Dumbbells: Newsletter Challenge (02/12/08)

What happens after the initially higher ball hits the ground ? I am way too scared to tackle that !

I don't think this is too scary. We can study the movement of the CG, as if it was a bouncing ball, i.e. synthesize a vertical free fall motion along with a horizontal steady one. The issue is only to calculate the initial conditions, which are the vertical and horizontal component of the velocity at the moment the upper ball hits the ground, and the displacement along x-axis that this happens. Then, it's easy: the dumbbell's CG just moves at a constant speed towards x-axis while the upper ball bounces up and down, never reaching a vertical position over the lower one, though! (Some dynamic energy has been converted to kinetic)

These initial conditions can be calculated by noticing that at the moment the upper ball loses contact with the wall, the dumbbell doesn't experience any horizontal force, therefore it's CG's velocity has a constant horizontal component from then on. This velocity is simply half the one given as answer to this challenge (i.e. the velocity of the lower ball). Similarly we get the vertical component, by halving the velocity of the upper ball at that moment of detachment. All these are easily obtained. Then, by using energy preservation we can go on and do some school math to see where the dumbbell hits the ground and at which velocity.

You will find that this was already covered, but marked as "off topic" because it was not part of the original problem. On that basis both your contribution and this comment are also off-topic.

In my humble attempt I have tackled the problem by considering M1 as a mass rotating about M2, and the forces that come into play are centripetal and centrifugal, and the arrows emanating from M1 being there representation.

FIG 1, shows the centripetal force at maximum and the centrifugal force at minimum. As gravity 'g' acts, the masses move to the position as shown in FIG 2, where the masses are at 45° to each other, in this position it can be seen that 'g' is balanced between the centripetal and reactive force with the wall, and the centripetal force is g/2.

and the force acting horizontal at M2 is g/2 x 0.0707. As the centripetal force at this point is greater than the centrifugal force the energy in M2 will cause M1 to shortly there after to come away from the wall, as M1 accelerates under the force of gravity, so does the centrifugal force increase to the point as shown in FIG 3, where they are balanced?

Pass this point the centrifugal forces increase beyond the the centripetal force and de-accelerates M1, where upon at the point shown in FIG 4, centrifugal force is at maximum and centripetal at zero. It is assumed that the centrifugal force in FIG 4, cancels out the centripetal force shown in FIG 2, leaving the acting force of g/2.

Therefore my answer is M2 rate of ecceleration at the point when M1 is level with M2 is: velocity = g/2 x 0.707 x t², where 't' is the time taken for M1 to fall.

Regards JD.

I think you may be assuming constant angular velocity around the point at which the dumbbells part company with the wall. However, the angular velocity is increasing all the time (under gravity), so the zero-horizontal-force point is delayed. What surprised me was that it came so close after the 45-degree angle (slightly less than 3.2^O later).

Fyz

Yes I agree there are a few assumption, and also I seem to have misread the question, not for the first time, I had at first thought that M2 would be doing more work than M1 if it moved away from the wall, but that is erroneous thinking also.

Regards JD.

This challenge has a few holes in it. 1. If the planes are indeed frictionless, that would mean that if the mass of the entire dumbell is greater than my mass, as I try to move it, the dumbell's inertia would exceed mine, and I would pull myself to the dumbell. 2. If I understand the intent of the question, when the top end looses contact with the vertical plane, the horizontal speed should be 0, but because the horizontal plane is frictionless, and nothing opposes the horizontal speed generated by the effect of the top mass falling from heighth L.....the horizontal speed of the bottom mass would be only what is generated by the vertical speed of the falling top mass. So by that convoluted reasoning, I would say that the horizontal speed would be the same as that generated by any mass being dropped from a heighth (l).

Nothing says you have to be on the smooth surface, but if you were, you'd pull yourself towards the dumbbell, but it too would move. (You might also fend off against the wall)

What is the vertical component of the motion when "the horizontal speed would be the same as that generated by any mass being dropped from a height (l)", and where does the additional kinetic energy for that come from?

My point about the floor being frictionless, is that if it truly so... that would mean that once the upper weight reaches the floor, the bottom weight has achieved a certain lateral (horizontal) speed. If the floor is truly frictionless, then the lower weight would not deccelerate from top speed.

