Consider this arrangement of pulley:
The problem is to find the acceleration of $m_1$.
This is my solution:
The problem explicitly wants $\ddot y_1$.
So starting by applying Newton's second law we get:
$$T_1-m_1g=m_1\ddot y_1$$
$$T_2-m_2g=m_2\ddot y_2$$
Where $T$'s are tension forces acting on the masses with the same subscript.
We need two more equations. First is obtained by geometry of the system:
$$l_2=y_3+(y_3-y_2)+\pi R=2y_3-y_2+\pi R\implies 2\ddot y_3=\ddot y_2$$
$$l_1=h-y_1+h-y_3+\pi R=2h-y_1-y_3+\pi R\implies \ddot y_1=-\ddot y_3$$
$$\ddot y_2=-2\ddot y_3$$
But I still need another constraint to relate $T_1$ to $T_2$.
How to write that?
Best Answer
Since the pulley is weightless, according to Newton's second law the net force acting on it should be zero. so: $$T_1=2T_2$$