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Topic: Infinite fast ladder (Read 5233 times) |
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BNC
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Infinite fast ladder
« on: Oct 12th, 2003, 3:55pm » |
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A ladder with length L leans against a frictionless wall. The wall and floor are at right angle. Mark the horizontal distance (on the floor) as X, and the vertical distance (on the wall) as Y. Then, we get X2+Y2=L2.
Now, the ladder is being pulled horizontally away from the wall (pulled by the bottom part) at a constant speed v.
Since both X and Y change, mark x=x(t), and y=y(t) [x(0)=X, y(0)=Y]. Naturally, x2+y2=L2.
From that, it is trivial to get:
y=[sqrt](L2-x2)
Differentiating with respect to t, and marking x'=dx/dt ; y'=dy/dt :
y'=-(xx')/[sqrt](L2-x2)
x'=v. Replace:
y'=-(xv)/[sqrt](L2-x2)
lim x->Ly'=-Lv/0 -> inf
Therefore, as you pull the ladder, its top falls infinitely fast.
Or does it?
[edit]Corrected typo pointed by aero_guy.[/edit]
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| « Last Edit: Oct 13th, 2003, 1:15pm by BNC » |
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towr
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Re: Infinite fast ladder
« Reply #1 on: Oct 12th, 2003, 11:48pm » |
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on Oct 12th, 2003, 3:55pm, BNC wrote:
Therefore, as you pull the ladder, its top falls infinitely fast.
Or does it? |
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Not in reality 
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BNC
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Re: Infinite fast ladder
« Reply #2 on: Oct 13th, 2003, 1:34am » |
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on Oct 12th, 2003, 11:48pm, towr wrote:
Not in reality
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I'm sure you realize that even though it was not specifically stated, the puzzle is really to point out the fallacy.
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towr
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Re: Infinite fast ladder
« Reply #3 on: Oct 13th, 2003, 5:15am » |
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yes, but I couldn't find it before I had to leave
Still can't find it to be honest.. 
Th only other thing I can think of is it depends on x'(t), which might be a function so that y'(t) is f.i. constant, or anything else that doesn't go to infinity
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| « Last Edit: Oct 13th, 2003, 5:24am by towr » |
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aero_guy
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Re: Infinite fast ladder
« Reply #4 on: Oct 13th, 2003, 8:03am » |
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Two problems: first I don't think y'=dy/dy, but that was just a typo.
The real problem is that:if you make x'=the constant v, it must instantaneously jump to zero when the ladder is horizontal (your limit) or the ladder seperates from the wall. Thus, a look at the graph of y'(t) will show exponential growth towards infinity with a sharp vertical line back to zero when the ladder is horizontal.
Also, your problem intrinsically assumes nonformable massless ladders that are attached to a slider against the wall so it will not separate.
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BNC
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Re: Infinite fast ladder
« Reply #5 on: Oct 13th, 2003, 8:59am » |
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aero_guy,
If I understand you correctly, you're saying that the top won't reach infinite speed, but may reach, say 100c ?
Now, that's an easy way to conduct relativity experiments in your garage! 
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aero_guy
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Re: Infinite fast ladder
« Reply #6 on: Oct 13th, 2003, 10:05am » |
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As I said, we are assuming that the thing has no mass or else it would be impossible. When we get to the ladder nearly flat on the floor we could start using small angle assumptions. If we did that though, we would see x remains nearly unchanged for any reasonable change in y, or that a modest change in x produces a huge change in y.
So, mathematically there is no problem with the riddle. The problem occurs in practical application.
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BNC
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Re: Infinite fast ladder
« Reply #7 on: Oct 13th, 2003, 1:21pm » |
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aero_guy,
I do beleive you have the answer (or at least are on the right track), but just to clear things up:
The ladder is not conneced to the wall. Hence, if it was without mass, pulling the bottom part would cause it to separate from the wall immediately, keeping the original angle. It stays on the wall only due to gravity.
Except for the "frictionless wall" that is there to keep things simple (not really necessary), everything else is under normal conditions.
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Icarus
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Re: Infinite fast ladder
« Reply #8 on: Oct 13th, 2003, 7:58pm » |
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But your mathematics DOES assume a massless ladder. It takes no account whatsoever of the inertia of the ladder. It assumes that as you pull the bottom out, the top is able to respond instantly to maintain its lean against the wall. This is not the case. Inertia of the ladder prevents it from keeping up as the angle approaches zero.