If you imagine a very large scale dumbell...at one second, the top weight is moving at 10m/sec (yeah I know 9.8-bear with me). at 2 sec, speed is 20;3 sec, 30 etc etc. If you do a distance covered (area under the velocity graph) the distance covered in the 1st sec is 5m, 2nd is 20m, 3rd is 45m, 4th is 80m, and 5th is 125m. Because the length of the dumbell remains the same, use trig to find the distance the bottom weight moves laterally each second. sec 0 to 1 the lateral distance moved is 35m; sec 1 to 2 - 32.8m; sec 2 to 3 - 28.2m; sec 3 to 4 - 20.6m; sec 4 to 5 - 8.4m. Without do any lengthy calculus...if the floor is truly frictionless, the speed when the dumbell loses contact with the wall, the speed would be the same as the max lateral speed which is around 35m/sec in my example. If there is some minimal realistic friction that will stop the bottom weight from shooting across the floor... do the calculation as the distance of the top weight to the floor approaches 0, and the speed will be 0. In other words, at the instantaneous moment when the top weight reaches the floor, the bottom weight has stopped moving away from the wall.

Once the dumbbells leave the wall, tension in the rod will cause horizontal deceleration of the bottom mass and acceleration of the top mass. However, the horizontal velocity component of the centre of gravity will be constant.

I failed to understand sufficient of what you meant in the second paragraph to give a relevant answer - but in any case I think it is unhelpful to introduce friction to this specific problem until the lossless case is sorted.

I believe (not completely certain) that it will be possible for there to be sufficient floor friction both for the rod to leave the wall and for the bottom to stop before the top hits the floor - but I suspect the range is quite narrow (it's a significantly more involved problem than this one). (Would you have identical coefficients of friction for both the floor and the wall? - and no, I'm not volunteering to do this).

By the way, when the top weight reaches the floor, the rod is horizontal, so the horizontal component velocities of the top and bottom masses are identical

"I believe (not completely certain) that it will be possible for there to be sufficient floor friction both for the rod to leave the wall and for the bottom to stop before the top hits the floor - but I suspect the range is quite narrow" What a load of rubbish! If there is floor friction the movement won't start unless the rod is at a suitably large angle to the vertical - in which case rod will fall to completion regardless of how high the (finite) coefficient of friction might be. So there's a wide range over which it stops - but not with the "small angle" specified.

If you imagine a very large scale dumbell...at one second, the top weight is moving at 10m/sec (yeah I know 9.8-bear with me). at 2 sec, speed is 20;3 sec, 30 etc etc

I use 10 much of the time, too. (In fact I would favor a movement to change the official value! That may involve changing where we think the planets are, or revising the value of a kilogram... but that's the price of progress.)

In fact, the top mass cannot accelerate at this rate. It is prevented from doing so by the lower mass, which would have to accelerate at a very high rate in the first few milliseconds, because the mechanical advantage of the upper mass over the lower is so disadvantageous. From the starting position, just pull the lower weight back toward the wall a tiny amount, and no acceleration at all would occur (assuming nobody sneezes, absolute zero, etc, etc.) By your figures, the lower mass moves 35 m in the first second, implying an average acceleration of 7g, which would require a very high force. However at this instant, the mechanical advantage is such that the force must instead be very low.

If the top weight could move along the wall, vertically all the way to the bottom, then the lower weight would have to decelerate so that it is stopped when the upper weight reaches the floor. However there is no way to transmit the required force to decelerate the lower mass: the upper mass would have to locked onto rails.

ok but the challenge implies frictionless surfaces. Without any friction against the lateral movement of the lower weight, wouldn't the top weight be in free fall?

The lower weight is resting on the horizontal surface with a force greater than its weight due the bar linking the two weights.

No.

The lower mass can slide without friction, but it still has inertia. It will accelerate according to A=F/M. The force starts out extremely low, because the mechanical advantage is so poor. For example, if the bar is one degree from vertical, a 1kg upper weight will have a downward force (i.e., weight) of 9.81 newtons. The horizontal component of that force is only 1.7% of 9.81 newtons. Therefore, the lower weight will accelerate very slowly at first. Because the lower weight refuses to get the heck out of the way, the upper weight cannot free fall.

ok but..the weights are in contact with frictionless surfaces. Imagine the dumbell standing perpendicular to the floor without the wall. If you let it go, the top weight should describe an 90degree arc at a predictable rate-shouldn't it? I dont know the expression for figgering the time it would take to fall, but the length of the dumbell divided by that time should be the speed of the bottom weight when the dumbell is horizontal if the floor is frictionless and doesn't oppose the acceleration of the bottom weight.

