Centrifugal forces ' appear ' to act opposite to gravity . How is this possible?

[swansont]

SwansonT,

 

So what direction are you moving when you are going in a circle?

Tangent to the circle.

 

 

If I draw a line depicting my motion tangent to the circle, for every degree of my turn, I get 90 different directions that look sort of like a fan. The lines all go radially outward from where you would hold the fan.

Then you did it wrong. The velocity vectors will be perpendicular to the radial direction. They never intersect the center.

 

 

During the turn your momentum is heading toward 0, 359, 358, 357...273, 272, 271 and finally 270 degrees (West). The lines all trace back to 90 origins on a small 90 degree arc of the circle that the turn is part of. If you draw a circle with a center where the 0 degree tangent line, and the 270 degree tangent line meet, all the directions inbetween can be seen as radial lines drawn out from this point.

 

motorcycles.jpg

 

While there is no arrow taking the rider East, there is an arrow eminating in each direction between North and West. And each direction can be seen as being in the same direction as a radial line drawn from the center of the curve. And in each case, following the line, in the case of gravel or ice causing the tire to loss traction with the road, will take the rider and bike off the road, to the right, to the outside of the circle of the turn. The momentum will take the rider and bike, radially outward.

Nope. I don't know what you're doing here, but you aren't properly drawing velocity vectors or calculating accelerations if you conclude there is an outward acceleration.

 

http://dev.physicslab.org/Document.aspx?doctype=3&filename=CircularMotion_CentripetalAcceleration.xml

[tar]

SwansonT,

 

I noticed the vectors never intersect the center of the circle.

 

That is why I concentrated on the point outside the arc of the bike, where the North (360/0) vector and the West (270) vector intersect.

 

If you draw a square, centered at the center of the circle, with the upper right corner of the square being the point where the North and West vector intersect, all the vectors intersect all the other vectors somewhere to the upper right of the circle and to the lower left of this corner point.

 

The "center" of the turn therefore may be considered the average position of these vector intersecting points, which by rough estimation is about 80% the way between the center of the circle and the square corner point. This as opposed to the roughly 70% of the way out that the bike is, halfway through the turn, when it is directly between the center of the circle and the square corner point.

 

Each of the directional vectors found by drawing tangents to the circle at each degree position of the bike and rider has its parallel conterpart going radially outward in the NW quadrant of the circle.

 

The rider leans into the turn. She leans left, toward the center to counteract something carrying her and the bike outward.

 

Regards, TAR

But, while I admit I am doing something wrong here, and the actual centripedal acceleration is being caused by the reaction of the tire and the road, and the rider is leaning merely to maximize these forces, allowing the change in direction to take place, I am still "speculating" as to why it "appears" that there is a radially outward "thing" going on during the turn.

Interesting to note that if you follow a tangential line away from the circle it intersects all 90 of the degree radials to its left (or right if you are going the other way around the circle.) These intersections are rapid at first and get more and more spread out as you go. But as Mike pointed out, each of these intersections is further "out", radially speaking.

Edited April 19, 2015 by tar

[swansont]

SwansonT,

 

I noticed the vectors never intersect the center of the circle.

 

That is why I concentrated on the point outside the arc of the bike, where the North (360/0) vector and the West (270) vector intersect.

 

If you draw a square, centered at the center of the circle, with the upper right corner of the square being the point where the North and West vector intersect, all the vectors intersect all the other vectors somewhere to the upper right of the circle and to the lower left of this corner point.

 

The "center" of the turn therefore may be considered the average position of these vector intersecting points, which by rough estimation is about 80% the way between the center of the circle and the square corner point. This as opposed to the roughly 70% of the way out that the bike is, halfway through the turn, when it is directly between the center of the circle and the square corner point.

You don't get to redefine what is meant by a circle. The center of the circle is the center of the circle. The acceleration vector points to there. If you have a different answer you are simply wrong. There really can be no debate about this.

 

Each of the directional vectors found by drawing tangents to the circle at each degree position of the bike and rider has its parallel conterpart going radially outward in the NW quadrant of the circle.

 

The rider leans into the turn. She leans left, toward the center to counteract something carrying her and the bike outward.

The rider leans to facilitate a force pushing her in, allowing her to move in a circle. If the rider does not lean, she goes straight.

 

 

But, while I admit I am doing something wrong here, and the actual centripedal acceleration is being caused by the reaction of the tire and the road, and the rider is leaning merely to maximize these forces, allowing the change in direction to take place, I am still "speculating" as to why it "appears" that there is a radially outward "thing" going on during the turn.

