so far in these lectures we've talked about mass, aboutacceleration and about forces, but we never used the word"weight," and weight is a very nonintuitiveand a very tricky thing which is the entire subjectof today's lecture. what is weight? here you standon a bathroom scale. gravity is acting upon you,the force is mg, your mass is m. the bathroom scale is pushingon you with a force f scale
and that f scale--which in this case if the systemis not being accelerated is the same as mg-- that force fromthe bathroom scale on you we define as weight. when i standon the bathroom scale i could seemy weight is about 165 pounds. now, it may becalibrated in newtons but that's, of course,very unusual.
if i weigh myself on the moon where the gravitationalacceleration is six times less then i would weigh six timesless-- so far, so good. now i'm going to put youin an elevator and i'm goingto accelerate you upwards and you're standingon your bathroom scale. acceleration isin this direction and i will call this "plus"and i will call this "minus." gravity is acting upon you, mg
and the bathroom scale ispushing on you with a force f. that force,by definition, is weight. before i write down someequations, i want you to realize that whenever, whenever you seein any of my equations "g" g is always plus 9.8. and my signs, my minus signstake care of the directions but g isalways plus 9.8or plus 10, if you prefer that. okay, it's clear thatif this is accelerated upwards that f of s must belarger than mg;
otherwise icannot be accelerated. and so we getnewton's second law: f of s is in plus direction... minus mg-- it's in thisdirection-- equals m times a and so the bathroom scaleindicates m times a plus g. and i have gained weight. if this acceleration is five meters per second squaredin this direction i am one and a half timesmy normal weight.
if i look on the bathroom scale,that's what i see. seeing is believing--that is my weight. if i accelerate upwards, with 30meters per second squared 30 plus 10 is 40-- i amfour times my normal weight. instead of my 165 pounds, iwould weigh close to 700 pounds. i see that--seeing is believing. that is my weight. now i am going to put youin the elevator-- here you are-- and i'm going to accelerateyou down.
this is now a. and just for my convenience i call this nowthe plus direction just for my convenience--it doesn't really matter. so now we have here mg--that is gravity acting upon you. and now you have the forcefrom the bathroom scale. clearly, mg must belarger than f of s; otherwise you couldn't gobeing accelerated downwards. so if now we write downnewton's second law
then we get mg minus f of smust be m times a. this holds for acceleration down and so i get f of sequals m times g minus a. this is one way of doing it and you putin positive values for a. if a is five metersper second squared you get ten minus five is five--your weight is half. you've lost weight. being accelerated down,you've lost weight.
you could alsohave used this equation and not go through this trouble of setting upnewton's law again. you could simply have said "okay, this a is minusin this coordinate system" and so you put ina minus five and a plus ten-- you get the same answer. so you have lost weightwhen you accelerate downwards. suppose now i cut the cable...cut it.
then this a is ten metersper second squared if we round it off. you go down with ten metersper second squared so g minus a is zero. you are now weightless,you are free-falling. you have no longer any weight. you look at the bathroom scale and the bathroom scalewill indicate zero. you're floating, everythingin the elevator is floating.
if you had a glass with water you could turn it over andthe water would not fall out. it's like havingthe shuttle in orbit with the astronautsbeing weightless. there is a great similarity between the astronautsin the shuttle and a free-falling elevator. the only difference is that the elevatorwill crash, will kill you.
in the case of the shuttle it never hits the earthbecause of its high speed. we'll talk about this much later when we deal with orbitsand with kepler's law. what exactly is free fall? free fall is when the forces acting upon youare exclusively gravitational. nothing is pushing on you; no seat is pushing on you,no string is pushing on you.
nothing is pulling on you,only gravity. i will returnto this weightlessness very shortly in great detail but before i do that, i wouldlike to address the issue-- how could i determineyour weight if i hang you from a string? so now, instead of standingon a bathroom scale you are here. here is a string.
you might even havein the string a tension meter as we have seen earlierin lectures. and you are holding desperatelyonto that string. just like that. the system is not beingaccelerated, gravity is mg and so there must be tensionin the string, t which is pulling you up which, if there is noacceleration, must be mg. i read the scaleand i read my weight.