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BNC
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Re: Infinite fast ladder
« Reply #9 on: Oct 13th, 2003, 10:40pm » |
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Of course that's the problem, but not due to "massessness". A massless ladder would have kept its original angle all the time. It's true that inertia was ignored, though.
The falacy, as I see it (given the clarification of regular ladder), is in the seemingly innocent transition from the correct X2+Y2=L2 to the incorrect x2+y2=L2.
[edited to add]
I do seem to remember (not sure) from my old days learning physics that "gravitational mass" and "inertial mass" (probably wrong terms) are not fundamentally identical -- but are the same under "normal conditions" (on earth, Newtonian physics).
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| « Last Edit: Oct 13th, 2003, 11:41pm by BNC » |
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Re: Infinite fast ladder
« Reply #10 on: Oct 14th, 2003, 12:42am » |
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That's probably the difference between weight (g*m or a*m) and (resting) mass (m0)..
I don't think the ladder would have kept it's angle if it were massless though, there's other effects like airresistance, attraction between wall and ladder etc.
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BNC
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Re: Infinite fast ladder
« Reply #11 on: Oct 14th, 2003, 1:21am » |
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on Oct 14th, 2003, 12:42am, towr wrote:That's probably the difference between weight (g*m or a*m) and (resting) mass (m0)..
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I checked a little...
Inertial mass is defined by Newton's law F=ma
Gravitational mass is defined by the force of gravitation F=Gm1m2/r2
The two masses stem from different rules (so they should be allowed to be different) -- but are equal to each other as far as we can measure.
As a side note, when searching the web on this, I found this site that claims to show us all that superstitious `scientists' (a quote from the site) are trying to hide from us.
According to the site, there exists a mistrious force called UEF that pushes all things together everywhere in the universe. That force is faster than light, and can be used to explain all physical phenomena, including those "NASA cannot explain".
For example, gravitation is an imaginary force. There is no "pulling" force anywhere in the universe -- only the one "pushing" force. Gravitation is a result of one body "shadows" the other body from this push, so the two bodies are "pushed" (rather that pulled) together.
I was blind -- but now can I see!!!!!!
Actually, I don't have the time to look into the "theory" thoroughly enough -- if anyone out here takes the time and find the falacy (I suspect it should be easy enough) -- do let us know...
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Re: Infinite fast ladder
« Reply #12 on: Oct 14th, 2003, 7:33pm » |
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The equivalence of intertial mass and gravitational charge has long been a subject for debate. However two scientists have offered quite compelling explanations.
Mach argued that inertia is a residual effect of the combined gravitational attraction of the entire universe. I am not familiar enough with the argument to reproduce it properly here, but consider that in an otherwise empty universe, there is nothing to distinguish an inertial frame from one which is not. Mach argues that in a universe with a single massive object, that object can never be accelerated because it alone determines what will be inertial. Thus it can never be "spun", as we discussed in January in this thread. By Mach's argument, inertial mass is equivalent to gravitational charge because that is where it comes from.
The other scientist is of course Albert Einstein. The Principle of Equivalence, which states that gravitational action is indistiguishable locally from inertial action, is the heart of General Relativity. In Einstein's view, gravity is "an illusion" created by our imposing a Euclidean viewpoint on a non-Euclidean spacetime. That is, things behave as they do because in the Non-Euclidean spacetime, the paths they take are inertial ones, until they bump into each other. These paths appear to be accelerated to us because we are misinterpreting spacetime as flat. The curvature of spacetime is proportional to the stress-energy tensor, which in turn is usually proportional to mass. Why this relationship should hold is not explained in General Relativity itself, but it has been theorised that inertia is the result of spacetime resisting changes in its curvature.
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rloginunix
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Re: Infinite fast ladder
« Reply #13 on: Nov 26th, 2014, 4:57pm » |
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Before you guys start yelling at me for bringing back to life an eleven-year-old corpse please read what follows. I hope you will enjoy reading it as much as I did solving it (and you wont hear me curse when the current crumbled sheet of paper flew into the bit bucket).
Before posting "A Cat on a Ladder" I decided to search the forum to make sure I am not introducing a duplicate. This puzzle came up, I saw the connection between the three puzzles, decided to get to the bottom of this and hence the dead are rising from the grave.
