Vertical acceleration physics help

"In other parts of this problem expressions have been found for the vertical acceleration of the cylinder a_y and the angular acceleration alpha of the cylinder in the z_unit direction; both expressions include an unknown variable, namely, the tension T in the vertical section of string The string constrains the rotational and vertical motions, providing a third equation relating a_y and alpha. Solve these three equations to find the vertical acceleration, a_y, of the center of mass of the cylinder.

Express a_y in terms of g, m, r, and I; a positive answer indicates upward acceleration."

I have already found that: Omega = v/r, ma_y = T-mg, and I(alpha) = -Tr.

I am so stuck right now that I don't know what to do, I am running out of attempts to get the right answer and I need someone to explain how to find the vertical acceleration a_y.

You almost have the answer you need with

ma_y = T - mg

except that they don't want T to appear in your answer. Therefore we need to find an expression for T in terms of g, m, r, and I. We can get it from

I*alpha = -Tr
Tr = -I*alpha
T = -I*alpha/r

But they also don't want alpha to appear in the answer. From

Omega = v/r

you can derive

alpha = a_y/r

T = -I*alpha/r
T = -I*(a_y/r)/r
= -I*a_y/r^2

ma_y = T - mg
ma_y = -I*a_y/r^2 - mg
ma_y + I*a_y/r^2 = -mg
Distributive law
a_y(m + I/r^2) = -mg
a_y = -mg / (m + I/r^2)

w= 0mega
T.r = I*a1 (a1 = alpha)
a = a1*r

T.r = I*a/r
T.r^2 = I*a ......(1)
T- mg = ma
T = m(g+a) put in (1)

mr^2(g+a) = Ia
mar^2 - mgr^2 = I*a
(mgr^2 = Ia - mar^2
mgr^2 = a(I - mr^2)
a= mgr^2(I - mr^2)

So , what's the answer ...
or give figure