1. If the planes are indeed frictionless, that would mean that if the mass of the entire dumbell is greater than my mass, as I try to move it, the dumbell's inertia would exceed mine, and I would pull myself to the dumbell.

In fact both you and the weight would move, but by different amounts. But it is reasonable to assume that you are not in the friction free zone. That zone could consist of just a wall and a floor each perhaps a square meter, and you simply reach in to position things.

2. If I understand the intent of the question, when the top end looses contact with the vertical plane, the horizontal speed should be 0,

... the horizontal speed of the top mass, that is. When the top mass leaves the wall, its horizontal speed would be zero at that instant, but the bottom mass will have already accelerated to its maximum horizontal speed.

So by that convoluted reasoning, I would say that the horizontal speed would be the same as that generated by any mass being dropped from a heighth (l).

However, at the moment addressed in the question, the weight has not fallen through a height of l. It has only fallen 1/3 of the way to the floor at the time when it (the top weight) is pulled away from the wall.

Initially, the force applied to the lower mass is infinitesimally small. Therefore, the initial horizontal acceleration of the lower mass is very slight. Not only its speed but also its acceleration are non-linear with time.

I think someone got up too early in the morning to post that answer. ;-)

"....that consists of a rod of length l.... "

Does the change of wording mean we have to start again ? If so, I'd like to plagiarize Fyz and claim kudos for answering the revised question.

Good answer.

Shame we can't rate it as such.

So what's the answer? It says the answer is posted but it's obviously not. All I see is the same question posted twice.

How about an answer instead of a restatement of the problem?

"It is what it is."

And apparently what it is, is Zen physics.

What is the sound of one dumbell sliding?

Where is the answer?

The answer, apparently, is in the question. Or is it the other way around? I can never remember! Perhaps the question is in the answer. I thought that Fyz did an excellent job in establishing a theoretical basis for a proper answer. The answer we received, no matter how eloquent, was not a proper answer. Come on, CR4, give us a proper answer to the question. Either that, or let Fyz take over. I think I'll have another drink.

On a completely different note, here is a picture of my back yard in January.

We have two mountain ash trees and a birch tree in our back yard. Every year, about this time we are visited by hordes of Bohemian Waxwings who feast on the orange berries of the mountain ash trees. In a single day, the berries are totally wiped out. Amazing!

The little structure on the right is a bird feeder which I fill with sunflower seeds. Chickadees love sunflower seeds. Just to the left of that is a bird bath, currently covered with snow. If you look very carefully at the photo, you will see the waxwings, although I must admit my photography skills are less than perfect.

And therein lies your answer. But I have forgotten...what was the question again?

mgl=2mv²/2

v=sqrt(gl)

Upon reading this, I feel compelled to second the motion of whomever it was who proposed that you be named the CR4 poet laureate!

You do me an honor, sir. But wait...you can't second your own motion under Bob's Rules of Order.

I think you will find a footnote in Bob's Rules. I don't have a copy in front of me, but I seem to remember something like this:

¹ Of course, in cases of premature senility, in which the second motioner can not be certain of the identity of the first motioner, then, that old fool can second any motion he darn well pleases.

Also from Bob's Rules, I seem to remember something about an overlap having to be established outside the two boatlength distance from the mark... unless there is an obstruction such as a falling dumbbell.

Very well, then, tell Bob that I will second that motion.

We have a motion and a second on the floor. All in favor, show of hands, please!

The answer appears to be remarkably similar to the question!

At the moment in question bottom end will stop, so speed is equal to zero.

Oh dear (New) official answer, how do I love thee?

Let me count the ways.

I love the disappearance of l, and the appearance of r.

I love the novelty of measuring from the floor to the center of the upper weight, rather than the tediously conventional center-to-center distance.

I love all the numbers and stuff.

I love how long it took me to figure out this: "We can express gravity as a function of the Vx" (Do we mean the gravity of the situation?) (I fear I used to think of gravity as being something of a constant, around where I live – something that would not necessarily vary with the velocity of a dumbbell.)

