The Breakdown: Defying Death at the Gym

Amazing, and compelling, sample of conservation of momentum. The video I want to see would be the one with the attempts made to hit that ball with just the right force and angle. Had to be some good wipeouts:) Great video.

Adam
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Allow me to tickle your brain!
www.ThisHoldsMyAttention.com
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I wouldn't call that "defying death"

This discusses physics which can be applied to a person bouncing on a trampoline or even just a bouncing ball.
It does nothing to explain this remarkable and unique trick.

Wow, I had to write a comment on this one.....as a teacher of statics and dynamics, the explanation is down right WRONG. Yes it helps bring a little technical jargon to the masses, but it is still wrong. The first question that needs to be asked should be, is the collision elastic or inelastic (think pool balls and play dough). In the inelastic case, energy is transferred from velocity and angular momentum into mechanical deformation (swashing of the play dough ball), Kinetic energy is not conserved! When analyzing an inelastic collision, you need to look at the coefficient of restitution (Cr), which is the ability of the object to return to it's original shape. V1 = [(Cr+1)(M2V2 + V1(M1 - CrM2)]/(M1+M2) where Cr of 1 is an elastic collision (pool balls) and a Cr of 0 is a perfect inelastic collision (classic bullet hitting a block of wood, the bullet comes to rest inside of the block, and they both move together). When talking about a Cr between 0 and 1, this deformation energy is released back into the system through velocity and angular momentum as the ball returns to it's original shape. The key to inelastic collisions is that kinetic energy (movement) is NOT conserved. In the video shown, both the ball and the guy are inelastic objects. The guy jumps, starts to rotate in the air, lands on the ball and squashes the ball, the ball then rolls underneath the guy and releases some of that deformation energy back into the system by imparting vertical velocity into the guy as it returns to it's original shape. He slows his rotation, not by increasing the rotation of the ball, but by straightening out his body to a standing position (a compact objects require less momentum to rotate). Not to mention that the momentum of the ball is negligible compared to the guy. So how can the author use an elastic analysis "mguy vguy + mball vball (before collision) = mguy vguy + mball vball (after collision)" on an inelastic problem. Has the author ever taken dynamic? Down right poor Pop-Sci........

Cool video though :)

To AreYouKidding,

Thank you for you comments, however you have made several errors in your critique that I have to point out. First of all you seem to be confusing the concept of kinetic energy with that of momentum. While it is true as you point out that kinetic energy is not conserved in inelastic collisons, momentum IS conserved. In the absence of external forces momentum is conserved in any and all collisions. Look that up in any introductory physics text book.

Also I did not say that his rotation rate decreases as a result of his interaction with the ball, I said it increases. AFTER the collision, he is able to slow down his rotation once in the air specifically because conservation of angular momentum holds. It's very unclear to say that "compact objects require less momentum to rotate". What I think you mean is that compact objects tend to have less rotational intertia than extended ones making it easier to change their angular momentum with a given external torque.

In the case of straightening out and slowing the spin rate the angular momentum of the guy is in fact constant (there us no external torque acting on him). Because he changes the distribution of mass he must slow down in order for momentum to be conserved.

As far as energy being transformed from kinetic energy into elastic energy in the ball and then back to kinetic energy that is all true. It does not however contradict conservation of momentum in the collision.

Now all that being said let's clarify an important point. Analyzing the system assuming no external forces is a simplification, and to be more accurate I should have included the Earth/ground as part of the system, because clearly the ground is exerting forces during the collison. If we acknowledge that, then any difference in changes in momenta of the guy and the ball would be accounted for by a change (basically impossible to detect) of the Earth itself.

Oh Snap!

and men wonder why women get fed up so fast.....

Is there formula for the two guys leaning on the horse in the background? Thats what I want to know! (How do you pronounce Weiner?) It must have been rough. sorry, just playin.

Well its all good, you scientists proved a point, but now as a Mechanical Engineer I want to know how can we utilize this in a practical way?

Otherwise this stay a very interesting, but useless piece of knowledge! :) Now THATS the real challenge!

Aviation is the proof that given the will, we are able to accomplish the impossible.

Useless piece of Knowledge????? NO KNOWLEDGE IS USELESS
Wait, I'm studying Sociology....Nevermind

*poof* troll vanishes into the depths of the interwebs

To adamweiner

Does the following equation apply to inelastic collisions?
"mguy vguy + mball vball (before collision) = mguy vguy + mball vball (after collision)"

The answer is no..... So back to my main question.
"So how can the author use an elastic analysis...on an inelastic problem?"

Oh and last time I checked, momentum is directly related to kinetic energy! (Kinetic Energy) = (P^2)/(2*m) where P is momentum and m is the mass of the object. And in an inelastic collision, kinetic energy (and momentum) are NOT conserved. Sorry not confused on my end.

Don't get me wrong, I think it is great that topics like this are discussed on Pop-Sci, and I like your column, but I think it still needs to be scientifically accurate.

