S.H.M. Tough one!!!

A uniform cylindrical pulley of mass M and radius R can freely rotate about the horizontal axis O. The free end of a thread tightly wound on the pulley carries a dead weight A. At a certain angle α it counter balances a point mass m fixed at the rim of the pulley. Find the angular frequency of small oscillations of the arrangement (refer figure).

Attempt-

Let the mass of the dead weight be m1
Considering rotational equilibrium about O, we have
m1gR = mgRsinα
i.e. m1 = msinα

Let the pulley be displaced through a small angle ∂ in the clockwise direction.

The Total Energy of the system after rotation through the small angle is-
E = mgRcos(α + ∂) - m1gR(α + ∂) + Iω²/2 + m1ω²R²/2

As the system is conservative, the time derivative of energy is 0
dE/dt = 0

on solving the above equation,
I got dω/dt = mgRsinα + m1gR/(I + m1R²)

Now m1 = msinα and I = MR²/2 + mR² ( I did not include m1R² as I already took its kinetic energy as m1ω²R²/2)

on solving further,

dω/dt = 2mgsinα/[(1+sinα)mR +MR/2]

which is not proportional to ∂!!!
and hence not a S.H.M.

The answer given is ω² = 2mgcosα/[MR + 2mR(1+sinα)]

Please help!!