So what do "Off the Edge of the Earth", "A Cat on a Ladder" and this puzzle have in common? Ponder as I proceed.
There is more than one way to resolve this paradox. I have found two so far - 1). as a logician who knows a little bit of physics and 2). as a physicist who knows a little bit of logic. Math being a common denominator.
1). Paradox Resolution as a Logician.
It is obvious that the velocity of the ladder's top edge hitting the floor can not be infinite. The sticky point here, however, is not the fact that an infinite velocity means an infinite amount of energy. It is also not the fact that according to Einstein no material object can reach the speed of light.
The sticky point here is the brutal simplicity of the problem statement and the childish simplicity of the experimental disproof. You do not need a Large Hadron Collider deep under ground that takes untold millions of Euros and thousands of geeks to operate. I took my kids' wooden ruler, leaned it against the wall on a hardwood floor and watched the top edge of the ruler hit the floor at a very finite speed.
So what gives?
Before I play along and arrive at the paradox via simple trigonometry let us introduce a definition of a rigid body on which we will lean heavily in the upcoming discussion:
A body is rigid if a distance between any of its two points stays the same over time regardless of the forces, external or internal, acting upon it, while the body is at rest or in motion.
Let us logically snap the ladder into a single rigid body rod of uniform density of length L and mass m - from now on this is the meaning of "a ladder" unless otherwise noted. Let us assume further that the ladder started its motion from the vertical position after a gentle nudge and we are observing it when it forms an angle  with the floor, "V" is the given constant velocity of the ladder's bottom edge touching the floor (point B), "u" is the unknown velocity of the ladder's top edge touching the wall (point A):
The fact that the ladder is rigid means that the projections of the velocity vectors from both edges of the ladder onto the ladder itself must be of equal length (marked with the same color in the drawing above). Otherwise different portions of the ladder will move at different speeds introducing ladder deformations which are not allowed:
BE = AD
V*Cos() = u*Sin()
u = V*Cot() (1.1)
The angle varies from 90 to 0 degrees. When reaches 0 degrees Cot(0)   . How do we resolve this? As logicians we shall use the approach popularized by Mr. Holmes: when all the impossible is eliminated, whatever remains, however improbable, is the truth.
Einstein, for example, noticed that wearing socks makes holes in them. His solution? Eliminate the socks.
Logically the fallacy of the infinite velocity comes from our assumption that the ladder forms a right triangle with the floor and the wall during its entire trip, including the instance when its top edge hits the floor. It follows then that our only choice is to eliminate the right triangle.
The right triangle is formed by three objects - the wall, the floor, the ladder. The wall and the floor are permanently fixed. Eliminate them. The ladder remains. It has two edges. Hence at some point during the ladder's motion: 1) the entire ladder is lifted up by some mysterious force, 2) the bottom edge is lifted off the floor by some mysterious force (while the top edge stays in contact with the wall all along), 3) the top edge is pushed away from the wall by some mysterious force (while the bottom edge stays in contact with the floor all along).
Choices 1) and 2) look unreasonable while choice 3) looks plausible. This is one of those typical problem solving situations - staring in the face of the unknown and being psychologically willing (or not!) to accept it. We go for choice 3.
Now the problem becomes more manageable as we break it into two parts. Part one is the right-triangle-motion prior to separation. Part two is the wall-free fall. We are given L and V. The (finite) answer then, based solely on the units of measurement, must be a function of V and what else? We need to somehow manufacture meters per second out of just meters (L). But we are not given any time. Gee. I mean g. It is meters per second squared so, purely mathematically, L must appear in the answer as  (gL) or some such.
