Speeding Box: CR4 Challenge (12/22/09)

And the other end of the spring must be rigidly mounted, and the spring must be undamped, and the box must hit the spring head-on and there must be no air resistance, and...

No quarrel with your reasoning and solution -- didn't even check the working. However, I can suggest one more necessary assumption.

Since any given spring constant can be achieved with an arbitrary combination of dimensional parameters, we need to ensure that a long and slender spring doesn't buckle while it gets compressed to its final length as demanded by the problem. Or else it must be provided with an internal or external frictionless guide, which can be a tricky practical design problem in itself!

I don't think anyone has visualised the spring to be anything other than the common helical coiled round wire metallic compression type with squared and ground ends. One could exercise the imagination and conceive of other kinds of springs having a constant 'rate', which is all that the question calls for. Some appropriate changes in the assumptions may be required.

But let me leave all that to come up in the discussion, if folk wish to take time off from their Christmas and New Year festivities.

Season's Greetings all round, =TeeSquare=

With all the usual assumptions, I make it 3.33 cm

Codey

All other conditions mentioned aside the question says the spring is in the surface.

It doesn't compress if the spring is in the same surface. The leading edge of the spring in the surface would act as a stop if embedded. If not embedded would depend on how far into the surface it is. The back edge acting as pivot would cause the spring to roll up decreasing the compression of the spring as it redirected the force applied.

Before we get into Newton's 3 laws of motion maybe the "frictionless" surface should be scrutinized eh? As far as I know there isn't a surface on earth that is "frictionless". Using the Electron microscope one can see space/air between any two objects and as we all know air creates friction. Therefore since man cannot create the "perfect" vacuum one would have to go into space for a hands on demonstration but what do I know....

For all you know the experiment is performed inside a UHV chamber, the box contains a magnet and the horizontal surface is superconducting. With a small enough box and cold enough surfaces we might even get to the point of zero friction between the two (for the duration of the experiment).

Since it's now been way over 1276 seconds since I was last pedantic (and because I want to steal a march on LynLynch), I must point out the box does not stop. It's velocity reverses and goes through zero, but at no time is there a discontinuous function in that motion.

No. "Stop" is incorrect, or perhaps deliberately misleading. Fernandotasso has the right answer of course, so I'm just piddling with this to see how else you could look at it. When you think about the box not stopping, you realize this can be modeled as a simple harmonic oscillator and that gives another solution (same answer, different solution). To see the meaning of my statement, watch the pendulum on a grandfather clock. It doesn't stop; it just reverses. Well, it could stop if the clock spring ran down.

You should perhaps try post #3. It's a much more interesting problem, and one that conforms to your interpretation without in any way conflicting with the challenge as posed.

It should really be marked "on topic".

My "interpretation" does not conflict with the challenge as posed. It seems obvious that the challenge was not clearly written (note the spring "in the same surface") but that doesn't mean it's wrong. But, in the absence of a nonconservative force(s), or some rather unexpected external forces, the mass does not stop and remain stopped. Now, you could make this a more interesting problem if you were to pin the mass when it reaches zero velocity and then ask what happens?

If you are allowing a second solution in the opposite direction don't you have to add the postulate that the box adheres to the spring?

Surely pinning the mass is a trivial addition if the spring is massless and lossless as assumed in the existing solution.

No comment on post #3, though?

Only if you want to repeat.

Pinning the mass is not trivial. Somthing very unexpected happens.

I did comment on number 3. The mass will not come to a full stop unless you come up with something extra.

I agree that the mass will never come to a stop but for a different reason. Unless the center of mass of the box is moving directly in line with the direction of action of the spring, there will be some component of the box's motion that will not be stopped. This could be a linear component and there would still be some motion when the spring was compressed its maximum amount. There could also be some rotational motion imparted to the box.

Thanks,

Jim

I'm not certain what you mean here - unless you are referring to a "single-point" pinning of the mass, in which case any equilibrium will of course be unstable (unless the pinned point of the box falls within the spring).

It refers to the problem of saying the box is "stopped" at the end of travel. It is not since only the velocity, not the derivatives of velocity, is zero. Pinning would show that dramatically.

Yes, pinning would reduce the acceleration instantaneously to zero (from its maximum value). But that is not at all unexpected.

Perhaps to you. Care to place a wager on how many responders would find that unexpected? That is the difference between stopping and reversing without stopping.

The box never stops. It's a FRCTIONLESS surface!

Think outside the box

0,138461538461538461538461538461538mm or

0,00138461538461538461538461538461538m

?

OK!

0,138461538461538461538461538461538mm or

0,000138461538461538461538461538461538m

the root of this = 0,0191715976331360946745562130177515m = 19,1715976331360946745562130177515mm

if you want to know the exact value - couldn't you calculate it?0,013867504905630728050458543336425m=13,867504905630728050458543336425mm

Your answer is different from post #1, which appears correct to me.

You suggest I calculate - but what would my calculation be based on, as you give no method or formula. (In any case, I'm not interested in more significant figures, only in having something meaningful).)

However, I note that your "corrected" answer is different from #1 by a factor of 1.2. This suggests to me that you misread the initial speed of the box. However, in the absence of supporting data there's no way we can be certain.

the spring constant is different from 650N/m to 660N/m in the answer is; maybe there's a problem with the windows calculator?

That's true - but it only results in a 0.8% error - nothing like the 20% difference proposed above.

recalculated with windows calculator: 0,01664100588675687366055025200371m = 16,64100588675687366055025200371mm

Finally the numerical answer is approximately right (more digits on an irrational number doesn't change that). But a late "correct" numeric following loads of wrong ones still looks silly - particularly when no explanation is forthcoming.

9 eleven?

In order for the box to reverse directions, at some point there is neither forward nor reverse velocity . . . i.e. there is, in point of fact both 0 forward velocity and 0 reverse velocity. Would that not be an alternate definition of the term "stopped". In which case, the problem is simply stated in a manner that requires one to understand the term "stopped" in reference to forward and reverse velocities.

And, an interesting complexity is what we do with momentum. I don't know of any way to just assume conservation of momentum out of the way.

Won't the standard simplifying assumption of an effectively infinitely-massive constraint do the business?

it's a box, not a ball

So? Anyway, if it helps, it's a spherical box.

Hello TVP45

I don't see a problem with momentum. The spring puts a force on the box for a certain time and integral of force x time = change of box momentum. The other end of the spring causes an equal and opposite change of momentum of the Earth (which doesn't change Earth's velocity much!). The same thing happened the other way round when the box was accelerated in the first place.

Cheers...........Codey

There's not a problem, but there is the consideration. In the harmonic motion solution without damping, the average momentum remains constant. I don't mean to be pedantic (except when I'm needling - and here I'm not), but it is never OK to just assume away conservation of momentum; that one is the gold standard of mechanics.

moving transversal?

Since it's on a spring, it moves transversal...reversal...transversal...reversal...tra

box or ball