I am not concerned about forces acting applied to the string.

It should be possible to draw a freebody force diagram, concerned only with the ball, replacing any interaction across the freebody diagram with an equivalent force.

I have showed you such a force diagram which is unbalanced. I did not say it was an equilibrium diagram, I put a question mark against horizontal equilibrium.

The ball is not in static equilibrium. There is a resultant force. This is a necessary condition of there being an acceleration.

It is a convenient mathematical fiction called quasi-equilibrium to supply a ficticious force to balance the resultant, for calculation purposes - because it works.

We return to what I said at the beginning about the equivalence of force and inertia. We only see (need) this ficticious force because our frame of reference is accelerating relative to the ball. Whenever this happens the calculations must supply a ficticious force determined by that acceleration. If we chose a frame of reference that was accelerating with the ball we would not need it, but the other real forces would still be acting.

This concept is known as Galilean relativity. This form of relativity was known long before Einstein.

 

What was the purpose of this question?

What did you want to show to us?

 

What was the purpose of this question?

The question actually arose in the first of these two threads and spilled over into the second. Several doubtful statements were made about Newton's Law.

http://forum.allaboutcircuits.com/showthread.php?t=23354

http://forum.allaboutcircuits.com/showthread.php?t=23716

The subject didn't really belong in either.

 

What did you want to show to us?

Any body which can generate a force that acts on itself could use this force to accelerate itself, thus creating perpetual motion.

Only Superman works like that.

In the real world the centre of mass of the system does not accelerate when a rocket flys.

 

So, Steve, you are suggesting that it is possible to for a body to apply a real force to itself?

Every force has an equal and opposite reaction. This means that a body can not apply a force to itself. If I push on my chest, my chest pushes back with equal force, so the net force on my body is zero and I don't accelerate or move.

 

My favorite author, Robert A Heinlein, made a similar point about orbits (as in satillites and moons). In these cases there is no centrifical force, the objects are in free fall. They are falling towards the earth, and forever missing.

Another thing the schools tend to teach wrong. I can forgive them though, it helps get the concept across, and is corrected by education more in depth. Kind of like AM modulation and sidebands.

 

Every force has an equal and opposite reaction. This means that a body can not apply a force to itself.

Exactly.

The ball's inertia will try to resist any change in velocity and so there is a reaction force outward. This is referred to as a reactive centrifugal force, and it is very real.

So I ask again. Where does this 'very real force' come from since it cannot come from the ball?


Eric, you need to be very careful using vectors in Newton calculations.

For instance a vector form of Newton's second law is

mr'' = F where bold denotes vectors.

Now momentum p = mr',

but we should avoid using p'' = F leading to

mr'' + mr' = F

a momentum conservation equation that can lead to incorrect answers in some cases eg the rocket system.

 

This force is not coming from the ball itself but from the motion of the ball.

It is like the electric field created by an electron but the magnetic field is created only when the electron moves.

 

Hello Mik, this is becoming a bit of a repetitive dialog, where you keep offering explanations for the origins of a mysterious fourth force.

There is no force.


Let us look at it another way, but introduce this mythical fourth force, call it F4

Vertically applying Newtons 2nd Law

Force = Tcosθ - W = mass x acceleration = 0

Since the mass of the ball is not zero, acceleration = 0
and everything vertical is hunky dory.

Horizontally, applying Newtons 2nd Law

Force = Tsinθ - F4 = mass x acceleration

What value do you place on F4?

One further question;
Would you accept that any real force must exist in any coordinate system I use?

 

Exactly.

So I ask again. Where does this 'very real force' come from since it cannot come from the ball?

Reaction centrifugal force does come from the ball. However, it is not from the ball to the ball. Instead, it is from the ball on to the string.

Reaction centrifugal force comes from the same principle: "Every force has an equal and opposite reaction force". If the string is applying an inward force to the ball, then the ball is applying an equal and opposite (i.e. outward) force to the string.

The issue is not just to make a free body diagram for the ball, but to also consider that every line element in the string needs its own free body diagram. The last line element (of the string) that connects to the ball is being pulled inward by the string tension. The other line elements have string tension on both sides. This continues all the way down the string until the first line element that attaches to the center. The central point then has a force to. The string pulls on the center, and the center pulls on the string.

