Physics talk

[mossy1200]

oh no.

10char

Post edited and was out of line sorry.

[HenryDorsetCase]

AC bitch.
(shit don't flow.)
[/COLOR]

then anodising doesn't work, and neither does a lead acid battery.

thanks for clearing that up.

[Spuds1234]

Im no physics major, nor have I taught physics before so this maybe a bit rough, but Im going to attempt to explain this like your 5 (dont be offended). If I make an assumption about what you may or may not know then I might miss something out that will be important to you understanding this.

Centripetal force is always directed toward the center of the curve, centrifugal force is always directed away from the center.

However centrifugal forces are fictitious forces. They don't exist. They appear to be there because if human perception though.

Imagine for a moment that you're on a merry-go-round that is spinning very fast. In order to not fall off you need to hold on. Your applying a force. The direction of that force that your exerting on the bar of the merry go round is actually inwards with respect to the center of the merry-go-round. Now imagine that you let go of that merry-go-round and you fall off. You would fell like you were pulled off by a mysterious force. However an observer that is on the same merry-go-round would see that your flying radially away from the merry-go-round. But an observer on the ground in a stationary reference frame (the dude on the merry-go-round is in a spinning reference frame) would see you fall of tangentially.

People who have done physics know that velocity is a vector (it has magnitude and direction) and in this case, is always at right angle to the centripetal force, we can therefore say, from the viewpoint of the observer in a stationary non-rotational reference frame that the person is really flying away from the merry-go-round due to his inertia, because inertia is the resistance of change in motion, in this case, the person's inertia has overcame the centripetal force, and thereby he will fly tangentially away from the merry-go-around.

The reason that centrifugal force is called a fictitious force is because that it only agrees with the definition of a force (a push or a pull) when the observer is in the same rotating reference frame as the object; while an observer in an non-rotational stationary reference frame does not need to be equipped with the concept of centrifugal force. The existence of centrifugal force is really a matter of fact that which frame of reference that the observers are in.

[Spuds1234]

Go take a weight on elastic out into your back yard and spin it in circles and come back in and tell me the elastic didnt stretch.

Ignoring the elastic completely for a moment, I'll get back to it shortly.

A moving object tends to move in a straight line right? Unless a force acts on that object. When you throw a stone at something, if there wasn't gravity that stone would just fly off into the distance never to land again. Gravity is constantly acting on the stone to make it curve and come back down to Earth. A similar thing is happening when you spin a rock on a string.

To spin the rock on the rope you have to first get it moving. When you apply a force to the rock to make it move its actually moving in a straight line initially (probably for only a very very short distance if the rope is already taught). In order to make the rock move out of a straight line and into a circle, you have to keep pulling on the rope (you can't spin something around without your hand being slightly ahead of the rock). This inward force that your applying to the rope is what centripetal force is. It's an inward force because you're always pulling the rope "down" with respect to the rock (if inward is a negative direction and out is a positive direction, down would be a negative direction).

Your pulling on the rope is applying the same force to the rock. Pulling it inward/down. If you let go of the rope you would see the rock fly away tangentially (ever play with a fox tail as a kid?)

It's that inward force that your applying to the rope with your hand which the rope is then applying to the rock, the force on the rock is inward.

Of course, your hand feels the rope pulling back on it (Newton's 3rd law and all that). That's "centrifugal" force. For the same reason as on the merry-go-round it doesn't exist. If your hand could think, it would certainly think it was there (centrifugal force), but your hand is in the same reference frame as the rock -a spinning reference frame. You know better because you can see with your eyes (which are hopefully not spinning to much at this point) what is happening if you let go of the rope. The rock would fly away tangentially. So there is no centrifugal force (as explained in my other post).

Now if the rope isn't getting any longer (or shorter) when your spinning the rock, it means that the inward force the rope is applying to the rock is equal and opposite to the reaction force that the rock applies in the string rope which must be outward right? So are we back to centrifugal forces being real? No, not in the sense everyone is talking about in this thread with the exception of a few posters.

From Wikipedia:

A reactive centrifugal force is the reaction force to a centripetal force. A mass undergoing curved motion, such as circular motion, constantly accelerates toward the axis of rotation. This centripetal acceleration is provided by a centripetal force, which is exerted on the mass by some other object. In accordance with Newton's Third Law of Motion, the mass exerts an equal and opposite force on the object. This is the reactive centrifugal force. It is directed away from the center of rotation, and is exerted by the rotating mass on the object that originates the centripetal acceleration.

This conception of centrifugal force is very different from the fictitious force. As they both are given the same name, they may be easily conflated. Whereas the 'fictitious force' acts on the body moving in a circular path, the 'reactive force' is exerted by the body moving in a circular path onto some other object. The former is useful in analyzing the motion of the body in a rotating reference frame; the latter is useful for finding forces on other objects, in an inertial frame.

This reaction force is sometimes described as a centrifugal inertial reaction, that is, a force that is centrifugally directed, which is a reactive force equal and opposite to the centripetal force that is curving the path of the mass.