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rloginunix
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Re: Infinite fast ladder
« Reply #14 on: Nov 26th, 2014, 5:08pm » |
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Let us examine the motion of the ladder's Center Of Mass (simply CM from now on) and this is where the connection with "A Cat on a Ladder" puzzle comes in. From "A Cat on a Ladder" puzzle we know that CM is moving along the circular arc of radius OC = L/2:
As such CM's velocity vector VCM must be perpendicular to OC ( OCG = 90 degrees) and by the earlier rigid body argument we also must have:
CD = BE
VCM*Cos( ) = V*Cos()
= 90 - OCB = 90 - (180 - 2) = 2- 90
= 2- 90 (1.2)
because the triangle OCB is isosceles, since OC = CB (OC is half the diagonal of the AOBM rectangle), and the internal angles at the base of the isosceles triangle are equal:
V*Cos() = VCM*Cos(2- 90) = VCM*Sin(2) = 2VCM*Sin()*Cos()
V = 2VCM*Sin()
VCM = V/(2Sin()) (1.3)
The above (1.3) expression is the (scalar) absolute value of CM's velocity. Its projection on the OX axis VCMX is:
VCMX = VCM*Cos( )
We find the angle from the 180-degree angle centered at the point C:
180 = OCB + + + = 180 - 2 + + +
= - = - 2 + 90 = 90 -
Here we took from (1.2) and taking VCM from (1.3) we get:
VCMX = (V/(2Sin()))*Cos(90 - ) = (V/(2Sin()))*Sin() = V/2
VCMX = V/2 (1.4)
From this we conclude that the ladder's CM is moving horizontally with a constant speed - without any acceleration.
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rloginunix
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Re: Infinite fast ladder
« Reply #15 on: Nov 26th, 2014, 6:25pm » |
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But CM is moving along the circular arc, it must have some acceleration whose projection on the horizontal must be zero. Our only choice here then is to point the vector of CM's acceleration strictly vertically and downward so that its projection on the radial axis OC is (sub index r below stands for "radial"):
aCMr = V CM/OC = 2VCM/L
aCMr = V/(2LSin()) (1.5)
Here we took VCM from (1.3). We find the absolute value of the CM's acceleration as a function of from the CKN triangle:
aCM() = aCMr/Cos(NCK) = aCMr/Cos(OCB/2) = aCMr/Cos(90 - ) = aCMr/Sin()
aCM() = V/(2LSin ()) (1.6)
From the right triangle CPG we find the vertical component of VCM:
VCMY = VCM*Cos(PCG)
PCG = 90 - OCB/2 = 90 - 90 + =
VCMY = (V/(2Sin()))*Cos()
VCMY = (V/2)*Cot() (1.7)
Here we took VCM from (1.3). But from (1.1) we have that u = V*Cot(). Putting that into (1.7) we get:
VCMY = u/2 (1.8)
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rloginunix
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Re: Infinite fast ladder
« Reply #16 on: Nov 26th, 2014, 6:25pm » |
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Now we have done all the legwork and we only need to draw the parallel with "Off the Edge of the Earth" puzzle. In that puzzle we have calculated when a material point flies off the sphere. Here we observe that when the same event occurs - the ladder's CM flies (outward) off the circular arc - it corresponds to the ladder's top edge breaking contact with the wall. This is exactly what we want.
Why would the ladder's CM fly off the circular arc outwards? Because the ladder is a rigid body and its CM can not possibly get "under" the arc, no deformations.
We must be careful in carrying the "Off the Edge of the Earth" equations over here. We can not do it blindly because over there we had a material point and over here we have a rigid body. We must reduce the the latter into the former. The way we will do that is by using the energy, kinetic and potential. This reduction is a purely mathematical equivalence.
Not exactly the same but a similar method of equivalence was mentioned by Icarus when Einstein postulated that gravity and acceleration are indistinguishable. We do not really understand gravity but we think we understand acceleration so we use one to deal with the other.
A similar deal is going on over here. We do not really understand why the top edge of the ladder must lose contact with the wall but we noticed an equivalent situation which we have resolved already and we want to exploit it to obtain a solution for this particular problem. We will use energy as an equivalence bridge. Adjusting for the hemisphere in the "Off the Edge of the Earth" the kinetic energy of a material point was T = mv/2 and its potential energy (adjusting for the alternate angle) was U = mgRSin().