So let me ask a question too. If the centrifugal force is not real, how does the string have tension? Doesn't a string need to be pulled from both sides to have tension?

 

I agree with Steveb.

 

If the string is applying an inward force to the ball, then the ball is applying an equal and opposite (i.e. outward) force to the string.

Yes perfectly true. And my equations reflect this.

But we are not considering the equilibrium or otherwise of the string. We are considering the equilibrium or otherwise of the ball.

Since you have joined forces (smile) with mik can you between you put some figures to this force I have called F4 so we can put it into Newton's Law and see what happens?

I maintain that F4 = 0 so that Newton's Law produces the correct inward acceleration as noted by Eric.

 

Yes perfectly true. And my equations reflect this.

Your equations may reflect this, but your statement about centrifugal force does not.

"There is no such thing as centrifugal force in the known universe.
CF is a mathematical fiction, albeit a convenient one."

 

Is the ball also in horizontal equilibrium, since it neither gets nearer to nor further from the centre line?

To answer your original question. The ball in not in horizontal equilibrium. The horizontal motion ( call it the x-direction or the y-direction, or left-right backward-forward) is sinusoidal. This is similar to the Lissajous pattern on an oscilloscope where a sine and cosine wave on the x and y axis produce a circular pattern.

 

Steve you are wriggling.

Yes it is a 3D problem.

Acceleration is a vector quantity, capable of being resolved into vertical (V), horizontal (H) and right angles to the plane (Z) directions.

The forces acting can similarly be resolved and Newton's first, second and third Laws be separately applied in each of these directions.

I am asking you to apply the second Law in the H direction, it should be simple enough I have done most of the work already.

If you really must, apply it to both the string and the ball (and even the beam if you like) but the result will be the same.

 

I was taught to use the term "centripetal acceleration" instead of "centrifugal force."

 

Steve you are wriggling.

Well, not really wriggling. I just thought the verbal description was essentially a valid answer. At least for people familiar with Newtons's Laws, circles and sinusoids, as most of us are.

If we observe the motion and use that as a starting point. The position of the ball in the horizontal plane is described by the following:

x=Rcos(wt); y=Rsin(wt) where R is the radius

This is just a Lissajous pattern that most people here are familiar with.

The velocity, acceleration and force all are easily calculated as the first derivative, the seond derivative and the then F=ma, respectively.

Thus:
Fx=-mRw^2cos(wt) and Fy=-mRw^2sin(wt)

Hopefully this is correct. I'm just doing this off the top of my head.

 

I was taught to use the term "centripetal acceleration" instead of "centrifugal force."

Absolutely correct sir!

I have some business to transact for the next couple of hours but,

Since I think it is essential that this educational website does not promote fallacious viewpoints I will post a zipped extract showing Newton's method which only uses real forces and side by side with D'Alembert's method which introduces an imaginary (his words) inertial force to solve the problem.

Centrifugal force is imaginary.

Centripetal acceleration is real.

Meanwhile you might like to look up centrifugal force, centripetal force/acceleration and D'Alembert on Wikipedia. They certainly support my view there.

 

I am not a teacher of this subject so the best way is to let those who are speak.

The following extract contains some theory and two examples, similar to mine, one simpler one more complicated.

Centrifugal force is clearly shown to be imaginary and the conditions for its introduction and use laid out.

The student is particularly warned not to mix the methods, as this is a frequent source of error.

Since inertial electronic navigation systems are at the heart of nuclear submarines and aircraft it behoves eletronic engineers to know something about the theory presented.

Mik, you might like to know that the authors are, or at least were, lecturers at the University of Newcastle.

Good luck in your studies.

 

I was taught to use the term "centripetal acceleration" instead of "centrifugal force."

Bingo! I've just started reading this thread, and thought surely someone would bring up this term before 36 posts. The inertia of the ball tries to keep it moving in a straight line at constant velocity (per Newton's First Law). The string exerts a force on the ball, pulling it away from its straight-line path toward the center of the circular path the ball is forced to follow. The horizontal component of the string's force is the centripetal force on the ball. The acceleration that accompanies this force arises from the ball's direction being changed. Since its velocity vector is changing direction, dv/dt is not 0, hence there is a nonzero acceleration.