The above might seem a bit convoluted but never the less its true. What the general population thinks centrifugal force is isn't true, the the people that actually understand it cant explain it for shit.

Now back to the problem.

Keeping things simple and assuming that the rock is traveling at a constant speed and your spinning it in a circle above your head.

The rock on the rope is held in circular motion by the tension in the rope. It is this tension that is causing the circular motion.

So what happens to the elastic? Why does it stretch? Simply because until the elastic is fully stretched the inward force the elastic is providing (from your pulling) is less than the reaction force the rock is applying (this force comes from outside factors like gravity and wind resistance, and also things like the rocks inertia which I wont get into). When the outward reaction force on the elastic is greater than the inward force that the elastic provides, the elastic must stretch until both forces are equal and opposite.

If any of this isn't clear, then feel free to ask questions, although I suspect at this point most of what I've said will have gone in one ear and out the other.

Dave- do you have anything to add? You seem pretty clued up on all this shit.

[mossy1200]

Ok so I understand.

But what are we calling the tyres inability to maintain it shape as the revolution speed becomes greater.
I understand that the exit line of an object is straight but the angle of exit is still distancing the object from the axis. So surely this means a retained object has force outwards appearing to be centrifugal force.

Question. What are you calling the resulting stretch?

[oneofsix]

Ok so I understand.

But what are we calling the tyres inability to maintain it shape as the revolution speed becomes greater.
I understand that the exit line of an object is straight but the angle of exit is still distancing the object from the axis. So surely this means a retained object has force outwards appearing to be centrifugal force.

Question. What are you calling the resulting stretch?

Centrifugal force. Whilst this may be a fictional force to the physicist to the rest of us it is the easy way to say and understand it, and the kicker is they also understand you even though they know it is a simplified expression of a complex concept.

Just like saying water freezes at 0C and boils at 100C, only works when it is pure and the pressure is right etc etc etc, but fucked if we care it is close enough for general use.

Gotta have a degree of tolerance, and back to speed ...

[Milts]

Just to help confuse things.

(From http://xkcd.com/123/)

[Dave-]

Dave- do you have anything to add? You seem pretty clued up on all this shit.

By the time I finished study last night (actually it was technically this morning) my brain wasn't much in the mood for explaining tangential, I started a post and wrote "Tangential" and then my brain quit, figured I'd leave it till the morning, plus this stuff is fresh for you, you sat 101 last semester? I sat it like 2 years ago haha.

I would add would be in terms on a tyre on a rim, imagine you lay the wheel down on the ground, and then like a pizza cut it up, except you've got a lot of friends and so you have to make a lot of cuts (luckily this is science so our cuts are infinitely thin) and we keep cutting until we get smaller and smaller pieces of pizza, we keep going until they're arbitrarily small, it's easier if I stop here and you just imagine they're in sections of about 5cm, this wee 5cm cross section piece of tyre has a certain mass, it's very small, maybe 20 grams? 50 grams? it's very small compared to the mass of the whole tyre.

Lets also look at the definition of acceleration, which is change in speed, acceleration is a vector, which means it has both magnitude and direction, in physics to properly identify an objects motion you must specify how fast it's going and in what direction, this is the same for velocity (there was a hilarious joke in the despicable me about vectors - top film) so lets say then that you spin a wheel up to a speed, and you hold it constant at that speed, now think about what your little pizza slice of wheel is doing, it's spinning around, but its also constantly changing direction, therefore its acceleration is non-zero (because it's vector is magnitude AND direction) even though the tyre isn't spinning any faster or slower, it still has acceleration.

It is this acceleration, with the mass of that wee pizza slice, that distorts the shape of the tyre at a tangent, but because the tyre is joined all the way around, all the other wee pieces of tyre do the same thing at exactly the same time, and it looks as if the tyre is moving away from the rim.

In reference to the tyre getting smaller, that was a more real world situation where it's on a road and you have many more forces coming in to play.

So you have gravity pulling your vehicle down at 1G, on a drag car this is higher cause the wing presses it down some more. I theorised that you have your tangential force pressing the tyre out at some force, the acceleration is huge, but the mass is tiny, so the force is fairly normal, if there was no weight on the axle then the tyre would be allowed to expand, would raise the ride height of the vehicle, and change the gearing, BUT we have a vehicle pressing down at 1G, if the force pressing down is greater than the force pressing "out" from the wheel then the tyre will be more squashed than it is expanded, therefore the gearing would go down, I may be wrong on this though.

And then I spoke with a mechanical engineer friend, he explained that as the tangential force hits the road the tyre deforms and the radius gets smaller again, lowering the gearing more substantially, you can sort of see it in this video.

