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

[swansont]

Surely there is a conflict here ,between what we see and feel . And what some mathematical structure somehow says " there is nothing there!" It's fictitious . When our senses tell us something that we know and feel is happening to us . If we were in a cardboard box we would feel ourself moving toward the edge of the. Roundabout, as if some other child was pushing the box . Surely by the law of equivalence , then it is so!

Mike

That's a reason why we generally don't use personal observation in place of instruments in science.

[Acme]

I don't get it ! ...

Mike

That's the crux of the matter Mike. Your getting it has no bearing whatsoever on whether or not it is true. Clearly others do get it and yet you have the temerity to argue with them. After all this time and many threads it appears that you never will get it; just let it go.

[swansont]

!

Moderator Note

Posts about earth deformation and gravity have been split

http://www.scienceforums.net/topic/88621-nonspherical-earth-split-from-centrifugal-forces/

[Mike Smith Cosmos]

That's a reason why we generally don't use personal observation in place of instruments in science.

Not sure what you mean by instruments ( do you mean Newton meters measuring scale or do you mean mathematical model )

Because I would have thought if observation showed one thing and maths contradicted it by its model . Surely that shows a good case the maths model could be wrong .

 

If you mean , measuring instrument ( instrument in science ) says one thing , whereas observation says another , then that is different.

 

I have the feeling there is a ( from which perspective you view things ) . If all my life from roundabouts to swings to bicycles to motor bikes to wall of death , and a lot more besides ,I have seen and felt the force , ' centrifugal force '. Then I am now wrestling with a maths model that says " there is no centrifugal force pushing out like I feel it is , it's fictitious " . Then my reaction is to say " I just don't get it " . This conflict of inertial straight line motion to circular motion , to me seems to generate , this from the centre force feeling ( centrifugal force feeling ) . As regards measuring it . In the school physics labs they have a bob that you can swing around fast . It is connected to a string , this goes vertically down ." Inside is a spring with a needle indicating newtons . You twirl the weight around , it pulls outward on the string. Goes down the tube to a simple Newton meter. Shows a few newtons of centrifugal force .

 

So you feel it you measure it .

 

 

Mike

Edited April 14, 2015 by Mike Smith Cosmos

[studiot]

 

mscosmos

it pulls outward on the string. Goes down the tube to a simple Newton meter. Shows a few newtons of centrifugal force .

 

 

 

Or the string is pulling Bob inwards.

 

 

Much more satisfactory.

[MigL]

Take one of those little self-propelled toy cars, Mike. The kind that only go in a straight line.

Now wind it up ( or whatever ) and set it down about a foot away, and offset, from a pole in the ground.

The car will travel in a straight line, past the pole.

 

Now introduce some form of attraction between the pole and the toy car, like gravity, or any other centripetal force.

For our case we will use a string tied between the car and the post. What happens when we repeat the previous experiment ?

 

We now see the string go taut and the toy car is constrained to travel a circular path around the post.

 

From your mistaken point-of-view ( and any occupant of the toy car ) there is a force which is making the string go taut.

Everyone else on this forum is telling you that this centrifugal force, which seems so real to you, is fictitious and only apparent in that accelerated ( spinning ) frame of reference.

In the inertial frame of the person who sets the tied off car in motion, there is only the toy car's inertia and the string's centripetal force.

Edited April 14, 2015 by MigL

[swansont]

I meant that humans are, in general, poorly calibrated and easily fooled

[Mike Smith Cosmos]

 

Or the string is pulling Bob inwards.

 

 

Much more satisfactory.

Yes but surely that is just ' arbitrary ' .

 

My remembering the device . The thing gets cranked up. Spinning . Then by a rhythmic imperceptible synchronous wobble the rotation is up to operating speed. You can see the top of the spring , being tugged at. The spring is stretched. The cause , the link to the bob, the bob itself is pulling at this linkage . It can only be pulling one way , otherwise the linkage would not be taught . It can't push in , it would all buckle up . But it can and does pull outward , away from the centre , centrifugally. No ?

