Centrifugal Force: Is it Radially Outwards?

[D H]

There are two forces acting on the stone: The outward horizontal centrifugal force Fhoriz; and the downward gravitational force Fvert. When you twirl a stone around your head in a Goliath sling, the downward slope of the string is about tan(Fvert/Fhoriz).

Regardless of the status of the nomenclature imbroglio over whether there is such a thing as a reactive centrifugal force, there is no real centrifugal force acting on the rock, period. The net force acting on the rock *must* be inwards for the rock to undergo circular motion.

 

[rcgldr]

Regardless of the status of the nomenclature imbroglio over whether there is such a thing as a reactive centrifugal force, there is no real centrifugal force acting on the rock, period. The net force acting on the rock *must* be inwards for the rock to undergo circular motion.

But that's the point of calling the foce the rock exerts on the string a "reactive" force,: it's the inertial reaction due to the acceleration caused by the real net force. When the real forces don't cancel, then there is a net real force, and the acceleration corresponds to the net force / the inertia (mass in the linear case), and the "reactive forces" are simply a means to describe the reaction to the net forces in compliance with Newtons 3rd law about forces only existing in pairs. It's like a method of accounting to get the forces to sum up to zero, the real forces and the reactive (to acceleration) forces.

 

[Dadface]

But that's the point of calling the foce the rock exerts on the string a "reactive" force,: it's the inertial reaction due to the acceleration caused by the real net force. When the real forces don't cancel, then there is a net real force, and the acceleration corresponds to the net force / the inertia (mass in the linear case), and the "reactive forces" are simply a means to describe the reaction to the net forces in compliance with Newtons 3rd law about forces only existing in pairs. It's like a method of accounting to get the forces to sum up to zero, the real forces and the reactive (to acceleration) forces.

Newtons third law does have the forces occurring in pairs but it also requires the forces to be :
1.of the same type
2.equal in size
3.opposite in direction
The above three criteria are not met when considering the centripetal force on the stone and any pull on the string.

 

[elect_eng]

But that's the point of calling the foce the rock exerts on the string a "reactive" force,: it's the inertial reaction due to the acceleration caused by the real net force. When the real forces don't cancel, then there is a net real force, and the acceleration corresponds to the net force / the inertia (mass in the linear case), and the "reactive forces" are simply a means to describe the reaction to the net forces in compliance with Newtons 3rd law about forces only existing in pairs. It's like a method of accounting to get the forces to sum up to zero, the real forces and the reactive (to acceleration) forces.

The only points of talking about the reactive force (for me) is to acknowledge one of Newton's laws and to name a force reponsible for internal stresses in a rotating system. The outer parts of the rotating system place a force on the inner parts. I won't even use the "bad" word here. Just acknowlege that the force is there and is sometimes important.

However, there is no balancing of outward and inward forces on the stone in this example. There is NO centrifugal force on the stone! This makes sense because there is no equilibrium in the horizontal plane. The vertical forces are in equilibrium so you want the upward component of string tension to balance gravity. But you do not want the horizontal force to balance, otherwise the stone wouldn't move and we would all be wondering why the stone was just floating there.

 

[D H]

But that's the point of calling the foce the rock exerts on the string a "reactive" force,: it's the inertial reaction due to the acceleration caused by the real net force.

This gets to the heart of the problem. The net force is not necessarily a "real" force, defined as a force with a causative agent. The only case in which the net force is a real force are the trivial case in which zero or one real forces act on an object.

Newton's third law applies to real forces with causative agents, and only to these real forces. Pretending that the net force is a real force can get you in trouble. It might, for example, make you think there exists a third law reaction against it. This particular problem exemplifies that. The third law reactions are the rock pulling the Earth upwards and the rock pulling the string outwards and downwards. There is no horizontal reactive force.

 

[Cantab Morgan]

Newtons third law does have the forces occurring in pairs but it also requires the forces to be :
1.of the same type
2.equal in size
3.opposite in direction
The above three criteria are not met when considering the centripetal force on the stone and any pull on the string.

What does "of the same type" mean? I've never heard that before.

 

[D H]

What does "of the same type" mean? I've never heard that before.

A specific example: Some people say that the upward force exerted by a scale on an object atop the scale is a third law reaction to the gravitational force exerted by the Earth on the object. This is erroneous. For one thing, the object on the scale is undergoing uniform circular motion about the Earth's rotation axis (the Earth is rotating, after all). The net force on the object on the scale cannot be zero, and thus the upward force exerted by the scale is not equal but opposite to the gravitational force.

Even more basic, the two forces are not of the same type. The downward gravitational force results from gravity. The upward force by the scale is something else: It is a manifestation of the electrostatic force. This is not a third law reaction force. It is something else.

 

[BobbyBear]

You mean with your hand (holding the string) below the stone?

Not possible unless you're cheating.

Um, so how exactly does one cheat? I mean, when we see that happening, what's really happening? o:

 

[tiny-tim]

Um, so how exactly does one cheat? I mean, when we see that happening, what's really happening? o:

Dunno … computer graphics, maybe?

Have you ever seen it happening?​

 

[BobbyBear]

Dunno … computer graphics, maybe?

Have you ever seen it happening?​

Um, I thought it was commonplace to see it happening, swinging a lasso over your head, like the cowboys do . . . in fact, I'm going to try doing it right now:P:P

-goes to swing rope around head-

-back-

Uhuh, to be honest I can't do it very well, the rope looks to be pretty much horizontal, but I guess that's because I'm trying to swing it very fast :P

So it's impossible to do? I mean, I thought when you said it would be 'cheating', I thought you meant that it could be done but by doing something we were not considering in our model, some addtitional effect we weren't considering, like some sort of upward force our wrist is exerting upon the rope (which is not present in the rotating shaft that has a string with a mass tied to it). I'm just asking, when we swing a rope with an upward tilt (if it's possible, lol, now I'm not so sure), we must be exerting an upward impulse or something? I was just wondering how you'd explain it, and how you'd model it . . . as an upward force?

 

[tiny-tim]

See if you can find a video of it …

I don't think you'll find one of the rope staying above the level of the hand.

 

[Dadface]

As you whirl the rope faster and faster it approaches the horizontal but never quite reaches.The mass on the rope has two main forces on it mg, and the tension in the rope(T).Tsin theta supports the weight and T cos theta provides the centripetal force(theta =angle rope makes to the horizontal).We can write:
Tsin theta=mg...1
Tcos theta=mv^2/r...2
Divide equation 1 by equation 2 to get a value for tan theta and you will see that for theta to become zero and the rope to become horizontal v must become infinite which is impossible.