this scale indicates,in my case, 165 pounds. while i'm hanging,i can see my weight. so you see,it makes very little difference whether i am standingon a bathroom scale and read the force with which the bathroom scalepushes up on me or whether i hang from a scale extend a springand read that value. it makes no difference.
the tension herewould indicate my weight. there is a complete similaritywith the bathroom scale except in one case,something is pulling on me; in the other case, something ispushing on me from below. now let's accelerate this systemupwards with an acceleration a-- and i call this plus. then, of course,this t must grow; otherwise youcannot be accelerated. newton's second law,t minus mg must be ma.
the tension in the stringequals m times a plus g. ah! we've seen that before. no difference with the elevator. you accelerate the system,the tension will increase and you will see that,you will read that on the scale. your weight has increased,you weigh more. needless to say, of course, ifyou accelerate the system down that you will weigh less-- wejust went through that argument. and if i cutthe cable completely
you go into free fall. t will go to zero, a becomeminus ten plus ten is zero. you're in free fall. the scale reads zero,you are completely weightless. if we accept the idea of weight being indicatedby the tension in a string then there is a veryinteresting consequence of that. i have here a pin whichis completely frictionless and i have on both sidesa string
and this string hasnegligibly small mass. now, just assumethat it is massless. and there is here an object m1and there is here an object m2 and i am telling youthat m2 is larger than m1. so we all knowwhat's going to happen. the system is goingto accelerate in this direction. m2 will be accelerated downand m1 will be accelerated up. what comes now is important,that you grasp that. i claim that the tension onthe left side must be the same
as the tension in this stringon the right side. t left must be t right. why is that? it is because the pinis frictionless and it is becausethe string is massless. take a little sectionof the string here a teeny-weeny little section. if there is a tension on it-- that is, a forcein this direction
and there is a forcein this direction-- these two couldnever be different because thenthis massless string would getan infinite acceleration. so there can never bea change in tension from this side of the stringto the other. if you take a little sectionof the string here-- there it is,teeny-weeny little section so there is tensionon the string
and there is tensionon the string-- this one couldnever be larger than that because this little pieceof string so because there isno friction on the pin and because the stringsare massless-- only because of that must thetension be everywhere the same. if there is friction in thepin-- which we will do later-- then that's not the case. given the factthat the tension left
and the tension rightare the same i must now conclude that thesetwo objects have the same weight because didn't we agree that tension is an indicationof weight? so these objects have nowthe same weight. and some people may say "oh, that's a lot of nonsense,you must be kidding. "if m2 is larger than m1 this must havea larger weight than that."
well, they are confusingweight with mass. it is true that m2 isa larger mass than m1 but it is equally true that the weight of thesetwo objects is now the same according tomy definition of weight. let us calculatethe acceleration of this system and let's calculate the tensionand let's see what comes out. i first isolate hereobject number one. this is my object number one.
i have gravity, m1 g,and i have a tension t. nonnegotiable. t better be larger than m1 g. otherwise it wouldnever be accelerated up and we knowit will be accelerated up. so what do we get? we get t-- i will call this plus direction,by the way-- minus m1 g equals m times a. so the tension equalsm1 times a plus g.
hey! we've seen that one before. this one is beingaccelerated upwards. notice it gains weight. that's the tensionand this is the acceleration. i have one equationwith two unknowns so i can't solve it yet. but there is another one,there is number two here. for number two,we have a force, m2 g and we have the tension up.
this one better be largerthan that one; otherwise itwouldn't be accelerated down. let me call this direction plus. the reason why i now switchdirections and call this plus-- as well as this--is a good reason for it. it's not so arbitrary anymore. i know that this acceleration is going to bea positive number. because it's goingin this direction, it's a given.
if i called this negative, i would get herea negative acceleration for the same thingfor which i get here a positive. that's a pain in the neck. i don't want to have a plusand a minus sign there, have to think about thatit means the same thing. so the moment that i decide todefine this the plus direction will also come out to bethe same sign as this one. so i flip the signs there.
