Rolling, Rolling, Rolling...: Newsletter Challenge (June 2014)

I think you have the correct answer and a GA from me.

Torque is a factor, and for a solid cylinder I = 1/2 MR².

Where a ring of equal mass is I = MR².

If both mass and radius are the same, then:

mgh = 1/2 mv² + 1/2 Iw²

Note, that I is either 1/2 MR² or MR² depending on the body.

To answer another ambiguous question that others have made a comment on, I think negligible means just enough friction for rolling to commence.

Distinguishing between air resistance and surface friction might have helped. Since rolling was mentioned, I read it as air resistance being negligible. In the case of sliding, europium is of course correct.

I remember enough of these problems from high school and college to know that the crux of the problem is concerned with fundamental classical mechanical physics. Going beyond that was the realm of advanced degrees and theses.

"... I think negligible means just enough friction for rolling to commence."

In other words, "'negligible' can mean anything we want it to mean so long as it supports our foregone conclusion." That, in a nutshell, is what you are saying.

How can friction be considered 'negligible' when its presence or absence is fundamental to the outcome?

Well, re-reading the question states the cylinders are allowed to "roll", so sliding should not be a factor and the impact of any other forms of friction are to be discounted (negligible).

Sometimes a rose is just a rose.

I would have liked to have been in your physics class in school. I bet you gave the teacher a hard time!

My high-school Physics teacher, Dr. Norton, was a retired physicist from Sandia National Laboratories. It was he who gave us a hard time! Dr. Norton was very rigorous and, were he here, he would have ripped this Challenge Question (and most of the Challenge Questions posted here) a new ____hole. I'm merely continuing his long and notorious tradition of questioning hidden assumptions, identifying experimental biases, not leaping to foregone conclusions and ... running against the grain.

I expect that if the Challenge Questions specified every detail precisely, it would give away the solution.

In this case, they could have left out any mention of friction. The difference in the moment of inertia to weight ratio between the two cylinders would be more significant than differences in rolling friction.

It sounds a bit like Galileo at the Leaning Tower of Pisa.

Contrary to popular belief, "...physics professors don't like working in frictionless vacuums after all."

"The framers of this problem are expecting those cylinders to rotate."

Really? Maybe because the 'problem' clearly states "...allowed to roll to the bottom."

I think this is another 'done in one' challenge. Tornado provided the obvious correct answer. Perhaps if the challenge also included a powerful magnetic field, or other complicating factor, the answer would have required more thought.

Yes, they certainly are allowed to roll and so we have the implicit demand that they must, and so, in one fell swoop, Physics flies out the window.

They're also allowed to reach the bottom of the ramp at different times, and therefore they must. The only difference is that the framers stopped just short of saying so, and so the real Challenge Question is "Forget what we wrote. Who can guess our hidden assumptions most accurately?" Physics has little to do with it. Scientific method even less so.

If this had been an early 20th-century CR4 Challenge Question about the difference in the speed of light as measured by two arms of a cross-beam interferometer, the obvious 'correct' answer (cough) would be the speed of the lumeniferous aether.

Fortunately for Science, Michaelson & Morely got it 'wrong.'

Michelson, Morley, luminiferous.

These challenge questions are often framed incompletely or ambiguously.

Michelson, Morley, luminiferous.

Thank you. Yes, I noticed also - 16 minutes after posting.

"Challenge Questions ... incomplete... framed ambiguously."

It is surprisingly difficult to frame such questions to exclude hidden assumptions. That fact that the assumptions are *hidden* makes it so. The problem is further compounded by the fact that natural language is, by its very nature, ambiguous.

What I think would have made a better Challenge Question here would be to pose the question as framed along with the 'correct answer', and ask

1) "What hidden assumptions are we making that lead to this particular solution?"

2) "If one were to take our problem definition at face value, how would the outcome be different, or would it?"

3) "If different, why different? If not, why not?"

For my part, I think that would stimulate some very interesting dialogue!

I completely agree. I offered my scenario of the flattened lead pipe sliding on the ramp while the aluminum rod rolls just to point out that a valid but unlikely scenario meets the stated conditions and produces the antithesis of the expected result. This is another poorly phrased problem in giving us a unique answer.

However, the very same ambiguity of the given conditions (negligible friction effects) has sparked a great discussion here at CR4 that has been missing lately. IMHO the parsing of this ambiguity and the subsequent variety of valid results from how one interprets this ambiguity has been far more informative than just getting a simple dry result answer.

This is exactly why I presented an alternative solution: to highlight the hidden assumptions implicit in the problem definition.

It is obvious from the wording (and the title) what the *expected* 'correct' answer should be, but the answer, were we to take the problem definition *strictly at face value,* that is, as being complete unto itself and containing all the information necessary to arrive at a correct solution, is something altogether different.

The problem definition, as worded, is in fact unambiguous. What *is* ambiguous are the hidden assumptions the author made whilst formulating the definition. What these assumptions are will be made clear when the author reveals what he or she considers to be the 'correct' answer.

