Infinite pulleys problem, why doesn’t this technique work?

[Hiero]

The problem statement can be seen here http://www.feynmanlectures.caltech.edu/info/exercises/infinite_pulleys.html

Since each pulley is presumably massless, it must have no net force on it and so the tension of each rope is half of the one above it. If we let T be the tension acting on m₀, then the tension acting on m_i will be T/2ⁱ

Thus the acceleration of m_i (downward positive) is
a_i = g - T/(2ⁱm_i)

Now to solve the problem, I thought I had a nice idea, but it leads to a wrong answer. The idea is to let a short time dt go by, then equate the kinetic energy with the work done by gravity.

Thus we get the equation:
Σm_i(a_idt)²/2 = Σm_iga_i(dt)²/2
Where the sum is over all i. From this we can solve for T.

I double checked my steps and cannot find mistakes but the answer comes out wrong, so I am asking is there any flaw in the reasoning of this approach?

 

[BvU]

Hello hero,

You overlook a trivial solution for your last equation

What is the total mass hanging from the left side of the top pulley ?

[edit] wow, apparently, so am I ! there's several pages of solution pdf.

I think your problem is that gravity does more work on m_i than you assume. There should be a double sum on the righthand side. And I think a sign error is lurking too.

 

[Hiero]

Hi BvU, thanks for the reply

You overlook a trivial solution for your last equation

There is a trivial static solution (all a_i = 0) if we take that equation alone (not sure if that’s what you meant) but the accelerations should obey the first equation a_i = g - T/(2ⁱm_i)

What is the total mass hanging from the left side of the top pulley ?

The total mass on the left (the sum from i = 1 to infinity of m_i) equals m₀, but we can’t just replace that subsystem with a mass of m₀. It’s not the same since the masses are able to accelerate at different rates.

I think your problem is that gravity does more work on m_i than you assume. There should be a double sum on the righthand side. And I think a sign error is lurking too.

As for the signs, I chose a_i to be positive if downwards, so positive a_i corresponds to positive gravitational work.

As for the double sum, I am not seeing it. The only work done by gravity that I can see is m_ig(dy) = m_ig(a_i(dt)²/2)

Can you elaborate why you think gravity does more work than this?

 

[BvU]

Oh boy, I'm making a mess of this aren't I ?

However, the trivial solution

accelerations should obey the first equation a_i = g - T/(2ⁱm_i)

is holding up: for a=0

T₀ = m₀g = g/(1-t) and
T₁ = T₀/2 has to be m₁g = g
from which t = 1/2 ... phew...​

And for the rest I take it you've seen Chandra's solution. I find it ugly, because I expected something smart and elegant from Feynman. But perhaps he wanted to be ugly for once.

but we can’t just replace that subsystem with a mass of m₀. It’s not the same since the masses are able to accelerate at different rates.

I agree. I hope I didn't suggest that but I definitely did not exclude it

Your signs are clearly established; my bad (2).

Double sums remark probably nonsense too since that's already in a_i (3)

Brings us to your approach and why it may be or not be correct

Σm_i(a_idt)2/2 = Σm_iga_i(dt)2/2

I recognize ##{1\over 2}mv^2 = mgh## but I don't see where it goes awry.

I double checked my steps and cannot find mistakes

I believe you, but all I can think of now is to ask that you show the steps -- by now I kind of feel obliged to check.

The alternative is that we call for expert help ! @haruspex, do you like this kind of exercise ?

 

[Hiero]

from which t = 1/2 ... phew...

I didn’t think about that until now, but yes I agree t = 1/2 should give the static solution.

I believe you, but all I can think of now is to ask that you show the steps

Ok.
(A reminder, T₀ ≡ T)
Starting from gΣm_ia_i = Σm_ia_i² if we put in the a_i equation then it becomes... (all sums from i = 0 to infinity)
gΣm_i(g - T/(2ⁱm_i)) = Σm_i(g - T/(2ⁱm_i))²
gΣ(m_ig - T/2ⁱ) = Σm_i(g² - 2gT/(2ⁱm_i) + (T/(2ⁱm_i))²)
The term Σm_ig² cancels from both sides leaving the equation
-gTΣ1/2ⁱ = -2gTΣ1/2ⁱ + T²Σ1/(4ⁱm_i)
Adding 2gTΣ1/2ⁱ to both sides (and then dividing through by T²) and noting that Σ1/2ⁱ = 2 we get the equation
2g/T = Σ1/(4ⁱm_i)
Now since m₀ doesn’t fit the pattern, I will separate it from the sum. So now the summation will run from i = 1 to infinity and the equation becomes
2g/T = 1/m₀ + Σ1/(4ⁱm_i)
2g/T = (1 - t) + Σt/(4t)ⁱ
2g/T = 1 - t + 4t²/(1 - 4t)
2g/T = ((1-t)(1-4t)+4t²) / (1 - 4t)
2g/T = (1 - 5t +8t²) / (1 - 4t)
Thus we find the tension acting on m₀ to be
T = 2g(1 - 4t) / (1 - 5t +8t²)
Then we can find a₀ from
a₀ = g - T/m₀
a₀ = g - 2g(1 - t)(1 - 4t) / (1 - 5t +8t²)
a₀ = g(5t - 1)/(1 - 5t +8t²)
Which is my final answer and incorrect. This is at least the third time I’ve worked through it, always getting this same answer.
(This does not give zero when t = 1/2)If you do take the time to check then thank you very much! It really bothers me that this method isn’t working

 

[Hiero]

One thing I did not mention though is that when t ≤ 1/4 then the sum ∑1/(4m_i)ⁱ diverges and so the formula I used no longer applies. Hence our final formula should only apply to t > 1/4

 

[andrewkirk]

So now the summation will run from i = 1 to infinity and the equation becomes
2g/T = 1/m₀ + Σ1/(4ⁱm_i)
2g/T = (1 - t) + Σt/(4t)ⁱ
2g/T = 1 - t + 4t²/(1 - 4t)

I think there's a sign error in the last line of this excerpt. It should be
2g/T = 1 - t - 4t²/(1 - 4t)

Note that the sum in the last term of the previous line is of all positive terms, so the sum must be positive. But the denominator (1 - 4t) in the last line is negative because t > 1/4.

 

[Hiero]

Oh wow I finally see the error I was mistakenly thinking that Σ1/(4t)ⁱ = 4t/(1-4t) (that would be Σ(4t)ⁱ) but it’s actually (1/(4t)) / (1-1/(4t)) = 1/(4t-1) isn’t it?
(Sums from one to infinity)

 

[Hiero]

I think there's a sign error in the last line of this excerpt. It should be
2g/T = 1 - t - 4t²/(1 - 4t)

Don’t you think it should instead be,
2g/T = 1 - t + t/(4t - 1)
?

 

[Hiero]

Fixing that though, I end up with
a₀ = g(2t-1)²/(6t-4t²-1)
which at least is zero when t = 1/2 but it’s still not the given answer.

I must be making another mistake, but I cannot see it cause apparently I like to repeat the same mistakes over and over

 

[andrewkirk]

Don’t you think it should instead be,
2g/T = 1 - t + t/(4t - 1)
?

Yes it should. I noticed the sign problem and so just re-wrote the line with the sign changed. I did not check the other details of the summation.

 

[haruspex]

@haruspex, do you like this kind of exercise ?

Yes, it's fascinating, but I'm a bit busy right now and it looks like good progress is being made.