so now i apply newton's law. i get m2 g minus t equals m2 a. and so i get t--i'll write it here-- equals m2 times g minus a. two equations with two unknowns. well, that shouldn't be so hardto solve these two equations. you can immediately eliminate t,by the way. if you add this onewith this one, you really-- i call this equation one,you call this equation two--
you immediately lose your t andyou get that the acceleration, a equals m2 minus m1divided by m1 plus m2 times g. and you substitute that "a"in that equation and you'll find that the tension equals 2mgdivided by m1 plus m2. this is very easy for youto verify. let us look. this is m1, m2... 2m1, m2--i lost one m-- 2m1, m2. let's look at these equations,let's scrutinize them a little.
let's get some feeling for it rather than accepting themas being dumb equations. let's first take the case that m2 equals m1,and i'll call that "m." notice that a becomes zero and notice, if you substitutefor m1 and m2 "m" here that you get 2m, you get mg. so t becomes mg. that isutterly obvious.
if m1 and m2 are the same,nothing is going to happen. they're going to sit there,acceleration will be zero and the tension on both sides-- which is always the same,we argued that-- is going to be mg. clear. now we're going to makeit more interesting. suppose we make m2much, much larger than m1 and in a limiting casewe even go with m1 to zero.
let's do that. what you see now,if m1 goes to zero this goes away, this goes away,a goes to g and t goes to zero. if m1 is zero, t goes to zero. that is obvious! because if i make m1 zero,m2 goes into free fall. and if m2 goes into free fall its weight is zeroand so the tension is zero-- that's exactly what you see--
and you see thatthe acceleration of that object is g, which it better be,because it's in free fall. so you see, this makes sense. this is exactly consistentwith your intuition. and if you wanted to make m1much, much larger than m2 and you take the limiting casefor m2 goes to zero you'll find again that a goesto g and that t goes to zero except that now the accelerationis not this way... (makes whooshing sound )
but now the accelerationis this way and now this object will gointo free fall. and therefore there is notension in the string anymore. m1, if i return to the casewhich we have there-- that m2 is larger than m1-- m1 is beingaccelerated upwards. that's nonnegotiable,so it must have gained weight. m2 is being accelerated down,so it must have lost weight. just like being in an elevator,there's no difference.
they each weigh the same-- one loses weight,the other gains weight. they each weigh the same,and so i can make the prediction that if this is m2 g,which was its original weight and this nowis the new weight, t that m2 g must be larger than t. m1 gains weight,so t must be larger than m1 g. m2 loses weight,so t must be smaller than m2 g. that's my prediction--it has to be.
and we can... i can show youthat with some easy numbers. let m1 be 1.1 kilogramsand let m2 be 1.25 kilograms. frictionless system, and thestring has a negligible mass. what is the acceleration "a"of the system? i get m2 minus m1-- that is 0.15 dividedby the sum, which is 2.35 and that is approximately0.064 g, approximately 0.064 g. it's about 1/16th ofthe gravitational acceleration. it's a very modest acceleration.
what is the tension? well, i substitute my numbersfor m1 and m2 in there. you can take, for g,10, if you like that and you will findthat the tension equals 1.17 g. and now lookat what i predicted. they both weigh 1.17 g,that's nonnegotiable. that is my definitionof weight-- the tension in both sidesis the same. that's my definition of weight.
this is their weight. this one had a weight 1.25 gwithout being accelerated. you see, it has lost weight,because it accelerated down. this one had a weight of 1.1 g. you see, it has gained weight,because it has accelerated up. so you see, the whole pictureties together very neatly and it's importantthat you look at it that way. i now want to return to the ideaof complete weightlessness and i want to remind you,a few lectures ago
how i was swinging you at theend of a string in the vertical. i was swinging you like this. and i was swinginga bucket of water like this. and i want to return to that. i want to look at you when youare at the bottom of your circle and when you areat the very top of that circle. you go around a circlewhich has radius r. here is that circle. there's a string here,you're here.
and there's a string here and at some point in time,you're there. and you're going around...let's assume that you're going aroundwith an angular velocity omega and for simplicity,we keep omega constant. but that's reallynot that important. okay, this is point pand this is point s. let's first lookat the situation at point p. you have a mass and sogravity acts upon you, mg.
