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Conservation of Angular Momentum


Delburt Phend

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The only thing that makes motion is Force: F.

 

Newtons explanation for the quantity of motion was F = ma

 

Acceleration is velocity over time: so, F = mv/t

 

Multiplying both side by t we get Ft = mv;

 

Newton's explanation of motion was Ft or mv;

 

Now the professors of the world of physics say that mv can't be used to evaluate motion. That is identical to saying motion can't be used to evaluate motion.

 

Instead the inappropriate use of two other formulas have taken momentum's place. In an absolute sense Newtonian Physics has been thrown out.

 

If the motion of a balanced flywheel is caused by 20 newtons applied for 20 seconds; then that is the quantity of motion that the flywheel has. It is wacky to say that it is zero.

 

High school students on spinning chairs and ice skaters are not experiments. To qualify as an experiment you would have to know the radius and mass and the rate of spin. These events are innuendo and I think they are the only things ever cited for angular momentum conservation. There are no angular momentum conservation experiments done in the lab.

 

Linear Newtonian momentum conservation will also increase rotation rates for students on chairs and ice skaters; so nothing is proven by these events. So you are pitting zero experiments against a plethora of experiments for linear Newtonian momentum conservation.

 

Energy conservation acknowledges losing heat; so that leaves that out.

Edited by Delburt Phend
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The only thing that makes motion is Force: F.

 

Newtons explanation for the quantity of motion was F = ma

As the equation clearly indicates, force causes acceleration, not motion.

 

Now the professors of the world of physics say that mv can't be used to evaluate motion. That is identical to saying motion can't be used to evaluate motion.

That's news to me.

 

Instead the inappropriate use of two other formulas have taken momentum's place. In an absolute sense Newtonian Physics has been thrown out.

 

If the motion of a balanced flywheel is caused by 20 newtons applied for 20 seconds; then that is the quantity of motion that the flywheel has. It is wacky to say that it is zero.

Who says it's zero? The rotation will depend on the force and where it was applied.

 

High school students on spinning chairs and ice skaters are not experiments. To qualify as an experiment you would have to know the radius and mass and the rate of spin. These events are innuendo and I think they are the only things ever cited for angular momentum conservation. There are no angular momentum conservation experiments done in the lab.

You've been in every lab, everywhere, at all times? How does that work?

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Wow: F = ma is no longer mainstream physics.

That really depends on the metric. The equation f=ma is valid. Though under GR there is no acceleration in freefall motion. As the metric of GR describes this as spacetime curvature that follows a geodesic equation.

 

End result until you have relativistic factors ie time dilation the two are equivalent to extremely good approximation.

 

Don't be fooled though Newtons laws apply even in GR though it may not be obvious as its shrouded in the mathematical treatment.

 

A better way to think of it, is Newtons laws f=ma is valid until one must factor in [latex]\gamma [/latex]. All equations are to a reasonable approximation. Whatever one considers reasonable.

 

For everyday situations f=ma is incredibly close. Close enough to cover any non relativistic kinematic motion.

 

There is no need to use complex GR to describe a thrown baseball. Though one can use GR to describe such, one can also use action from the "Principle of least action" These different treatments don't invalid one another. They simply help define the limitations in accuracy

Edited by Mordred
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The only thing that makes motion is Force: F.

 

Newtons explanation for the quantity of motion was F = ma

 

Acceleration is velocity over time: so, F = mv/t

 

Nope.

 

[math]a=\frac{v_1-v_0}{t_1-t_0}[/math]

 

Acceleration is change of velocity in period of time.

Constant velocity = no acceleration, as both v1 and v0 are equal, and reduce to 0 in the above equation.

You can multiply either velocity by mass:

 

[math]F=\frac{m*v_1-m*v_0}{t_1-t_0}[/math]

 

[math]F=\frac{p_1-p_0}{t_1-t_0}[/math]

 

and force is equal to change of momentum in period of time.

Edited by Sensei
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Wow: F = ma is no longer mainstream physics.

 

 

One section of correct physics is not enough to make a post acceptable - however, one section (and yours had plenty) with erroneous assertions is enough to make a post unacceptable.

 

You were way off course by the third line of your post; acceleration is the change in velocity over the time taken for that change for simple cases as Sensei has elucidated above; in most cases it is dv/dt as the acceleration is not constant.

 

Most school have and will experiment with gyroscopes - perhaps not quantitatively - can you explain gyroscopic precession? Or even why a linear force on the string of a top will allow it to stand on its point?

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Haven't a clue what that video is meant to show. But your equation is still foolish - it only works for constant acceleration, what you have as time (t) is change in time (delta-t), and it doesn't always work; F=ma and the other equations of motion apply in all classical mechanics at non-relativistic speeds - yours falls over very quickly.

 

If you apply 5 newtons of force to 2 kilograms of mass for 4 seconds you get a velocity of 10 m/sec. So the formula works as printed. Ft = mv
OK - so take the mass going 10m/s and apply 0 newtons of force for ten seconds, or apply 10 newtons for zero seconds.
You could - if you really wanted to - write F * (t1-t0) = m * (v1-v0); but this cannot deal with, for example, an acceleration which varies with time. So what we actually write is F=ma or F=m (dv/dt) or F= dp/dt
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F and v are vectors in that equation. You are applying 5 N at one side of the cylinder and -5 N at the other, for a resulting force of 0 N, as evidenced by the fact that (ignoring the fall) the cylinder remains at the same spot.

Each part of the cylinder has a linear momentum mv, but adding them all together in a vector sum, the resulting linear momentum is 0.

 

In fact, for a small particle, the linear and angular momentum are interchangeable. Just multiply your equation with the radius r and you get

[math]Frt=mvr[/math]

split up v

[math]Frt=mr^2 \omega[/math]

And rewrite with torque M and moment of inertia I

[math]Mt=I \omega[/math]

 

Or where did you think the formula's for calculating angular momentum or moment of inertia come from, if not by deriving them from the formula's for linear motion? I suggest at least reading the wikipedia article if you are genuinely interested in the subject. You will never convince anyone if you don't even know the basics.

 

Of course, the last equation is more practical for rotating objects because integrating all the little parts doesn't yield 0. (in case you care, it is because in vector-notation all the vectors of angular momentum are now pointing in the same direction, orthogonal to the original force)

 

Also, the correct abbreviation for seconds is s, not sec.


Edit: in case you are unfamiliar with summing vectors, look at this image:

16.jpg

a) shows what happens if all vectors (eg forces) act in the same direction. b) shows how to add vectors with different direction: you can draw them end to tail, and the resulting vector is going from the start to the end. c) is the situation we are discussing: the end of the last vector arrives at the start of the first, so the resulting vector is 0. If this is not clear, you are welcome to ask more detail.

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