Because humans are not well-calibrated instruments and are pretty easily fooled. We are used to inertial systems, and when we aren't in one, we often get confused.

[tar]

SwansonT,

 

Circles are confusing.

 

I was not so much trying to redefine the circle, as trying to figure if it made any sense to consider a relationship between moving radially outward and moving tangentially to the circle, since the one direction becomes the other, 90 degrees later.

 

Instantaneously speaking you might be able to say a direction is either tangent to the circle or normal to the tangent line, but in terms of "feeling" a force, the feeling comes over time.

 

I understand that if an inertial body maintains its speed but changes direction an acceleration must be occurring. In the case of the bike rider this force is applied radially inward, turning her momentum left, in the example. But that impulse to keep going forward, that is being conteracted by the force, the reaction of the tire against road, is a real impulse. The inertia is real. And this inertia starts heading to 360 and winds up heading to 270. The inertia lines at each point in the turn are parallel to 90 degrees worth of radial lines. If you would make a complete circle a couple of times and draw lines in all the directions tangent to the circle, and all the directions radially inward from the rider and bike, and all the directions radially outward from the rider and the bike, you would get three identical sets of directions.

 

A vector is a direction and a magnitude, you can not get far unless you take some time to do it. An instantaneous figuring does not allow the human instrument the time it takes the human instrument to sense and remember and compare. It might "feel" like you are being pulled outward, because in total, you are being pulled outward. Regardless of the forces causing a change in your direction of movement when figured instantaneously.

 

Regards, TAR

 

The spring in Mikes spinning bob example is pulling inward. But something is stretching the spring. Which "looks like", over time, an outward force.

Edited April 20, 2015 by tar

[swansont]

Instantaneously speaking you might be able to say a direction is either tangent to the circle or normal to the tangent line, but in terms of "feeling" a force, the feeling comes over time.

The it's wrong. Acceleration is the instantaneous change in velocity per unit time.

 

 

I understand that if an inertial body maintains its speed but changes direction an acceleration must be occurring. In the case of the bike rider this force is applied radially inward, turning her momentum left, in the example. But that impulse to keep going forward, that is being conteracted by the force,

The momentum is perpendicular to the force. They do not "counteract" each other.

 

the reaction of the tire against road, is a real impulse. The inertia is real. And this inertia starts heading to 360 and winds up heading to 270.

Yes, and the change in inertia is directed toward the center of the circle.

 

The inertia lines at each point in the turn are parallel to 90 degrees worth of radial lines. If you would make a complete circle a couple of times and draw lines in all the directions tangent to the circle, and all the directions radially inward from the rider and bike, and all the directions radially outward from the rider and the bike, you would get three identical sets of directions.

Because that's how geometry works, not because there's an outward force. Radially out and radially in lines will overlap. But the net force is toward the center.

 

A vector is a direction and a magnitude, you can not get far unless you take some time to do it. An instantaneous figuring does not allow the human instrument the time it takes the human instrument to sense and remember and compare. It might "feel" like you are being pulled outward, because in total, you are being pulled outward. Regardless of the forces causing a change in your direction of movement when figured instantaneously.

Except you aren't. The net force is inward, and it must be so.

 

You are arguing a lost cause here. No amount of hand-waving is going to overturn the math.

[tar]

SwansonT,

 

You

 

The it's wrong. Acceleration is the instantaneous change in velocity per unit time.



The momentum is perpendicular to the force. They do not "counteract" each other.


Yes, and the change in inertia is directed toward the center of the circle.


Because that's how geometry works, not because there's an outward force. Radially out and radially in lines will overlap. But the net force is toward the center.


Except you aren't. The net force is inward, and it must be so.

You are arguing a lost cause here. No amount of hand-waving is going to overturn the math.

Except in our example the acceleration is not a change in velocity, its a change in direction.

 

The force is counteracting the momentum of the body, which would take it in a straight line tangentially away from the circle.

 

The net force is toward the center but the net movement is in a path along the circle, which is the resultant path of the tangential (outward) momentum and inward force of the tires against the road.

 

I am not trying to overturn any math. Just trying to explain why a person feels like they are being pulled away from the center of the circle.

 

Regards, TAR

[swansont]

Except in our example the acceleration is not a change in velocity, its a change in direction.

