Re: Gyroscopes - Usenet Physics FAQ - Reference frames

From: Tom Roberts (tjroberts_at_lucent.com)
Date: 07/29/04 

Date: Thu, 29 Jul 2004 13:25:13 GMT

JM Albuquerque wrote:
> "Tom Roberts" <tjroberts@lucent.com> wrote:
>>When I spin a rock around my head, without rotating my head
>>or body, then I have no need of "centrifugal force", F=ma
>>applies and the only force is the tension of the string.
>
> Only tension of the string?
> If so the string tension appears by magic then.
> I don't believe in magic. Something creates the string tension.

Yes, the only FORCE here is the string tension, and it points inward at
the rock's end of the string, and outward at my hand's end. What
"causes" it, of course, is the inertia of the rock (that's a naive use
of causality, however).

>>A "force" that disappears in one frame and appears in another
>>cannot possibly be "real" in any useful way, because physical
>>phenomena cannot possibly depend upon the frame or coordinates
>>one uses to describe them (Nature uses no coordinates, they
>>are pure fictions of the human mind used for descriptive
>>purposes only)..
>
> The force doesn't disappear.

Yes, it does. In the inertial frame of my non-rotating body, if there
were a "centrifugal force" equal and opposite to the string tension,
then the rock would move in a straight line (no net force => no
acceleration => straight-line motion wrt an inertial frame). In the
inertial frame there is no "centrifugal force", and the only force here
is tension on the string, as above.

In the rotating frame, as I said before, one can imagine Newton's law
F=ma to hold by assuming there is a "centrifugal force" that is equal
and opposite to the string tension; in this frame, of course, the rock
remains at rest, so one must have no net force IN THIS FRAME.

> And the most important subject is the force, not the frame of reference.
> Forces are the important subject and the real effect. Frames of reference is
> what I call a fictitious subject, since they are not real at all.

OK. But frames of reference are what we use to describe the motion of
objects. And fictitious "forces" like "centrifugal force" depend
explicitly on one's choice of frame (no rotation => no "centrifugal force").

This is very basic Newtonian mechanics, and you have it wrong. You seem
to be confused by the elementary freshman error of confusing centripetal
force with "centrifugal force".

> Do you know any string that can have a compression strength?
> I repeat for you. A string only conveys normal traction strength,

Which we normally call "tension".

> not
> transverse force, not shear force, not torque or torsion, not normal
> compression strength, not bending stress.

Sure. So what? nothing I have said disagrees with that.

> BTW, tension is force over an area.

In elementary discussions like ths we commonly and usually call the
force conveyed from string to rock the tension of the string. The area
of the string is not important, and we often/usually assume it is
negligibly small (and the string is perfectly flexible, has no
elasticity, etc. -- all the usual idealizations we make because they are
IRRELEVANT to your underlying misconceptions about "centrifugal force").

The "centrifugal force" on an object of mass m at location r in a frame
rotating with angular velocity w is:

        F_centrifugal = m (w x r) x w
        [c.f. any elementary mechanics textbook]

Here r is the 3-vector location of the object wrt the origin on the
rotation axis, and w is the 3-vector angular velocity (its direction is
along the axis of rotation and its magnitude is the angular velocity in
radians per second). x is the usual 3-vector cross product.

The fact that F_centrifugal is proportional to m means that in F=ma the
mass falls out, and one is left with a relationship between
purely-coordinate quantities. That is, the centrifugal force at a given
location is independent of the object and dependent only on the choice
of coordinates. Moreover, the "centrifugal force" vanishes in a
non-rotating frame (w=0).

        That's why we call it a "fictitious force" -- it was invented
        to keep the FICTION (of F=ma being valid) alive in a rotating
        frame. In fact, F=ma is really only valid in inertial frames.

> Yes, the force the string exerts on the mass is toward the center.
> That leaves you with a mass and a force applied on the mass toward the
> centre. So, why the mass doesn't fall to the center?

It _DOES_ fall in toward the center, PRECISELY as F=ma predicts. But as
it also has a tengential velocity, and because the tension on the string
adapts appropriately to the motion of the rock (in both orientation and
magnitude, as does the hand spinning it), the rock moves in a circle
around the hand. So it falls TOWARD the center, but never "to the
center" -- it never gets there. If there were no force of tension, no
CENTRIPETAL force, then the rock would not move in a circle, it would
move in a straight line (F=0 => a=0 => straight-line motion). Similarly
if the CENTRIPETAL force were balanced by an equal and opposite
"centrifugal force", then again F=0 => a=0 => straight-line motion --
BUT THE ROCK MOVES IN A CIRCLE (here I'm working in the inertial frame).

> Any mass with a force applied will accelerate. Here we have a mass that
> any given time wants to fly away in a tangential direction. So, you have a
> mass moving in a direction perpendicular to the said force and the said
> force applied toward the center. Now you have the two vectors defined.

Right. One vector is the rock's velocity and the other vector is the
rock's acceleration; they remain perpendicular as the rock moves,
becuase the string adapts. As acceleration is the time derivative of
velocity, one integrates the equation of motion (F=ma) to determine the
path of the rock, which is a circle.

You REALLY need to go back and review elementary mechanics.

> What do you do with them. The "dot product"?

Not at all -- one integrates the equation of motion. See above.

There's no "magic" here, simply the elementary application of basic
mechanics.

        [This is getting repetetive; don't expect me to respond unless
         and until you actually STUDY basic mechanics.]

Tom Roberts tjroberts@lucent.com

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