torque

Ask Ethan #9: Why everything rotates

esiegel — Fri, 01 Nov 2013 18:39:04 +0000

Ask Ethan #9: Why everything rotates

"Galileo got it wrong. The Earth does not revolve around the Sun. It revolves around you and has been doing so for decades. At least, this is the model you are using." -Srikumar Rao

It's the end of the week, so that means its time to take on another one of your questions from the question/suggestion box, and continue our ongoing Ask Ethan series! Even though there's a backlog of hundreds of questions, you should keep sending the new ones in, as all questions are fair game for any segment. This week's question comes from reader Brian Mucha, who asks us:

Where did the sun and planets get their angular momentum resulting in their rotation. I am not asking about the orbits but the actual rotation. I understand the ice skater analogy where bringing in the extended arms increase the skaters rotation due to the conservation of angular momentum. But the skater starts with spin. IF the skater is standing still they can extend and retract their arms all day and they wont spin.

So when the planets and the sun started to form how was their initial angular momentum achieved?

Ahh, the old question of rotation, and why everything does it.

Image credit: Alicia of http://surrenderingallofme.blogspot.com/.

It's easy to make something spin faster once it's already going: you just change its moment of inertia.

What does that term mean, moment of inertia?

Image credit: PDFcast and Utah Electronic High School

You know Newton's second law: the one that tells you force is equal to mass times acceleration (F = ma). Technically, it's a little more accurate to say that force is how another quantity -- momentum -- changes over time. It means if you apply any external force to a mass, its momentum -- or how it's currently moving -- will change, and it tells you exactly by what amount it will change. And if you don't apply an external force to something, its momentum cannot change.

And if everything in the entire Universe only consisted of point masses along the same line, we'd never need anything else. But in the real Universe, masses-in-motion are distributed in more than one dimension.

Image credit: Greg L at the English language Wikipedia.

And whenever you have that, your system has not just momentum, but also angular momentum. And while momentum changes are dependent on mass, angular momentum changes are dependent on a combination of the mass and how that mass is distributed. That combination of factors -- mass and how it's distributed -- is what makes up moment of inertia. So yes, Newton's second law relates how objects change their momentum (i.e., how masses experience changes in their velocities), and there's an equivalent law that relates how objects change their angular momentum, or how moments of inertia experience changes in their rate of rotation.

Images credit: Markus Pössel, Einstein Online Vol. 3 (2007), 1011.

How the figure skater who pulls her arms and legs in spins faster is one example of this: as her mass becomes distributed closer to the axis-of-rotation (and her moment of inertia gets smaller), her rotation rate increases to compensate. If your mass-and-how-it's-distributed changes (goes up or down), your rate of rotation will also change (go down or up) to compensate. But just like Newton's second law tells you that you can change a system's momentum by applying an outside force, you can change a system's angular momentum by applying an outside torque.

Image credit: Wikimedia Commons user Yawe.

And a torque is just a force applied in such a way that it causes an object's rotation to change.

Image credit: NASA, ESA, and the Hubble Heritage Team (STScI/AURA).

Now here's where it gets interesting: every star system in the Universe once began as a cloud of gas-and-dust. These clouds may have been thousands or millions (or in some rare cases, maybe even larger) of times the mass of our Sun, but they were once incredibly diffuse, and spread out across many hundreds or thousands of light years. If these gas clouds had (or the ones we see today have) any sort of global rotation to them, it's far too small to be detectable, as it would take billions of years for such a gas cloud to make even one complete rotation.

Image credit: NASA/ESA and The Hubble Heritage Team (STScI/AURA)

But gas clouds -- like all objects in the Universe -- don't exist in isolation. They exist in the presence of all the other matter-and-energy in the Universe, all of which is subject to the laws of gravitation. And whenever any two masses in the Universe are in relative motion to one another, so long as they're not moving exactly and directly towards-or-away-from one another, the gravitational force they exert on each other causes a torque.

Illustration credit: Bill Saxton, NRAO / AUI / NSF.

This phenomenon is known as a tidal torque, and was first theoretically understood by Jim Peebles -- my Ph.D. advisor's Ph.D. advisor (or my grand-advisor) -- back in 1976. (So, you asked the right person!) It's why pretty much every mass that exists in this Universe, whether it was born with non-zero angular momentum or not, has one now, 13.8 billion years onward. That includes every gas cloud, including the one that gave rise to our Solar System. We can break these large gas clouds up further, into the regions that give rise to the individual stars and star systems that came into existence.

Each one of these regions that eventually result in a star/star system, with whatever angular momentum they have inside, are typically distributed in shapes known as triaxial ellipsoids. A triaxial ellipsoid is a fancy way to say that they're like spheres, except inevitably if you draw three perpendicular axes on them (X, Y, and Z, for example), one of the axes will inevitably be the shortest of the three. When a region gravitationally collapses, it's going to collapse along the shortest axis the fastest, and because normal matter -- the stuff all stars and planets is made out of -- interacts (i.e., collides) with itself, that means it's going to go "splat," like a pancake. (In fact, the scientific word for this process is known as pancaking.)

Image credit: NASA / JPL-Caltech.

But along the other two axes, you'll have a disk-like distribution, which is going to have an overall, net rotation in the direction of whatever its angular momentum is! It's the reason why -- in our Solar System -- all the planets revolve around the Sun in the same direction (counterclockwise, looking downwards from north of the Sun's north pole), the Sun rotates in that same direction, almost all the moons revolve around their planet in that same direction (with notable exceptions explained here), and finally, why practically all the planets rotate about their axes in that same direction, too.

Image credit: Calvin Hamilton, and click for a huge version!

There are only two major exceptions to the rule: Venus, which hardly rotates at all (but does so in the opposite direction), and Uranus, which rotates practically on its side. Both of these worlds are thought to have had their angular momentum significantly changed by the intervention of an outside body, most likely a significant collision a long time ago. That is to say, their rotation was changed by the influence of an external torque!

Image credit: Bhavana Jagat, via http://bhavanajagat.com/.

So that's the story of why planets, moons, stars and star systems revolve and rotate the way they do! Thanks for a good question, Brian, and to anyone else who has a question or suggestion for me, drop me a line!

esiegel Fri, 11/01/2013 - 14:39

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ask Ethan

gravity

Physics

Solar System

Could be my fault... but I'm not convinced by what I've read here Ethan.

I'm still rather amazed that a solar system resolves into a 2D disk from [I assume] a random set of particles in 3D. I think I would be happier if I had some idea the range of torques that could fall out of a *typical* solar system

I would assume that a typical pre-solar system should have zero spin on average, but obviously I'm wrong

Michael

The initial spin is just gravity. If you have an amorphous blob of stuff, pick any point in the stuff, and every other bit of mass in the stuff is going to exert gravity on it. Now repeat that for every point within the blob of stuff. These forces are not going to be equal, hence it will accelerate (rotate). Unless the initial form were a perfectly uniform sphere, (I think) this will happen.

I'm guessing that it would eventually just begin rotating about the centroid of the inital blob of stuff.

As gravity continues to form more cohesive masses out of the stuff, this is where change in mass distribution and conservation of angular momentum kicks in and compels it to form a disc, (and spin faster?). (spin a water filled balloon about an axis and it would have the same effect).

I think I have that more or less right, though I'm just going off of a year of University physics, and several years of reading Ethan's blog here... so ... yeah.

Its the invisible hand giving us the invisible finger and telling us "Rotate".

The Oort cloud rotates in many difference axises, and that is unstable since they cross each others path a lot, causing all these comets to bomb into us.

But that far out, they're very diffuse so interactions are not common.

Closer in, those off-ecliptic orbits were destroyed by much more frequent interactions and they bombarded the inner solar system. See the Moon's backside.

Once crashed, they aren't orbiting any more, so they don't exist much.

(note: only one method for the weeding out. There's still a preferential axis that everything would have "just picked" because perfect isometry is impossible and the biggest mass there (the sun) would be displaying a rotation around that axis more than any other.)

Remember too that zero spin is mathematically impossible, an infinitesimal chance.

For a start, that part of the dust matter closer to the centre of mass of the galaxy would lose velocity when "pulled" back out by in-falling to the new sun, and that further out would gain velocity when "pulled" back in.

But even absent that, zero has a 0/1 chance (infinitesimally small) of occurring.

@Michael Fisher #1: I thought Ethan explained it relatively clearly, but he did skip some steps in the argument (citing external papers instead).

Addressing your questions in reverse order.

1) Why did the presolar nebula have non-zero angular momentum initially? The answer to that is _tidal_torque_. Consider the simple case of two separated blobs of extended matter (gas clouds, dust clouns, galaxies, whatever). Suppose these are the only things in the Universe. What happens? Tides! Since the blobs are extended, there will be different gravitational forces on the near vs. far sides, and also on the "left vs. right" sides (unless both blobs just happen to be perfectly spherical). If the blobs are moving relative to one another, then there will in general be a non-zero impact parameter between their centers (that is, they will zip past one another, not collide head on). Put those things together, and you should be able to work out that the two blobs extert a net torque on one another, with the result that each one -- even if initially irrotational! -- will start rotating.

2) Why does a non-spherical 3D matter distribution evolve into a 2D disk? "Pancaking." Start with a distribution, which can be perfectly spherical or not, with net angular momentum about some axis Z (see item (1) above). Gravity alone would pull that 3D distribution inward radially, making a smaller and smaller blob, until pressure or solidity or whatever made it a little sphere (this is the "hydrostatic equilibrum" referred to in the IAU definition of a planet!). But we just said the blob is spinning. For the part of gravity pulling bits together along Z, the angular momentum doesn't change. But for the part of gravity pulling perpendicular to Z, moving bits closer to the axis means either (a) angular momentum must disappear; or (b) tangential speed must increase to conserve angular momentum. But increasing the tangential speed of a bit about the center pushes that bit to a _larger_ orbit, even though gravity is trying to pull it to a _smaller_ one! The net result? In the plane perpendicular to Z, stuff stays roughly where it started, but along Z, stuff collapses down. So you go from a 3D blob to a 2D-ish pancake (2D up to a thickness corresponding to pressure-balance).

Does that clarify the situation?

Michael #6 and all: I think pancaking can be shown very effectively by putting a small ball of water to rotation somewhere within the ISS. We've seen several watery experiments on the ISS, but I don't recall pancaking being one of them. I think that would be a cool experiment to see...

I think I see why solar systems spin the way they do. Because gas and dust clouds that don’t will never condense into solar systems. The material that starts to form a disk has to rotate or it falls out of the sky and, in the process, builds up the central mass that will evolve into a sun. If enough matter does, rotation of the amorphous blob develops into revolution (in the same direction) of discrete objects. The ones that survive as planets or asteroids are the ones that are moving at the right speed and distance to balance gravitational forces.
But it’s the persistence of the link between rotation and revolution that puzzles me.
I’m not quite clear what causes planets to rotate in the same direction as they revolve. I can see the sun tugging against counter-rotation, much as Earth has pulled the moon’s rotation to a standstill, but planets generally spin faster than they revolve. And I wonder if Venus and Uranus will ever have to pay a price for their waywardness.
If we zoom out, our solar system behaves toward the galaxy much like such a rogue planet – revolving the same way the Milky Way rotates, but at a wicked angle. Is this a big mistake on our part? Could that play a role in our predicted fate of crashing into Andromeda? Not that there’s much we could do to change things at this point. Let’s just try to enjoy the ride, I guess.