 

Velocity is a vector, so a change in direction is a change in velocity. This is a concepts that trips up many students when they encounter the concept. Speed (a scalar) does not change (which is because the force is perpendicular to the velocity, and does no work. It's all self-consistent)

 

The force is counteracting the momentum of the body, which would take it in a straight line tangentially away from the circle.

Force and momentum are different beasts. Force changes momentum, but it does not "counteract" it. The change in momentum is directed toward the center of the circle, thus the force is directed there.

 

 

The net force is toward the center but the net movement is in a path along the circle, which is the resultant path of the tangential (outward) momentum and inward force of the tires against the road.

Tangential is not outward. It is tangential.

 

I am not trying to overturn any math. Just trying to explain why a person feels like they are being pulled away from the center of the circle.

It's because people are not well-calibrated instruments and can be fooled. In this case, because they are in a rotating frame of reference and that's not what we're used to. Not because of any real physical effects going on.

[Strange]

I am not trying to overturn any math. Just trying to explain why a person feels like they are being pulled away from the center of the circle.

 

I suppose that is quite an interesting question. I assume (in addition to swansont's comments) that it is because our brains are designed to predict certain types of motion very accurately - this is why we can throw and catch balls, jump across obstacles, etc. We obviously don't do this by solving the equations of motion. But those same predictive heuristics fail when faced with situations that would not have been encountered earlier in our evolution: accelerating in a car at a good proportion of 1g, being thrown round a corner at high velocity, and so on.

[swansont]

 

I suppose that is quite an interesting question. I assume (in addition to swansont's comments) that it is because our brains are designed to predict certain types of motion very accurately - this is why we can throw and catch balls, jump across obstacles, etc. We obviously don't do this by solving the equations of motion. But those same predictive heuristics fail when faced with situations that would not have been encountered earlier in our evolution: accelerating in a car at a good proportion of 1g, being thrown round a corner at high velocity, and so on.

 

Exactly the point I was trying to make. Being in a rotating frame fakes us out. But the physics tells the truth.

[tar]

SwansonT,

 

My mistake. A change in direction at a constant speed IS a change in velocity.

 

But I am still stuck on tangential and radially outward, in terms of the fact that both directions take you away from the center of the circle. Not that I don't realize that the two directions are orthogonal to each other, but if you put a lifesaver on the ground and travel at 30 miles an hour tangentially to it, after a minute, you will be half a mile radially outward from the center of the lifesaver.

 

Our being fooled by the motion in a circle, might be related to the fact that tangential motion results in a separation from the center as surely as radially outward motion does. Just at a different rate of separation. Significantly different rate at first, but as you travel a distance greater than the diameter of the circle, the two directions both take you away from the center of the circle at an increasingly similar rate.

 

In the round-about example you wind up rolling/running/sliding away from the thing when you fall off, and in the bike example, you wind up off the road to the right. In either case you can draw a radial line, from the center of the circle, to where you lay on the ground. You can also draw a tangential line from where you lay to where you left the circumference of the circle, but where you would have been, still upright on the round-about or bike, had you not fallen, is on the radially outward line, where you lay.

 

Regards, TAR

Perhaps this "where you would have been" is important in explaining why the tangential launch, feels like a radial motion. Watching the kid you were crouching next to when you fell off, stay between you and the center of the thing, as you increase you distance from the center as you run/fall off, makes it seem like a radially outward manuver.

Edited April 23, 2015 by tar

[swansont]

But I am still stuck on tangential and radially outward, in terms of the fact that both directions take you away from the center of the circle. Not that I don't realize that the two directions are orthogonal to each other, but if you put a lifesaver on the ground and travel at 30 miles an hour tangentially to it, after a minute, you will be half a mile radially outward from the center of the lifesaver.

True but inconsequential.

 

Our being fooled by the motion in a circle, might be related to the fact that tangential motion results in a separation from the center as surely as radially outward motion does. Just at a different rate of separation. Significantly different rate at first, but as you travel a distance greater than the diameter of the circle, the two directions both take you away from the center of the circle at an increasingly similar rate.

 

In the round-about example you wind up rolling/running/sliding away from the thing when you fall off, and in the bike example, you wind up off the road to the right. In either case you can draw a radial line, from the center of the circle, to where you lay on the ground. You can also draw a tangential line from where you lay to where you left the circumference of the circle, but where you would have been, still upright on the round-about or bike, had you not fallen, is on the radially outward line, where you lay.

That radial line will not have been your path. Motion and force are not the same thing, and the presence of either one does not imply the other is present.