I love how the calculations end with the square of the velocity, rather than the velocity.

I love how it shows that there is more than one way to skin a cat – the answer, although a bit squared, it still pretty good.

I'm not sure it's as good as the first official answer, though.

This analysis is very nearly completely correct. However, the implicit assumption of y=(r^2-x^2)^1/2. in equation (1) is that the bells on the dumbbells are point like masses. This is not what the problem states nor is it what the free body diagram shows. This means that the continuous integral of the sine and cosine functions of the angles expressing the relation radius of the contact points for the planes must be computed. This is necessary to correct for the continuously varying amount of angular offset between the point of contact for the falling mass and true center of mass for that portion of the rigid body.

The last time I attempted to solve that particular type of problem was several decades ago and it took me several days of hard work to find a first solution. In checking my work I stumbled across a vastly more elegant solution based on a ratio equivalence in pythagorean theory. I believe that this solution became the basis for the algorithm that machine tools currently use to compute tool radius offsets on the fly.

However, there is no set of equations of which I am aware that can be used to compute this integral without using the radius of the bells which is not given in this problem. So, I contend that the complete and accurate solution of the theoretical problem is non-determinant.

I am rather unwilling to devote the couple of days it would take to develop the software to solve this problem so at this point i can not prove my claim that there is an actual difference in the result when you assume a non-zero radius for the bells however, I did a highly similar computation involving the cantilevered slide in a crankshaft, piston rod, piston, cylinder wall assembly and in that case the degree of misalignment between the piston rod pivot point and the piston skirt which slides against the cylinder wall was a small but note worthy factor. If you have ever wondered why there is typically wear and scoring of the piston skirt in this area, the offset load between the piston and piston rod when the charge is exploding is the answer.

In that instance it was necessary to include fudge factors to allow for the action of friction in the mechanism in order to achieve a useful model of the real world.

In this problem the upper bell will separate from the frictionless wall somewhere when V(horizontal, lower bell) > V(vertical, upper bell. This condition will be met very nearly around the 45 degree angle point. If allowances are made for the intrusion of real world into the problem then the action of friction would delay the moment of separation to a later time and a lower angle.

Sincerely,

Mr. Gee

PS In engineering the "correct" answer is the one that most closely approximates a real result! If there are some enterprising students out there who would like to determine what the true correct answer is then affixing two wheels to each end of a ladder will approximate the frictionless condition of the problem and filming a falling rod, bar, or ladder while falling past a tape measure should give a useful approximation of the "correct" answer.

Increasing the drag in the wheel bearings would assist in modeling friction and measuring and or mathematically estimating the degree of inclination before the device free falls would make for a rather nice experimental test of the validity of whatever mathematical model or models you choose to evaluate.

The greatest difference between a working engineer and an individual who professes to teach engineering is that the professor believes that the correct answer is the one the text book endorses and the working engineer knows from sometimes bitter experience that the correct answer is the one that best matches reality.

Furthermore, If you wish to engineer really good systems then consider dynamic systems that make predictions, measure results, and correct the model better match the results.

Men have been hurling projectiles at each other for several centuries. Your course work did, or does include formulas for ballistic projectiles and these formulas marked a milestone in man's progress in the art and science of killing each and so they are much celebrated in our educational literature. But, in the real world a brave individual known as a a forward spotter is often forced to sneak behind enemy lines so that he can observe the actual point of impact of the artillery round. The aiming is adjusted based on his observations and then verified, (miscommunication is a common problem) and only then does the battery fire for effect. As the barrels heat the steel in them is weakened and subsequent rounds fall short of first rounds so an experienced gunner will automatically raise his sights and continuously adjust for Kentucky windage. Inexperienced gunners will often tragically not adjust and lay down a devastating barrage of "friendly fire" on their own troops.

As you may have guessed by now, I have a genuine distaste for challenges of the type expressed in this problem. If you make unrealistic assumptions about the behavior of the real world like frictionless planes and such a then maintain that these fantasies are "correct" you perpetuate the myth that mathematical correct answers are also valid answers. Which is simply not true often enough to warrant the sort of faith we tend to put in these types of results.

Mathematics are the servants of man, but they are the slaves of reality. They do not determine reality, at best, they imperfectly reflect it.

Fyz's solution and the official answer only ignore the MIs (Moment of Inertia) of the end masses.

