Dumbbells: Newsletter Challenge (02/12/08)

Looks like someone's first year homework in dynamics. Too simple. See you all next week.

Not up to the usual standard.

I didn't set this problem, but I agree that all the principles should lie well within the scope of a first-year engineering** course. However, there's plenty of scope - both for errors of principle and for careless mistakes. As evidenced by the lack of correct solution so far.

Of course, for the case with point-masses and massless connecting rod this is the same as the first maximum in the velocity of the horizontal mass. However, as they are not explicitly specified, you could also add in the effects of mass dimensions (shape?) and rod mass.

I haven't double-checked my sums, but a first-go-through for zero-diameter spheres and massless rod gave an answer of 2/3.sqrt(2.g.l/3).

Fyz

**I don't know the present situation, but at one time all the necessary tools would have been in place at least a year before leaving high school.

"2/3.sqrt(2.g.l/3)"

Can you help me understand how you came to that equation?

Positions = l.sin(theta) and l.cos(theta) respectively. theta=f(time), and differentiate for velocities. KE=m/2.sigma(v^2). Equate KE with change in PEs. Substitute to give v(horizontal) as an expression of theta,m,l,g. Find maximum velocity.

Unless someone else does it first, I'll write it out in full when I've a bit more time.

Fyz

I vaguely remember a similar problem in college dynamics class where there is a string anchored on one end and the other end is connected to a weight. With the string tight and horizontal, the weight is released and accelerates through a perfect arc. The problem was to find the speed of the weight when the string reaches vertical. The solution only involved the length of the string and the acceleration of gravity.

I still think that you need to consider where the center of gravity is though.

I do not think it is a trivial problem. Have you tried to find the solution? I mean the exact solution?

Abe

I agree. In fact I found this far more interesting than the average challenge question. To each his own, I suppose.

v=√(g x l)

Sounds simple but the determining "L" makes it tricky. "L" is the distance that the center of gravity drops. So "L" would be the difference of 1/2 the lengthof the dumbbell and 1/2 the dumbbell diameter. We also need to know the dumbbell's geometry to determine this.

When I do it the bottom end slides away from the wall, the top slides down the wall and never loses contact...

That is what I first though, but the wall and floor are frictionless.

It is not stated, but I will assume that l is the total length of the system and I am going to think of the ends as two spheres with zero size to make the calculations easier.

The velocity of the horizontal component will be equal to the vertical component just before the vertical component stops.

Since the whole system is frictionless, then the answer should be the velocity for a given distance over length l and the acceleration of gravity... But it is not!

The problem is that this is not a true free falling body. Only a portion of the body is free falling (The upper sphere).

The mass that slides horizontally is not the free falling portion, so it has mass and thus momentum. Since it is the rest mass, the kinetic energy from the free falling mass must be used to move the rest mass. We can calculate that kinetic energy as:

KE = (mV^2)/2

I am assuming that the object can be thought of as two spheres of equal mass the distance l apart and that the rod between them can be thought of as zero mass. So remember to use only 1/2 the total mass which actually drives the object and the other half resists it by momentum. That means m = m/2.

We can solve for KE for the top sphere of the dumbbell (use the first equation for a free falling object and the force of gravity's acceleration) and determine the energy imparted on the second sphere and the top sphere! The top sphere has zero horizontal velocity, so that needs to be factored into the final equation. It isn't until the lower sphere has reached maximum velocity that the upper sphere hits the floor. The moment that the upper sphere begins to move horizontally it becomes an anchor for the lower sphere that now has maximum horizontal velocity. We do know what the lower sphere's KE is because it will be the same as the PE of the upper sphere, but now the mass suddenly doubles at the point when the upper sphere hits the floor, which affects the final velocity of the system.

Again KE = (mV^2)/2 is the equation to use, but m = the full mass of the system now. We recalculate solving for V = sqrt(KE/m).

We should be able to shortcut the process for calculating the KE. KE = PE of the upper sphere only, which has mass m/2 and height l.

PE = mgh

where:

m = the mass/2

g = Force of gravity (9.8 m/s^2)

h = height of the upper sphere = l and the sphere is a point mass.

M = total system mass of the dumbbell in the equation below...

Now we can solve for V = sqrt(PE/M). Tada!!!

"no vector force to shove it across the floor"?? Even in the presence of friction, if you lean ordinary dumbbells against a wall at a suitably flat angle, the top will fall and the bottom mass to move outwards. In the absence of friction there is no restraining force, so it must eventually start to pull the top mass away from the wall.

If you try it with real dumbbells, it will obviously stop its movement, but the probability of the dumbbell actually touching the walls when it comes to rest is quite small (though it could be quite close - if so, just start the sliding from a different angle).

Ahhh! No! No! Please don't join -- you'll fall off the edge!

If you look at my crude little picture in post 12, you'll see that the cg of the dumbbell starts off following an arc, if the top end stays in contact with the wall. But you can also see that there is a point in that arc where the horizontal component of the velocity must start to decrease, if the cg continues to follow that arc.

