I'm trying to understand how a movable pulley of mass $m$ and radius $r$, that rotates and doesn't slip actually works.
Consider the picture, if a force is applied at the free end of the rope, are the following equations valid?
$\begin{cases} F+T_1+T_2=m \frac{a}{2} \\ (T_1-T_2) r=I \alpha=I \frac{a}{r} \end{cases}$
Where I is the moment of inertia of the disk.
If I'm asked to determine both the tensions of the rope and the acceleration, how would I do? Do I have enough informations to do it?
Let $T$ be the tension in the part of the rope that’s tied off to the roof or ceiling. Then the tension in the applied force side of the rope is $F$.
There are 2 accelerations: $a_t$ the tangential acceleration of the rope as it makes and breaks contact with the pulley and $a_y$ the vertical acceleration of the pulley. So there are 3 unknowns: the tension $T$ in the rope, $a_t$ and $a_y$. They can be obtained in terms of the pulley mass $m$ and the applied force $F$.
We need 3 equations: They are:
$F + T –m*g = m*a_y$ (1)
$(F-T)*r = I*\alpha = \frac {m*r^2*a_t} {2*r}$ (2)
$a_y = \frac {a_t} {2}$ (3)
Solving for $T, a_t, a_y$ gives:
$T = \frac {m*g} {2}, a_t = \frac {(2*F – m*g)} {m}, a_y = \frac {(2*F – m*g)} {2*m}$
If the applied force $F$ is equal to $\frac {m*g}{2}$, there is no acceleration of the pulley.
This is the only solution that makes any sense to me.