@Tihomir #7: I don't think you can use water blobs in microgravity to show pancaking. The problem, as I see it, is that they don't have an attractive potential (i.e., something trying to pull them toward their center).

They have a surface tension, which is going to pull tangentially everywhere, which gives you a "vertical" pull along the rotation axis down at the equator, but a "radial" pull _toward_ the rotation axis up at the poles.

What you might get with a well-formed but spinning water blob is a sphere with an equatorial rim.

In any event, seeing the experiment directly would still be awesome.

Your explanation and the comments that followed, all explain why the universe evolved into what we see today and I understand this, but it does not explain why the gas left over after inflation, just after the big bang would have any angular momentum at all. Without that initial angular momentum, the gas should have simply collapsed back into a singularity. Obviously, it did not, so some initial angular momentum was present, no matter how small. Also the distribution of mass within the cloud of gas was not uniform. I have read that the non-uniformity of mass can be explained by quantum fluctuations, but no explanation of the initial angular momentum has ever, to my knowledge, been suggested. Quantum fluctuations do not seem to work as an explanation either.
I think that this was the actual question posed here. Not how did the objects in the universe end up spinning the way they did, but where the initial momentum came from that allowed that spin to evolve.

Paul, precisely zero has a chance of happening of 0/1. I.e. infinitesimally small.

Only one organisation of matter, pf the entire 10^80 particles of matter, will give you that.

That's not a likely scenario.

Okay Folks, why would some thing begin to exert gravitational force when its a particle, under what unknow circumstances whould these particles move or change direction when everything to begin was in rest, and please how in this world can you come up with past billions of years of story when real time any idea or theory is just few years old...ITS LIKE AND ANT ON ROAD TRYING TO WRITE A THEORY ON A CITY WHEN HE HAS NO MEANS OF PHYSICALLY /MENTALLY KNOWING WHAT N WHERE TO LOOK ...
If you still trust Newton laws....then at all times things are in Equilibrium. why would a change to begin with ...Why should gravity of A should intervene the gravity of B in space.

@Paul #10: This is the same question Michael Fisher asked, and which both Ethan and I have tried to answer. Non-spherical blobs of matter exert and respond to gravity with both "central forces" (the usual F = -GMm/r^2) _and_ with torques. In this case, the torque arises because there are different forces acting on different parts of the non-spherical blob.

If the blob is holding itself together (either graviationally or with chemical/EM forces), then those differential forces will _cause_ a rotation of the object, even if it wasn't rotating before.

Thinktank, if you're going to berate everyone for being dumber than you, you'd best stop for a bit, calm down and collect your thoughts.

Because SOMETHING is making your post full of bad English.

Then work on what it is you're trying to say. Because I'm sitting here thinking "Well what the hell do you mean there?" and having to interpret it to something that is coherent.

With that out of the way:

"why would some thing begin to exert gravitational force when its a particle"

Because it has gravitational mass and therefore transfers energy by the force of gravity to things within range.

"under what unknow circumstances whould these particles move or change direction when everything to begin was in rest"

Who says it was at rest? This isn't an explosion where the bomb was sitting still, someone lit the fuse and it went BOOM.

"and please how in this world can you come up with past billions of years of story when real time any idea or theory is just few years old"

By looking at the physics of distant stars and determine that the physics hasn't changed appreciably in that time. E.g. the value of the fine structure constant is a combination of several fundamental constants and if they change, the laws of physics change.

There's no change.

"ITS LIKE AND ANT ON ROAD TRYING TO WRITE A THEORY ON A CITY WHEN HE HAS NO MEANS OF PHYSICALLY /MENTALLY KNOWING WHAT N WHERE TO LOOK"

No it isn't.

"If you still trust Newton laws….then at all times things are in Equilibrium"

The latter does not follow from the former.

"why would a change to begin with"

A what? I think you a word.

"Why should gravity of A should intervene the gravity of B in space."

I don't think that word "intervene" means what you think it means.

Paul, what Michael is saying is that for a collapse not to AUTOMATICALLY result in increased rotation would require the gravitational field over the entire size of the collapsing cloud to be flat. Absolutely flat.

Otherwise by pulling it out of its current location it would experience a force that will cause it to deviate from the straight line to the CoM because it's in a different gravitational potential field.

Excellent post and some good comments as well. Thanks!

This is my favorite version of Wow. I like this one better than the mean one. Argument destroyer, not self-esteem destroyer, although the latter does have it place in the very rare exception. Just my opinion though.

@ Paul

the universe was never static. Even at earliest stages. There was some intrinsic motion (however small) emparted to matter from leftovers of inflation. Then as you said youself, the small non-uniformities caused local attractions (gravity)

cd2, we have different opinions when it's deserved.

I’m not quite clear what causes planets to rotate in the same direction as they revolve.

I haven't seen anybody else answer Chuck's question @8, so I'll give it a try.

Gravity is trying to pull mass toward the center of the disk, but in order to move inward any given bit of mass would have to lose orbital angular momentum. So if a bunch of that mass starts spinning in the same direction as the orbital motion, that spin angular momentum can absorb some of the orbital angular momentum and allow that bit of mass to fall inward. To get some mass to rotate in the opposite direction, you would have to increase its orbital angular momentum to compensate, and gravity opposes that. So while it's not impossible to get retrograde rotation, it requires external torques, as Ethan points out in the original post.

Regarding spin rates vs. orbital rates, keep in mind that that is only one of the two factors which go into angular momentum, the other being the mass distribution. Planets orbit a star at a distance much larger than their radii, so despite the much lower angular velocity the orbital angular momentum can be comparable to, or even larger than, the spin angular momentum.

@10:

it does not explain why the gas left over after inflation, just after the big bang would have any angular momentum at all.

It doesn't have to. It does, for reasons people have already given, but we can start with a universe without any and get galaxies, solar systems, and planets that rotate.

A layman version goes like this: the interactions between particles or masses converts gravitational energy into momentum. Start with two sationary balls apart, and gravity, and they will start slowly moving together. Right? That's very intuitive. The exact same statement is true - but less intutive - for torque. Start with two objects without any torque, plus gravity, and (with a couple of wierd exceptions), they will acquire torque.

The wierd excetions are point masses and perfectly spherical masses. But otherwise, your two objects are not only going to start moving towards each other, but different parts of each object are going to gain microscopicaly different momenta. That results in a tumbling motion and leads to rotation.

So is the entire universe considered as a single entity also rotating? Or does the expansion somehow prevent that?

@Rich T #21: That's a bit of an unanswerable question, kind of like asking, "What is the velocity of the universe considered as a single entity?" In order to measure a net rotation of the whole observable universe, you would need to be able to observe it with respect to some external reference frame. "External" and "whole observable universe" are mutually exclusive.

Rich T:
If the universe were rotating, that would be in relation to what? The picture is that there is no space "beyond" the universe, nor is there a time according to which a dynamic process like rotation could be said to occur. So there's nothing to rotate in.

If there's a multiverse and some intervening space between our universe and another, then we might be able to talk about rotation. But that doesn't seem like a promising scenario, because if it's the same space as our three dimensions, I don't see how we're talking about a "universe" instead of "part of a universe." And if it were rotation in some other dimension, it seems that whatever it is must not be ordinary angular momentum at all.

@Michael

Could one still measure net angular momentum of the universe if it was rotating strongly in one direction? My understanding is that the Cosmological Principle would state that the universe is isotropic, and a universe with a preference for a certain spin wouldn't be isotropic, I feel like that'd be a significant symmetry break.

It's my understanding that the while random solar systems or galaxies might have a net angular momentum, there is no reason why galaxies couldn't spin opposite directions, so the overall universe would have zero, or nearly zero net angular momentum.

I could, of course, be very wrong, as I'm but a BsC. PS, I had a prof from SLAC, she was awesome, seems like that's a great place to work.

Michael and others:

So why doesn't pancaking happen to planets and the sun just like the whole (solar) system? I know they do flatten a little at the poles. Do the multiple moons around some planets rotate in one plane from pancaking, and if not, why not?

"If the universe were rotating, that would be in relation to what?"

In relation to "spinning around, therefore not being an inertial frame".

So why doesn’t pancaking happen to planets and the sun just like the whole (solar) system? I know they do flatten a little at the poles.

In planets and stars, the atoms are also interacting via the electromagnetic force, strong force, and weak force. These counteract gravitational pancaking. So, for example, the Earth can't pancake because the atoms on the surface are being electromagnetically pushed "up" (away from center) by the atoms underneath them. For stars and in particular neutron stars, the strong force does the same job.

Do the multiple moons around some planets rotate in one plane from pancaking, and if not, why not?

I believe the answer to this question is: the pancaking was a solar-system-wide effect that imparted approximately the same directional momentum to pretty much all the atoms in it. That's why pretty much everything in our solar system rotates more or less in the same plane and direction. The exceptions (moons orbiting in the wrong direction, etc...) are a result of collisions or non-local objects being captured by a local system, or both.

Q to the cosmologists - for really big spinning masses like galactic center black holes, does relativistic frame dragging reinforce the pancaking effect?

Thinking over the shape of the equations, eric, I rather think that frame dragging would retard the effect. Except that frame dragging would tend to be where there's already a lot of rotation going on, so it'd still be "Gotta pancake" overall.

And the volume which even a super massive massive black hole would have its effect even in the dense core would have it include very few actual stellar bodies.

I'd look more toward photon pressure there keeping things spherical more than frame dragging.

I don't know that I've got anything at home that'd help me be more accurate or precise, mind.

First Ethan et al great questions and answers

specifically, #21, 22, 23 about the observable universe rotating.

"in Euclidean geometry.. three categories of objects, points, lines, and planes.. are not defined but assumed to be intuitively given." Herman Weyl, pg 3, Philosophy of Mathematics and Natural Science, 1949

then Weyl, pg 105, "Incidentally, without a world structure, the concept of relative motion of several bodies has, as the postulate of general relativity shows, no more foundation than the concept of absolute motion of a single body."

Then we have, "After the Godel discovery of rotating model universe, several other rotating cosmological models were discovered with interesting properties different from the Godel model universe." pg 243, Gravitation and Inertia, Ignazio Ciufolini and John Archibald Wheeler, 1995

So it seems we have a kind of a chicken and an egg situation regarding the possibility of a observing our visible universe as a rotating universe. we need clear concept is our visible universe rotating relative to space or time or relative to extra dimensions or relative to its own space time or what? I defer to the experts on what a more correct testable question might be?

But then Ciufolini and Wheeler suggest, "for example, one can put limits to the rotation of the universe using upper limits to the large scale anisotropy of the cosmic blackbody radiation.. Test of non rotation of the universe.. for example, using, VLBI and local gyroscopes" pg 251 Just which definition of rotation are they using? I don't know but I assume a closed GR universe.