But given that the surfaces are frictionless, what force could cause a decrease in the horizontal component of the velocity? The piece of linkage that was on my roof bolters does not exist in this case, so the dumbbell would be free to leave the wall, given its horizontal momentum, and the CG should continue moving with that horizontal speed, until it hits the person who was so foolish as to stand it up against a frictionless wall and on a frictionless floor. (I suppose the person doing so travelled via jet pack.)

If Fyz is right, then the angle of the dumbbells, at the point where wall contact is lost, would be that angle which has a cos of .666. <<--- That number alone should be a sin () that this problem is the work of the devil.

In fact, I suppose that you could say that you will find sin in many problems that use trig. Repent!

Naturally, if you do this correctly and round upwards, the beast will disappear. As Flaubert wrote: "le bon Dieu est dans le detail".

I essentially agree with your answer, but since the ends of the dumbell are circular, the potential energy difference between the initial position of the dumbell and the final position can be given by mg(Δh) where Δh is (l/2 - r), l is the given length of the dumbell, and r is the radius of the circular ends. When you set the kinetic energy (1/2)mv² equal to the difference in potential energy, the mass terms cancel.

So v = √¯(2g[l/2 - r])

Quibble: The question says the dumbell is in the corner at the intersection of 2 planes, but counting the two walls and the floor, it should be 3 planes.

Anonymous guessed,

I suggest you concentrate on the problem, rather than carping about your misreading of the detailed wording.

Your solution makes an unstated assumption about the point at which the upper mass loses contact with the wall; unfortunately, either the assumption or the subsequent calculations are wrong.

The two intersecting planes that make the corner are clearly defined as being the wall and the floor. Given the definition of the problem, it makes no difference if you choose to place the top of mass in the corner between two walls - but that is not in the problem as stated, nor is it necessary.

Similarly, the masses are defined as points; according to any description of a dumbbell, the masses have greater diameter than the connecting rod - so the rod must have zero diameter. That makes the radius you say needs to be to subtracted equal zero.

Anonymous calculated

Point A, I didn't 'carp'. I said it was a quibble, a minor point. The person who formulated the question could have said the dumbbell was sitting against a wall, but he (or she) did not; he used "standing in a corner"; "corner" sounds like 'corner' to me, not 'wall'. So my quibble is simply that the questioner ought to use clear language.

Point B, my initial assumption was that it did not lose contact with the wall. Therefore my initial thought was that the speed it achieved was the speed when the upper part of the dumbbell contacted the floor. Subsequently it occurred to me that the dumbbell would lose contact when the maximum distance between the wall and the center of the dumbbell had been achieved. This distance, s, is given by the pythagorean formula, and requires a real length (l) and radius (r). This distance, s, is given by s = √¯[r² + (½ l)² ]

Point C, your comment that the masses are 'points' does NOT occur in the description of the problem, nor does your statement that the 'masses have a greater diameter than the connecting rod'.

I didn't write the question, and you are sure carping now, even if you weren't in the first instance. But the question clearly states that the corner is "formed by two frictionless planes, the wall and the floor". You are right about the masses not being stated to be points (which would mean that in principle we need to include the effect of moment of inertia). However, the question does state "dumbbells" which are of a known shape as stated.

As you supposedly follow on from anonymous hero, and he describes the outer surface as two spheres, the centres of the spheres will move along vertical and horizontal lines respectively. Under these conditions, it is incorrect to use Pythagoras with the radii of the spheres.

Even for the situation you claim to have calculated (velocity of upper mass when it hits the floor), your result must be wrong - if the upper mass stays in contact with the wall, the lower mass will be stationary when the upper mass hits the floor. So the velocity of the upper mass would be sqrt(2gl) (plus any modifying terms for shape. It was the factor of two error that made me think you had at least started to address the problem (my mistake)
It should be obvious that the lower mass must pull the upper mass away from the wall as soon as it (the lower mass) starts to decelerate. We can show that this must happen as follows: if the upper mass were to stay in contact with the wall, the lower mass would start stationary, and be stationary again when the originally upper mass hits the floor. So the lower mass must accelerate and then decelerate - and immediately it starts to decelerate it will pull the upper mass away from the friction-free wall.

I've thought about my reply, above, and realized that the Dumbell will leave the wall when s is a maximum, which is given by the pythagorean relation, s=√¯[(l/2)² + r² ] , where l is the length of the dumbell and r is the radius of the ends.

Dumbell

At this point the change in potential energy is given by the amount by which the center of mass has fallen. This difference in height would be Δh = l - l cosθ - r sinθ.

The angle θ is given by θ = atan(r/(½l))

If the length of the rod is to the centres of the spheres, there will be no length modification - the centres of the spheres will just move along planes that are displaced by the radii from their respective planes. If the rod is joined to the ends of the spheres, the effective length will be (l+r). If you are working with a cylinder, the directions of motion are altered as well as the potential energy. Perhaps get the basic version right first?

Am I alone in reading the question as meaning that each of the masses has value 'm' (not that this makes any difference to the final answer)?