[YOUTUBE]v-DMkO3g2SI[/YOUTUBE]


edit: I just had a thought, what's going on in this video isn't exactly as I've described, what I've described is wiped out by the bob and sway of the car going over bumps and braking etc, but the bumps and braking etc gives a good impression of what I'm describing, you can see that as he goes over a bump and the force pushing down on the axle increases the base of the tyre deforms more and more.

now while that's only happening to the bottom of the tyre, the rest of the tyre is expanding sure, but you'll find the gearing is based on the average (halfway), as it turns out from the mech student, the tyre gets more smaller due to this deformation than it gets larger due to tangential force, therefore at speed the tyres gearing goes down. and the speedo should reflect this.

But because they don't use an active filter on modern digital speedos you wont see this cause it's lost in the signal noise.

I feel we've achieved something here today. Back to network protocol design!

[Spuds1234]

By the time I finished study last night (actually it was technically this morning) my brain wasn't much in the mood for explaining tangential, I started a post and wrote "Tangential" and then my brain quit, figured I'd leave it till the morning, plus this stuff is fresh for you, you sat 101 last semester? I sat it like 2 years ago haha....

Na did all my physics stuff last year, so a lot of what I wrote above I had to look up. For all intents and purposes this year has been 3/4's maths and fuck all physics. Next year will change that though I think.

As for the tyre shrinking or growing thats well out of my area of expertise, but I suspect if the tyre was off the ground (like that video of the RC car a page or two ago) that what your seeing is much like the rock on the rope. The tyres sidewall acts like a stretchable rope, and the part of the tyre that contacts the road acts like a rock. So the tyre would expand in this case, but here we're completely taking out the affects of the cars mass and the force that the tyre was exerting on the ground.

What happens when the tyre is on the ground with the weight of the car pressing down on it? Im not sure. Its well out of the realms of what I know and I cant find any material online that even begins to explain it.

[Dave-]

What happens when the tyre is on the ground with the weight of the car pressing down on it? Im not sure. Its well out of the realms of what I know and I cant find any material online that even begins to explain it.

There's a book in the EPS library on tyres, there'll be something in there about it, you're probably sitting less than 100m from it

[Spuds1234]

Speaking to my flatmate, he reckons that the radius of the tyre shrinks because of plastic deformation caused by the forces the tyre exerts on the road (due to both the tyres rotation and the mass of the car) and then at speed, the tyre does not have enough time to come back to its original shape in one rotation.

So basically if you look at a tyre on your car, you notice that the bottom part of the tyre is squashed more than the top part. Now what happens when you spin that tyre really really fast like what is happening in the video Dave- posted?

All those small parts of the tyre (that Dave- mentioned) get squashed a bit each time the tyre goes through 1 revolution, but now the tyre is rotating so fast that those small parts of the tyre cannot expand back out to their original position before they are squashed again. So it looks like the tyre is contracting.

This seems to make no sense intuitively when you first think about the problem because when you think about it its hard to imagine a tyre going that fast. Its not something you look at every day. Unlike the situation when your spinning a rock on a rope (because who didn't do that as a kid in some form or another).

This seems to make sense as you'll notice in that Bathurst video when the car speeds up the tyre gets deformed and squashed and the radius of the tyre seems to shrink. Then when it slows down or the weight on the tyre decreases (when it lifts off the ground) the radius of the tyre increases. When the tyre is off the gound we're back to a rock on a stretchy rope kinda problem and when the car slows down each small section of the tyre is allowed to return to its original shape because its not immediately getting squashed again.

[mossy1200]

Why do you think the bike gets forced down heavier onto the road at speed when a bike has no wings to force it onto the road?
The front wheel of bikes I have riden seem to get lighter the faster you go.
I also dont understand why we are comparing a race car wheel braking and moving over bumps to a motorcycle front wheel at 200 or more.
Idd say the only time a motorbike wheel has compressed for extended time is front during braking and rear during acceleration.

[Akzle]

thanks mann!

[Dave-]

Why do you think the bike gets forced down heavier onto the road at speed when a bike has no wings to force it onto the road?
The front wheel of bikes I have riden seem to get lighter the faster you go.
I also dont understand why we are comparing a race car wheel braking and moving over bumps to a motorcycle front wheel at 200 or more.
Idd say the only time a motorbike wheel has compressed for extended time is front during braking and rear during acceleration.

you're entirely right.

a bike doesn't get heavier, but the physics involved means the tyre shrinks. also a few guys mentioned drags cars and I believe the original discussion was in a car/truck to do with cruise control/limiters....

the front wheel will get lighter during acceleration due to a torque at the back wheel.

the race car wheel video is an analogy, there aren't any videos on youtube with a motorcycle wheel at speed which show a representation of what I'm describing to you.

braking yes, also cornering on a cambered corner.

I see we have our own physics thread now, I was considering starting one so next time someone talked about centrifugal force I'd just link them to the thread.

Anyone happen to know what the maximum lateral G a bike can pull in a corner is? I bet you'll be surprised how low it is.

[mossy1200]

Hey. I never started a new thread. I just took the original thread off topic. Id rather just have an infraction than be responsible for this thread.

When everyone thinks your stupid dont open your mouth and confirm it.