Beads on a thread - What stops the acceleration?

[Alettix]

That's the right step. Can you see how to rewrite that as m_N= sum of a series? You can easily sum that series and find the exact algebraic form of v_N.

Should it be rewritten: ## m_N = m_1 + m_2 + m_3 ## ?
No, that can't be right, can it? I am sorry, I can't see which sum ## m_N ## equlas.
Should I go back to some of the "basic" equations and use v instead of w to find a sum?

 

[haruspex]

Should it be rewritten: ## m_N = m_1 + m_2 + m_3 ## ?
No, that can't be right, can it? I am sorry, I can't see which sum ## m_N ## equlas.
Should I go back to some of the "basic" equations and use v instead of w to find a sum?

No, persevere with ##m_{N+1}=m_N+2kN##.
This tells you what to add to go from m_N to m_N+1. So what do you need to add to go from m₁ to m_N? If it's still not clear, find m₂ in terms of m₁, then m₃, etc.

 

[Alettix]

##m_N = m_1 + (N-1)N *k ##
is what I find, is that right?
Now, I tried finding the difference between ##m_{N+1}## and ##m_N##
--> ##m_{N+1} - m_N = (N+1)N - (N-1)N = 2kN ##
Now, using ## m_N = w_N * N^2 ## , we can obtain:
##w_{N+1} = \frac{2kN + w_N * N^2}{(N+1)^2} ##
if now N → ∞ , we can apporoximate: ##w_{N+1} → \frac{2kN+w_N * N^2}{N^2} = 2k/N + w_N → 0 + w_N = w_N ##
because ## w_N = v_N^2 ##, this proves that ## v_{N+1} → v_N ## when N → ∞. Or am I wrong?

 

[haruspex]

##m_N = m_1 + (N-1)N *k ##
is what I find, is that right?

Good, but the next step is to clean up that m₁. The whole system starts from rest, so m₁=0.
m_N=N(N-1)k.
Use that to find v_N² as a function of N and k.

 

[Alettix]

Well in that case, I get:
## v_N = (k - k/N)^{1/2} ##
If then N → ∞ ## v_N = k^{1/2} = (\frac{Fd}{m})^{1/2} ## which is the correct answer.

Thank you very much for your kind help and patience Sir!

 

[haruspex]

Well in that case, I get:
## v_N = (k - k/N)^{1/2} ##
If then N → ∞ ## v_N = k^{1/2} = (\frac{Fd}{m})^{1/2} ## which is the correct answer.

Thank you very much for your kind help and patience Sir!

You got it.