[Mike Smith Cosmos]

I am out on the road doing real live experimental science on my 250cc Madison Italian motorbike .

 

I am going in straight lines. ( I can report first hand on the observation of Inertia . At slow and fast speed , and no speed )

I am going around circular paths ( I can report on leaning , either the bike only or the bike and me. . The feelings are distinctly different )

 

Report later :

 

-

 

Three observations.

 

1) Starting from being still. Stationary. The bike is heavy , and anything other than straight upwards is hard to keep balanced. Also very difficult to hold it upright if I lean a little.

At any speed , slow, medium or fast.

 

 

Mike

Edited April 23, 2015 by Mike Smith Cosmos

[Endy0816]

How much of this force would continue to feel if you jumped off while making the turn?

[Robittybob1]

How much of this force would continue to feel if you jumped off while making the turn?

In what direction are you going to jump? Please for me jump straight up and down.

[Mike Smith Cosmos]

I am out on the road doing real live experimental science on my 250cc Madison Italian motorbike .

I am going in straight lines. ( I can report first hand on the observation of Inertia . At slow and fast speed , and no speed )

I am going around circular paths ( I can report on leaning , either the bike only or the bike and me. . The feelings are distinctly different )

Report later :

-image.jpg

Three observations.

1) Starting from being still. Stationary. The bike is heavy , and anything other than straight upwards is hard to keep balanced. Also very difficult to hold it upright if I lean a little.

At any speed , slow, medium or fast.

Mike

The results , observation , of the straight line (test 1 ).

As stated above , when stationary ( a ) Stationary. The bike is heavy , and anything other than straight upwards is hard to keep balanced. Also very difficult to hold it upright if I lean a little. ( quite a heavy bike )

( b) At any speed , slow, medium or fast. Here the inertia ,phenomenon kicks in . The bike definitely and sensed in/ by me the driver , wants to go straight . If I attempt to lower the bike from its up right condition , as when I am stationary , it does not want to lean over. The faster I go , the harder it gets to lean over. In fact at high speed it feels like all it wants to do is ' FLY' . In other words , it's as if I could raise the wheels like a plane does on take off , it would fly. (Above the ground). This is no doubt, why motor cyclists get such a bus in driving a motor bike fast , up the motorway ( to car motorists disgust ) . It gives you the motorcyclist a real buzz, as if you are somehow safe at speed. Of course this is not true , but nonetheless less , that is the feeling . ( children gets a similar experience , when they first learn to ride a bike . And often fall off , and graze their knees ) we all remember that, I still have the scars)

 

What I don't understand , is why we do not feel this while stationary , and can't even balance , because after all , we are travelling at approximately 1000 mph on the earths surface , as it spins ( once every 24 hours) . We must carry that inertia , why do we not have the same feeling I get when on the motor bike at say 70mph now possibly 1070 mph if same direction of spin of earth? Is the inertia only relevant to the nearby Earth or is it absolute? If so where is the absolute ? If not where is the Reference? The ground, the gravitational field just nearby ?

 

Observations ( 2 & 3 ) traveling on the motorcycle in a curve around part of a circular motion circle .(TO FOLLOW )

 

Mike

Edited April 25, 2015 by Mike Smith Cosmos

[swansont]

 

What I don't understand , is why we do not feel this while stationary , and can't even balance , because after all , we are travelling at approximately 1000 mph on the earths surface , as it spins ( once every 24 hours) . We must carry that inertia , why do we not have the same feeling I get when on the motor bike at say 70mph now possibly 1070 mph if same direction of spin of earth? Is the inertia only relevant to the nearby Earth or is it absolute? If so where is the absolute ? If not where is the Reference? The ground, the gravitational field just nearby ?

 

It's relative. We don't notice our motion because everything is moving at that speed. Except for large-scale phenomena (weather, or things moving many miles in a short period of time) the fact that we're rotating is not apparent to us, and we are effectively in an inertial system.

[Mike Smith Cosmos]

It's relative. We don't notice our motion because everything is moving at that speed. Except for large-scale phenomena (weather, or things moving many miles in a short period of time) the fact that we're rotating is not apparent to us, and we are effectively in an inertial system.