I might also point out that hidden assumptions, biases and foregone conclusions are the arch-nemesis of Good Science.

It's very much like Galileo's experiment, except that in an experiment like this, where they're rolling down an incline (barring slippage and other complications) a hollow cylinder will always flag behind a solid one regardless of any other difference between them. They could differ in radius, mass, length, density and/or composition. It doesn't make a bit of difference; as long as they both have uniform radial density, the solid cylinder will always win.

rofl - I am still going with the pic you posted of the cylinders in an `x` formation.Some variations come to mind, but that is a brilliant way of busting how a question is phrased.

I said nothing about how Gailileo dropped his balls, so to speak. The thing was leaning over before even completed, and does any of the sources from history specify, here to mess with argumants, surely MR G would be near the top of the list.

Might the low strength of the lead make a deformation in the lead tube. Thus more friction will be needed for the lead tube to start rolling. So if the aluminum cylinder does roll with only negligible friction (an oxymoron scenario but given) and the lead tube slides due to a deformation and insufficient friction for rotation then the lead tube should reach the bottom of the ramp first since no energy or momentum is going into rotation.

The frictional effects are negligible, but not zero...I might add that the lead cylinder is not likely to be 100% balanced, and will have a wobble.....so the aluminum cylinder will arrive at the bottom first.....

"The frictional effects are negligible, but not zero"

More accurately, "The frictional effects are negligible - except where they're not"

"I might add ..."

We must be very careful not to add to (nor subtract from) the problem definition. When we do so, we are no longer answering the same question but a different question altogether; one that wasn't asked.

Just my two cents' worth.

As ever, it`s all very much in the wording. This one will be a hoot when the `official` answer is given.

"Forget what we wrote. Who can guess our hidden assumptions most accurately?"

This accurately describes every test I've ever taken.....

What about the effects of eddy currents resulting from motion within the Earth's magnetic field?

the aluminum will be the first down...if you think about two bags one filled with stone and the other with sand,,,which will be heavier?...the bag of sand of cause because less air is allowed in...hence the bag of stone will be lighter.. :)

If the shape of the sand grains and the stone/s are the same (e.g. spheres) then the amount of air in each bag will be the same - for a big enough bag. Bag of sand would have more rolling hysteresis so stones reach bottom first - Pom kit wha.

Unless you just bag up one big rock

Who said the ramp had a bottom?
Or that the cylinders started at the top?
What if their ends are touching, they will stay together.. ah, but there's no friction. It's what we call a silly question.

Del

Oh I said "bottom"

They are not saying there is negligible friction, rather that its not different between the two cylinders. and imagine that it all takes place in a vaccuum so air resistance is not an issue to worry about. Now lets get out our pencils and paper and think about how we are going to solve this problem.

What you wrote: "They are not saying there is negligible friction ..."

vs

What they wrote: "Assuming the frictional effects are negligible ..."

You can redefine the problem statement into a new problem statement if you like, but then your solution is no longer a solution to the original problem, but to yours.

The equations were already posted in Post #10.

You will need to rearrange the equations to solve for t, but I will leave those gymnastics for others.

Yes, but the torque on both cylinders is negligible, this being the result of negligible friction and resulting in negligible rotation. The cylinders will arrive at the bottom of the ramp, one having a negligible speed advantage over the other and so, any solution involving rotation will have a negligible advantage over one that does not. As Mr. Spock so aptly said on one STOS episode, "Any difference that makes no difference is no difference." (yes, I'm being pedantic)

Ah, not so. I suggest an experiment.

Lie down on the top of a hill and report back when you roll to the bottom. :-)

"... imagine that it all takes place in a vaccuum"

Oh, but in a vacuum things tend to stick together (good bluddy luck getting all that Moon dust off your pants before entering the LEM). Van der Waals forces and all that, no air acting as buffer between objects to reduce the effects of electric fields at the atomic level. In a vacuum, rolling friction increases dramatically and so does stiction - a portmanteau of 'static' and 'friction.' It is why the flap actuators on high-performance aircraft intentionally vibrate at a (relatively) high frequency, even when stationary, so that the flaps can respond quickly to commands without having to be 'unstuck' first.

You don't want to put this in a vacuum. No sir! A low-pressure (~1 Torr) helium atmosphere would be far more preferable for this sort of experiment.

Nothing was said about the environment the test was conducted in. If this was done in zero gravity, none of your assumptions would be right as both cylinders would have just stayed put! LOL

"... none of your assumptions would be right as both cylinders would have just stayed put! LOL"

'Stayed put' with respect to what? The galactic center? Pluto? Alpha Centauri? Yuba City, California?

LOL

Well, California is always on the move anyway.

Since friction and windage are neglected the acceleration component is proportional to gravity and ramp angle. Therefore whatever mas will reach the same speed after a time t (v=1/2a*t^2).

Both reach the bottom at the same time.

.....but wouldn't the solid gain a head start and remain just a little in front?