there is tensionin the string, t. there must be--this is nonnegotiable-- a centripetal accelerationupwards. otherwise,you could never do this. remember, from the uniformcircular motion. so there must be herecentripetal acceleration which is omega squared r or, if you prefer,v squared divided by r if v is the speed,tangential speed at that point.
it must be there. let's look here. right there,gravity is acting upon you, mg. let's assume this stringis pulling on you. let's assume that for now,so there is a tension. the string is pulling on you. therefore, nonnegotiable, whenyou make this curvature here there must bea centripetal acceleration and that centripetalacceleration
must be omega squared r. that is nonnegotiable,it has to be there. let's now evaluatefirst the situation at p and i will call this plus and i will call this minus. so what i get now is that t minus mg must be m timesthe centripetal acceleration so t must be m times thecentripetal acceleration plus g.
hey! that looks very familiar. it looks like someone is beingaccelerated in an elevator-- almost the same equation. if the centripetal accelerationat this point for instance, were 10 metersper second squared then you would weigh twiceyour normal weight. the tension herewould be twice mg. if this were five metersper second squared then you would be1â½ times your weight.
let's now lookat the situation at s. at point s, i'm going to callthis plus and that minus. i'm going to find that t plus mg must be m timesthe centripetal acceleration-- newton's second law. so i find that the tension thereequals m times a of c minus g. hey! very similarto what i've seen before. this object is losing weight. let us take the situation
that a of c is exactly10 meters per second squared and we discussed that last time when we had the bucket of waterin our hands. if a of c... if the centripetal accelerationwhen it goes through the top is 10, then this is zero. so the string has no tension,the string goes limp and the bucket of waterand you are weightless. if the centripetal accelerationis larger than 10
then, of course,the string will be tight. there will be a force on you and whatever comes out of herewill indicate your weight. if a of c is smaller than 10,that's meaningless. the tension cannever be negative. a string with negative tensionhas no physical meaning. what it means isthat the bucket of water would never have made itto this point. if you try to swing it up--
as someone triedin the second lecture-- but didn't make it to that point the bucket of waterwill just fall. you end up with a mess,but that's a detail. so the bucket of water,when it is here... if the acceleration there,the centripetal acceleration were exactly 10 metersper second squared then that bucket of waterwould be weightless. so i said earlierthat when you're in free fall
all objects in free fallare weightless. it's like a spacecraft in orbitor an elevator with a cut cable. it also meansthat if i jump off the table that i'm weightless whilei am in mid-air, so to speak. it means this tennis ball... while it is in free fall,it has no weight. now it has weight. now the weight is even higherbecause i am accelerating it and now it has no weight.
the tennis ball is weightless and i assume, for now,that the air drag plays no role. if i jump off the table i will be weightlessfor about half a second. this is about one meter. if i jump from a towerwhich is 100 meters high i will be weightlessfor 4â½ seconds ignoring air drag. i prefer todaythe half a second.
i am going to jumpoff this table with this water in my hand. and i'm going to tell youhow i can convince you that as i jump, that i will,indeed, be weightless. here is the bottle. there is a gravitational forceon the bottle. my hands are pushing upon this bottle. my hands are beinga bathroom scale. i feel, in my muscles,the need to push up.
in fact, i might even be ableto estimate the weight playing the roleof a bathroom scale. it's a gallon of water,it's about nine pounds. now my own body...gravity is acting upon me but i am being pushed up,right there. suppose we jumped. there would be no pushingfrom me on the bottle anymore no pushing there on me,the table. only gravitation would act uponus and we would be weightless.
how can i show youthat we are weightless? well, if i don't have to use my muscles to pushon this bottle upwards i might as welllower my hands a little bit during this free fall. and you will see that the bottlewill just stay above my hands without my having to push up. therefore,being the bathroom scale i no longer have to push on it.