So the question of what it might mean to even ask and then to answer, is our visible universe rotating (and relative to what)?; seems a valid scientific question.

And the best answer to date seems to be http://www.earlyuniverse.org/is-the-universe-rotating/ " we have studied models of rotating universes, the so-called Bianchi models, in order to test the isotropy of the Universe... So, is the Universe rotating? Well, probably not... However, only very simple Bianchi models have been compared to the data so far. There are more sophisticated Bianchi models that more accurately describe the physics involved and could perhaps even provide a better explanation of CMB observations. We’re looking into it!"

So it's a difficult question. Very difficult mathematics (because "only very simple Bianchi models have been compared") and very difficult physics (because a century of observation needs to be explained in the detail).

But there are, "two types of scientific development, normal and revolutionary.. normal science is what produces the bricks that scientific research is forever adding to the growing stockpile of scientific knowledge... Revolutionary changes are problematic. they involve discoveries that cannot be accommodated within the concepts in use before they were made." Thomas Kuhn, the road Since Structure, pg 14

In my opinion the question of whether the universe is rotating or not (and what that means) is tied up with other questions such as what is dark matter and energy, are there extra-dimensions, is there only one dimension of time, etc.

If we are to consider "revolutionary scientific development" and not just "normal scientific development"; then we need to be open to understanding (even in very fuzzy terms) what the range of possible universes are, i.e. some very crazy speculative ideas.

That is, if we consider that "revolutionary cosmology" development is still possible and not just "normal stockpiling of more cosmological bricks" from here on out.

Yep, yuk, yuk, just dumb luck we got it all basically right in the 20th century.

This amateur suggest http://vixra.org/abs/1307.0056 A Toy Universe "replaces the large unseen dimension of time of the real universe, cti ∈ I, with a tiny curled unseen dimension of imaginary space, yi ∈ I." well actually "three imaginary number curled dimensions y1i, y2i, y3i ∈I " (Just crazy speculative nonsense. Right? What does 3-dimensions of T-duality time mean? Yikes!)

This mathematically sophisticated physicist suggests http://arxiv.org/pdf/1303.3044v2.pdf Stable Cosmic Vortons "We present for the ﬁrst time solutions in the gauged U(1) U(1) model of Witten describing vortons – spinning ﬂux loops stabilized against contraction by the centrifugal force... We construct explicitly a family of stationary vortons characterized by their charge and angular momentum. Most of them are unstable and break in pieces when perturbed... until recently it was not clear whether cosmic vortons were stable or not... Interestingly, a very similar conclusion was made for the 'spinning light bullets' which share many ideas with vortons." (Nice try, seriously crazy enough. And even almost ready to be explained in everyday English. Maybe with vortons we get a rotating universe. Probably not that easy. Do we have to go back to the Bianchi models?)

Well anyway or NOT, I personally expect to see "revolutionary cosmology" in my lifetime not just ""normal stockpiling of more cosmological bricks."

OKThen, the book "The 4% Universe" has a few sections on one professor's attempts to measure and define whether the universe as a whole is rotating.

However, being female she was not accorded the recognition of that work.

Worth a read.

Hey OKThen, haven't seen you for a while! :) welcome back

Wow,
Thanks, I'll look at it.

Sinisa Lazarek
Thank you. Been busy in a good way. Life, you know.

Well anyway or NOT, I personally expect to see “revolutionary cosmology” in my lifetime not just “”normal stockpiling of more cosmological bricks.”

Good lord, man! In the last human lifetime we've gone from the steady state model to simple big bang to inflationary big bang. We've gone from thinking we understand just about everything in the universe to thinking we understand about 5% of its mass and energy.

This IS a revolutionary period in cosmology. You're living in one right now. Cosmological sciences 1930-2030 is what a scientific revolution looks like.

Heads up Ethan, the majority of your images have stopped loading for me. I don't know why, your pictures always worked before.

Alissa,

They are working for me in all my browsers (Chrome, Firefox and Safari); can anyone else verify that images aren't loading?

So, Ethan, what you're saying is that you were the padawan of the padawan of Jim Peebles? ;)

Fortissano,

That is true. Unfortunately for me, when my advisor (Jim Fry) was Jim Peebles' student, Star Wars Episode IV came out. When I was Jim Fry's student, Star Wars Episode II came out.

The Universe isn't always fair.

What is Arnold made of?

rallain — Mon, 21 Jun 2010 11:55:45 +0000

What is Arnold made of?

Of course I am talking about Arnold Schwarzenegger. After looking at how many bullets he carries in Commando, I remembered this scene (also from Commando) (warning: maybe some not great language and some killing. You have been warned)

If you don't want to watch that clip, here is a shot (sorry for the quality).

Clearly Arnold is strong, but there is more than strength involved here. Oh, don't bring your "he did it with wire stuff". I am not buying that. Also, I am talking about THE Arnold - he is real. I am not talking about the character in the movie (not real). Now for some physics.

Let me look at the combo of the two men (Arnold and Sully) and look at the forces acting on this "system". Picture time:

If I assume Arnie is holding Sully just to the point where they are both going to fall, then there would just be one force on the Arnie-Sully system from the ground. It would act on the system at the edge of the cliff. The other two forces are the gravitational force on Arnie and the gravitational force on Sully.

If they are in equilibrium (and from the movie, it looks that way), then two things must be true (for this case):

There are no forces in the x-direction, so that condition doesn't really matter. For the torque equation, if it is not rotating about a particular point, it will not be rotating about any point. So, you can pick any point to look at the torques. Here is another picture with some distances in it. I have replaced the stick-figures with balls.

Where ra is distance from Arnold's center of mass to the edge of the cliff and rb is the distance from Sulley's center of mass to the edge of the cliff. If I use the edge of the cliff to add up the torques, I get the following two equations (oh m_c is the mass of Arnold-commando and m_s is the mass of Sulley)

If I know the mass of Sulley and the two distances, then I can solve for the mass of Arnold:

Now for some values. First, Arnold's arm length. Wikipedia says that he is 1.88 m tall. Using that measurement and bing.com, I made this diagram.

Using that diagram, I can scale the image with Sulley hanging over the cliff. This is what I get (I added the red line to show the vertical line up from the edge of the cliff).

From that (and one estimate of Sulley), I get:

r_b = 0.44 m
r_a = 0.15 m
m_s = 68 kg (150 pounds)

Putting these values in, I get:

199 kg is like 440 pounds. So? I am going to calculate Arnold's density. According to Wikipedia's Arnold page, his (obviously fake) weight is listed at 250 lbs. Clearly, whoever made this fake page used Arnold's volume and the density of a normal person (about 1000 kg/m³) to calculate an expected mass. This is what they used.

The volume of Arnold is more difficult to disguise. So, if I assume they used his correct volume then the following should be true:

Using this, I can solve for Arnold's real density.

Using his "fake weight" of 250 pounds and his real weight of 440 pounds (since I only need the ratio, it doesn't matter if I use the ratio of weights instead of the ratio of masses), I get that Arnold's density is 1.76 times the density of a human.

A good value for the density of a human is about 1000 kg/m³ (about the same as water - humans just float). This would give Arnold a density of 1750 kg/m³. So, what is he made of? Aluminum has a density around 2700 kg/m³ and titanium has a density of 4500 kg/m³. I don't know. Maybe he has some lower density stuff along with some titanium or something. He obviously isn't pure titanium. He also is obviously not human.

I knew it.

rallain Mon, 06/21/2010 - 07:55

Tags

arnold Schwarzenegger

Well done! Only problem is that his center of mass doesn't seem (to me) to be accurate in your assumption. Throughout the video, Arnold's center of mass looks like it is directly over the normal force (unless, of course, Arnold's rear-end is much more dense than the rest of him).

Well, you just proved that he isn't made of solid pure titanium. But his internal metal skeleton may be made of titanium metal, and he still could have a lower density, since the volume may be filled with air or something.

Are you so smart that it makes you blind? you can clearly see the wire off his ankle at around 55 seconds.

@Lace:
DAMN YOU FOR DESTROYING MY ILLUSIONS!!!

Oh, and before I forget it: this post is pure awesome. If my physics teacher at school would have used examples like this, I would have fallen in love with physics at a much younger age. This is great!

Oak Trees are awesome

rallain — Fri, 02 Apr 2010 06:00:17 +0000

Oak Trees are awesome

In this part of the world, we have oak trees. Technically they are called live oaks - but I don't get it. Of course they are alive. I was at a soccer game and this is the tree I always look at.

Look how far those limbs extend horizontally. That branch is about 12 meters long. Why is this amazing? Have you ever tried to hold an 8 foot 2 x 4 board horizontally by holding one end? Pretty tough. How about I calculate the forces needed to hold that branch in place? I will do a simple model and then maybe later I can make it more complicated. Suppose I replace that limb with one straight uniform limb that looks like this:

In this replacement limb, I am going to say it is a cylinder that is 9 meters long and 30 cm in diameter. Let me assume that this limb is connected at two points to the tree (the white dots). So, for this limb to stay there, the following must be true:

The first two equations say that the total force must be zero. The last one says that the torque about any point must be zero (since it is in rotational equilibrium about any point). First for the forces. There is the gravitational force. This pulls on all parts of the limb, but I can represent this as one force pulling on the limb at the center of mass (long ago, I said I would explicitly show this - but I haven't yet). Then there are two other forces. Let me pretend like there are two pins that hold the limb to the tree. Each of these pins can exert a force in the vertical and horizontal direction I will call these F_1-y F_2-y etc...where the top pin will be 1. That is 5 forces.

For these forces and the first two equations, I get:

So, already I have some constraints. The horizontal components of the forces from the two pins must be equal and opposite. The vertical components of the forces from the pins have to add up to the weight of the log. Now for the torque, I am going to add up the torques about the lower pin. Let me draw a distorted view of the log so that the important distances can be seen.

What is torque? Torque is like a rotational force. Here is an example, what if you try to open a door by pushing near the hinges? It is much harder than pushing near the handle, right? When rotating about some axis, the torque is:

Here F is the applied force, r is the distance from the point where the force is applied to the axis. Theta is the angle between F and r. I will call torques that would make a rotation counterclockwise positive (really, torque is a vector). So, what is the some of torques about axis O (that passes through point O)? First, there are some forces that have zero torque. Both of the vertical pin forces have either theta = 0 or r = 0 so that the torque is zero. The same is true for the horizontal force on the bottom pin. This just leave two forces that have non-zero torques:

Now that I have the horizontal force on the top pin, the bottom pin has the same value (but in the opposite direction). I don't have an expression for the two vertical pin forces. Let me just say that each has a force equal to have the weight of the limb.

How about some values? First, I need the mass of the limb. If this is a cylinder of wood (with a density rho) then the mass is:

So the magnitude of the two horizontal forces on the pins would be:

If I use my values from above and an estimation of the density of wood, I get:

Wow. Oh, I know I made some estimations but even 50,000 Newtons would be huge. Impressive, most impressive. I salute you mighty tree.

rallain Fri, 04/02/2010 - 02:00

Tags

They are called "live" oaks because they keep their leaves
during the winter, thus still looking alive when their
northern cousins look dead, dead, dead.