 

Mike

[studiot]

You have already described the correct interpretation in the first part.

 

Think about what happens before you start and during the 'crank up'.

 

Which way do you direct the bob when you start it spinning and which way do you jiggle it to keep it spinning?

 

That is the key and equivalent to Mig and his toy car.

[Mike Smith Cosmos]

Take one of those little self-propelled toy cars, Mike. The kind that only go in a straight line.

Now wind it up ( or whatever ) and set it down about a foot away, and offset, from a pole in the ground.

The car will travel in a straight line, past the pole.

 

Now introduce some form of attraction between the pole and the toy car, like gravity, or any other centripetal force.

For our case we will use a string tied between the car and the post. What happens when we repeat the previous experiment ?

 

We now see the string go taut and the toy car is constrained to travel a circular path around the post.

 

From your mistaken point-of-view ( and any occupant of the toy car ) there is a force which is making the string go taut.

Everyone else on this forum is telling you that this centrifugal force, which seems so real to you, is fictitious and only apparent in that accelerated ( spinning ) frame of reference.

In the inertial frame of the person who sets the tied off car in motion, there is only the toy car's inertia and the string's centripetal force.

I am not sure I could have , or would even now , trust ,believing there was not some outward pushing force during rotation.

When I played on those circulating playground rotary roundabouts. No way could I let go while it spun fast . I would be thrown off , to disastrous consequences ( this is survival ) .

 

Similarly , when I am zooming along at 70 mph on my motor bike , no way , when seeing a corner coming up , going to think ' there is no force acting outward away from the centre of the corner'. I know , having come off to my cost in the past, that there is a dangerous real force away from corner centre, so I lean in desperately toward the centre , in a hope there will be enough inward pull by my levered angle , to combat that outward push . I cannot change this mental paradigm now, it's survival stuff. I will have a bad accident if I give up this mental paradigm. ( this is survival ) .

 

I think it must be the constant change, or ' battle with ' inertia ' ( away from straight line to circular ) , that circular motion causes this effect ?

 

Mike

Edited April 14, 2015 by Mike Smith Cosmos

[pzkpfw]

Cool, a motorcycle analogy.

 

Imagine you are zooming along a straight piece of road, say, heading due North. You come to a corner, turning sharp to the left (West).

 

Oh no- there's ice. You can't turn. Which way will you go? You'll go straight ahead, not West, not East, but North.

 

If there was instead no ice, and your tyres were able to make your bike turn to the West - yes, you do seem to feel a force. But that's just the "desire" of your body to continue going the way it was. North. To make your body instead go to the West, your bike is exerting a force on you, towards the West - inwards, relative to the circle of your turning path.

 

There's no force East, or, outwards relative to the circle of your turning path (yes, other than your body resisting the turn, and wanting to go straight ahead)

Edited April 14, 2015 by pzkpfw

[MigL]

Motorcycle analogies always seem to work best.

[swansont]

Yes but surely that is just ' arbitrary ' .

My remembering the device . The thing gets cranked up. Spinning . Then by a rhythmic imperceptible synchronous wobble the rotation is up to operating speed. You can see the top of the spring , being tugged at. The spring is stretched. The cause , the link to the bob, the bob itself is pulling at this linkage . It can only be pulling one way , otherwise the linkage would not be taught . It can't push in , it would all buckle up . But it can and does pull outward , away from the centre , centrifugally. No ?

Mike

It doesn't matter.

 

F = ma Do you understand that equation?

 

The acceleration of an object is dependent on the net force acting ON the object. The force exerted BY the object DOES NOT MATTER. There is no outward force acting ON the bob.

[Mordred]

Key word NET.

[Mike Smith Cosmos]

Key word NET.

Do you know I think I have it!

 

It doesn't matter.

F = ma Do you understand that equation?

The acceleration of an object is dependent on the net force acting ON the object. The force exerted BY the object DOES NOT MATTER. There is no outward force acting ON the bob.