i no longer... my musclesdon't feel anything and the bottle istherefore weightless. the bottle is weightlesswhen we jump; i am weightless and eventhis bagel is weightless. we're all weightlessduring half a second. there is no such thingin physics as a free lunch. you have to pay a price for thishalf a second of weightlessness. what happenswhen i hit the floor? i hit the floor with a velocityin this direction
which is aboutfive meters per second. you can calculate that. but a little later,i've come to a stop. that means during the impact there must bean acceleration upwards. otherwise my velocityin this direction could never become zero. therefore, i will weighmoreduring this impact-- there is an accelerationin this direction.
the five metersper second goes to zero. if i make the assumption that it takestwo-tenths of a second-- that's a very rough guess,this impact time-- then the average acceleration will be five meters per seconddivided by 0.2; that is 25 metersper second squared. that means the accelerationupwards is 2â½ g. that means i will weigh3â½ times more.
remember it is a plus g, so a is 2â½ g upplus the g that we already have; that makes it 3â½ g. so insteadof weighing 165 pounds i weigh close to 600 poundsfor two-tenths of a second. so we get four phases. right now, i'm my normal weight if i standon a bathroom scale. i jump for half a second,weightless
hit the floor for abouttwo-tenths of a second maybe close to 600 pounds. and then after that i will havemy normal weight again. now, you're going to haveonly half a second to see that this bottle, as i jump,is floating above my hands. i will pull my hands off so you will see thati no longer have to push it. that means it's weightless. are you ready? i'm ready.
three, two, one, zero. did you see it floatingabove my hands? we were both weightless. now, i have been thinkingabout this for a long, long time. i have been thinking whether perhaps this could not be shownin a more dramatic way perhaps evena more convincing way. and so i thought of the idea
of putting a bathroom scaleunder my feet tying it very loosely so that itwouldn't fall off when i jump and then show you that while iam half a second in free fall that the bathroom scaleindeed indicates zero. and don't thinkthat i haven't tried it. i've tried it many timeswith many bathroom scales. i made many jumps. there is a problem,and the problem is the bathroom scalesthat you buy--
that you normallyget commercially-- they indeed want to go to zero. it takes them a long time. they have a lot of inertia,their response time is slow. but even if they make it to zeroby the time you hit the floor then immediatelythe weight increases because you hit the floor and your weight comes upby 3â½ times. so it begins to swingback and forth
and it becomescompletely chaotic and you can no longer seewhat's happening. and it just so happened thatabout six months ago, dave... i had dinnerwith professor dave trumper and i explained it to himthat it is just unfortunate that you cannever really show it that you jump off the table,have a bathroom scale under you and see that weight go down tozero when you are in free fall. and he said, "duck soup--i can do that."
he says, "i can make you a scale "which has a response timeof maybe 10 milliseconds "so when you jump off the table in 10 milliseconds you will seethat thing go down to zero." and he delivered,he came through. he built this wonderful device which he and i are goingto demonstrate to you. let me first give yousome reasonable light for this. and i would like to show youon the scale there
what this scale that he builtis indicating. here is the scale,i have it in my hands. and on top of this scaleis a little platform just like on your scale. this platform weighs4â½ pounds. and you can see that,it says about 4â½. now, you will say "hmm! i wouldn't wantthat kind of a bathroom scale. "i mean, if i want to seemy bathroom scale
"i want to see a zerobefore i want to go up. "i'm heavy enough all by myself. i don't want to getanother 4â½ pounds." the manufacturer has simplyzeroed that scale for you but obviously also your bathroomscale has a cover on it. once you have seenthese demonstrations you will be able to answer foryourself why we don't zero this why we really leave thisto be 4â½. that's the actual mass which ison top of the spring.
but it's not really a spring-- it is a pressure gauge,but think of it as a spring. 4â½ pounds. here we have a weight which is a barbell weight,which is 10 pounds. is this from oneof your children, dave or were you doing it yourself? 10 pounds...we put it on top here. what do you see?roughly 14â½ pounds.