Your density is way too low for green live oak. Even seasoned live oak has a density of nearly 90 g/cm3 - it's one of the densest woods around. I'd expect its green density to be nearly double that. This gives the green density as 90 pounds per cubic foot, whatever the hell that is in real units.

Oh, and it's called "live" oak because it's evergreen.

@Ahcuah, @Dunc,

You might be right about the density - I estimated the density from a stick I found near the tree. Who knows how old it was. If the density was higher, then the forces would be higher.

Thanks for the info about "live" - that makes sense.

I estimated the density from a stick I found near the tree

Yeah, I looked - a stick that size is practically all sapwood, but the heartwood is much more dense.

This non-physics major thanks you for this post. I am really going to be looking at trees through new eyes, ones more appreciate of the way in which they are an impressive example of natural engineering. (On a literary note, the photograph of the live oak makes me think of the Party Tree in Tolkien's The Lord of the Ring.)

@Dunc,

thank you tree expert. I am glad that I underestimated the density rather than overestimate.

I'm nowhere near an expert, but thanks anyway. ;)

Wow. Oh, I know I made some estimations but even 50,000 Newtons would be huge. Impressive, most impressive. I salute you mighty tree

Is there a typing error in the line for the net y-force?

I know it's off-topic, but how about a post on kitesurfing and the forces on a kitesurfer. What can a kitesurfer do to travel faster? What role does his mass play?

Trees are amazing. We just had some 65mph wind gusts last eevening and the tops of the trees in my backyard swayed as much as 20 or so feet at the top. Over the last year we had winds in the neighborhood of 106 mph and I didn't lose one tree. My backyard gate didn't fare so well.

I love trees. I ued to climb them and sit in their branches and listen to them in the breezes. I'd hug them. I don't have as much of my innocent connection to them but in being a gardener I care for the trees in my yard ad have planted/nurtured many a smaller shrub, brush trees for visiting songbirds. But I do love to visit and walk through parks with old growth stands of oaks and walnuts, maples etc.

I don't think the density could be higher than 64 lbs/ft^3, because I'm pretty sure oak floats and water is 64 lbs/ft^3.

Hmmm. From my experience, live oak wood will float, but barely. If you really want to see a tree that looks like it is physically impossible to stay upright, check out the mesquite. They tend to grow sideways farther than they grow up, and sometimes all limbs on one side.

Unit conversions and decimal points ...
90g/cm^3 is 5 times denser than anything on earth, I guess 0.9 g/cm^3 is more realistic? That gives 56.25 lb/cft.

Probably find that the tree isn't so much "strong" as cleverly dare I say it? "designed"
I.e the load is spread much farther than at those 2 points, possibly opposite branches are counterbalanced and are effectively self anchored around the trunk? Load could be spread
in the trunk, suspended above and butressed below the branch line with the tree fibers acting as cables and supports but running deep into the vertical trunk, Sorry no equations, just a day dream thought experiment? :)

Apolo Ohno Physics

rallain — Wed, 17 Feb 2010 19:28:07 +0000

Apolo Ohno Physics

It is winter Olympics time and time for physics. The event that I always gets me thinking about physics is short track speed skating. It is quite interesting to see these skaters turn and lean at such high angles. All it needs is a little sprinkling of physics for flavor.

Check out this image of Apolo (apparently, it is not Apollo).

How about I start with a force diagram?

I know what you are thinking...F_cent....what force is that? Yes, I am going to use the centrifugal force in this case - but remember that sometimes fake forces are awesome. In short, if I want to pretend like Apolo is not accelerating then I need to add the fake centrifugal force (which is in the opposite direction as the actual acceleration). Remind me later and I will re-visit this problem without using fake forces. Anyway, the centrifugal force will have the magnitude:

Here v is the speed that Apolo (or any skater) is moving and r is the radius of the circle he (or she) is moving in. I drew this vector for the centrifugal force as acting at the center of mass of the skater. This isn't exactly true. The problem is that different parts of the skater are moving in circles of different radii. However, the difference probably (but I will look at it later) not that large that it matters.

For the other forces, notice that the ice exerts two forces (well, one force that I broke into two components). There is a component parallel to the ice. This is a static friction force where the skate blades cut into the ice. Also, the ice pushes up perpendicular to the ice. This is the normal force. I assumed that the kinetic friction force (which would be into the page opposite the direction of motion) is small enough to be ignored. Really, that is the cool thing about ice skating. Ice needs to be low friction in the direction the skate moves and high friction perpendicular to the blade.

Back to the force diagram, there are two things to consider. The forces must add up to the zero vector (because I am assuming the reference frame of the skater is not accelerating). Also, the torque must be zero about any point. For this case, I will choose the point where the skates touch the ice. This will give three scalar equations (two for the forces an one for the torque). Forgive me, but I am not going to go into all the torque details for now. Wait - I forgot one more parameter - the distance from the point where the skates contact the ice to the center of mass. I will call this distance s (for no particular reason).

Now I can make a substitution for both the centrifugal force (which I wrote above) and the frictional force. I will assume a coefficient of static friction of mu_s. I will also assume that the skater is just at the point of slipping. This means that the static frictional force is the greatest it can be (so there will be an equal sign and not a less-than-or-equal sign)

Substituting in for the friction and the centrifugal force in the x-direction force equation:

And again for the y-direction substituting for the centrifugal force:

There is another important relationship here. I am going to assume that the sum of the frictional and normal force must be directed towards the center of mass. This means that:

And now, using this in the x-direction force equation to eliminate mu. I get:

This gives me the speed of the skater in terms of the angle he (or she) is at and the radius of the circle the skater is moving in. It turns out I get the same thing if I solve the y-direction force equation (and that would have been a little simpler). Does this result seem reasonable?

Do the units work? If g is in N/kg (same as m/s²), then g*r will be m²/s². When I take the square root of this, I get units of m/s - that is good.
If r is constant, what should happen as theta gets larger? This should be slower speed. It is sort of difficult to see from that function, so let me make a quick plot.

That plot looks pretty good. For an angle that approaches 90 degrees, the skater's speed would be smaller. A skater wouldn't have to lean at all if the skater was stopped. As the angle gets smaller (approaching zero), the skater would have to be going faster and faster. That is just what that graph shows.

So, let me see if this works. What is the radius of a short-track? According to the ISU (International Sk8ing Union) the inner radius must be ~~25-26 meters~~ 8 - 8.5 meters (see correction in comments). What about the angle? From the picture of Apolo above, I get about 33 degrees (0.6 radians). Using these values, I get a speed of:

note:

This is the part where I discuss why the speed was too great. I originally used the 25 meter radius and calculated a speed of around 42 mph. I will leave this part in here even though it is incorrect.

Clearly, this is way too fast. Apolo's best time on the 500 meter race is 41.5 seconds. This gives an average speed of 12 m/s. Is the angle the problem? I don't think so. Looking at the plot of the function above, the angle would have to be around 50-60 degrees for the speed to be 12 m/s. Is it because he is pushing on the ice with his hand? Again, I don't think this is the case because sometimes they don't touch the ground. What about the radius? Even moving the radius down to 23 meters doesn't make that big of a difference.

The problem must be with one of my assumptions. I suspect the assumption that the "center of the centrifugal force" was at the same location as the center of mass. This would make a difference. Now I guess I will have to calculate that.

Update

You should give me some credit for knowing something was wrong. According to commenter Milan, the radius is around 8.5 - 8 meters. You can take off some points in my internet looking-up scores. This gives a speed of 24 mph - that I am much happier with. I will fix the figures above.

rallain Wed, 02/17/2010 - 14:28

Tags

Might it have something to do with part of the non-accelerating assumption or the kinetic friction force? He has to do an Olympian level of work climbing towards the center in order to maintain that 12m/s.

The 25-m radius is for the regular track, I think -- short track has a smaller radius.

Found it -- the radius for the short track is around 8.5 m. That gives you a speed of about 11 m/s.

Not related to your math here, but one of the talking sports heads mentioned that short track blades are actually slightly curved and not asymmetrical on the bottom of the skate.
Here's a link I just found:
http://www.nbcolympics.com/short-track/insidethissport/equipment/newsid…

It says the asymmetry is to allow the skater to lean more without the boot touching the ice, but it must also move the blades slightly closer to the center of gravity during the leans.

The curved blade probably decreases friction in the desired direction of motion and adds more surface area and friction perpendicular to the blade.

There must be some fun calculations behind the blade locations and curvatures.

The inner radius of the turns on a 111 meter short track course is 8.0 m although the "measured track" radius is 8.5 m. Top competitors try to hug the 8.0 m track edge, but that's very difficult at their highest speeds, and in timed trials they often swing wider on turn entry and exit, approximating a wider turn radius.

Blades are offset on the boot to provide clearance on turns, and are both ground to a slight radius (tips higher than middle) and curved slightly longitudinally (center to right of tips) to maximize edge on ice on turns.

@Milan,

I was thinking about the radius last night. After watching the olympics, I was sure I had the wrong value - thanks for your input.

Wondered if the skater changing leg alters the amount of time friction is excerted on ice - on flat blade and forward. Look what happens when they change leg. You know when you see slow motion of animals running, at some point they are air born - less friction so more area covered over time. In this senese, is the curve of the skater also a result of the changes of friction by each change of leg on the ice surface?

I meant that path of the curve the skater takes, not the angle of the skater.

Please do curling. The most I could find on the Internet was an explanation of why a curling stone curls in the opposite direction to, say, an upturned tumbler on a glass surface. But I don't understand where the sideways force on the upturned tumbler comes from either....

Concerning the discussion about why the skater makes a curve, doing Inline Speed Skating i'm experiencing, that the skate starts the curve as soon as it isn't orthographic to the floor.
Even with only one foot on the ground for the whole corner (thus on air time) you can make it around the corner (see short track skaters especially).
I assume that cornering is caused by slight drifting.

I love your work here, Rhett.

I have a suggestion for a future post, perhaps during the NBA finals: "Spudd Webb Physics."

Because who among us didn't ask ourselves about him "How Dat?"

On terminology: Simply call it "the centrifugal pseudoforce" and get on with it....the readers will understand what it means.

I have nothing of importance to contribute here, but I just stumbled across this and had to say how awesome it is.

I just stumbled across this page. Very cool!
As a former short track speedskater with a degree in Biomechanics I can confirm everything posted here.

For those who are still curious, I'll throw in some more data.

The radius of the actual track layout is exactly 8m radius.
Obviously they dont skate ON the layout. The actual radius that anyone skates ranges from as low as 8m to as high as 16m, all within the same turn. It is never consistent through the entire turn. So any scientific calculations could only be specific to the instantaneous sample. 12m/s is a very typical velocity for short track. The actual question to be asked with these calculations is "What radius is he skating at this exact point?"

The blades are both curved and offset towards the center of the corner. In most cases the heel will be centered. On the right foot the front of the blade passes approximately between the big toe and second toe. On the left foot, depending on the width of the skaters foot, it will pass +/- through the center of the second smallest toe.