I think I am getting the picture!

 

Cool, a motorcycle analogy.

 

 

 

Imagine you are zooming along a straight piece of road, say, heading due North. You come to a corner, turning sharp to the left (West).

 

Oh no- there's ice. You can't turn. Which way will you go? You'll go straight ahead, not West, not East, but North.

 

If there was instead no ice, and your tyres were able to make your bike turn to the West - yes, you do seem to feel a force. But that's just the "desire" of your body to continue going the way it was. North. To make your body instead go to the West, your bike is exerting a force on you, towards the West - inwards, relative to the circle of your turning path.

 

There's no force East, or, outwards relative to the circle of your turning path (yes, other than your body resisting the turn, and wanting to go straight ahead)

Something is sinking in !

 

Motorcycle analogies always seem to work best.

I believe I am getting ,what is going on here !

 

--------------------

 

If I think of what is really important to me . Me on the children's roundabout , or me on my motorcycle . And what could happen if I fall off the roundabout or fall off the motor cycle . This above all else , that is going on outside my circle , my bend , that is outside in some other sphere of operation , possibly what you say another frame . Absolute frame maybe.

 

I don't care if there are strait lines or tangents or fictitious forces , what I am worried about is flying off the roundabout , off my bike onto the Tarmac.

 

So!

 

I must work on the roundabout to keep myself in the circle of the roundabout , I hang on , pulling myself constantly to the centre. Centripetally . I do get an outward force centrifugal force equal and opposite to the inward force. So to me in my close world, the one I am most concerned about , am in a state of force equilibrium , about my immediate world of ( the edge of the roundabout, and the centre of the roundabout. By hanging on , in my little 'frame ' from the edge ,to the centre. The two forces centripetal and centrifugal equal and opposite , cancel out . There is no net force now ( provided I hang on ) on my rotating radius (frame) , so along the radius there is no net force therefore , no net acceleration ( from f=ma =0 , along the radius) .

 

Now out in the big wide world , things appear completely differently. A bigger frame , maybe absolutely big . Then from that perspective , everything looks completely different . Forces disappear, possible strait lines appear, things going down small radii are unimportant , more absolute things are important !

 

Similarly with the motorcycle , I do have to balance my forces , for me it is critical ,that I do not leave the circle , and slide along the Tarmac in a straight line, to my hurting , possibly critical cost. So again I must develop this outward force of centrifugal with the centripetal to keep me in equilibrium ( stay on my motor bike ) , keep it going in a circle , or at least for the corner. Then I can afford to let up , and go in a straight line along the straight road.

 

So if I have it right . If we want to stay in a circle , we must balance the centrifugal and centripetal forces . Closed frame of operation . No net acceleration along the radius , neither in or out . If there were an acceleration along the radius, there would be movement, velocity, and the radius would get smaller, I would move closer to the centre , no , I hang on , neither moving in or out along the radius , staying at the extremity of the radius ,

 

If we wish to consider the much bigger picture . Outside the circle , some large or absolute frame , then all these peculiarities that have given me so much grief , have to be considered.

 

Is that about it ?

 

Mike

Edited April 15, 2015 by Mike Smith Cosmos

[Mordred]

Yep your getting it change in momentum requires an imbalance of net forces upon an object.

[Mike Smith Cosmos]

Yep your getting it change in momentum requires an imbalance of net forces upon an object.

Can you say that again in words that an idiot would understand ! Namely me !

 

Mike

[Mordred]

An old trick , to teach the math USE a compass and a ruler. Take 9.8 metres/cm^2. Draw a line 90 degrees,measuring 9.8 cm. Now add another vector force. Use the bottom end of the above line. Draw that line according to its direction and acceleration. Then connect the top of the first line (gravity). With the end of the second line( random vector acceleration). Measure the angle and distance . You now have the Vector sum.

Can you say that again in words that an idiot would understand ! Namely me !

 

Mike

Let's start with an object on a flat surface. I push left 1 Newton and push right two Newton of force. The vector sum is 1 Newton in the right direction.