Del, not to be picky, but since the floor is a "frictionless plane", why wouldn't the dumbbell continue to slide away from the wall for an indeterminate distance? The increasing velocity of the falling top mass should drive the entire dumbbell a significant distance, which one of you guys who are smarter than me can tell us about. With no friction, why would the velocity of the moving dumbbell slow with distance from the upright frictional surface?

It depends upon whether the former upper mass bounces in an elastic collision with the frictionless floor or not. If it does, then the system slows to a halt at a new metastable position, with the former upper mass immediately above the former lower mass like before, though this time with the whole dumbell system shifted off away from the wall (someone else can do all that clever algebra-stuff and work it all out).

Of course, not having any friction in this strange world, it will just stay there metastabily, though a passing frictionless cat might disturb the non-frictionless air such that the impinging velocity of the stagnant streamline upon the upper mass and the lower pressure caused by boundary layer separation downwind of both it and the rod (length l - watch out for vortex-shedding downwind of it setting up lateral vibrations at its harmonic frequencies as it is not fitted with spiral strakes) causes the equilibrium to be disturbed such that the whole darn shooting match falls down upon the poor cat's head. If THIS collision is not elastic then the cat might be taking half a day out with the undertaker, having written off one of its nine lives.

Of course, if the first collision is inelastic, then the dumbell will continue to move in a hostile manner, frightening our poor cat out of the room until the inelastic collisions cause it to come to rest. Or at least, that's the plan. In reality, our frictionless cat will be unable to accelerate relative to the floor to get away from this system, be probably terrified by the action of this dumbbell thrunging about, and will lose another one of its lives as it dies of fright.

Who'd be a cat, eh?

Sounds like undertaking is a good industry to invest in.....

Assuming that nothing gets in the way, that the ground remains horizontal, and neglecting Coriolis effects, viscosity, etc:
elastic or not, "no friction" means that magnitude of the horizontal component of the velocity will remain unchanged once the top mass has parted company with the wall.

Good ! I agree !

I agree that the question is ambiguously worded "two masses (m) at each end" or is it "two masses (m), one at each end".

Anyway, if the rod is massless, and the wall and floor are frictionless, then because they're connected the velocity of the sliding mass will equal the velocity of the falling mass which will be as if it was free-falling from a height (h) equal to the length (l) of the rod (the basic equations of motion should do it). You're also probably right about the falling mass not leaving contact with the wall unless it is acted upon by some other force and bounces or something.

Cheers

Graduate of the "5 minute university"

Ah ...

Let's see ...

Dumbells of zero dimension operating in a frictionless world moving off to infinity (and beyond )

I think I've been there before.

Wait! That sounds like those years High School listening to Pink Floyd... :-)

I don't mean to upstage Del's drawings, I just want to illustrate a couple of points.

Even with frictionless surfaces, the dumbell will slide more easily with rounded edges.

Obviously, acceleration is provided by gravity acting on the upper mass. The inertia of the lower mass will resist this motion.

If the resistance of the lower mass decreases at the same rate the velocity of the upper mass increases, the dumbell comes to rest with the top mass still in contact with the wall. The problem would have been better stated to find the speed of the bottom end when the top end stops falling.

Now I need to ruminate on the math.

Hmmm.

In your second picture, the dumbbell would not slide down the wall. Its CG is to the right of the support point on the floor. It would rock back to the shaft vertical position.

The calculations would be easier, I think, if we assumed the dumbbell ends were points.

If we imagined the upper point to fall straight down, then in the last increment of time, when the lower point is moving fast to the right, imagine how fast the wall end would be moving. Let's consider the situation at .1 degree from horizontal. At that point, the vertical movement would have to be 572 times as fast as the horizontal movement.

I remember from my coal mining days a linkage (on a roof bolter) that looked like this:

I should have labeled this, but the hash marks indicate that the piece of linkage with the arc at its end is equal to half the length of the piece that moves like our dumbbell. The joint between the two pieces would be at the CG of our dumbbell. As the CG moves through that arc, there comes a point where the horizontal component of the CG's movement would slow. Being that our situation lacks the linkage of the roof bolter, there would be nothing to keep the dumbbell upper end from pulling away from the wall at that point. Then the CG could no longer follow the arc that would lead to the upper end ever touching the wall/floor intersection.

Another way of looking at this is that if the upper end wound up in contact with the wall, that means that at that instant, the other end could have no horizontal velocity (for the corollary of the reason that the inner end moves 572 times as fast as the outer end in that last .1 degree.) So, I think the dumbbell will leave part company with the wall before it hits the floor.

Yes, in the ideal case the top is still at 2.l/3 when it parts company with the wall - much higher than I would naively have expected.

Just trying to visualize in my mind, it appears to me that the downward acceleration of the upper dumbell would be controlled by the force and time required to accelerate the mass of the lower dumbell. That force, as transmitted through the link will continually vary with the angle at which the link acts upon the lower dumbell. At some point as that angle decreases, the horizontal speed and inertia of the lower dumbell will cause the link to pull the falling dumbell away from the vertical surface.