all right, we are goingto tape it down. there we go. and we're going to drop it from about 1â½, two meters and we drop it in here,well-cushioned because we don't want to breakthis beautiful device. when we drop it,the response is so fast that you will see, indeed,that pointer go to zero. now, keep in mind,when it hits the cushion
that the weight will go up. for now, i want youto concentrate only on the thing going to zeroand not what comes later. we will deal with thatwithin a minute. okay... 14â½ pounds. you know why the thingis actually jiggling back and forth? i can't hold it exactly still and so i slightly accelerate itupwards and downwards
and when i accelerate itslightly upwards it weighs a little more and when i accelerate itdownwards, it weighs less. it's interesting. you can see i'm nervous. that's my nervoustension meter there. okay, we're ready? look and... don't look at me,now, look at that pointer. did you see it go to zero?all the way to zero.
now comes somethingeven more remarkable. he said to me, "i can also makethe students see the response on a time scale of abouta fraction of a second." by the way, this is the herowho made all this stuff. he's fantastic. (class applauds ) lewin:he can show you the weighton an electronic scale and this weight you will seeas a function of time. i will put the ten poundsback on again...
tape it a little tighter and so the level that you seenow is 14â½ pounds. this is 14â½ pounds and this iszero, this mark is zero. i'm going to hold it in my hand. and notice,if i can hold it still you're backto your 14â½ pounds. now i'm going to drop it. you will see it go down to zero. it will hit the floor,the cushion.
it will getan acceleration upwards. it will become way heavierthan it was before and then it will even bebounced back up in the air and it goes againinto free fall. we will freeze that for you,and you will be able... we will be able to analyze it,then, after it all happens. so, 14â½ pounds...three, two, one, zero. and now professor trumperis freezing it for you. now look at this, lookat this incredible picture.
this is truly an eye-openerfor me, when i saw it. the physics in hereis unbelievable. here is your 14â½ pounds. tick marks from here to hereare half a second. it was half a secondin free fall and it goes to zero,that's no weight. now it hits the floor,the cushion and its weight goes up in somethinglike a tenth of a second.
look, this isabout one, two, three... it's about 3â½ timesits weight now. so the 14â½has to be multiplied by 3â½ or four which is exactlywhat we predicted-- that it would be much higher. but now it's being... it bounces off, becauseit's a very nice cushion. it throws it back up.
so it goes back into the air so it goes immediatelyto weightlessness again and then it oscillatesback and forth. and then here you would expect that this level, 14â½ pounds,would be the same as this. and the only reasonwhy that's not the case is there's a little cablethat fell with it which is pushing a little bit up on the upper... on the upperdisc that is there
so it's making ita little lighter. isn't it incredible? you see here in front of youthe weightlessness and you see the extra weightwhen it hits and again followedby weightlessness. dave, a-plus,you passed the course. there is a great interest in doing experimentsunder weightless conditions. nasa was very interested in it.
and if you would jump100 meters up in the sky you would only benine seconds up. you wouldn't even be weightlessbecause of air drag. however, if you could jump up way near the topof the atmosphere-- where the air dragis negligible-- then you would be weightlessfor quite some time. and that iswhat people have been doing for the past few decades.
professor youngand professor oman here at the aeronautics department have done what they call"zero gravity experiments" from airplanes-- and i willexplain that in detail-- but first i want youto appreciate that "zero gravity"is a complete misnomer. "zero weight," yes--"zero gravity," no. if you have an airplaneanywhere near earth, flying whether the engines are onor whether the engines are off
or whether it is free-fallingdoesn't matter. there is never zero gravity. there is always gravity--thank goodness. but if you are in free fall,indeed, there is no weight. apart from that, they call them"zero gravity experiments" and why not? maybe it sells better. they fly an airplane,which is the kc-135 and they do these experiments
at an altitudeof about 30,000 feet. if i could clean thisas best as i can... the plane comes inat one point in time at an angle of about 45 degrees. there's nothing specialabout that 45 degrees. it's just...that's the way it's done. you have to also thinkof the convenience-- convenience for the passengers. the speed is thenabout 425 miles per hour
so the horizontal componentis about 300 miles per hour and the vertical component isalso 300. the air drag is very little. let's assume, for the sakeof the argument that the engines are cut and the plane goesinto free fall. it's no differentfrom this tennis ball-- the same thing. you're going to see a parabola.
and so this planeis going to free-fall and comes back to this level. and let's analyze this arc,this parabola. right here at the top, clearly there will still be300 meters per second in the absence of any air drag. you should be able to calculate with all the toolsthat you have available how high this goesfrom this level.
