Pulley connects 2 masses, problem with inertia to accelleration calculation

[satisf]

Homework Statement

Given is the setup in the attachment, I think the values speak for itself. Additionally, there is a friction coefficient of 0.36 that works in on the 2 blocks.

So in short: 2 masses (1 one a slope), connected through a pulley, that does have friction and to complete it, the ground has friction to.

Asked:
a) Accelleration of blocks (0.309m/s²)
b) Tension in both cords (7.67N and 9.22N)

I actually got quite far, I even know what I am missing in my own derivation, a factor of 2 (will explain below). But I can't manage to see why. It's in the derivation for a, a to b Is no problem.

Homework Equations

F=ma

\tau=RFsin(\theta)

\sum \tau = I\alpha

a_t=r\alpha

The Attempt at a Solution

Using free-body calculations I found expressions for T₁ (left block) and T₂ (right block), as follows:

T_1=m_1a+0.36m_1g
T_2=-m_2a+m_2g\mathrm{cos}(\frac{\pi}{3})-0.36m_2g\mathrm{sin}(\frac{\pi}{3})

I'm quite sure that these are correct, since substituting the solution for a in these gives me the correct tensions.

Now to bring in the pulley I started with this:
\sum \tau = I\alpha=I\frac{a}{R}=MR^2\frac{A}{R}=MRa

Then, knowing that \tau_1=RT_1\mathrm{sin}(\theta_1) and \tau_2=RT_2\mathrm{sin}(\theta_2) and both sinuses will equal one I get to:

RF_1+RF_2=MRa
or the more familiar
F_1+F_2=Ma

But this is where things go wrong, as far as I know it should be \frac{M}{2} on the right hand side. And yes, when I do put in a factor of 2 I get the correct solution.

So now my question is: where the hell in my derivation did I forget the factor?

P.S. while previewing this post the latex got messed up, If it's not solved in final submit, can some moderator adjust it?

 

[satisf]

Allright, the error was of course that for a disk I=\frac{MR^2}{2}. I relied to blindly on the formulaium i got. I=mr² is for point masses, so for a disk you get following equation for I:

[ŧex]I=int_0^\pi int_0^R \frac{M}{\pi r^2} r^2 r \mathrm{d}r\mathrm{d}\theta = \ldots = \frac{MR^2}{2}[/tex]

Note the extra r is of course because the equation is in polar coordinates and the division by \pi r^2 is of course to get the uniform mass distribution.

If you use this formulah in my original post, everything is correct.