If I am in error and the falling dumbell remains in contact with the vertical surface until it contacts the horizontal surface, the the horizontal speed of both dumbells will be zero, again due to the link.

I think you are correct.

For the dumbell to remain in contact with the wall, something would need to slow the horizontal motion of the lower mass. With no friction, there is nothing to do that.

Assuming it slips at all, an inflexible rod must eventually part company with the wall**, though it may fall back under some circumstances - but it would have to fall back "just right" for there to be no gap between the wall and the end of the rod in the final position.

**After all, as the orientation tends to horizontal, the required velocity of the lower mass to avoid this tends to zero - and if the lower mass is stationary the lateral force to keep it from going through the wall tends to infinity.

At some point as the dumbbell falls, there will be a maximum distance between the wall and the center of the dumbbell. In your drawing, if l is the length of the dumbbell and r is the radius of the ends, then this maximum distance, s, will occur when s = √¯[r² = (½l)²].

Once this maximum distance has occurred, the dumbbell will continue to slide away from the wall, so there should be a gap between the wall and the dumbbell in the bottom view in your sketch.

The dumbell would pull away from the wall as soon as the angle was less than 45 degrees due to momentum of the lower mass, and the center of gravity moving away from the wall. The velocity of the lower mass would be at maximum at that point. Lower mass velocity would be equal to upper mass velocity at the moment it reached the 45 degree angle.

The sum of the kinetic and potential energies is constant and equal to mgl. If we define theta as the angle between the rod and the corner, x as the distance on the bottom mass from the wall, and y as the distance of the upper mass from the floor, x= l sin(theta) and y = l cos(theta). Take first derivatives to get linear velocities and second derivatives to get linear accelerations of the masses. When the linear acceleration of x drops to zero, the force with the wall drops to zero and contact is lost. Do a little algebra. I get cos(theta) = 2/3 or about 48.2 degrees. The bottom mass is moving at sqrt(g l) times (2/3)^1.5.

It's reassuring to see the same answer as in #6. And by a slightly different method (see #14).

Isn't an associated mass necessary for a calculation of velocity?

Or is the lack of friction nullifying the need for this component?

I am used to dealing with thing in a friction based world so sometimes the hypothetical world confuses me.

Also I am use to using the kinds of dumbbells that you add and remove the weight from so I have an added supposition.

Wouldn't the protruding end dig into the wall as it fell or would the Frictionless coating that seems to apply in this magical room prevent that from happening.

I only ask this because when I was younger and much more ignorant I left my barbells precariously perched against the wall on my moms beautiful hardwood floor which was massively scarred as well as the wall due to the afore mentioned sliding effect.

Misplaced reply moved.

The mass is unnecessary. The principal of dropping a bb and a bowling ball applies: they both accelerate at the same rate. I think for this problem you need to assume at least spherical shapes for the ends of the dumbbell. Also making them points seems to make the visualization a little easier, because then the length is clearly the overall length and not the shorter center-to-center length of the spheres.

What is wrong about this reasoning?

The problem is to determine the speed.

L= Length of rod

g = accl of gravity

t = time

Velocity = distance/time

In this case the 'time' is = square root of L/g

In this case the 'distance' = L/2

Ergo V = [L/2] / [Ö L/g] = (L* Ö L/g) / 2

The rod length times - the sqrt of the rod length divided by g – all divided by 2 equal the speed

My sqrt symbols have been transformed into gibberish

Assuming point masses and a mass-less rod, here's my try:

The only force acting on the system is gravity. Both masses will move length l in same amount of time, therefore linear velocity of each mass along it's respective plane is equal. Since x=1/2at² we can say that l=1/2gt². Since v=at, we can say v=gt. Solving the first for t gives t=√(2l/g). Substituting this t expression into second equation gives v=g√(2l/g).

That seems too easy - what am I missing?

You are missing the detail of when the top loses contact with the wall. The displacement is the same, but the velocity varies over time. The upper mass is accelerating it's entire length but the lower mass is accelerating only until the the upper mass reached approx the halfway point when the horizontal force vector approaches 0. Since it is frictionless it will actually be a bit before that. I belive a 48 degree angle was calculated earlier. Incidently, the acceleration of the upper mass is not constant due to inertia of the lower mass. At the point that the upper mass loses contact, acceleration increases for the upper mass and disappears for the lower mass.

I haven't seen this answer yet so I'll have a go

I'm going to go for the easy option of ignoring the actual shapes of the weights assuming that they are point masses of no size.

First notice that the locus of the centre of mass of the dumbbell conveniently follows an exact circle (until it leaves the wall).

At first I thought it was as easy as working out when a sliding mass would fly off the circular slide, but I was forgetting the rotational energy of the weights.

So we need to work out the kinetic energy of the two masses separately. Let the top mass be y from the floor and the bottom mass be x from the wall, and, let the two velocities be V₁ and V₂.