 

Yes .ok. We are in an inertial system , travelling somewhere at approx 1000 mph . Why do I not experience the same symptoms I feel on a motor bike ? In other words , I must be pointing in some direction or other . Why do I not feel like mainly going in that direction , rather than all the other 359 degrees of compass directions ? Also why do I not feel that lightness I feel on the motor bike , like I was nearly flying ? And why do I not feel , it is not going to be easy ,to be able to start a turn , left or right , as I do on the motor bike when I am going comparatively fast at 70 mph? Yet now I am going 1000 mph ?

 

Mike

Edited April 25, 2015 by Mike Smith Cosmos

[Robittybob1]

Yes .ok. We are in an inertial system , travelling somewhere at approx 1000 mph . Why do I not experience the same symptoms I feel on a motor bike ? In other words , I must be pointing in some direction or other . Why do I not feel like mainly going in that direction , rather than all the other 359 degrees of compass directions ? Also why do I not feel that lightness I feel on the motor bike , like I was nearly flying ? And why do I not feel , it is not going to be easy ,to be able to start a turn , left or right , as I do on the motor bike when I am going comparatively fast at 70 mph? Yet now I am going 1000 mph ?

 

Mike

It must be because everything is moving along with you at the same rate whereas on the bike you are moving (relatively) and everything else is not.

Edited April 25, 2015 by Robittybob1

[swansont]

Yes .ok. We are in an inertial system , travelling somewhere at approx 1000 mph . Why do I not experience the same symptoms I feel on a motor bike ? In other words , I must be pointing in some direction or other . Why do I not feel like mainly going in that direction , rather than all the other 359 degrees of compass directions ? Also why do I not feel that lightness I feel on the motor bike , like I was nearly flying ? And why do I not feel , it is not going to be easy ,to be able to start a turn , left or right , as I do on the motor bike when I am going comparatively fast at 70 mph? Yet now I am going 1000 mph ?

 

Mike

 

When you're on a bike, you have to cut through the air that surrounds you; that's one way you know you're moving, unlike in a car, train or large boat/ship traveling under calm conditions, where you are possibly shielded from the elements. On a bike you have to balance, because you are in a state of unstable equilibrium, and that's easier to do while moving. And when you turn you notice it because that means you are no longer in an inertial frame. Acceleration is not relative.

[tar]

Mike,

 

The stability when moving on the bike has to do with the gyroscopic effect the spinning wheels have. Here another inertia is taking place and the spinning mass of the wheels wants to stay in the orientation they are in. I have had a small gyroscope since I was young that I still play with. Pull the string and get it spinning fast and you can "feel" it trying to keep its orientation as you try and tilt it . A body in motion...

 

Regard, TAR

With my little gyroscope the resistance the spinning thing has to my attempted tilt, "feels" like a force resisting my attempted tilt. That the inertia of a mass, does not count as a force, when it feels so much like one, is the thing that is at the crux of the question of this thread.

As an aside, I had a thought 15 or 20 years ago that you could design a flywheel system made up of larger and larger rings where you could store large quantities of inertial energy if you got the whole complex spinning really fast...then I figured...wait a minute, the vehical wouldn't want to turn, because you have this giant gyroscope going. Then I thought you could have three flywheels at different orientations counter acting each other...then I figured...wait a minute, then the thing wouldn't want to move.

The spinning space station creates a radially outward "force" that feels like gravity pulling in the direction of a spot on the outside circumference directly opposed to the axis of rotation, or radially outward.

[swansont]

 

The spinning space station creates a radially outward "force" that feels like gravity pulling in the direction of a spot on the outside circumference directly opposed to the axis of rotation, or radially outward.

 

No, it doesn't. You feel a force radially inward — the space station pushing on you.

[tar]

SwansonT,

 

Then why, when I wheigh myself, is it not the scale pushing radially outward on my feet?

 

Regards, TAR

[swansont]

SwansonT,

 

Then why, when I wheigh myself, is it not the scale pushing radially outward on my feet?

 

Regards, TAR

 

 

It is the scale pushing out on your feet. The centripetal force of you on the surface of the earth is much smaller than the gravitational force, so there must be another force present, and acting in the opposite direction. If you were moving faster, the scale reading would drop, and go to zero if you achieved the orbital speed.

[tar]

SwansonT,

 

Definitionally, how close to the Earth's equator can you be an object in geosynchronis orbit, before you are an object standing on the rotating surface?

 

Regards, TAR

[Mordred]

Here is how to calculate a geosynchronous orbit.

 

http://m.dummies.com/how-to/content/how-to-calculate-the-period-and-orbiting-radius-of.html

Edited April 29, 2015 by Mordred