What was/is the angle of the ramp? At 180 degrees, the ramp itself would do little to affect the outcome and they would hit bottom simultaneously. :-)

DUH! The first one gets there first!

Let me throw another wrinkle into the ring ( I like mixing metaphors and stuff). The diameters are not given. The ramp ends when it intersects with the level ground. A large diameter cylinder does not have to travel as far down the ramp as a smaller one before it touches the ground, when the slope and the ground are simultaneously tangential to the cylinder.

"Two cylinders of equal length and radius"

They are both the same size, so it doesn't impact the results.

An elliptical cylinder wouldn't roll very easily. Actually, I believe any closed curve bound by two parallel planes counts as a cylinder. The word "radius" implies that the Jordan curve at the ends of the cylinder will be a unit circle. What if these are instead oblique cylinders and not right cylinders. The ends intersecting the two parallel planes could still be a circular but not parallel to the planes defined by the rolling axes. The rolling plane will intersect the cylinder in an ellipse thus making this an elliptical cylinder. The unspecified slope of the ramp may not be able to provide enough force at the assumed one g acceleration field to lift the center of mass the difference of the two axes for either cylinder to roll with the aid of any friction. Neither cylinder rolls if friction exists, if it doesn't then they slide down the friction less ramp in unison.

What if there was peanut butter uniformly smeared over the outer surface of each cylinder?

Bet you didn't think of that?

The ramp could be composed of whipped cream...in which case the lead cylinder, being the heaviest, would sink to the bottom first....

Genius!

so higher the moment of iniertia so lower ist the acceleration of the cylinder.

the 3/4-hollow cylinder has an inertia moment of nearly 3/4 (1-81/256) of an massiv cylinder; the density of lead is nearly 4 times the density of aluminium - the massiv aluminium cylinder will reach the finish line as first.

What if the ramp has an angle of 89°?

.....and maybe the ramp is 24,000 miles long!

That reminds me of some astronaut vandalism.

Two XKCD links in one thread.

That's one hell of a ramp.

And what if it has only one side? And one edge? And is made of bacon? (nom nom)

No one has discussed warped space.

For that you need a warp drive...

... and beer.

Beer in a Klein bottle!

Now that's my kind of party!

Oh god. Not another DWDT (Driving Whilst Doing Topology)

It should have been, "Step out of the jar, sir."

I still love these speakers. Even the follow up.

Yes, they are over the top.

what if the ramp is a rotating cylinder and ketches the falling cylinder allways on the some angle of his surface?

Are the cylinders side by side and both start together?.....or one behind the other?....

Yes.

'...frictional effects are negligible...'

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At least they avoided calling it 'frictionless'.

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There are a number of other conditions that were not stated that would definitely affect the results. I'm not sure it is always possible to specify all the conditions that might possibly effect the outcome.

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What about temperature? The original question states the aluminum is solid but does not comment on the stated of the lead. The temperature might be 1000° F, giving an advantage in many situations to the lead sample.

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What about the ramp? A ramp is an inclined plane, right? So a flat plane that intersects another essentially flat ground plane, at 'the bottom of the ramp'. If that is the case, and assuming the temperature is below the melting point of lead, neither cylinder ever 'reached the bottom of the ramp', since the bottom would be a line at the sharp intersection of two planes that could not be touched by the circumference of either cylinder.

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What about the original alignment of the cylinders placed on the ramp? This was not specified. There are numerous ways a cylinder can be placed on an inclined plane, some that allow rolling, that sill can have significant effects of how quickly the mass proceeds downhill.

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What about the marmosets in the room? This was not specified. There are numerous ways that .....

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It is good to be aware of what has not been specified. It is also good to be able to work with what was intended, even while making suggestions about improving the way in which something was asked.

Should you not have had:

I of cylinder = 1/2 x (M) x ((R outside) **2) - (R inside **2)).

Then substitute R inside for 3/4R outside

and solve.....

Are you sure?

Let

I = moment of inertia,

a = inner radius,

b = outer radius,

L = length,

M = mass

and

ρ = mass density:

Let's derive this starting with the moment of inertia for a hoop with radius R:

I_hoop = m · R².

Divide the cylinder into thin concentric hoops of thickness dR.

Density = Mass per unit volume

Density = dm / dV

where:

ρ = Density

dm = Mass of a hoop of radius R and thickness dR

dV = Volume of a hoop of radius R and thickness dR

Lets assume the height of the cylinder is h.

we have

We can obtain moment of inertia by integrating over all these hoops; in general for the first case and, in the second case, between radii a and b:

Assume the cylinder has uniform density, ie, ρ = constant

The volume of this cylinder is

and its mass M is

and since

,

the moment of inertia for hollow cylinder is therefore

Ahhh..... you are right.

I am looking up how to do the fonts for maths so I can show another way to do this.

can anyone tell me. (Like a**2)

Do NOT get LaTeX using the TeX Live link to the Windows installer. It contains a trojan. I've contacted the site admins.