When you want to ad vector angles the ruler compass trick helps.

Edited April 15, 2015 by Mordred

[studiot]

 

Take 9.8 metres/cm^2.

 

Is this a typo for 9.8 metres per second² ?

[Mordred]

Lol yeah oops

[tar]

Thread,

 

So with the motorcycle going North and the road turns left the cyclist leans left to put the bike between him and the direction his momentum is taking him. There is, as everybody is saying, no force pulling him East, but the force pulling him North, his momentum, his inertia, becomes a force to his right, as soon as he points the bike a little left of North. The Northward force becomes a force to the right. Once he is headed West the Northward momentum he is carrying that is pushing the tire into the pavement is overcome by the solid friction with the road, and the momentum that was heading North, is redirected to the West. If there is gravel on the road, or ice as in the example, the bike and rider will not be able to redirect the momentum to the West, and the bike and rider will continue in a Northern direction, which will be 90 degrees to the right of West, or radially outward from the center of the turn.

 

Regards, TAR

[swansont]

Thread,

If there is gravel on the road, or ice as in the example, the bike and rider will not be able to redirect the momentum to the West, and the bike and rider will continue in a Northern direction, which will be 90 degrees to the right of West, or radially outward from the center of the turn.

 

In that case there is no force, and north is not "radially outward". At the beginning of the turn, "radially outward" would be east. North is only radially outward after the rider is going west, 90 degrees into the turn

[tar]

SwansonT,

 

So what direction is the rider's momentum carrying him in, if the turn would be a hairpin one and the road would start turning to the West and wind up heading South? Northern momentum would pull him first to the right and increasingly to the right as he got further into the turn. As the Northern momentum was redirected to the West, the Western momentum would be redirected to the South as the rider's lean and the friction of the tires conspired to change the direction of the momentum of the moving mass of bike and rider.

 

Radially outward would be the direction the bike and rider would want to go if the rider was in the barrel at the county fair. This radially outward force would allow the bike and rider to ride up the wall against the acceleration of gravity.

 

Regards, TAR

SwansonT,

 

Once, before I hurt my knee, during the summer between my 11th and 12th grade years, I ran in a field behind the highschool at world class speed. I had been second fastest in my 11th grade class, and ran faster than I had ever run before, by considering that a body in motion tended to stay in motion, unless acted upon by an outside force. I started sprinting and once I had established my normal top speed, I kept my stride and increased the pace of my footfalls just a bit. Once I gained the new momentum, I increased my stride a bit by concentrating on the push as the ball of my foot hit the ground. As I reached a new speed by increasing my stride, I ran faster again by maintaining the stride, but increasing the pace of my footfalls. I was flying along so fast I thought I would hurt myself if I fell, and I started to run out of good field, and got into an area that was a little rougher and less well kept than the playing fields, so I slowed to a stop.

 

I asked myself why I could not use this technique to run faster than everybody else on the track. The answer was, because the technique only works in a straight line. Tracks turn. The momentum you can add to, while running in a straight line is your friend. On a track you have to use strength and energy to turn your momentum to the left. The technique does not work on a track.

[swansont]

SwansonT,

 

So what direction is the rider's momentum carrying him in, if the turn would be a hairpin one and the road would start turning to the West and wind up heading South? Northern momentum would pull him first to the right and increasingly to the right as he got further into the turn. As the Northern momentum was redirected to the West, the Western momentum would be redirected to the South as the rider's lean and the friction of the tires conspired to change the direction of the momentum of the moving mass of bike and rider.

 

Radially outward would be the direction the bike and rider would want to go if the rider was in the barrel at the county fair. This radially outward force would allow the bike and rider to ride up the wall against the acceleration of gravity.

 

The momentum is always in the direction you are moving. That is not radially outward.

 

Draw a diagram. Do the math.

[tar]

SwansonT,

 

So what direction are you moving when you are going in a 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.

 

Regards, TAR

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.

 

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.