Then the gain in K.E = the loss in P.E..

i.e. ½mV₁² + ½mV₂² = mgh = mg(l-y)

so V₁² + V₂² = 2g(l-y) ...............(A)

Now I still think that the bar leaves the wall at the same time as the centre of mass would fly off the locus (at the velocity calculated from above), but, I'm open to correction. So the centrifugal force needs to equal the forces holding the mass on the slide.

The velocity of the centre of mass is just the vector sum of half V1 and half V2

So the centrifugal force is 2m((V₁/2)² + (V₂/2)²)/(l/2) = m(V₁²+V₂²)/l

We can think of the opposing forces in two different ways (which both fortunately give the same result). Either the full 2m is acting on the imaginary slide, or, the upper m is acting with mechanical advantage of 2, and the lower weight is supported by the floor.

i.e. force = 2mgCos(ß) = 2mg(y/2)/(l/2) = 2mgy/l

So when the centrifugal force equals the opposing force

m(V₁²+V₂²)/l = 2mgy/l

i.e. V₁²+V₂² = 2gy ..................(B)

That's lucky combining (A) & (B)

2gy = 2g(l-y)

i.e. y=l/2 (that is when the bar is at 30° to the floor)

At this time the horizontal velocity of the bottom mass equals twice the horizontal velocity of the centre of mass. Let the velocity of the centre of mass be V

Now from (B) V² =gy/2

so substituting for y: V = ½√(gl)

so velocity bottom weight = √(gl) Cos(ß)

= ½√(gl)

I agree with many of your statements - though I would find this a hard way to do the modelling.

I first come adrift with your "centrifugal force". What you seem to have is the unsigned sum of the forces on each mass; however, these are in opposite directions and add up identically to zero (this is a result of the definition of the centre of gravity, of course).

Nonetheless, where you lose me completely is "or, the upper m is acting with mechanical advantage of 2, and the lower weight is supported by the floor" which appears to simultaneously consider the mechanical advantage over a central mass of 2m, and the balance of vertical forces on the right-hand mass of value m.
I'm also unable to follow apparently equating of a generalised centrifugal force to external forces. I assume you are saying something equivalent to "the compression/tension in the rod is zero at this point, as is the force from the wall, so the inward component of force due to gravity on the top mass must be exactly sufficient to account for its centripetal acceleration". That should be true.
BTW, I probably have personal issues, because I find (true) statements such as "At this time the horizontal velocity of the bottom mass equals twice the horizontal velocity of the centre of mass" confusing (only because it is unnecessarily narrow - the statement is true at all times preceding loss of contact).

Fyz

Although I've had problems following your detail, one of your observations has triggered the following:

You noted that the middle of the rod followed a circle with a radius of half the length of the rod. This is exactly the same path that would be traced by the middle of the rod if the top was disturbed away from the wall instead of the bottom. Following this further, we can find that the vertical velocity of each part of the rod (at a given angle to the vertical**) will be unchanged; on the other hand, the horizontal velocities are swapped between the top and the bottom.

Moreover the base of the rod will move away from the wall when the rod reaches precisely the same angle as in the original question.

That wasn't immediately obvious (at least to me). It was interesting to 'discover' this (though I doubt I'm the first). I've no idea if it's interesting in any other way.

**The orientation being reversed, of course

I was thinking all this looked pretty good, and then I saw this:

So deceleration starts when (1-cos(θ))).cos²(θ)

It has a certain "oddness" to it.

Very nice, clear, explanation. If I could vote for "good answer" twice, I would.

So deceleration starts when (1-cos(θ))).cos²(θ)

I did wonder whether to differentiate the whole thing to find the zero acceleration point. But that would produce a load of untidy terms, and is only equivalent to finding the peak of the velocity. (The final answer would be identical, and it would be a nightmare to present via a CR4 interface).

Actually, I was being silly. The oddness applies only to the number of parentheses.

Yes, cut and paste is great for these things - but I forgot to double check for residual braces.

Now for the sole "insight" of the process: ....

Well done! This is the "insight" I missed in my effort... Myself, I tried using cartesian coordinates until I stuck at this critical point. I now completed it and post it just to have it as an alternative solution:

Let v₁ the velocity of lower mass and v₂ the velocity of the higher one. Also, let the (0,0) being at the floor corner. Then v₁=dx/dt, v₂ = -dy/dt. The binding condition between these two masses (during the time the upper one has contact with the wall) is that it must always be x²+y²+l². By differentiating with regard to time, we get: 2xv₁-2yv₂ = 0 => v₂ = v₁.x/y (1)

Now, by using energy preservation theorem, we get:

1/2mv₁² + 1/2mv₂² = mg(l-y) => v₁² + v₂² = 2g(l-y) (2)

(1) and (2) yield: v₁ = y/l.(2g(l-y))^1/2 .

By using Physicist's insight, we seek to find the maximum value of this velocity. That means, we seek the y which maximizes l.(l-y)^1/2 or better: l²(l-y)

By differentiating and setting to zero we get y=2/3.l, which leads to the same result.

but...by your reasoning, at the moment that the dumbell loses contact with the wall, the dumbell must be absolutely horizontal, and hence no longer moving away from the vertical wall. Speed = 0.

