR MOTION IS ACCELERATED MOTION.
An INWARDS force causes that accleeration. [Centripetal Force]
There is no outward force to balance the inward force.
If there was, the net Force would be zero, and the motion would be in a straight line.

tiny-tim Mar9-12 05:17 AM

welcome to pf!
 
hi jason! welcome to pf! :smile:
Quote:

Quote by jason.farnon (Post 3803434)
where a car negotiates an unbanked turn, I usually see that the centripetal force is supplied by friction. My understanding is that friction applies in the direction opposite motion

this really is only to do with the difference between sliding and rolling

if we ignore air resistance etc, the car will keep going in a circle (even on an unbanked surface) without any power from the engine

no forward friction is needed :wink:

and there's no forward relative motion between the rolling wheel and the road

the point of contact is stationary

(nor is there any tendency to forward relative motion)

so there's no forward friction
there is a tendency to move sideways (the wheel can't roll sideways! :wink:), and so the friction is opposing that :smile:

cmb Mar9-12 06:22 AM

Re: outward radial motion in unbanked turn
 
Which comes first - the slide or the friction?

Actually, in automotive chassis-dynamics it is often said there is no grip without slide. I don't buy that, I think there is such a thing as purely static friction, but I don't think it is a completely unreasonable argument to make either way.

PeterO Mar10-12 03:14 AM

Re: outward radial motion in unbanked turn
 
Quote:

Quote by cmb (Post 3806217)
Which comes first - the slide or the friction?

Actually, in automotive chassis-dynamics it is often said there is no grip without slide. I don't buy that, I think there is such a thing as purely static friction, but I don't think it is a completely unreasonable argument to make either way.

Based on the example of applying an ever increasing force to a block on a rough surface, I would say the friction comes first.

Suppose μN for a block is 10N [μN is the maximum possible friction]

Apply a force of 1N, you get Friction of 1N
Incresase to 2N - friction is 2N
keep going to 10N - friction becomes 10N

Finally get to 11N and we get the slide.

cmb Mar10-12 03:17 AM

Re: outward radial motion in unbanked turn
 
I think I should've used the word 'slip' rather than 'slide'.

'Slip' is the term the tyre dynamicists use to describe this concept. I'm with you, I'd say 'the friction' comes before 'the slip'.

jason.farnon Mar10-12 10:51 PM

Re: welcome to pf!
 
Hi Tiny-tim, Thanks I had not even considered the difference between rolling and sliding. But I am still not clear.

Quote:

Quote by tiny-tim (Post 3806159)

this really is only to do with the difference between sliding and rolling

if we ignore air resistance etc, the car will keep going in a circle (even on an unbanked surface) without any power from the engine

You mean no additional power, after whatever gives the car an initial velocity? I can visualize that, e.g., a satellite rotating around a planet, where gravity replaces the friction of our example. But for gravity there is a law that says it always acts radially between two bodies in the direction of the segment between their centers. With friction it only acts in the direction opposite motion. This is my question, what motion is it opposite to.

Let me try this way. Tell me where I mess up. 1. When a car is driving in a circle with uniform speed, friction supplies the centripetal acceleration. 2. Friction is a force between two surfaces in contact that acts in the direction opposite to the direction of motion between the two surfaces that would exist in the absence of friction. 3. Therefore in the car example friction opposes motion in the outward radial/centrifugal direction. 4. Therefore there is a centrifugal force.

PeterO Mar11-12 03:33 AM

Re: welcome to pf!
 
Quote:

Quote by jason.farnon (Post 3808996)
You mean no additional power, after whatever gives the car an initial velocity? I can visualize that, e.g., a satellite rotating around a planet, where gravity replaces the friction of our example. But for gravity there is a law that says it always acts radially between two bodies in the direction of the segment between their centers.
Gravity only acts radially on a body orbiting in a circle!!! Similarly the friction you are encountering her acts radially because the car is travelling in a circle.
Gravity doesn't "look" at you, when you leave a diving board and say "Ah, he is just going to fall so I will act down - or look at a rocket being launched and say "Oh, that rocket is trying to get away, so I will pull down and at least make it little bit harder for it" or look at a satellite in orbit and think " I may as well be kind here and supply a radial force to keep it in orbit - gee, I hope that is what it wanted!!"
As common, there is an attraction - centre of mass to centre of mass regardless


With friction it only acts in the direction opposite motion.

Opposed to motion or opposed to a potential motion.
When a box slides across the floor, the friction opposes that motion. When a box is placed on a gentle slope, friction opposes the potential for the box to slide down the slope


This is my question, what motion is it opposite to.