To "collapse" a ladder into a suitable material point energy-wise we will use the kinetics Konig's theorem (wiki link) which says that a kinetic energy of a ridig body is the kinetic energy of its CM plus its rotational kinetic energy:
T = mVCM/2 + Trot
To find Trot (since the ladder also rotates around its CM) we can use yet another equivalence idea. Let us squeeze the ladder into a massless rod carrying two material points of mass m/2 on each end, point A moving at V and point B moving at u. In that case we have a kinetic energy of just two material points which is easy to calculate - it is just a scalar sum of the energies of two points. However, which method we use to calculate the total energy of a system - via a collection of material points or via a rigid body - does not matter. The net result must be exactly the same in both cases:
(m/2)V/2 + (m/2)u/2 = mVCM/2 + Trot
Trot = (m/2)V/2 + (m/2)u/2 - mVCM/2
But VCM = VCMX + VCMY and we already calculated the horizontal and the vertical components of CM's velocity in (1.4) and (1.8), that is why we did all this work. Put these values into the above equation:
Trot = (m/2)V/2 + (m/2)u/2 - m(V/4 + u/4)/2
Trot = (m/2)V/4 + (m/2)u/4 = (m/2)VCM = mVCM/2 (1.9)
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rloginunix
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Re: Infinite fast ladder
« Reply #17 on: Nov 26th, 2014, 6:26pm » |
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In other words in this case the rotational kinetic energy of the ladder equals the kinetic energy of its CM so the total kinetic energy of the ladder is:
T = mVCM/2 + mVCM/2 = mVCM = (2m)VCM/2
Note that we mathematically rearranged the above formula for it to look like a well behaving kinetic energy of familiar form. It means that our reduction of a rigid body in to a suitable material point yielded a virtual material point of mass 2m:
T = MVCM/2, where M = 2m.
Now that we equated the kinetic energies let us do the same for the potential energies. In our case the potential energy of the virtual point of mass 2m is (remember that CM's acceleration vector "a" points down):
U = 2ma(L/2)Sin(), take L/2 as R.
And in the "Off the Edge of the Earth" puzzle it is:
U = mgRSin()
Equating the two, still purely mathematically, we get:
2ma(L/2)Sin() = 2maRSin() = mgRSin()
a = g/2
From which we conclude that our collapsed ladder of mass 2m is moving over the hemisphere of radius L/2 under weak gravity of g/2. Now we can blindly apply the equations of the "Off the Edge of the Earth" and say that the point flies off the hemisphere when all the "other" forces, whatever they are - reaction and what have you, in that instance are zero and the projection of the weakened force of gravity 2mg/2 must be equal 2maCMr only, we already calculated aCMr in (1.5), so the expression for the angle of escape is:
2m(g/2)*Sin(e) = 2mV/(2LSin(e))
Sin(e) = V/(gL)
Sin(e) = (V/(gL))1/3 (1.10)
That is the value of when the top edge of the ladder loses contact with the wall. Sin() is  1. The values for V and L are independent. So if V gL then V is large which means that the ladder is yanked away from the wall right away and it does not even have a chance to touch the wall. If V is small enough then our problem "works".
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Re: Infinite fast ladder
« Reply #18 on: Nov 26th, 2014, 6:27pm » |
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Since we are interested in the velocity of the ladder's top edge, which is now in a semi-free fall, let us keep the ladder as a massless rod with two material points of mass m/2 on each edge. We move into the Inertial Reference Frame (IRF) of the point B since B is moving at constant velocity with no acceleration. In that IRF the ladder's top edge is rotating counterclockwise about a fixed point B and has two velocity components.
One component is obviously the -V vector which is equal in absolute value to B's velocity V but points in the opposite direction. The second component must be directed in such a way that the projection of the resulting A's velocity vector onto the ladder is zero - again, the rigid body argument, no deformations. In other words the resulting (total) A's velocity vector must be perpendicular to the ladder:
Without the B's IRF, in the wall-floor IRF, a force variable in time is applied to B to make sure it moves at constant velocity. It may not be apparent now but we will discuss this in greater detail in the second type of the paradox resolution. So in the wall-floor IRF the energy is not conserved as this force does a non-trivial work. To avoid this complexity we move into B's IFR where the variable force is applied to a non-moving object. As such it does not do any work and the energy is conserved:
(m/2)ue/2 + (m/2)gLSin(e) = (m/2)uy/2
Here ue is A's escape velocity which is V/Sin(e) and uy is the velocity with which A hits the floor in B's IRF. Here m/2 cancels out and we keep the square of uy:
uy = V/Sin(e) + 2gLSin(e) (1.11)
To calculate the total resulting velocity of A when it hits the floor we move out of B's IRF back into the original wall-floor IRF. In the drawing above I show these two frames separately.
