# Rotating movable pulley [closed]

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.

• Thanks for the answer! But I have a similar exercise on my book and for $F=mg$ it gives in the solutions $T_1=mg$, $T_2=\frac{2}{3} mg$ and $a_t =\frac{2}{3} g$. I did not get why the tension in the side of the rope where F is applied must be equal to F or it is not considered at all Commented Mar 7, 2016 at 13:37
• @Gianolepo: You're welcome. How can the tension on the applied force F side be anything other than equal to F? How similar is that other exercise? Commented Mar 7, 2016 at 15:04