The "rocker" of a blade ranges from 8m to about 12m radius. The "rocker" is the normal curve in the blade you would expect. A few skaters will have a consistent rocker of only one radius. Most elite skaters have a compound radius, flatter in the center and more round at the front and back.
Flatter glides better in the straights. Rounder allows tighter corners and pivots. Tweaking the ankle at different points of the track positions the force through a different part of the blade and thus a different rocker value.
The "Bend" of the blade is a curvature that allows for more blade contact while cornering. Both the right and left blades have a slight concave bend as referenced from the center of the track radius. As the skater leans over, more blade contacts the ice. This curve ranges from about 18m to around 26m radius.
These two parameters combined result in some very complex relationships with the ice. In real life the numbers are far to complex. Thus elite skaters spend most of their careers going through trial and error to find the best setup. Once they find something they like it becomes a hugely guarded secret. As an aside, if you look closely at any pictures that show the bottom of someones skate you will see tape next to where the blades attach to the boot. This tape has little pen marks to help the skater remember past positioning. In other words, all of these things are constantly being tweaked. EVERYONE has a wrench with them while on the ice during practice.

To comment 14, Craig,

Hey thanks for that. All a bit clearer. I wasn't suggesting that the skater becomes air born when changing leg - the animal analogy was to highlight what more can be seen when slowing down a process.

By the way, I posted my first comment on this page a few days ago here, thinking I was still on the New York Time Science section and commenting on that - I was reading an article on NYTS then clicked here thinking it was still the same site. So, now accidently quite glad I saw this kewl site for the fisrt time.

Thanks,

Claire

l like ice cream

I am a short tracker myself. I just wanted to note the hand touches the ice only when we are near top speed and need extra support

I don't know if you know this but Apolo has angled cups on his skates. the blade isn't quite perpendicular to his foot which means that the angle your measuring might be off.

The angle remains the same since what is being measured is the angle that is created from the line that starts at the point of contact on the ice and passes through the body's center of mass.

Angled cups don't effect this measurement. In other words, they don't change the angle of lean. Just the contact angle of the blade to the ice.

Nice breakdown of the basic forces

But I think you overlook some things
1- The lean angle is not the same all the way thru the corner- you fall in to a maximum lean at the apex- the pop backout the exit- it's an orbital tube - think of a swept volume tracing where all the body parts have been over time.

2. The COM is not fixed as the body moves - as you say the diff body libs have different orbital radii, and it really is a big difference - and all have different moments of inertia, especially the parts that swing- like legs and arms.

3. You ignore the corriolis force ( yes psudo forces are awesome, and are very real) especially on trailing limbs with reference to the first person frame of ref.

4. You ignore the Newtons law of conservation of angular momentum, Iw = Iw - this is a key equation for objects that change their orbit, it's what gives good skaters that "slignshot " effect when they tighten their track. It's also why core stability is so important, without it you get no efficient conversion of angular momentum to radial spin/torque.
Newton says " In a rigid system angular momentum is conserved" So rigidity is important,even when you are moving a lot, you have to keep some rigidity /core stability

Also there are many internal torques in the body that act to store and release rotational and compression energy (like plyo effects from muscles and tendons), when skaters fall they sometime spin out anti clockwise- as the rotations are contained by the centripedeal force of the blade path, then released when they fall

I'd say there is a bunch of Euler transforms involved and you would need a bunch of Quaternion math to solve it, espcially the blade dynamics.

The rotational dynamics are very complex, and are the hidden forces that can be quite non linear, as the blade behaves dynamically depending on angle of lean, blade profile and torque and vert loading applied to it.

this might be interesting to u

http://www.freepatentsonline.com/5320368.pdf

Great thread- I have some other writing on this I will dig out.

Cheers
Bar

Found this on the web- can't remember where . .
Generating Angular Momentum
An object does not just typically have angular momentum. Recall Newton's first law that an object in motion tends to stay in motion. Well, if a figure skater is just skating straight down the ice and then needs to perform a spin or jump with several rotations in the air, he or she needs to generate angular momentum. Angular momentum is generated by the skater applying a force against the ice. The ice then applies a ground reaction force on the skater. This ground reaction force causes gives the skater angular momentum.
The point of application and line of action of this force is critical. If the line of action of the force is directed through the skater's axis of rotation, then he or she won't spin. The force must cause a torque, or moment, which means it must be applied some distance from the axis of rotation AND have a line of action which does not go through the axis of rotation.

The larger the force or the farther the force is from the axis of rotation, the larger the torque. The larger the torque, the greater the angular momentum.
Another key consideration in generating angular momentum is the object's moment of inertia. The larger an object's moment of inertia, the more angular momentum the object can obtain. For example, if a figure skater wants to generate a lot of angular momentum, they should have their arms spread wide, which increases their moment of inertia. In this position, while the skater will have to have a large torque to start rotating, his or angular momentum:

will be larger due to the large I. A skater who starts spinning with his arms at his side, with the same angular velocity will have a smaller angular momentum. Moreover, this skater will not be able to increase his speed in the spin, because he will not be able to reduce his moment of inertia to increase his angular velocity. Two animated figures are provided to illustrate this idea.

The larger the moment of inertia the more torque it takes to start the object spinning. Thus, there is a trade-off between moment of inertia and angular velocity when generating angular momentum. In figure skating, the skaters do not usually have a problem with having producing large enough torque to start spinning. Accordingly, it is to their advantage to start every spin, or rotational trick, with a large moment of inertia. They accomplish this by having their arms and free leg held away from their body.

Some skaters reach rotation speeds of 7 rev/s during a jump. This corresponds to 420 rpm (revolutions per minute). This is as fast than the idling speed of the engine on some cars!!!!

Given the following moments of inertia and angular velocities of the skaters initiating spins, calculate their angular momentum and answer the questions that follow. Note that when calculating angular momentum, it is important to convert any angular velocity to readians/s before performing the calculation.

Also this is interesting, and often overlooked in ST technique.

----

The Physics of Ice Skating - Angular Momentum
Â Â Â Â Â Â Â The angular component of linear momentum is angular momentum. When an object rotates around a fixed axis, the force acting on the object is called the centripital force. This force points inward, toward the center of the circle traced by the rotation. The velocity of the object points tangential to the circle traced. This is illustrated by swinging a ball on a string around your head (donât hit any lamps though). If the ball becomes detached from the string, it goes flying in a straight line.
Â Â Â Â Â Â Â The vector for angular momentum points perpendicular to the velocity and force vectors. It goes according to the âright hand rule.â This is just a simple way of remembering where the angular momentum vector is pointing. Angular momentum is represented by the equation L=I where I equals the moment of inertia and is the angular rotation or the period of rotation divided by 2 . The moment of inertia depends on the mass of an object and also the distribution of that mass around the axis of rotation. So a skater can have a different moment of inertia based on whether their arms are extended or not. This can be compared to linear momentum where p=mv or linear momentum equals mass times velocity.
Â Â Â Â Â Â Â Angular momentum is conserved when no outside torques act on an object. As say, the moment of inertia decreases, the angular rotation has to increase to keep the same angular momentum. This is most evident when a figure skater spins. A skater starts the spin with arms outstretched (a large moment of inertia).Â As the skater brings the arms in (decreasing the moment of inertia), the rotational speed increases. This is how those incredible spins skaters like Paul Wylie, Todd Eldridge and Kristi Yamaguchi are accomplished. Along with many long years of practice.
Â Â Â Â Â Â Â Most of the spins done by world class figure skaters are edge turns, meaning they are spinning while remaining on an edge. For beginners, often the first spin learned is the two-footed spin. A skater rides a large curve with most of their weight on an outside edge. As the curve spirals into the center, the skater rises up on the flats and begins to spin. One of the most important aspects of a spin is how to âcenterâ a spin. This refers to the property that the spin should stay in one place and not travel all over the ice (which is quite hazardous). This requires converting all of the linear momentum into angular momentum. (Another conservation law)
Â Â Â Â Â Â Â Another example of conservation of angular momentum occurs when a massive star (meaning several times the mass of our sun) dies. As the star, which is already rotating, begins to collapse, it becomes a smaller sphere which decreases its moment of inertia. Since the star is an isolated system with no forces acting upon it, the angular momentum must be conserved and the rotational period of the star increases. If the star (known now as a neutron star) is emitting a beam of radiation, its rotational motion makes this beam appear to us like pulses. These stars are known as pulsars.
Back to the Physics of Ice Skating
by Karen Knierman and Jane Rigby

The Physics of Ice Skating - Torque
Â Â Â Â Â Â Â Torque is a rotational force, in fact the word itself comes from the latin for âto twistâ. Torque, in a sense, causes rotation about an axis. Torque involves both the force applied to an object and the distance from the rotation axis you apply the force. Perhaps some examples will help. In order to open a heavy door, you need to apply a force. But force alone will not do the job. Where the force is applied and in what direction is also important. If you apply a force close to the hinge of the door rather than out by the doorknob, it is much harder to move the door. Thatâs why doorknobs are located at the opposite side of the door to the hinges; itâs much easier to move the door out there.
Â Â Â Â Â Â Â The definition of torque is the product of the distance from the axis of rotation (often called the lever arm) with the force that is perpendicular to the lever arm. (If you pull on a door parallel to the plane of the door, you do not rotate the door.) Another example of torque occurs when you turn a screw or bolt. Using a screwdriver (the non-electric kind) is often hard and time consuming since you must apply a large force in order to turn the screw (small lever arm). However, if you use a wrench for tightening bolts, you only need to apply a small force since you have a long lever arm. That's why wrenches used to turn large bolts have much longer handles compared to those that turn small bolts. This enables the user to use less force since they have a long lever arm. Of course, the user must apply that force over a longer distance. Â So there's a tradeoff between force and distance.
Â Â Â Â Â Â Â So how do we get from tools to ice skating? A skater, in order to rotate, must exert a torque on his body by pushing against the ice. In edge spins, the skater pushes one foot against the ice to start the turn. You also see this in multiple rotation edge jumps. In these jumps, the skater takes off from the ice, turning the skate as he does so, which creates a torque. Â Thus, the skater spins!

Â You can think about the physics of moving objects in two different and equally acceptable ways: in terms of forces, or in terms of energy. Which you use depends on which is more convenient; both will always work. Â In this section, we'll look at energy in ice skating. Â It's an energetic section!
Â Kinetic and Potential Energy Explained
Â Â Â Â Â Â Â Kinetic energy is the energy of motion ("kinetic" means motion). So an ice skater flying across the ice has lots of kinetic energy. When he slams into the boards, that is transferred abruptly into thermal energy, sound, and work done on the skater (ie compressing his chest, rearranging his face, etc.) Â Seriously, to measure kinetic energy (KE), just measure the mass of the object and its velocity. KE = 1/2 mass*velocity*velocity or 1/2 mv^2. Â
Â Â Â Â Â Â Â If the movement is rotational instead of straight-line, then the equations are similar. (See Karen's section on rotation.) Â The idea is the same - how fast the skater is spinning is directly related to her kinetic energy.
Â Â Â Â Â Â Â It's easy to "see" kinetic energy in the motion of an object. But what about potential energy? Potential energy is stored energy. Energy can be stored in chemicals, by compressing a spring, or by doing work against gravity (ex: by placing an object on a higher shelf.) Â Muscles act like springs, so the chemical potential energy of muscles is converted, by the muscle applying a force, into kinetic energy of motion.
Energy Conversions in Jumping
Â Â Â Â Â Â Â In a jump, the skater uses chemical potential energy (muscle power) to gain speed across the ice. When she jumps, she's also converting her chemical potential energy into kinetic energy. As she flies upward, her kinetic energy is converted into gravitational potential energy; as she slows, she gains height above the ice. At the top of her jump, she has no kinetic energy (for a moment!)Â Â There, all of her former kinetic energy is now gravitational potential energy. As she falls back down, her potential energy is converted back into kinetic energy. At the moment she hits the ice, all the gravitational potential energy she had at the jump's peak is again kinetic energy, so she hits pretty hard. If you measure the height of her jump, you can determine how hard she pushed off, which is also how hard she smacked down.