in other words, what is the time that the velocity in the ydirection comes to zero? you can calculate that and then you knowhow much it has traveled. very crude number,this is about 900 meters. and it will take about 15seconds to reach this point so it will take about 30 secondsto go from here to here and in those 30 seconds the horizontal displacementis about 3â½ kilometers.
and all these numbersyou should be able to confirm. right here,the engines are restarted. during this free fall, everyonein the airplane is weightless including the airplane itself. now the engines start,and the engine is sort of... the plane is going to pull up,it goes into this phase and then the plane flieshorizontally for a while. during this phase,as we just discussed it's like hitting the floor.
you need an accelerationin this direction. there will be weight increase so there is herean acceleration upwards. and during this time,very roughly people haveabout twice their weight. and then here,they have again normal weight. and then the planepulls up again and here it goesand repeats the whole thing again going into free fall.
so again here, people havemore than their normal weight. zero weight,more than normal weight normal weight, morethan normal weight, free fall. and the whole cycle takesabout 90 seconds. you can imaginethat it is very important when you are here in free fall,when you have no weight that when your weight comes backand your weight doubles-- and professor oman told methat this change from zero to twice your weight takesless than a second--
that you better know where yourfeet are and where your head is because if your head is down and you all of a suddendouble your weight you crush your skull,so you have to be sure that you are standingstraight up in the plane when your weight beginsto double and we will see thatvery shortly, how that works. i want to show you first someslides from these experiments. so here you see the situationthat we just described.
let us start here, that iswhere i started with you. the plane turns the engines off. this is the parabola. here the engines are restarted. this is the free-fall period. this is about 30 seconds. the engine is restarted,and during this time there is an acceleration upwardsand they call it "2g peak." well, they really mean 1g.
what they really mean,that my weight doubles. they call that "2g" but, of course,they call this "0g" which is equally incorrect. it's not 0g--you have noweight. this is weightless,here your weight is double here your weight is normal,here your weight roughly doubles and you go intoanother free-fall period and the cycle from here to hereis about 90 seconds.
now, the irony has it that the reasonwhy these flights are done is to study motion sicknessunder weightless conditions. astronauts were complainingabout motion sickness. and so professor youngand oman have done lots and lots of experimentswith airplanes and later, also, in the shuttleto study this motion sickness. i find it rather ironic because if you and i werepart of these experiments
we would get terribly sickbecause of the experiments. just imagine that you gofrom weightlessness into twice your weight,back to weightlessness. we would be puking all day! how can you study peoplewho are sick? how can you study the sicknessdue to weightlessness? well, they musthave found a way. they do thisabout 50 times per day. and now i want to show yousome real data
which were kindly given to meby professor young where you see them actuallyin the plane. i believe i have to put thison one and start the... can you turn offthe slide projector? so here you see themin the plane. they are not weightless,they are climbing up. i think this is professor young. the guys lying on the floormust be a bit tired. the light will shortly go on,and when the light goes on
that's an indication thatthe weightlessness is coming up. it already went on, i must havemissed it, i wasn't looking. and there they gointo weightlessness. see, this person isupside down here. you better get straight upbefore your weight doubles because you'll crashinto the floor. (class laughs ) lewin:and now it takes 60 seconds because the whole cycleis 90 seconds
and in these 60 seconds they get readyfor the next free fall-- for the next weightlessness. and you will see very shortly the light will go on again,and that will tell them that the weightlessnessis coming up and then they will be weightlessfor another 30 seconds. the sound that you hear isobviously the engines of the plane.
there you go-- light goes on, they get a warning, theytake their headphones off and everythingbecomes weightless. they may not like that and so they put their headphonesin a secure place. you see that hereprofessor young takes his off. and there they go again...swimming in mid-air. 30 seconds weightless. lewin:and the planein which this happens...
lewin:yeah, these things happen. i'd like to show youa last slide of the plane that they dothese experiments from. this is the planewhile it is in free fall. about 45-degree angle and these people have donea tremendous job in indeed makinga major contribution to the airsickness dueto weightlessness. all right, see you friday.
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