It is the potential motion that it is opposing.
When the front wheels of the car are angled (you turn the wheel) there is the potential for the tyres to now slide sideways (to some extent - we usually don't turn them through 90o
Friction opposes that potential to slide sideways by creating a force at right angles to the motion - and so an acceleration at right angles to the motion - a centripetal acceleration


Let me try this way. Tell me where I mess up.
1. When a car is driving in a circle with uniform speed, friction supplies the centripetal acceleration. Yes, because the front wheels have been angled
2. Friction is a force between two surfaces in contact that acts in the direction opposite to the direction of motion between the two surfaces that would exist in the absence of friction. That's right - the tyres are trying to slip sideways; but remember, sideways to an angled tyre may be straight ahead to a moving car. Many drivers discover this problem each winter when a snap freeze means ice on the road and the car can no longer turn or stop
3. Therefore in the car example friction opposes motion in the outward radial/centrifugal direction. The motion is not radially outward. When the driver in winter slips off the road, it is only that the road was constructed with a radius of, say, 50m but the limited friction means the car travels in a circle of , say, 100m.
When suddenly placed in an unexpected non-inertial (accelerating) frame - the human will immediately invent a fictitious force to "explain" the problem. You don't need understanding to make things up. The fictitious force everyone invents even has a name - centrifugal force - and once a force gets a name it must be real - NOT

4. Therefore there is a centrifugal force.

Exactly - as soon as everyone agrees on the same name for a piece of fiction it must exist, mustn't it? That's what physics is for - to show that there is no such thing!!!

tiny-tim Mar11-12 05:30 AM

(PeterO, do you know you're typing in red? :redface:)
hi jason! :smile:

(just got up :zzz:)
Quote:

Quote by jason.farnon (Post 3808996)
You mean no additional power, after whatever gives the car an initial velocity?

assuming no air resistance, friction in the bearings, rolling resistance, etc yes :smile:

(of course, the front wheels must be angled, or the whole vehicle must be leaning, like a motorbike)

Quote:

With friction it only acts in the direction opposite motion. This is my question, what motion is it opposite to.
it's static friction, so it's opposite the way the car "wants" to slide, ie outwards
you've got that anyway, in 2 and 3 below! :wink:
Quote:

1. When a car is driving in a circle with uniform speed, friction supplies the centripetal acceleration.
2. Friction is a force between two surfaces in contact that acts in the direction opposite to the direction of motion between the two surfaces that would exist in the absence of friction.
3. Therefore in the car example friction opposes motion in the outward radial/centrifugal direction.
yes :smile:
Quote:

4. Therefore there is a centrifugal force.
no, no such thing as centrifugal force for an inertial observer

the friction can be called a centripetal force :wink:

(though if the road was banked, the friction wouldn't be centripetal, would it? :biggrin:)

PeterO Mar11-12 05:45 AM

Re: outward radial motion in unbanked turn
 
Quote:

Quote by tiny-tim (Post 3809275)
(PeterO, do you know you're typing in red? :redface:)
hi jason! :smile:
I type in red when I respond within the previous post rather than after it - so it is obvious which part I am commenting on
(just got up :zzz:)


assuming no air resistance, friction in the bearings, rolling resistance, etc yes :smile:

(of course, the front wheels must be angled, or the whole vehicle must be leaning, like a motorbike)



it's static friction, so it's opposite the way the car "wants" to slide, ie outwards
you've got that anyway, in 2 and 3 below! :wink:

yes :smile:

With description given, I am afraid it is NO :frown:

no, no such thing as centrifugal force for an inertial observer

the friction can be called a centripetal force :wink:

(though if the road was banked, the friction wouldn't be centripetal, would it? :biggrin:)

But would the friction by up the bank or down the bank ?:smile:
Perhaps no fiction at all, even if the track was "rough"!!

PeterO Mar11-12 05:51 AM

Re: outward radial motion in unbanked turn
 
Quote:

Quote by tiny-tim (Post 3809275)
(PeterO, do you know you're typing in red? :redface:)
hi jason! :smile:

(just got up :zzz:)


assuming no air resistance, friction in the bearings, rolling resistance, etc … yes :smile:

(of course, the front wheels must be angled, or the whole vehicle must be leaning, like a motorbike)



it's static friction, so it's opposite the way the car "wants" to slide, ie outwards …

And importantly the car will NEVER slide out, it just may turn in less sharply than the road/planned route turns in. ie the car may be travelling in a circle of radius 100m when the road was constructed with radius 70m. Under no circumstances, on a flat surface, will a car go right when you turn left - it just may not go left as sharply as you hoped for! It may even continue straight ahead if the surface is ice.
you've got that anyway, in 2 and 3 below! :wink:

yes :smile:


no, no such thing as centrifugal force for an inertial observer

the friction can be called a centripetal force :wink:

(though if the road was banked, the friction wouldn't be centripetal, would it? :biggrin:)

.....

tiny-tim Mar11-12 05:58 AM

Quote:

Quote by PeterO (Post 3809291)
I type in red when I respond within the previous post rather than after it - so it is obvious which part I am commenting on

but then it appears in italics :frown:
it's much easier to read (and to quote!) if you simply divide the previous post into separate quote-blocks (eg as in my last post) :smile:

PeterO Mar11-12 06:09 AM

Re: outward radial motion in unbanked turn
 
Quote:

Quote by tiny-tim (Post 3809297)
but then it appears in italics :frown:
it's much easier to read (and to quote!) if you simply divide the previous post into separate quote-blocks (eg as in my last post) :smile:

What is the problem with italics - I would concentrate more on correcting your suggestion there is a radially outward motion - even a potential for it - in the situation mentioned in the OP.

tiny-tim Mar11-12 06:15 AM

Quote:

Quote by PeterO (Post 3809308)
What is the problem with italics

i find italics difficult to read :redface:
Quote:

in the situation mentioned in the OP.

i'm confused :confused:
can you quote it?


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