When A hits the floor the ladder is perfectly horizontal. Since it is a rigid body all of its parts must move at the same horizontal speed at that moment. And at that moment these velocities are "convenient" - all of them are equal to V. So from Pythagoras we get the resulting and finite! when-A-hits-the-floor velocity uf of:
uf = (V+ uy) = (V+ V/Sin(e) + 2gLSin(e))
We take Sin(e) from (1.10):
uf = (V+ VV-4/3(gL)2/3 + 2gLV2/3(gL)1/3)
uf = (V+ 3(gLV)2/3) (1.12)
The above expression shows the (finite) velocity with which the ladder's top edge hits the floor.
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towr
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Re: Infinite fast ladder
« Reply #19 on: Nov 26th, 2014, 10:46pm » |
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on Nov 26th, 2014, 4:57pm, rloginunix wrote:| I took my kids' wooden ruler, leaned it against the wall on a hardwood floor and watched the top edge of the ruler hit the floor at a very finite speed. |
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I don't think you'd notice even if it were infinite. Because it would take place over in infinitely short amount of time (in fact, velocity would be infinite only at one single moment in time, not even over an interval of time). And humans don't have infinite resolution in their perception.
Quote:| Logically the fallacy of the infinite velocity comes from our assumption that the ladder forms a right triangle with the floor and the wall during its entire trip, including the instance when its top edge hits the floor. It follows then that our only choice is to eliminate the right triangle. |
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So, then if I attach one edge to a vertical rail, and the other to a horizontal rail, so that it must always form a right triangle as the ladder moves. Then it would have infinite velocity?
on Nov 26th, 2014, 5:08pm, rloginunix wrote:| From this we conclude that the ladder's CM is moving horizontally with a constant speed - without any acceleration. |
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Did we need all that to arrive at that conclusion?
Every point on the ladder is moving with a constant velocity to the right, as long as the ladder forms a right triangle with the wall and floor. This follows from its projection on the floor being stretched with a constant velocity.
Since we can adjust the problem easily enough so that it remains a problem, by enforcing that both edges always touch the respective surface they're on, there has to be another issue.
I think that in a Newtonian universe, the problem would work as stated, except that there's a problem with the assumption of a constant velocity for the right edge, it requires an infinite force to sustain it (there's a horrible transfer rate along the ladder to accelerate the top edge).
In an Einsteinian universe, things just get weird either way, because there's an upper limit to velocity. A rigid connection between two points doesn't make sense because information cannot travel at infinite speed.
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| « Last Edit: Nov 26th, 2014, 11:28pm by towr » |
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rmsgrey
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Re: Infinite fast ladder
« Reply #20 on: Nov 27th, 2014, 8:10am » |
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on Nov 26th, 2014, 10:46pm, towr wrote:| Since we can adjust the problem easily enough so that it remains a problem, by enforcing that both edges always touch the respective surface they're on, there has to be another issue. |
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After the ladder is horizontal, if the top end remains in contact with the wall, the bottom end continues to move horizontally with constant non-zero velocity away from the wall, and the ladder is rigid, then you have a contradiction. As you approach the singularity at the point where the situation becomes theoretically impossible, one or more of the three conditions must be broken in practice - the top of the ladder comes away from the wall, the ladder ceases to behave as a rigid body, or the bottom of the ladder ceases to maintain a constant velocity. Once one of those happens, the apparent paradox falls apart.
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towr
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Re: Infinite fast ladder
« Reply #21 on: Nov 27th, 2014, 9:07am » |
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To avoid the immovable object vs unstoppable force issue, we can just let the ladder detach from the wall once it's horizontal. So in theory the problem remains.
In practice rigid bodies don't exist, and neither do unstoppable forces or immovable objects. So yeah, one of those three breaks.
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rloginunix
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Re: Infinite fast ladder
« Reply #22 on: Nov 27th, 2014, 11:07am » |
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Before we discuss the questions raised let me guys finish sharing my thoughts and finds with you as it may shed some more light on the issue as this part deals more with why.
2). Paradox Resolution as a Physicist.
In this particular case we will be examining the forces acting on the ladder's edges so it is more convenient to represent the ladder as a massless rigid body rod with two material points of mass m/2 at each end. However, it is pain in the rear to type these fractions all the time, in the end it does not really matter, we can always effortlessly scale the results back to what we want so let us take it that the end points have the mass m:
The force of gravity mg is the same for both points. The force NB is the force of reaction of the floor, it is perpendicular to the floor. The force NA is the force of reaction of the wall, it is perpendicular to the wall.