I think you are reading something into my text that is not there.

I say the ladder parts company with the wall as the lateral acceleration of the bottom end passes zero, not the lateral velocity. That means the lateral velocity is at its maximum.

I have not digested all the replies, but after giving it some through, have come to the following conclusion, At first I thought that the bottom mass would first accelerate then de-accelerate when the angle of the bar fell below 45°, I don't have an answer but pose the following question? If M¹v¹ = M²v² would it not follow the the bottom velocity would be the same as the falling velocity? as both masses are the same? or has this been taken to be obvious. Therefore would the velocity be half g?

Regards JD.

I couldn't resist trying the solution for a case that is closer to the one actually specified in the question – rather than the one implied (although I believe the implied one to be the one intended?).
We can allow masses that are not point masses, as well as a rod that does not need zero mass. You may find part of the result surprising.

The constraints here are as follows:
. The ends of the rod (length l) are at the centres of spheres or equivalent)
. The centre of gravity is at the middle of the rod
. The only forces acting on the dumbbells are gravity and the pressure due to the walls.

Purely for consistency with the original statement of the problem, I shall retain a total mass of 2.m
The distributed nature of the total mass is taken into account via the moment of inertia (about the axis perpendicular to the plane of movement). I shall write this as 2.m.R²
The angle between the rod and the vertical is defined to be θ
The horizontal distance between the centre of gravity and the initial position of the ends of the rod is defined to be x
The vertical distance between the centre of gravity and the initial position of the bottom of the rod is defined to be y
The acceleration due to gravity is gn
Derivatives w.r.t. time are written as θ' x' y'

x = l/2.sin(θ) , x' = l/2.cos(θ).θ'
y = l/2.sin(θ) , y' = -l/2.sin(θ).θ'

Change in potential energy (relative to stationary position = 2.m.(l/2-l/2).cos(θ)).gn
Kinetic energy = 2.m/2.(x'²+y'²)+2.m.R²/2. θ'² = m.((l/2)²+R²).θ'²

Equating change in PE and KE gives
l.(1-cos(θ)).gn = ((l/2)²+R²).θ'² , or
θ'² = l.(1-cos(θ))/((l/2)²+R²).gn

That gives the horizontal velocity of the mass as:
x' = l/2.cos(θ).(l.gn(1-cos(θ))/((l/2)²+R²))^½

The rod parts company with the wall when the horizontal velocity of the centre of gravity reaches its maximum – i.e. when
cos(θ) = 2/3
The angle at which the rod loses contact with the wall is independent of the distributions of the masses – I suppose that I should have expected this**, but I had not thought it through.
**on the basis of linearity – the accelerating force depends only on the position of the rod, and the fact that the (cos²(θ)+sin²(θ)) term is constant means that the resistance to motion is invariant with position (as is the moment of inertia).

Completing the problem as stated, the velocity of the bottom end of the rod when it first loses contact with the wall is:
2.x'_separation = l.(2/3).(l.gn.(1/3)./((l/2)²+R²))^½
= (2/3).(2.l.gn/3)^½/(2/(1+(2.R/l)²)

Why all the math? It maks no difference what the Mass is OR path of the fall.

A bullet shot horizontally from a gun at the same time an identical bullet is dropped to the ground - both bullet will hit the ground at the same time.

The question is SPEED. Speed is distance divided by time. Distance for top of dumbbell to hit floor is L or l of rod - time is sqrt of l/g .

That means the speed is [l]/[sqrt l/g]

However, the question is not asking what the speed of a mass dropping from a height of l would be. Per several explanations above, the mass on the floor must reach a maximum well before the upper end of the dumbbell reaches the floor.

The question is asking what the horizontal speed of the lower mass would be at the time that the upper end looses contact with the wall, (which occurs at 48.2 degrees from the vertical, i.e, with the upper mass 1/3 of the way down).

Here's a link to the "sliding ladder" question as provided by the physics department at Harvard. In a post above, I mentioned a roof bolter which has the same geometry as a sliding ladder (or sliding dumbbell). I'd remembered coming across this sliding ladder problem around the time I worked with mining equipment (decades ago) and remembered that the ladder would part company with the wall when the tip is 1/3 of the way down. Therefore Fyz's solution struck me as being right (and is still seems that it is), although I note that the Harvard teacher has come us with a different answer.

I agree with the Harvard folks that as long as the masses at the end of the dumbbell are the same, that you can work with the CG rather than the individual masses. Now, I'm perplexed, because the Harvard solution seems to make sense, but I haven't had time to compare it to Fyz's. I assume I'll come away completely befuddled, agreeing with both -- although Fyz's solution seems to get the lower weight ripping along pretty fast -- maybe too fast.

In fact, definitely too fast... I think -- it seems to suggest that a lot of the potential energy has been converted when the top weight is only 1/3 of the way down. Gotta go, but I'll come back at some point.