The Physics of Ice Skating - Isaac's Third Law
Â Â Â Â Â Â Newton's third law is one of the most-quoted in physics: Â Â for every action, there is an equal and opposite reaction. Â Lots of times, this is used out of context (to describe politicians, etc.)
Â Â Â Â Â Â The basic stroke in ice skating provides a good example of Isaac's 3rd law in action. When you "stroke" (the basic push in ice skating), you apply a backwards force to the ice. The ice applies an equal, forward force on you, so you go forward. Skaters stroke at an angle, so part of the stroke is wasted. You're pushing forward and to the side. The side push is resisted by the edge of your other blade. The forward push is resisted only by ice/blade friction, so you go forward.
Â Â Â Â Â Â Â Actually, this is exactly why sailboats go forward, not sideways. The wind pushes the boats forward and to the side. The sideways push is resisted by the long keel, but the forward push is relatively unresisted. Boats are designed to be aerodynamic (actually, "hydrodynamic) to forward motion and are intentionally unhydrodynamic to sideways motion.
Â Â Â Â Â Â Â Â Note that these forces apply only to direct pairs of objects. I push on the ice; it pushes on me. As I push in on the wall, it pushes me outward. Third law reactions never involve a third body.

For reference

http://www.youtube.com/user/Barraldinho#p/c/D2C4A989CEE5FFB1/1/j8Qrfo7A…

surprising that so much mental muscle has gone into accepting the fact that we lean round corners. Even my horse knows that (which can be daunting as he leans to go round a tree and near wipes you out). And the fact of maybe coming out of a turn faster than entry is clearly nothing to do with some imaginary sling-shot effect. Quite simply if your linear speed out > linear speed in then yo have inserted extra energy (more than reqired to ovecome resistive forces). which is what they do. QED

Of much greater interest - and perhaps not such risk of confounding by minutia - is the question of the "slalom" style of propulson of long track long distane (10km olympics for example) speed skaters. They constantly skate a curve path (to left and right) up their 100m straight. Putting aside the illusory sling-shot model, how do they insert forwards thrust into their bodies by swerving side to side. no other sport (motorcycle racing, even salom ski racers on their final straight swoosh to finish) - nobody believes that swerving increases speed.

Wow - thanks - this is great!

I stumbled across this photo looking for for images of skaters to put into a biomechanics exam that I am preparing. Funny that the photo I clicked on happened to be associated with the exact problem that I was formulating!

Yo-Yo: Rolling, sliding, pulling

rallain — Wed, 27 Jan 2010 15:20:31 +0000

Yo-Yo: Rolling, sliding, pulling

This is actually been sitting around for a while waiting for me to post it. Here is another short Christmas-toy demo. I am going to pull this yo-yo at different angles and on two different surfaces. Check it out.

What is going on here? Let me look at the first case where I pull the yo-yo and it slides without rolling. Here is a diagram.

Normally, I would just say - "hey - a free body diagram". And this is one, but you have to be careful. Normally, a free body diagram treats an object as though it were a point mass. You can't do that in this case because you have to consider rotation also (points can't really rotate). When I draw a diagram as a point, this is the key thing I am looking at:

Which I could break into 2 or 3 component equations such as:

Since this object can rotate, I must also consider that with:

I can't believe this, but I never really had a post just devoted to torque. Weird. Well, here is a post that basically goes over all the ideas of torque - Friction Demo with a meterstick. In short:

tau is the torque about some axis (labeled as O). You can think of torque as the rotational equivalent of force.
I is the moment of inertia of that object about the same axis as the torque. The moment of inertia can be a complicated thing, but in this case it can be thought of as the object's resistance to change in rotational motion. The moment of inertia depends on both the mass of the object and how this mass is distributed about the axis of rotation.
Alpha is the rotational (angular) acceleration.

Hopefully, you can see how similar this last equation is to the linear version (Newton's second law). Ok, I am proceeding onward. Back to the yo-yo. Really, I have three equations - the x equation, the y-equation and the rotational equation. I need to note a couple of extra things. First, I will call the radius of the inner part of the yo-yo r and the outer radius R. Also, the mass is m, and the coefficient of static and kinetic friction will be mu_s and mu_k. This gives the following:

A couple of notes:

I picked the case of the sliding and not rolling yo-yo because: the acceleration and angular acceleration are zero. The friction is kinetic friction. This means that I can determine its value. For static friction, I can only calculate the maximum friction. (here is a review of friction)
The acceleration in the y-direction is zero since the yo-yo stays on the table.
I can use the model for friction to get an expression for F_f (did you notice I changed F_friction to the shorter F_f?)
Also, I have shorter notation for the force from the table (F_N), tension (F_T) and the gravitational force (mg)
There are 4 forces. However, I only show two torques. The torque from the force the table exerts is zero about the axis since this forces points right through the axis. The torque due to the gravitational force is also zero. This is because gravity pulls on all parts of the yo-yo.

Here is the model for kinetic friction. Note that this an expression for the magnitude of the friction force - it is not a vector equation.

With this, I can replace all the F_f and I get:

Now, I will get an expression for F_T from the last equation:

And now I can substitute this in the other two equations. I get:

From the top expression, if F_N is not zero, then:

So, this says that the angle needed to pull the yo-yo so it doesn't slip only depends on the ratio of the inner and outer radius. Note that r would be smaller than R so that the ratio would be less than 1. This is good because the cosine function must produce a number less than one.

If you take the video above and analyze it with Tracker Video Analysis, I get that the yo-yo slides at an angle of about 53 degrees. You should notice that I repeated the experiment with the yo-yo on a different surface (WebKinz mouse pad) that was much slicker. The angle of the string was still 53 degrees. Since the coefficient of friction wasn't as much, I didn't have to pull as hard (for constant speed) but it was the same angle.

If you wanted to, you could measure the outer radius of the yo-yo and use this to calculate the inner radius.

The other two motions:

What happens if I increase the angle of the string above 53 degrees? The frictional force will be less. This is because if I pull at a greater angle with the string, then the normal force will be smaller (since it doesn't have to exert as large of a force to make the vertical acceleration zero). This smaller normal force means the frictional force will be smaller and thus a smaller torque from the friction. Both of these together make the torque larger in the direction that makes it roll to the left.

If the angle of the string is too small, the frictional force will be greater (basically because of the opposite of above).

I think the coolest part of this demo is that by pulling at different angles you can make the yo-yo roll right, roll left, or slide (not roll).

rallain Wed, 01/27/2010 - 10:20

Tags

Hi Rhett - love this problem. If you're interested in which way it does (or doesn't) roll, why not make the axis the contact with the table? Only one torque then - none if it's not rolling! That makes solving for the critical angle a one-liner, and drawing angles greater and smaller and finding the direction of the cross product show instantly which way it rolls.

Finding the critical angle for this problem was on the Caltech Physics placement test when I went there 20 years ago. I thought it was pretty neat then too.

Can you post how to find the linear acceleration of the yoyo. I think my solution is flawed.

RP 2: The Physics of Fantastic Contraption

rallain — Wed, 23 Dec 2009 09:08:32 +0000

RP 2: The Physics of Fantastic Contraption

One of my students showed me this game, Fantastic Contraption. The basic idea is to use a couple of different "machine" parts to build something that will move an object into a target area. Not a bad game. But what do I do when I look at a game? I think - hey! I wonder what kind of physics this "world" uses. This is very similar to my analysis of the game Line Rider except completely different.

Fantastic Contraption gives the unique opportunity to build whatever you want. This is great for creating "experiments" in this world.

The first step is to "measure" some stuff. The game includes three types of "balls" and two types of connectors. The balls are:

Clockwise rotating
Counterclockwise rotating
Non-driven

Connectors:

wood lines - these can not pass through each other
water lines - these can pass through each other, but not the ground

First question: Do the different balls have the same mass? This can be tested by creating a little "balance"

Now, I can test this by adding two of the same balls on each side (well, one on each side). It is still balanced. Now for two different types of balls:

Note: the blue ball does not spin and the yellow is a clockwise spinner. They look balanced. What about a blue and a conterclockwise spinner? Still balanced. So, it appears all the balls have the same mass.

What is the linear mass density for the two types of sticks? To measure this, I created a device with a ball at one end and the pivot NOT in the center, but it still balances:

Here you can see three forces acting on the device: the gravitational force on the ball, the gravitational force on the stick, and the pivot point pushing up. Since the stick is clearly not a point object, I have to draw it's gravitational force at the center of the stick. (I am not going to derive that right now, you will just have to trust me).

Newton's laws says that the forces must add up to the zero vector if the object is staying at rest. This means (in the y-direction, where y is up):

Here m_s is the mass of the stick and m_b is the mass of the ball. This would make the the gravitational pull on the ball -m_bg (notice it is the y-component, so I can have it negative). From all of this, I could solve for the force the pivot pushes on the balance, but what good is that? What I am really looking for is the mass of the stick. To do this, I need to consider torque. Here is the real definition of torque:

This definition is a little more complex than I want to go into (but I had to say it). The torque is technically a vector resulting from the cross product of a force and a vector from the point of rotation to the point the force is applied. The scalar version of torque can be written as:

Here, r is the distance from the point that you want to calculate the torque about (I chose the pivot point) and the point where the force is applied. Theta is the angle between the force and the distance to point about which to calculate the torque. In this case, the angle is 90 and sin(90) = 1. Another important consideration is the sign of the torque. I will arbitrarily call counterclockwise torques positive and clockwise torques negative.

So, how do I use torque? Well, I need to know the distance from the pivot point to the center of the ball and from the pivot point to the center of the stick. I can use my favorite free video anlaysis program, tracker, to do this. (even though it is just an image)

I will use the diameter of one of the balls as my unit (from the center of an attachment point circle to another one). Doing this, I get the distance to the ball and the center of the stick as:

Here I am using "U" as my distance unit - described above. To find the distance from the pivot to the center of the stick required some trickeration. I measured the length of the stick. I then used half that distance and measured from the one end of the stick to find the center. Knowing that point, I could then measure to the pivot point. Using these measurements in the torque equation:

Note that the torque due to the pivot does not contribute at all. This is because I calculated the torques about the pivot point. The distance from the pivot point to the pivot point is zero (thus zero torque).