To AreYouKidding,

In fact the answer is yes! The above equation DOES apply to inelastic collisions. It applies to all collisions (specifically between two objects for the equation above).

[If the collision is perfectly inelastic - meaning that the objects actually stick together - then the final velocities are the same and you can combine the terms on the right side of the equation.]

Kinetic energy is a scalar quantity while momentum is a vector. Just because algebraically you can write kinetic energy in terms of momentum doing so does not really have any physical meaning. An inelastic collision is DEFINED as one in which momentum is conserved but kinetic energy is not.

Again, don't take my word for it just look up collisions in any physics textbook. Momentum IS conserved in inelastic collisions, as it is for all collisions as long as there are no external forces.

I appreciate your interest in the physics, and your desire to make sure everything is scientifically accurate. And I am certainly open to corrections when I make a mistake, but in this case your statements about collisions and conservation of momentum are wrong.

Additional general note: The analysis in the article is simplified for the sake of illustrating a fundamental principle. I have treated the collision as one dimensional and we didn't have to delve into the vector nature of momentum for the purposes of illustrating the basic point. However, there is a vertical component in there as you can see by the slightly downward trajectory of the guy as he impacts the ball.

Learning whatever math is behind the trick won't enable anyone to do it themselves...

guy rolls the ball, runs a bit quicker than the roll to catch up with it, squashes the elastic ball as he jumps on it with rubber shoes, stops his own forward motion by rolling his body along the ball in the opposite direction, then stiffens his back just as the ball pushes back, flattens himself out in the air to slow his spin even more, sticks the landing...viola!

Neat trick... go teach your children.

Is 'the guy' actually available for parties? I will hire him. Leave the ball, bring the shorts.

Wow. Cool video. Cool Trick. But this has to come very strongly under the "do not try this at home" category. It's obviously a very carefully practiced trick to do it just right. I have no doubt that if I, or most others, were to try it we would simply accelerate the back of our heads into the floor hard.

@Adam & AreYouKidding:

Are both of you kidding??? ;-)

This is not a two-body collision. You are both ignoring the most important body in this system - the Earth! Not only is it significant mass, with momentum of it's own, ("jumper impacts ball impacts earth"), but gravity acting on the jumper contributes a significant and important force throughout.

BTW, Adam, I have to side with AYK regarding the angular momentum of the ball - it's negligible. I don't have the equations in hand to compare numbers, but given the relative mass and size differential, we're probably talking about a 100x difference in momentum given similar rotational velocities. I.e. 99% of the energy driving this system is coming from the jumper - it's just a question of how it is directed.

I would argue that the ball is best treated as an elastic, massless bearing - sort of like a trampoline on wheels - although its interaction with the jumper is, I suspect, a bit more complicated.

Thus, a proper analysis should look at how the jumper translates his initial velocity and [lack of] rotational momentum into his final velocity/momentum by interacting with the ground using a rather quirky force-directing device - a.k.a. the exercise ball.

Broofa,

I agree with you. It really is a collision between the guy, the ball, and the Earth. (I did mention that in the last paragraph of my first comment but clearly should have included that in the original analysis!)Only for the horizontal component of the collision can we ignore Earth and gravity. In oversimplifying this into a one dimensional collision you rightly point out that we miss a major part of what is happening here.

Including the Earth as part of the system we can still treat it as a collision using conservation of momentum, but as you suggest there is a greater change in momentum of the guy compared to the ball, which is compensated for by a change in momentum of the Earth (both linear and angular).

Another more complicated way to look at the interaction would be by looking at each object individually and analyzing the external forces acting on these - which is what AreYouKidding is essentially suggesting. That is certainly a legitimate way to look at it. My problem there was with an important mistake he was making regarding the principle of conservation of momentum.

Let's summarize!

"In the two dimensional collision between the guy, the ball, and the Earth the total momentum of the system is conserved. The change in linear and angular momentum of "the guy" as a result of the collision is equal in magnitude to the change for the ball and Earth. The total change in momentum for the entire system is zero!"

Thank you for your clarification/correction.

AREYOUALLKIDDINGMEDOTCOM!!!!!!!!!!!!!!!!!!!!!!!

Adam,

kudos on a GRRRREEAATT article, and yeah, some of us with actual applied scientific experience understand that you simplified the overall system/mathematics/science portion of your little demonstration a)for the benefit ot those who are not so technobabbically minded, and b)because over examining the stunt doesn't make it anymore entertaining.

BUT IT IS A R-E-A-L-L-Y G-R-E-A-T STUNT!!!!!
(luv to c the ouchtakes, tho :) )

Everyone who tries to explain E-V-E-R-Y-L-A-S-T-P-A-R-T-O-F-T-H-E-S-T-U-N-T [and, let's face it kiddies, that's what it is, an exercise in gymnastics] just don't get it.

It's a SHORT article!!!Stop trying to explain the H-U-H-W-H-O-O-O-O-O-L-E T-H-T-H-T-H-T-H-T-H-T-H-I-I-I-I-I-N-G!!!!!
and just enjoy the trick, for pete's sake!

Later, folks!