Sciencemadness Discussion Board - Mechanics brain teaser

Mechanics brain teaser

blogfast25 - 22-8-2016 at 06:56

Mass m₂ is connected to mass m₁ via a massless string (S) and perfect Hookean spring (with spring constant k) running over frictionless pulley P.

The static coefficient of friction between m₂ and the surface is μ.

At t=0 the spring is not extended (no tension) and everything is stationary. At that point m₁ is released.

Question:

What is the minimum mass m₁ for m₂ to start moving?

Answers via U2U, please.

[Edited on 22-8-2016 by blogfast25]

blogfast25 - 22-8-2016 at 10:30

One incorrect answer so far.

blogfast25 - 22-8-2016 at 12:05

Three more incorrect answers.

Didn't think you'd find it that hard!

blogfast25 - 22-8-2016 at 14:06

And three correct ones too:

- woelen
- metacelsus
- wg48

Well done.

Solution tomorrow.

[Edited on 22-8-2016 by blogfast25]

blogfast25 - 23-8-2016 at 05:12

One more incorrect answer. So a total of 3 correct answers out of 7 given.

Solution: Here's the easiest way of looking at it.

As m₁ is released, gravitational potential energy is converted to spring potential energy:

$$m_1gy=\frac12 ky^2$$
$$\implies y=\frac{2m_1g}{k}$$
As it falls, tension in the spring and string builds up:

$$T=ky=2m_1g$$

To make m₂ move:

$$T>\mu m_2g$$
$$\implies 2m_1g>\mu m_2g$$
$$\implies m_1>\frac12 \mu m_2$$

If this condition isn't met, then m₁ will enter into a simple harmonic oscillation and m₂ won't move.

Thanks for playing!

[Edited on 23-8-2016 by blogfast25]

wg48 - 23-8-2016 at 07:42

Quote: Originally posted by blogfast25

One more incorrect answer. So a total of 3 correct answers out of 7 given.

Solution: Here's the easiest way of looking at it.

As m₁ is released, gravitational potential energy is converted to spring potential energy:

$$m_1gy=\frac12 ky^2$$
$$\implies y=\frac{2m_1g}{k}$$
As it falls, tension in the spring and string builds up:

$$T=ky=2m_1g$$

To make m₂ move:

$$T>\mu m_2g$$
$$\implies 2m_1g>\mu m_2g$$
$$\implies m_1>\frac12 \mu m_2$$

If this condition isn't met, then m₁ will enter into a simple harmonic oscillation and m₂ won't move.

Thanks for playing!

[Edited on 23-8-2016 by blogfast25]

I can not agree with how you got your solution its at least misleading or confusing.

Your first statement is incorrect. The gravitational potential energy is converted to potential energy of the spring and to kinetic energy of m1.

In effect you assume the kinetic energy of m1 is zero with out justification which will/should be confusing to some. It will be zero at the maximum of y. Perhaps you were being deliberately obtuse to stimulate discussion. In which case I have kicked it off for you.

blogfast25 - 23-8-2016 at 08:50

Quote: Originally posted by wg48

I can not agree with how you got your solution its [sic] at least misleading or confusing.

Your first statement is incorrect. The gravitational potential energy is converted to potential energy of the spring and to kinetic energy of m1.

In effect you assume the kinetic energy of m1 is zero with out justification which will/should be confusing to some. It will be zero at the maximum of y. Perhaps you were being deliberately obtuse to stimulate discussion. In which case I have kicked it off for you.

Yes, m₁ acquires kinetic energy but we don't need to know that here, as my simple derivation clearly shows.

I'm in no way "assuming the kinetic energy of m₁ is zero" (it is not, until spring tension reaches T_max and assuming m₂ cannot be moved) but in my derivation knowledge of the kinetic energy of m₁ is simply not needed. That this may or may not confuse the easily confused is neither here nor there.

The simplest and valid explanation should always deserve merit.

I suggest you study the following, very similar problem and the upvoted answers. I'm 'Gert' and my answer got 6 upvotes (on a PHYSICS Q&A site, no less). See also the answer by 'dmckee'.

As regards:

Quote:

Perhaps you were being deliberately obtuse to stimulate discussion. In which case I have kicked it off for you.

Let me be very clear here: I take GRAVE offence to that remark. Think of that what you will. :mad:

Do you even know what obtuse actually means?

