I was trying to solve the following problem:
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.
Best Answer
Atwood machine with cylinder
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 $
results:
$$\ddot x_1=-{\frac {g \left( m_{{R}}-m_{{C}} \right) }{m_{{R}}+m_{{C}}+M}}\\ \ddot x_2={\frac {g \left( m_{{R}}-m_{{C}} \right) }{m_{{R}}+m_{{C}}+M}}\\ \ddot x_3=-{\frac {g \left( m_{{R}}-m_{{C}} \right) }{r \left( m_{{R}}+m_{{C}}+M \right) }} $$