I followed your clue to find your missing link.

So far as I can see, the Harvard answer is equivalent to mine - the "differences" are that they are considering a ladder rather than a dumbbell, and the velocity they give is that of the centre of gravity rather than that of the base. For the dumbbells, η = 1, so the horizontal speed of the C. of G. is
(2.g.(l/2)/(1+1))^½.(1-cos(θ))^½.cos(θ) = (g.l/2)^½.(1/3)^½.2/3
That makes the speed of the base of the ladder =
2.(g.l/2)^½.(1/3)^½.2/3 = (2/3.g.l)^½.2/3 as before

BTW, Harvard give a very elegant presentation (well worth looking at), including showing that the angle at which contact is lost to be independent of the (symmetrical) distribution of mass.

I wish I could claim I was drunk when I posted my previous comment: failing to actually make the link rather than simply referring to it as if it were there, could be used as evidence. Sadly, though, I was sober.

Despite the fact that CM is mentioned 9 times (let's say about 10 -- I could be wrong here too) in the Harvard answer, I somehow convinced myself they were talking about the speed of the lower end rather than that of the CM. And let's see... how many times does η appear?

Perhaps there is more to reading than simply visual perception.

I could while away the hours conferrin' with the flowers...

A ladder (distributed mass) invests slightly less of the P.E. in K.E. of rotation (for a given ω). Fyz's observation that the problem is exactly the same whether you disturb the top or bottom of the object works for both the dumbbell and the ladder and should give the two different results, much more easily.

Yes. I seem to have regained my senses... although for how long I cannot say.

I am now more confused than I was before I didn't know I was confused.

Rereading the puzzle it states that the bottom is moved away from the wall a finite distance to allow the top end to free fall.

So I loaded up Excel just to confirm what the prevailing thoughts are on the problem. I made the length an arbitrary 10.0000001 meters long so I can use round numbers and set the height to 10 meters so the base is is 1.41 mm away from the frictionless wall.

I then setup the spreadsheet to calculate the horizontal velocity of the bottom of the dumbbell per given drop in height of the top of the dumbbell.

When you look at the graph for the horizontal velocity you see an initial increase in velocity for a very short time, then a steady decrease. What I am looking for is the point where the velocity levels off just before it decreases again. Sure enough it does, but the point where the transition occurs is about 1.00 mm down from the starting height of the dumbbell.

I believe at that horizontal velocity transition point is where the top of the dumbbell will exit the wall horizontally. My rational is that initially there is momentum driving the increase in horizontal velocity of the bottom of the dumbbell. When the horizontal acceleration stops the inertia from the lower dumbbell will draw the dumbbell away from the wall. Since the surfaces are frictionless, the dumbbell leaves the vertical wall at the precise moment when maximum horizontal velocity is achieved (about 9.898 m/s) and maintains that velocity indefinitely.

This leads me to think that the dumbbell's top leaves the wall almost instantly, but how soon it leaves is totally dependent on the amount of distance that the lower end of the dumbbell is moved from the wall.

The puzzle states, "When you move the bottom end slightly away from the wall, the dumbbell begins to slide." The Excel sheet shows that the initial distance moved from the wall determines how soon the top end looses contact. Furthermore, I assume that the act of moving the bottom end does not impart a net initial velocity on the dumbbell. In other words, you move the bottom out at set distance, then hold it in place, then release so that all movement is from the force of gravity.

My calculations do not take into account inertia of the lower dumbbell that must be overcome by the free fall of the upper dumbbell, so the actual acceleration (g) will not be constant or equal to 9.81 m/s/s throughout the fall. Or will it? Does that mass really matter? Will the free falling mass cancel the horizontally moving mass's inertia?

A spreadsheet is only as good as the equations you enter. If you declare what these are, someone may be able to comment constructively. For what it is worth, while the ladder is almost vertical, the acceleration and velocity of the top are both tiny, so you can approximate the acceleration of the bottom as gn.sin(θ). So I can't see any scope for an immediate maximum of the velocity anywhere near there.

Obviously, the angle of lean will affect the leaving point - an initial angle will cause a slight delay due to the reduced angular velocity. To get the scale of this, for a 1.5-metre dumbbell and a 3-mm initial lean, the change in angular velocity near the parting point will be about 0.16%. Of course, if we don't care which way the ladder falls, we can rely on random thermal processes to start the fall one way or other, giving an effective initial velocity that corresponds to microns of initial offset.

I am not able to post the Excel table, but basically, I used the vertical downward velocity = SQRT(2*a*x) for the upper dumbbell.

x = the displacement from the initial height for the upper dumbbell.

a = 9.81 m/s/s

Given a length of 10.0000001 meters and an initial height of 10.0 meters for the upper dumbbell I used a^2 + b^2 = c^2 and solved for b = SQRT(c^2 - a^2), where b represented the horizontal displacement.