The force T is the internal tension force of the rod. It does not change the size of the rod (rigid body) but it affects the balance of forces at the ladder's edges. The reason it is directed that way is because at this point in time the rod is in the state "compressed".
The force FV is the mysterious force that insures that B moves at constant velocity. Later we will develop a better understanding of this force and prove that it must point that way.
Project the vectors of the forces on the X- and Y-axes:
Point A:
x: maAx = 0 = -TCos() + NA (2.1)
y: maAy = TSin() - mg (2.2)
Point B:
x: maBx = 0 = TCos() - FV (2.3)
y: maBy = 0 = -mg + NB + TSin() (2.4)
From (2.3) we see that FV must act in such a way as to "eat" the acceleration of B along the X-axis since otherwise the projection of the force T, TCos(), would have done its job - accelerated B. Why B must be accelerated prior to loss of contact with the wall will be proved a bit later.
To address towr's question about CM's horizontal motion. If we sum (2.1) and (2.3) we will see that T's cancel out and NA is equal to FV - another proof that CM moves horizontally at a constant rate.
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rloginunix
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Re: Infinite fast ladder
« Reply #23 on: Nov 27th, 2014, 11:07am » |
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Rewrite aAy in (2.2) as a second derivative of the vertical coordinate of A over time using the traditional tick notation - one tick is the first derivative over time (speed), two ticks is the second derivative over time (acceleration):
my'' = TSin() - mg (2.5)
Prior to A's loss of contact with the wall we do have a right triangle and hence at any point:
y = LSin() (2.6)
Differentiate (2.6) over time once to obtain the vertical velocity of A:
y' = L*'*Cos() (2.7)
Differentiate (2.7) over time once to obtain the vertical acceleration of A:
y' = L*''*Cos() - L*'*Sin() (2.8)
Put (2.7) and (2.8) back into (2.5):
mL(''*Cos() - '*Sin()) = TSin() - mg (2.9)
Our objective now is to express the square of the angular velocity 'and the angular acceleration '' via the given parameters L and V (and ). The way we do that is by using the right triangle for the horizontal coordinate x of B:
x = LCos() (2.10)
Differentiate (2.10) over time once to obtain the horizontal velocity of B, V:
x' = V = -L*'*Sin() (2.11)
Differentiate (2.11) over time once to obtain the horizontal acceleration of B, zero:
x'' = 0 = -L*''*Sin() - L*'*Cos() (2.12)
From (2.11) we right away find ' and its square:
' = -V/(LSin()) (2.13)
'= V/(LSin()) (2.14)
And from (2.12), knowing the square of the angular velocity already, we find the angular acceleration:
'' = -'*Cos()/Sin()
'' = -V*Cos()/(LSin()) (2.15)
Put (2.14) and (2.15) back into (2.9):
mL(-V*Cos()/(LSin()) - VSin()/(LSin())
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rloginunix
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Re: Infinite fast ladder
« Reply #24 on: Nov 27th, 2014, 11:07am » |
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Here we have multiplied and divided the second term by Sin() to obtain Sin(). One L cancels out, Cos() + Sin() = 1. We carry -mg over the equal sign and write the expression for T as a function of :
T() = (m/Sin()*(g - V/(LSin()))
We now see a mathematical proof that the internal tension force T must turn to 0 at some point, namely when:
g = V/(LSin())
Or in terms of the angle of escape e:
Sin(e) = V/(gL)
We obtained the same expression in the resolution of the paradox 1). We now take that expression and put it into (2.1):
0 = -TCos() + NA
NA() = TCos() = mCot()*(g - V/(LSin()))
which tells us that the force of reaction of the wall turns to 0 at the same moment as T turns to 0, at the same escape angle, which means that the top edge of the ladder breaks the contact with the wall.
Not only that. We now can calculate FV as a function of from (2.3):
0 = TCos() - FV
FV = mCot()*(g - V/(LSin()))
which tells us that the mysterious force FV also turns to 0 at the point of escape. And since T is 0 at that point it means that the ladder goes from the state "compressed" into the state "not compressed and not stretched" or "neutral". After the event of escape it will go into the state "stretched" and the vectors of the forces T must be flipped 180 degrees - they will now point not "away from" but "towards" the center of mass.
We obtained the same value of the angle of escape e as in the previous section and the rest of the solution can be taken from there.
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