So, I have the mass of the stick in terms of the mass of the ball. I can also get the linear mass density of the stick:

Cool - I should stop here. No!!!! I am on a roll. I will now calculate the linear mass density for the "water" stick. I can't do quite the same thing because the water would fall through the pivot. Instead, I will do the following. First, I will make a stick with two ball (one on each end) balance. Then I will replace one of the balls with "hanging" water so that it is still balanced. At this point, the mass of the water stick will be the same as the ball (I could have done this with the wood stick if I had thought of it then).

You may not be able to tell, but this is two overlapping full water sticks and one shorter one. I will have to combine the length of all of these. This gives a total length of water = 8.5 U. So, the linear mass density for water is:

Interesting. The linear density is half that of the sticks. Must be dense sticks. I tried putting a wood stick versus a water stick that was twice as long - they balanced.

Acceleration of falling objects

Do things accelerate? Is there air resistance? I created an engine that just kind of "flung" a ball up. I used copernicus to capture the video from the screen. Then tracker video to get position time data. Here is what I found:

This shows that it does indeed acceleration. Using the ideas from a previous post on graphing, the acceleration of the object is twice the coefficient in front of the squared term, this means that:

If this is on Earth, then this acceleration should be 9.8 m/s². With this assumption, I can find the conversion from U to m:

What is left?

Questions to answer:

Is there air resistance? From the above data, maybe not. To test this, I need to launch a ball with very high speed. If the horizontal velocity changes, then there is likely air resistance
Make a pendulum, does it oscillate at the expected rate (assuming the dimensions from here)? I already started to set this up, but there is CLEARLY some type of frictional force slowing it down.
Friction - what is the coefficient of friction? Does this game follow the model for friction where the frictional force is some coefficient times the normal force?
What kinds of torque are these rotating balls capable of
What is the moment of inertia of these balls? Are the cylinders or spheres?

I will probably answer some of these questions - but if someone ones to answer them first, I will gladly link to your results OR post them here.

RePost Note

Actually, I did look at Fantastic Contraption some more. Here is the other stuff I did:

rallain Wed, 12/23/2009 - 04:08

Tags

fantastic contraption

Physics

analysis

Angular Momentum Example

rallain — Thu, 03 Dec 2009 13:31:51 +0000

Angular Momentum Example

I showed this demo in class and I was surprised at how cool the students thought it was.

They actually thought it was some kind of trick. It is not a trick. Instead, this is an example of the angular momentum principle. If you want to try this yourself, I guess you are going to have to find some type of wheel. I attempted to get this to work with a small Lego wheel, but it wouldn't spin fast enough. You should be able to do this with one of those toy gyroscopes though. Anyway, here is the angular momentum principle:

Or, if you prefer it without a derivative it could be written as:

Here Tau is the torque on the object (about some point) and the vector L is the angular momentum as the object rotates around an axis through that point. Wow. I was going to link to a previous post where I talked about angular momentum and torque. However, it seems I have never done that. Ok - short version: Torque is like the rotational force, it can be defined as:

The second line is the magnitude of the torque. The vector r is a vector from the point of rotation to the point where the force is applied to the object. The angle theta is the angle between r and the force. Remember that torque (just like force) is a vector. Direction matters. If you don't know what the cross product is, just think of the magnitude expression. However, you do need to know the direction of the torque for this example. The torque must be perpendicular to both the force and the r vector. There will be two vectors that meet that criteria. Choose the one such that when the fingers of your right hand cross r and then F, your thumb is in the direction of the torque.

What about angular momentum? For an object rotating about a particular axis, the angular momentum can be defined as:

I is a scalar quantity called the moment of inertia. It is best to think about this as the "rotational mass". Just like normal mass makes it more difficult for a force to change the velocity of an object, rotational mass (moment of inertia) makes it difficult for a torque to change the angular velocity. The moment of inertia depends on the mass of an object and how that mass is distributed about the axis of rotation. Note: things can actually get much more complicated than this. The angular momentum can be in a different direction than the angular velocity - but in this case, I is not a scalar quantity.

Here is the short scoop on the angular momentum principle. If you recall the momentum principle, it basically says that forces change the momentum (linear momentum) of an object. The angular momentum principle says that the net torque changes the angular momentum of an object.

Now back to the bike wheel. Here is a diagram showing the wheel while it is spinning.

I left off the vector representing the torque because it would be out of the screen. Here is a diagram from above.

What if I look at what happens after a short time interval, delta t? In this case, I could write:

And I can represent this with the following diagram:

Ok, one problem. I have two angular momentum vectors, but the wheel is still rotating in the direction of the first angular momentum. Oh well, I think you get the idea. But, in case you didn't here are the keys:

Angular momentum is a vector
Torque is a vector
The net torque CHANGES the angular momentum
The new angular momentum will be in a different direction
This is really could be made way more complicated, but I left some stuff out.

The last point - why do people like this demo so much? I think it is because their everyday experiences say this wheel should fall and it doesn't. If you look at this in terms of forces, the tension from the string and the gravitational force add up to the zero vector, so motion of the center of mass doesn't change. If I have time, I will do a more detailed calculation of this demo.

Update:

As pointed out by Dr. Pion, I made a mistake. The two above diagrams showing the torque should be from BELOW the wheel.

rallain Thu, 12/03/2009 - 08:31

Tags

angular momentum

demo

torque

You know what's funny? I watched the demo and it didn't even occur to me that the "cool" thing about the demo is that the wheel doesn't fall (until I read the rest of the post). I guess I've just done too many demos.

The demo my students thought was the coolest was me sitting on a stool holding a spinning bicycle wheel and changing my direction and speed of rotation by turning the wheel.

@Chris,

My students also think the stool thing is awesome. I am with you, I have seen cooler stuff.

San Francisco's Exploratorium and (if memory serves) the Cabot Science Center in Balboa Park in San Diego both have these demos as hands-on exhibits.

It is amazing what some students think is cool. My 3rd period conceptual-level class was awed when I swiped an embroidery hoop from under a dry-erase marker, letting it drop straight down into a glass bottle. Some of my AP students are fascinated by a rubber band stretched between two support stands. That can keep 'em occupied for a good 10 minutes (unless I nudge them to get a move on and collect some data!).

A dramatic illustration of angular momentum:

On a rifle range, two paper targets are set downrange, one at twenty feet, the other twenty feet farther, directly in the line of fire. The shooter fires at the front target.

Observers expect the second target to be holed, like the first, but that does not happen. Instead, the second target remains intact.

Why? A high-velocity .22 centerfire, such as a .223 Remington, fired down a barrel rifled to one turn in 12 inches, exiting the muzzle at 3000 feet per second, is spining on its long axis at 180,000 RPM. When the bullet integrity is disrupted by contact with the fragile target paper, the bullet fragments, breaking outward, and whatever particles were near the axis and so had very little angular momentum, they have such high aerodynamic drag that they fall to the ground less than twenty feet from the impact.

Cooler in person. A bike wheel, a handle made from a rolling pin (the kind used for making pie), and an office chair (the kind that swivels) and they can do it themselves.

I think it would be very interesting to include the following in the demonstration: prevent the precession of the spinning bicycle wheel.
If you prevent the precession then the wheel will drop down just as hard as when it isn't spinning.

A better option than @7, which applies a second torque:

Include the case when L1 = 0 in your explanation. In that case, L2 is in the direction of tau which results in the wheel rotating around that axis -- hence "falling".

I'll put some comments in my own blog about how I fit this in with the demo and the full explanation. I find this (along with magnetic forces) to be the best way of showing that the cross product is "real".

Responding to @8:
I agree that preventing the precession involves applying a second torque.
However, the second torque is different: the torque from gravity follows the instantaneous orientation of the wheel; the second torque has a fixed orientation.
Preventing the precession from commencing has as side effect that the torque from gravity becomes simpler; it's orientation becomes stationary too.

Preventing the precession demonstrates that it's the precession that keeps the spinning/precessing wheel from toppling over.

Haha, so *that's* how that's supposed to work.

In Spectroscopic Elucidation, as we were discussing proton NMR and the precession of the protons involved in it, the professor tried to do this with a bicycle wheel as an example. He'd just taken the front wheel off of his bike, though, and lacked any sort of handle to get a good hold on it, so he couldn't get it spinning fast enough to do, nor have any way of holding one end of the axle strongly enough to support it like that.

@9:
I just saw your comment on my blog and came back here to see your observation. I think it explains why you don't see the relevance of the L=0 case.

The second torque, the one you apply, causes the wheel to "fall" (its center of mass rotates around a horizontal axis), not gravity. In the case of that applied force, rxF points down so the delta-L precesses L toward the floor. If you tried to "help" the precession, the wheel will rotate up (above horizontal).

PS to Allain:
The two diagrams are backwards if they are supposed to be from above. Tau is towards the wall, not away from it.

@CCPhysicist - holy cow! you are correct. Sorry about that.

@11
The following 37 second YouTube video (uploaded by Glenn Turner), offers a good display of spinning gyroscope observations.
http://www.youtube.com/watch?v=NQSYERWASE4

In the first 20 seconds you see Glenn gently moving the gyroscope wheel. He pushes with his fingertips, swiveling the gyroscope wheel clockwise and counterclockwise 10 or 20 degrees. Swiveling one way the gyroscope wheel pitches one way, swiveling the other way the gyroscope wheel pitches the other way. That is: to reverse the direction of pitch the direction of swivel must be reversed. Also you can see that the gyroscope wheel swivels readily, it just adds a motion component: to the swiveling motion a pitching motion is added.

Now the setup that CCPhysicist and I have been discussing (comments 8, 9 and 11):
Start with the setup displayed in Rhett's video. I'll call the torque from gravity 'the first torque'. Precession sets in. As demonstrated in the Glenn Turner video: the precessing motion gives rise to a tendency to pitch up. That tendency to pitch up keeps gravity from pitching the wheel down.

- What will happen when you apply a second torque, around the swivel axis, in such a way that the existing precession rate increases? Answer: the bicycle wheel will pitch up.
- What will happen when the second torque is applied in such a way that it decreases/nullifies the existing precession rate? Answer: then the precession-correlated-tendency-to-pitch-up is reduced/nullified, and the torque from gravity is free to pitch the wheel down.

CCPhysicist has argued that the second torque should be regarded as the cause of pitching down, and not gravity. I think the following analogy applies: think of a satellite in orbit, firing retro-rockets: the altitude of the satellite will decrease. I prefer to say that gravity is the cause of the descent, and that firing of the retro-rockets created opportunity for gravity to pull the satellite down.

Returning now to my original observation in comment 7:
If you prevent the precession from continuing then gravity will cause the bicycle wheel to flop down, just as fast as when it's not spinning.

Hi
please tell me what your branch of physics is ?
thanks alot.

Moonwalking on the moon

rallain — Mon, 29 Jun 2009 22:05:27 +0000

Moonwalking on the moon

Slate's Explainer has an answer and question post about moonwalking. Here is one of the very good questions answered there.