[Edited on 23-8-2016 by blogfast25]

Praxichys - 23-8-2016 at 09:50

The spring should be irrelevant. In the relaxed state, the spring will have elongated equivalent to F=mg of m₂.

So,

1/2 kx^2 = m₁g , solving for x.

But the tension is the same on both ends of the spring. Therefore the maximum tension m₁g can provide is if the spring were not there at all.

So,

μm₂g = m₁g

or

μm₂ = m₁

EDIT:

Wait, no, I'm wrong. If m₂ has accelerated, it will decelerate against kx² and that F will be > m₁g during the deceleration.

[Edited on 23-8-2016 by Praxichys]

woelen - 23-8-2016 at 11:36

First I want to say, blogfast25, keep up the good work! Good to see this kind of problems on sciencemadness.

I do agree, at least to some extent, with wg48. There is conversion of potential energy (in the height of the mass) to both kinetic energy and potential energy in the spring. The fact that your answer is correct is because it is reached at the point of maximum deflection of the spring. At that point the mass m1 is not moving. If, however, the problem were different and the mass were moving at your desired point, then the answer would be incorrect. That is why I mentioned in my U2U to you that the system, as long as m2 is not moving, is a harmonic oscillator and that the maximum force on m2 is reached when m1 is at its lowest point and not moving (hence has zero kinetic energy). In your simple solution I would add one sentence, telling about the kinetic energy and telling that it is zero at the point in which we are interested. That makes your explanation based on conservation of energy completely valid. Now, without this added explanation, there is room for discussion.

wg48 - 23-8-2016 at 12:11

Quote: Originally posted by blogfast25

Quote: Originally posted by wg48

Quote:

Perhaps you were being deliberately obtuse to stimulate discussion. In which case I have kicked it off for you.

Let me be very clear here: I take GRAVE offence to that remark. Think of that what you will. :mad:

Do you even know what obtuse actually means?

[Edited on 23-8-2016 by blogfast25]

Out of curiosity I checked the problem in the link you gave. It confirms my view. That derivation use ymax and explains that its the point when the kinetic energy of m1 is zero. So your suggestion that your not asseming any value is ridiculous.

I thought I was being kind in giving you a way out of your error. You could have just said "yes I left a few things out sorry" or You could have given the link to full explanation.

Sorry to have gravely offended you in pointing out your sloppy maths but an error is an error and its needs to be pointed out especially when it an explanation for several people.

wg48 - 23-8-2016 at 12:40

Quote: Originally posted by woelen

First I want to say, blogfast25, keep up the good work! Good to see this kind of problems on sciencemadness.

Snip.

and I second those sentiments.

blogfast25 - 23-8-2016 at 13:16

Quote: Originally posted by wg48

Out of curiosity I checked the problem in the link you gave. It confirms my view. That derivation use ymax and explains that its the point when the kinetic energy of m1 is zero. So your suggestion that your [sic] not asseming [sic] any value is ridiculous.

I thought I was being kind in giving you a way out of your error. You could have just said "yes I left a few things out sorry" or You could have given the link to full explanation.

Sorry to have gravely offended you in pointing out your sloppy maths but an error is an error and its needs to be pointed out especially when it [sic] an explanation for several people.

I don't accept your fake apology and your pointing to an 'error': yours is almost certainly a case of sour grapes as your own first attempt was wrong and I had to point this out to you.

You still don't seem to understand the problem. If, e.g. m₁ was considerably larger than the minimum (to move m₂):

$$\frac12 \mu m_2g$$

... then E_k would still not need to be known and would not need to become zero either. What matters is tension in the spring/string, not how one arrives at deducing it. No assumptions re. E_k were made, as none were necessary. Your continued claim to the contrary is a BLATANT LIE, which you seem to want to pass off as an 'apology'. Go try fool someone else!
<hr>

I put up these threads to inject a little variety and non-chemistry into SM. To be told 'you're wrong' when I'm not by twits like you is off-putting, to say the least.

So here's a kind request: stay off these threads and mind your own stupid business. :mad:

[Edited on 23-8-2016 by blogfast25]

blogfast25 - 23-8-2016 at 13:27

Quote: Originally posted by woelen

The fact that your answer is correct is because it is reached at the point of maximum deflection of the spring. At that point the mass m1 is not moving. If, however, the problem were different and the mass were moving at your desired point, then the answer would be incorrect.

