Infinite plus one cylinders

Let's look at k-th cylindar and write these equations for him:

$ma_k=mg+T_{k+1}-T_k$

$\alpha_k R=v_k-v_{k-1}$

$\frac{1}{2} m v_k^2 + \frac{1}{2} I \omega_k ^2=mg S_k+T_{k+1}S_k-T_kS_{k-1}$

Combinateing these equations one can find relationship:

$-a_{k-1}+3a_k-a_{k+1}=g$ (1)

And here comes the best part.

We can suppose that $a_1$ has form like this $a_1=\lambda g$ .

And now if we go in the first cylindars frame of reference the effective gravity will be $g'=g-a_1$ and the second cylindar became first one and now his acceleration $a_1 '=a_2-a_1=\lambda g'$ putting this in equation (1) you will get equation:

$\lambda^2+\lambda-1=0$

And the physical solution is :

$\lambda=\frac{\sqrt{5}-1}{2}$

I like this problem and hope you will too :)!

My solution is not the most beautiful, (it uses a lot of algebra) but it only uses basic Newtonian equations of motion (no accelerated frames of reference and no energy equations).

The first idea is this: There will be some mass, $M_{eff}$ which, when put in the place of the second pulley, will cause the first pulley to accelerate with the same acceleration as the infinite chain of pulleys caused.

The second idea is this: if you take the entire system and hang it from another pulley, the top pulley will behave in the same way as before (because the new system is identical).

These two ideas are illustrated by this drawing:

We can then solve the problem by describing the motion of both systems in my drawing, and then equating the accelerations of the top cylinders in both systems.

I have decided not to write the entire solution in detail, because I had to write 6 equations with 6 unknowns. Each equation was very simple (4 net force equations and 2 net torque equations) but it is a lot of algebra to eliminate all the variables.