As you imply, the equations are correct provided that the "length" of the dumbbell is the distance between the centres of the spheres.

NB that my extended answer (posting #73) provides an easy basis for the use of more generalised moments of inertia**, should the information to calculate their values become available.

**That would include incorporating the mass and MofI of the connecting rod as well as the MofI of the spheres

Fyz

As you may have guessed by now, I have a genuine distaste for challenges of the type expressed in this problem. If you make unrealistic assumptions about the behavior of the real world like frictionless planes and such a then maintain that these fantasies are "correct" you perpetuate the myth that mathematical correct answers are also valid answers.

Although I understand and respect your point of view, I certainly don't agree. I think this is among the best challenge questions we've had. Some pretty smart people have come up with potential solutions that are entirely unsupportable by generally accepted physics and math principals. I take that as evidence that a correct solution requires some clear thinking -- and that is, largely, the point of these exercises. Certainly, no one thinks that a frictionless world is "real," but many would agree that there is a mathematically "correct" answer to this problem. It is not a myth that mathematically correct answers are also valid answers.

Simplifying assumptions are part and parcel of both physics and engineering. In engineering, CFD and FEA are both used for prediction of continuous phenomena, but the techniques are discontinuous. They are simply approximations, and it's only fairly recently that these approximations have become close enough to the real world that some products can go directly from drawing board to manufacturing.

As a student, I used to tease my physics teacher when presented with textbook problems: "Did you want us to ignore Brownian movement in the gases surrounding the experiment area? How about local variation in gravitational field? Are the materials involved magnetic, and if so should we worry about possible eddy currents established as these things move, and if these eddy currents develop, should we feed the resulting temperature rise back into our Brownian movement calculations? ... etc., etc." Fortunately, my physics teacher had a good sense of humor.

Of course one is expected to ignore many real world effects, and most of the time, the nature of the question tells you which ones to ignore. If the question is ambiguous, (as all are to a degree) then you simply state the assumptions upon which your answer depends. The intent of the question is to get at the essential physics involved, and to come up with the math that can make a reasonable prediction of behaviour. It turns out, that in this case, if you use a ladder with wheels on both ends (as you suggest) Fyz's solution, and the few other correct solutions, accurately model real world behaviour. Several other solutions proposed do not, because they are not fully "thought out".

When I read through Fyz's solution (and the related solution from Harvard) I said "pretty cool," because they are masterfully presented: clear, concise, elegant, and appropriately complete. And when I started to think of Fyz's solution as being off by a factor of two... and then he patiently explained the cause of my (fortunately temporary) delusion... again, I said "pretty cool."

We might take into account 10 or so real-world phenomena to get a pretty accurate result for this thought experiment. Doing so, however, obscures, rather than elucidates the real issues.

Mathematics are the servants of man, but they are the slaves of reality. They do not determine reality, at best, they imperfectly reflect it.

Perhaps. Although one does not need to go more than a couple pages into a physics text before math comes in: it's not unreasonable to say that you can't understand reality without understanding math. Even simple concepts are very nearly impossible to appreciate without math. As soon as we make the "conceptual" comment that air resistance varies with the square of speed, we are already talking in mathematical terms.

Our ability to fly to the moon suggests that, although the math may imperfectly reflect reality, it is plenty close enough.

I have an objection to the experimental model suggested by Mr Gee (post158) using a ladder fitted with wheels. The rotational kinetic energy of the rolling wheels introduces extraneous terms which significantly alter the nature of the problem. One could perhaps get around it by having a heavy ladder compared to wheels of negligible mass (or moment of inertia), but aren't we then drifting back into the realm of idealistic assumptions in the same category as frictionless surfaces?

Does the 'good answer' categorisation in CR4 require further subdivisions to accommodate the off-topic posts which are sometimes more enlightening than the 'true' answers to stupid questions. I am in favour of more 'challenges' in which the answer remains a matter of dispute!

=TeeSquare=

Agreed on all fronts - even (sometimes) multi-answer challenges. (Posing usefully "open" questions is a rare skill).

Hi Fyz, thanks for your concurring remarks.

Allow me to add some qualitative observations which I feel are relevant to the topic as it has evolved so far. They were triggered by your reference to the Harvard solution in post 170, which I managed to skim through. But I don't expect you or anyone else to undertake any further analysis, as everyone would be in a state of fatigue from the dumbbell exercises by now.

