I was trying to solve the following problem: enter image description here

Basically we have a normal Atwood machine, the pulley is negligible and we have two masses, a square that falls normally and a cylinder that also spins. The rope goes round the cylinder so the rope doesn't do weird stuff. They both have mass m, and it has to be found the accelerations of both objects.

My approach was to consider the square and we see that $mg-T=ma_s$. Then considering the cylinder we can see it spins, so $I\alpha=I\frac{a_c}{R}=TR$ and thus $\frac{M}{2}a=T$. Substituting we get $m\frac{a_r}{2}=mg-ma_s$. The best shot I can think of is assuming $a_r=2a_s$ cause the rope acceleration has to split between both right and left. Thus $\frac{a_r}{2}=g-2a_s \Rightarrow a=\frac{2}{5}g$. But the solution given is that they are both $\frac{g}{2}$. I can see it has to be something about rototraslations but I can't get how to solve it, any helping hands? If possible I'd prefer dynamic approaches rather than energetic ones.

  • $\begingroup$ If the masses are equal the accelerations are zero ? $\endgroup$
    – Eli
    Dec 30, 2021 at 16:37

1 Answer 1


Atwood machine with cylinder

enter image description here

constraint equations:

$$x_1+x_2+\frac{\pi}{2}\,r=L_R\\ x_2+r\,x_3=0$$

where $~L_R~$ is the rope length

from here you can apply the Euler-Lagrange with the above holonomic constraint equations . you obtain the accelerations $~\ddot x_1~,\ddot x_1~,\ddot x_3~$ and the generalized constraint forces $~\mathbf\lambda $


$$\ddot x_1=-{\frac {g \left( m_{{R}}-m_{{C}} \right) }{m_{{R}}+m_{{C}}+M}}\\ \ddot x_2={\frac {g \left( m_{{R}}-m_{{C}} \right) }{m_{{R}}+m_{{C}}+M}}\\ \ddot x_3=-{\frac {g \left( m_{{R}}-m_{{C}} \right) }{r \left( m_{{R}}+m_{{C}}+M \right) }} $$

  • $\begingroup$ Hi thank you for anwering but the solution you provided doesn't solve the problem that was asked, I can somehow see you didn't made the cylinder spin as was hinted in the text and said in my solution, plus the solution isn't g/2 as was said in the question. Btw I asked people and found a solution, just for future reference I was very much close: $\alpha = 2a/r$ thus $ma=T$ from momentum considerations. Then we substitute in $mg-T=ma$ for the cylinder and obtain $mg=ma+ma \Rightarrow a=g/2$. $\endgroup$
    – sonod4
    Dec 30, 2021 at 19:05
  • $\begingroup$ Note: With equal vertical forces, both masses are accelerating down at g/2 as the cord unwinds from the cylinder. $\endgroup$
    – R.W. Bird
    Jan 1, 2022 at 14:28

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