Knowing the horizontal displacement rate from the Excel table you can calculate the horizontal velocity of b easily enough. So I just set up a number of columns as above and you can see the increase of velocity, where it levels off and then drops for the horizontal component of the dumbbell.

gn*sin(θ) is not what the data in the Excel chart predicts. There is clearly an initial horizontal acceleration (driven by the free fall of the upper dumbbell), but it swiftly reverses sense and decelerates about 13.5 ms after T zero.

Hi AH,

You might find this helpful (assuming you have not already looked at it).

If I understand your previous post re your spreadsheet, you are ignoring the inertia of the lower mass. You shouldn't do that, because it has a dramatic effect on the downward acceleration of the upper weight, particularly because the initial mechanical advantage is so poor (starting, if the dumbbell were truly vertical, at an infinitely disadvantageous value).

If you imagine that the dumbbell ends were not of the same mass, with the lower one being large and the upper one being small, you can see that the problem of getting the lower one to move expeditiously would be exacerbated.

Also, don't make my mistake (re the Harvard solution) of failing to realize that the horizontal motion they are referencing is at the center of mass. The lower weight, which is under consideration in our question here, would move twice as fast. Also, although you can think only in terms of the center of mass (assuming the dumbbell is symmetrical) to determine at what angle the dumbbell parts company with the wall, remember that the dumbbell is rotating, and that therefore distribution of mass affects the rotary acceleration.

Thanks!

No, I missed it before, but that helps a lot.

I happened to peek into CR4 again. Let's see if I can survive unscathed amidst the balancing bar bells and falling dumb belles.

This is a slightly different argument which leads to the same result as in the masterful exposition by Fyz (post 63).

Since we are dealing with a rigid body prior to any episode involving impact or deformations, we are not concerned with its elastic properties.

As far the laws of conventional mechanics go, we need only to define the external geometric shape of the object, and locate its centre of mass for studying its motion under external forces. The mass distribution should be immaterial* as long as we know the values for mass and principal mass moments of inertia (hereafter 'MI').

In our case we have two equal spheres with their centres separated by a distance 'l' and connected by a rod** of small radius to provide a dumb bell shape. To avoid confusion between 'ell' and 'one' let us define the half-length 'l/2' as 'h'.

We assume the centre of mass 'C' to be mid-way between centres, and the total mass is given as '2m' (could perhaps be interpreted as '4m' in the problem statement!). We also assume that the line joining the centres (hereafter just 'axis') is a principal axis , but the MI about this is not relevant for us. The MI about any transverse axis through C is 'I' (= 2 m h² ).

The body is initially balanced with the axis vertical, resting on a frictionless floor with both spheres touching a vertical frictionless wall. The floor contact point starts slipping away from the wall, and the body continues to 'fall' under constant gravity as the upper sphere slides down the wall as long as Newton will permit.

Any loss of potential energy due to lowering of C will appear as kinetic energy of translation plus rotation, since no work is done at the sliding contacts with floor or wall.

We can treat this as a two-dimensional problem, and if we take the origin of X-Y axes at the lower sphere centre, the initial co-ordinates of C are (0, h). In a subsequent arbitrary configuration if the axis is inclined at 'A' to the vertical, the following relations hold good, using the dash notation for time derivative (i.e. A' = dA/dt).

x = h sin A

x' = ( h cos A ) A'

y = h cos A

y' = -( h sin A ) A'

The constant energy condition is given by:

2m g h = 2m g y + (1/2) (2m) (x')² + (1/2) (2m) (y')² + (1/2) (I) (A')²,

which can be transformed as:

2 g h (1-cos A) = [ ( h cos A ) A' ]² + [ ( -h sin A ) A' ]²+ ( h A')².

2 g (1-cos A) /h = [ cos² A + sin² A + 1] (A')² = 2 (A')²

A' = dA/dt = [ g (1-cos A) /h ]^1/2

Separation at the wall will occur when the horizontal kinetic energy reaches a maximum value (beyond which some imaginary parameters may be required to satisfy the energy condition).

(1/2) (2m) (x')² should be a maximum, or

[ ( h cos A ) A' ]² should be a maximum.

Substituting for A' from the energy condition we get:

( h cos A )² [ g (1-cos A) /h ] should be a maximum.

The constants g and h (non-zero) can be removed, and we are left with:

cos² A (1-cos A) should be a maximum.

Since I've forgotten most of my calculus, I'll believe the guys who claim that it occurs at about 48.2 deg. Shaky calculus is also why I stayed clear of accelerations and wall reactions.

I anticipate some quibbling about my value for 'I', as I have taken the masses to be concentrated at the sphere centres. Now how long does it take for the axis to become horizontal, and how far will it be from the wall? Just adding to the challenge (I haven't worked it out!). Can someone give a good explanation for my somewhat intuitive 'maximum energy' assertion?

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* Not true for problems concerning impact or centre of percussion.

**The argument should hold in the more general case when the spheres intersect. The centres could even lie within both spheres. The only constraint is that no part of any connecting rod, whatever its shape, projects beyond the cylindrical envelope around the two spheres.

TeeSquare