I am limited to a very rough qualitative appreciation of quantum phenomena, as they are somewhat beyond my mechanical imagination. I got the drift of the argument minus the maths, but it set me thinking about the falling pencil and how it would hit the floor -- whether the kink at the cone/cyl interface would cause the tip to bounce, and what sort of motion would ensue subsequently (in the scenario of frictionless floor/wall and elastic impact).

Since impact is an additional complication (introduced by Kris and myself mainly) I feel we can avoid it altogether by assuming the 'body' to have a smooth convex shape all over (like an ellipsoid or prolate spheroid). After sliding some way down the frictionless wall it should just move off with a motion akin to a travelling rocking chair. Given the mass and MI, the mechanics ought to follow your equations in post 73. I'm not going to try working out the point of departure for an elliptic profile though.

If it was a free fall without the wall, then the top and bottom ends should keep exchanging positions, with each new fall being in a random horizontal orientation. The maths may be simpler if we consider an elliptical cylinder and treat it as a two-dimensional problem.

I just added these thoughts as a matter of interest. I'll remain satisfied with the qualitative picture, and hope there is no gross conceptual error in my thinking. Regards =TeeSquare=

Looks good to me - except that a generalised convex shape would be extraordinarily hard to analyse . There may be a few that I could handle analytically (at an appropriate fee!), but most would have me resorting to numerical tools.

Your sollution is presented well; however I have a problem with your equation

y = (r² - x²)^0.5

When x = 0 then y = r - ok!

When r = 0 then y = (-x²)^0.5 - Not ok?

Please enlighten me?

As the equations are only valid while the dumbbells are touching the wall, x² must always be less than r².
Does that solve your problem?

I see! You are not treating r as a variable but as a constant.

In these cirumstances then surely x is a function of y and the equation amended accordingly?

Wouldn't simplify matters ?

r is specified in the drawn restatement of the problem as the length of the dumbell (it was l in the original question). Personally, I would choose to express everything in terms of the angle θ between the rod and the vertical, as this is easier to generalise. Given that the required answer is the horizontal velocity, I agree that x would be a more natural choice for the parametric variable than y - but it makes little difference to the manipulations in practice

Sorry guys and gals, but I can't hold on any longer. The answers given are almost ALL incorrect. (Including the official answer.)

Conservation of energy. YES

Trajectory of the centre of mass. YES

Energy equations used in derivation of "point of separation from wall". ALL WRONG

There are THREE energy systems in play while the dumbell is ROTATING from one stable condition (vertical) to another stable condition (Sliding across floor with constant velocity.). The third is the rotational dynamics of the dumbell accelerating from a vertical rest with rotating velocity = zero, through a maximum and reducing back to zero when the dumbell is horizontal. The energy stored in the rotational momentum means that the "theory" calculation used by everyone for the dynamic movement of the dumbel are thus incorrect as they only consider conservation of energy in two of the three systems.

The answer to the problem is very simple when we use the two stable conditions and some logic. The transition (unstable) condition does not matter.

My answer with some elaboration.

Start condition. Two masses (each m) joined by a rigid rod. I'll call total length "L", (centre of mass at L/2.) with end masses each diameter "d". (With no rotational velocity/energy.)

End condition. Two masses (each m) joined by a rigid rod moving with constant velocity "v" across the frictionless floor. (Centre of mass now at d/2 from the floor, meaning both dumbell ends in contact with the floor.) (With no rotational velocity/energy.)

Logic to use: Once the dumbell loses contact with the vertical wall there can be no further change in horizontal velocity. (Since from that point on there is no "opposing force" available to change the horizontal velocity, so the end velocity is equal to the velocity when the dumbell leaves the wall.)

Now we use our conservation of energy formulae.

Delta Potential Energy (PE) = (L/2 - d/2) * g * (m + m)

[Centre of total mass falls from L/2 to d/2.]

Delta Kinetic energy (KE) = 1/2 * (m+m) * (v * v)

[Total dumbell is moving with constant velocity.]