That is not true. The question asks for the minimum m₁ but if m₁ was larger than that, the tension in the spring/thread would still be calculated the same way. The value of kinetic energy would still be irrelevant (and it that case it would be non-zero). Only:

$$m_1gy=\frac12 ky^2$$

... would be needed to calculate y and T and to ascertain whether m₂ moves or not. That would of course be the 'boring case'.

There's nothing to be gained from writing:

$$m_1gy=\frac12 mv^2=\frac12 ky^2$$

... if the middle expression will not be used.

The Hamiltonian:

$$m_1gy=\frac12 mv^2$$

... can be used to derive the equation of motion of m₁, if m₂ doesn't move of course. But that's a different problem.
<hr>

Thanks for the further encouragement, though.

[Edited on 23-8-2016 by blogfast25]

wg48 - 24-8-2016 at 13:03

Quote: Originally posted by blogfast25

Quote: Originally posted by wg48

Out of curiosity I checked the problem in the link you gave. It confirms my view. That derivation use ymax and explains that its the point when the kinetic energy of m1 is zero. So your suggestion that your [sic] not asseming [sic] any value is ridiculous.

I thought I was being kind in giving you a way out of your error. You could have just said "yes I left a few things out sorry" or You could have given the link to full explanation.

Sorry to have gravely offended you in pointing out your sloppy maths but an error is an error and its needs to be pointed out especially when it [sic] an explanation for several people.

[Edited on 23-8-2016 by blogfast25]

I have offered you no apology fake or otherwise for my criticism of your derivation. Your derivation remains at best sloppy and at worst simple wrong. Your anger, accusations and attempts at justifying it just make you look more of a twit. Your failure to understand that what you wrote was wrong and the fact you got the correct answer suggests you plagiarized the solution and produced your own sloppy version of the derivation. Not that there is anything wrong with using someone else’s solution.

I will post in any thread I choose but in general I will avoid directly interacting with you your way past difficult for me. In a mathematics thread of a science forum you should expect constructive criticism regardless of whatever your contribution to the forum is and nor does not absolve you from not behaving like a twit.

blogfast25 - 24-8-2016 at 14:55

@wg48:

It's obvious that you still don't really, fully understand the problem. Kinetic energy need not be invoked here, as shown also clearly in the answers to the question at physics.stackexchange. Even a basic numerical example shows that. m₁ does of course acquire kinetic energy as it falls against the restoring force of the spring, ky, but that isn't in any dispute: it just isn't necessary to calculate or invoke it here.

As regards 'plagiarism', that's just another deeply false accusation you're now adding. This answer here, as well as the one provided and 6 times upvoted at physics.stackexchange are both mine and mine alone. You really have no shame whatsoever.

False allegations do not constitute 'constructive criticism' and cause righteous anger. As do characterisations of being 'obtuse' or accusations of plagiarism.

Any other contributions I may or may not have made to SM are neither here nor there.

Again, I ask you to simply leave this and other threads on math/physics of mine alone.

[Edited on 24-8-2016 by blogfast25]

blogfast25 - 24-8-2016 at 15:08

What is of course even funnier that is in wg48's answer provided via U2U (screenshot provided, if interest warrants it) there's no mention of kinetic energy either (or even 'energy', for that matter), only of force (quite correctly). Here it is, "warts and all", titled 'Machanics [sic] brain teaser' :

Quote:

Ooops I dropped the m1 some where
so the correct answer is:

M2 will move if m2 x u < 2 x m1

I thought about it while watching the news. The average force on the spring must g x m1. the force varies sinusoidally and starts off at zero so the peak must 2 x g x m1.

The friction holding the block is g x m2 x u.

If the friction holding m2 is less than the peak force on the spring m2 will move. The g cancels.

I must have dropped the m1 some where in my long winded mathcad work sheet version.

Who on Earth needs a 'long winded mathcad work sheet for this kind of problem', BTW?

Note that if m₁ is significantly higher than the stated minimum, m₂ will start moving without m₁ ever entering into simple oscillation. The case where m₁ descends to its lowest possible value of y, creating maximum possible tension in the spring/string is only the limit case that we're looking for.

[Edited on 25-8-2016 by blogfast25]