Angular acceleration of a cylinder

After some long calculations, I found that the linear acceleration of a cylinder on an inclined plane (attached to a block over a pulley on another inclined plane) was 8.76238 m/s^2. How do I find the angular acceleration of this rolling cylinder?

Thanks in advance!

a=alfaxRadius
alfa=a/R
you should know the radius of the solid cylinder

Bonus equation:


Sigma Force (F)=ma
W-T=ma
T=W-ma
T=mg-ma...(1)
(*)m=actual mass (the hanging stuff)
T=Tension(string)
a=linear acc
W=Weight
g=9,8m/s^2

Sigma Torque = Ixalfa(angular acceleration)
T.R= Ixalfa(angular acceleration)
T.R={(1/2)xMR^2}x(a/R)
T={(1/2)xMR}x(a/R)
T={(1/2)xM}x(a)
(*)T=Tension(string)
a=linear acc=8.76238 m/s^2
m=actual mass (the hanging stuff)
M=Cylinder mass
R=Radius
alfa=angular acceleration=(a/R)=
I=Moment Inertia of solid cylinder
g=9,8m/s^2

equation (1) and (2)
T=T
{(1/2)xM}x(a)=mg-ma
mg={(1/2)xM}x(a)+ma
mg=[{(1/2)xM}+m]xa
a=(mg)/[{(1/2)xM}+m]
a=[m/{(1/2xM)}+m]x g

The relation ship is angular acc equals linear acceleration divided by the radius. U do the math :)