Delta rotating energy = Zero - Zero

Thus 1/2*(m+m)*(v*v)=(L/2-d/2)*g*(m+m)

Eliminating (m+m) and multiplying by two.

v*v=(L/2-d/2)*g*2

v*v=(L-d)*g

v=sqrt[(L-d)*g]

Think about all the other answers that say the dumbell leaves the wall before reaching horizontal situation. They all account the total PE as having been converted, but in fact unless the dumbell is completely horizontal, then the final PE condition has not been reached.

If the dumbell left the wall as described in many of the solutions, then they are implying the dumbell would move across the frictionless floor with constant velocity with what was the top end never falling any further.

If you really want to "crunch the math" then add in the necessary rotational inertia and such in the energy equations, do the integrations and get a headache.

Let me apologise in advance, as your courage in going against the crowd does not deserve a condescending response, and I fear that what follows may appear that way.

Conservation of energy applies continuously through time. So the amount of potential energy that has been removed from the top mass at any particular time must be present in some other form at that time. If you bear that in mind when you look at the equations in the "good answer" solutions, you will see that they do in fact conform to this requirement. [That is to say they do not convert "all" the available potential energy - just the amount that has been sacrificed at any particular time]

Now, the following may be a misinterpretation, but the only way I can see that you have reached the conclusion that the final position has to be horizontal is if you believe that the balls on the ends can only have constant horizontal individual velocities. That would be the same as saying that a rigid body moving freely through space cannot rotate. Please see what happens when you throw your ruler (gently now). What happens is that the centre of gravity must have constant horizontal velocity once the dumbbells have parted company with the wall; the ends can accelerate relative to the centre of gravity (equal and opposite if the dumbbells are symmetrical), because the connecting rod provides force along its length. It is this rotation of course that allows the top end to continue falling.

You also express discomfort with rotation not being considered as an additional degree of freedom; while we consider the total mass of the dumbbells to be located in just point masses at the ends, all the kinetic energy is indeed accounted for in the linear motion of the masses - provided you consider them separately. That is no longer practical if the rod has mass or the masses are distributed, in which case you need to consider the body in terms of a mass at its centre of gravity and its moment of inertia. This more general case was analysed in post #75; it was still a restricted case because the dumbbells were assumed to be centro-symmetric. (BTW, this formulation has the advantage of starting with just three degrees of freedom in the plane of movement, whereas the two-body solution starts with four, which then reduced to the equivalent by constraining the separation of the masses)

I think that covers all the issues you raise. If not, please clarify what is omitted.

Fyz

I hate to cause consternation between us engineers; I enjoy intelectual exercise along with the best of them, however not all the potential energy will be converted directly into horizontal kinetic energy. The velocty of the top dumbell will be at its maximum when it hits the ground. Its action would be normally through its own center of gravity, especially if its mass is signifcantly larger that the rod.

In reality the top dumbell would normally bounce when it strikes the ground. Presumably, when specifying the initial assumptions, you could have added "dumbells having no elastic properties"? But would this have any effect on the horizontal velocity because both plains are frictionless? If there is no friction between the dumbell and the surfaces, will it matter if the dumbells are in the air or not?

If the top dumbell does bounce it will cause a partial negative rotational effect; retarding the motion of the lower dumbell. The motion will therefore be of a non linear nature.

Life and engineering are not always as straightforward as they first appear.

I fear that you might have read the question incorrectly.

The upper end of the dumbbell looses contact with the wall long before it gets anywhere near the floor. Therefore, there is no bounce to worry about. Any bouncing would occur at a point in time after the instant referred to in the question. The nature of that motion could be a subject for another question, but it is not part of this one.

You write: "not all the potential energy will be converted directly into horizontal kinetic energy". I agree, and the proportions of the total kinetic energy that appear in the horizontal and vertical directions respectively (at any time before the top dumbbell loses contact with the wall) can readily be seen to be sin²(θ) and cos²(θ) (using the notation of post #63). In fact the horizontal component of the kinetic energy reaches its maximum just as the top of the dumbbells loses contact with the wall - and the horizontal component of KE stays constant so long as the dumbbells continue moving unrestrained across the frictionless floor, whereas the vertical component continues to increase until the upper mass hits the floor. We know nothing about the vertical component following impact, because we aren't told the coefficient of restitution.

BTW TeeSquare's (post #157) image of the "dumb bunny hopping along to infinity, nose never leaving the floor, but rear end bobbing up and down" is exactly right for lossless movement on an infinite plane (with minor modifications if you want a round Earth)