Would it be easier to moonwalk on the moon?

The Explainer says "absolutely not" and attributes this to the awkwardness of walking on the moon. The article gives an example of Earthly legs being too powerful as the "astronaut's hop". The explainer also says it is awkward because of the pressurized space suits.

I think the problem is almost entirely the pressurized suits. I believe that the astronauts do their moon-hop because it is difficult to bend their legs in the pressurized suits rather than because the gravitational field is smaller.

Suppose you had a space base on the moon so that you could be on the moon and NOT wear a space suit. In this case, I think moonwalking would be essentially the same as on earth. As I showed on my previous post on moonwalk, the move has a lot to do with center of mass and the ratio of the normal forces on the two feet.

On the moon, the ratio of the two forces would still be the same as on Earth. I could be wrong. The only way to know for sure is to build a moon base so that we can test this out. Well, I guess someone could go up in the vomit comet while they are simulating moon gravity.

rallain Mon, 06/29/2009 - 18:05

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The physics of Michael Jackson's moonwalk

rallain — Fri, 26 Jun 2009 08:28:51 +0000

The physics of Michael Jackson's moonwalk

Was the moonwalk fake? No, not the Apollo landings. I am talking about Michael Jackson's moonwalk. You got to admit, he had a big impact on a lot of stuff and this is my way to give him respect - physics.

I am sure you know about the moonwalk. Maybe you can even do the dance move yourself, but how does it work? First, here is a clip of MJ doing his stuff.

As a side note, I can't remember where I saw it but there was a great discussion of the history of the moonwalk. If I recall correctly, some were saying Michael didn't create this move. One thing is for sure, he made it popular. Now for the physics.

The key concept here is friction. Friction is actually uber-complicated, but a simple model works for many cases. Static friction is a force exerted on an object when it is in contact with some surface but those two surfaces do not move relative to each other. Kinetic friction is a force exerted on an object when the two surfaces are moving. Suppose I have a block at rest on a table and I pull it with a slowly increasing force. This is what it would look like:

Two key things from this graph. As you pull on the stationary block, the block doesn't move. If I pull with 1 Newton, and it doesn't move then the frictional force is 1 Newton. If I then pull with 2 Newtons and it still doesn't move, the frictional force is 2 Newtons. The static frictional force does what it can to make the thing not move - but not more than it can. This leads to the static friction model of:

In this model, the force is less than or equal to the product of some coefficient (that depends on the two types of surfaces) and the normal force (how hard the two surfaces are pushed together). The direction of this frictional force is parallel to the surface in the direction that prevents the object from sliding.

The other key feature in the graph is the small jump down when the thing starts to slide. This is because the coefficient of kinetic friction is typically smaller than that for static friction. Also, if the object is sliding, the frictional force is constant.

Back to Michael and the moonwalk. The key here is: how do you make one foot slide and the other not slide? If both feet are stationary, then this is dealing with static friction. I could make the frictional forces on these two feet different by changing my center of mass. Here is a free body diagram:

Since he is not accelerating up and down, the following must be true:

These are the forces in the y-direction. They must all add up to zero so that:

There is another condition that must be satisfied. Since he is not rotating, the total torque about any point must also add up to zero. If you want more info on torque, check out this post. But for this post I will just say that torque is like the 'rotational force'. It depends on the point about which you want to rotate and is essentially the force applied times the perpendicular distance to the point of rotation. For the free body diagram of Michael, I have chosen one of his feet to be the point about which he is not rotating (I could chose any point). This makes 3 of the forces have zero torque (N₂, F₂ and F₁ have zero torque because the perpendicular distance to point O is zero). Here I labeled the other important distances:

The only two forces that exert torque about O are the weight and the N₁ force. They have opposite directions of torque because they would cause rotation in different directions. This along with the previous equation gives:

Eliminating mg, and solving for N₁, I get: (I know the indices for the forces and distances don't match)

If his center of mass is in the middle, then r₂ - r₁ = r₁ and the two normal forces would be equal (as you would expect). If the center of mass is more towards the foot on the right, then r₂ - r₁ is less than r₁ and N₁ will be larger than N₂. This will make the frictional force on the foot to the right greater and the other foot slide.

Well, what if r₁ is greater than r₂? One of two things would happen. Either he would fall over, or there would have to be a force pulling the foot on the left down. This is similar to Michael Jackson's trick in "Smooth Criminal".

Here he used special shoes that connect to the floor so that he could do this. More details on this page.

Ok. So that is how Michael gets one foot moving. How does he keep one foot sliding and the other not sliding? It is really the same thing as above except that he can increase the force on the moving foot a little bit more since it is sliding. Sounds easy, but Michael could really make it look cool.

Finally, I just want to show another demo that is essentially the same idea.

Meterstick friction demo from Rhett Allain on Vimeo.

You can find more details on the meterstick demo in this blog post.

rallain Fri, 06/26/2009 - 04:28

Tags

The moonwalk will forever be remembered and so would the LEGEND himself,Michael Jackson!

It is such a great loss that a man with great talent like Michael Jackson dies. RIP King of POP

Michael Jackson is my favorite pop artist ever since i was a child. He is truly the King of Pop and i am saddened by this news.

michael jackson was, is , and will always be music history. I grew up to his music and loved it, now my 3 yr old is wanting me to play his music. im a real m.j. fan stop breaking his dancing down a real fan doesnt say his moonwalk is fake i believe he did it and its the real deal. R.I.P MJJ.

that demo with the meterstick - wtf ? people cannot be that stupid

Talent and efford are not the same. No matter what you are trying to do. Individuals with god given talents can not explain why they can do certain things, while others with endless amounts of practice will never crest the bell.

Talent is what most societies call this power. The beautiful thing is when a person can identify or discover his true god given talent like Michael Jackson. The moon walk was one of his birth talents.

However for you all needing practice...watch this video, click on the link below:

http://www.ourtown.com/snellvillega/article/2009/6/29/learn-how-to-moon…

Beautiful.

But I am told my demo is beautiful (even erotic !) and certainly instructive

www.youcanmoonwalk.com

Peace

R x

Michael Jackson will be the greatest pop singer in the whole world. Everyone will miss him.

| Michael Jackson is truly the King of Pop. i am a die hard fan of him and we are going to miss him now that he is gone.

i am already a great fan of Michael Jackson ever since i was just a little kid. i would really miss the King of Pop -

no one ever will do moonwalking as MJ did.....

all notable tributes and quotes on Michael Jackson from peers: http://www.tributespaid.com/quotes-on/michael-jackson

Michael Jackson is truly the King of Pop. He made a lot of great songs in the area of Pop Music. His death is a great loss to the music industry.

wdf was the meter stick all about like one hand stoppedd nd one hand kept goin we all arent dat stupid nd its just weird nd mj is de best i love him

Michael Jackson is one of the greatest singer in our time. He is really the King of Pop and we would really miss this great person,

Rotational Energy of the Earth as an energy source

rallain — Mon, 22 Jun 2009 17:46:39 +0000

Rotational Energy of the Earth as an energy source

Note: This is an old post from the time before my blog was in wordpress. I noticed there was some incoming link for this, and I never moved it over. Here it is in it's unaltered (except for this part) format.

I don't know why I even suggest a new energy source. Fusion power is only a few years away in the future (just like it as always been). This will replace any other sources of energy that we could come up with. But, I can't help myself, I need to share my idea and save the world. It's what I do. (call me a superhero is you want).

We can get all of our energy from the rotation of the Earth.

The basic idea is to use the rotation energy of the Earth to power our Nintendos and computers and stuff. How much energy could we get out of this and what would we lose? First - what we lose. If we use the rotational energy of the Earth, it would spin at a slower rate. I will start off assuming we make the day 1 second longer.

Currently, the Earth takes 23.9345 hours to rotate. It take the 24 hours for the Sun to be back in the same location. This is the difference between sidereal day and synodic day. I am concerned with the rotational rate of the Earth on its axis, so I need to use the sidereal day. Check the wikipedia link for a great diagram that shows the difference between the two days.

This leads to an angular speed of:

Now, suppose I want to increase the length of the sidereal day by 1 second, this would give a new angular speed of:

How much energy would this produce? (assuming it could all be turned into useful energy).
The energy of motion for an object rotating is:

Where (1/2) is (1/2), I is the moment of inertia about the axis you are rotating and ? is the angular speed. The moment of inertia is sort of like the "rotational mass". An object with a higher moment of inertia is more difficult to change its rotational motion. In this case, we are dealing with a spherically-shaped object (the Earth is mostly spherical). The moment of inertia for this can be approximated by assuming it is a uniform sphere, but it isn't. The density in the center of the Earth is much greater than on the surface. So, for this calculation, I will use someone else's determination of the moment of inertia of the Earth - here is Wolfram's Research value -

From this, I can calculate the change in energy by slowing the Earth down.

and putting in the values from above:

This is how much energy the Earth loses, so we could use this for other stuff. Is this enough energy?

Energy Usage:

Let me just look at the energy usage by the U.S.A. because they would be the ones to harness the rotational energy of the Earth (but really, because I found the data for US energy usage first). http://tonto.eia.doe.gov/ask/electricity_faqs.asp#home_consumption - this site has the data that I started with. It has a spreadsheet with average monthly usage for residential, commercial and industrial:

Residential: There were 122,471,071 consumers that used an average of 920 kilowatt hours per month.
Commercial: There were 17,172,499 users that used an average of 6,307 kilowatt hours per month.
Industrial: There were 759,604 users that used an average of 110,946 kilowatt hours per month.
NOTE: by "users" I mean companies or places or whatever the spreadsheet meant.

A kilowatt is a unit of power - or the rate at which energy is used. A kilowatt-hour is a unit of energy. Since 1 watt is a Joule per second, 1 kilowatt-hour would be:

and the monthly usage in the US would be:

that is per month.
So how many months (and years) would this energy last for the US assuming a steady energy usage? Actually, I will also assume that only 50% of the rotational energy of the Earth goes to useful stuff (like nintendos) and the rest is wasted. This would mean the amount of useful energy would be 2.5484 x 10²⁴ Joules. So

This is a long time. So the length of the sidereal day would only increase by 1 second over this time period (that way wouldn't have to store all this energy, but rather generate it as we use it).

Now for the details:

How exactly do I propose that this rotational energy be harnessed? I will leave that as an exercise for the reader - but I will give a hint: Magnets and wire (I have already said too much).

Also, this method produces no greenhouse gases.

rallain Mon, 06/22/2009 - 13:46

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And we are already slowing down by some fraction of a second per year (depending on the year) due to tidal friction with the moon and other factors. Nobody will notice that this extra increase happened.

Any harness I can think of for this kind of energy requires the first step to be "build space elevator" (or space tether, at least). And, as we've seen, that has some of its own issues.

So how many months (and years) would this energy last for the US assuming a steady energy usage? Actually, I will also assume that only 50% of the rotational energy of the Earth goes to useful stuff (like nintendos) and the rest is wasted. This would mean the amount of useful energy would be 2.5484 x 1024 Joules.

surely the earth has some reluctance so we just need brushes!?

John

i have a new idea about using energy of earth rotation