I have the following suggestions for your experiment:
First, I would make the two pulleys to be identically the same (this greatly simplifies the analysis because you won't to deal with two unknown moments of inertia). I have made a sketch of your experiment.
View attachment 258601
In the following I calculate the moment of inertia of the pulley in terms of the time it takes for the mass to descend from height h to the floor using the technique of Lagrangian dynamics. Lagrangian dynamics is usually introduced in the upper division physics sequence but you can (and should) do a force analysis to verify my result. The Lagrangian of a system is,$$\mathcal {L}=K.E. - P.E.$$where K.E. is the kinetic energy and P.E. is the potential energy. For this system,$$K.E.= I \dot{\theta}^2+\frac{1}{2}m \dot{y}^2\\
P.E.=mg(h-y)$$where ##\theta## is the angle of rotation of the pulley. The arc length for a circle is ##s=R\theta## and with a moment of thought you can see that ##\theta = \frac{y}{R}##. The Lagrangian is therefore,$$ \mathcal {L}=I\frac{ \dot{y}^2}{R^2}+\frac{1}{2}m \dot{y}^2-(h-y)mg$$I suggest you google Lagrangian mechanics to understand the following sequence of calculations,$$
\frac{\partial {\mathcal{L}}}{\partial{\dot{y}}}=2\dot{y}(\frac{I}{R^2} + \frac{1}{2}m)\\
\frac {d(\frac{\partial {\mathcal{L}}}{\partial{\dot{y}}})}{dt}=2\ddot{y}(\frac{I}{R^2} + \frac{1}{2}m)\\
\frac{\partial {\mathcal{L}}}{\partial{y}}=mg\\
2\ddot{y}(\frac{I}{R^2} + \frac{1}{2}m)=mg\\
$$We now integrate the acceleration twice assuming the initial velocity is zero, the initial displacement is zero, and the final position is h.$$
\ddot{y}=\frac{1}{2}(\frac{mg}{\frac{I}{R^2}+ \frac{1}{2}m})\\
\dot{y}-\dot{y_0}=\frac{1}{2}(\frac{mgt}{\frac{I}{R^2}+ \frac{1}{2}m})\\
\dot{y_0}=0\\
y-y_0=\frac{1}{4}(\frac{mgt^2}{\frac{I}{R^2}+ \frac{1}{2}m})\\
y_0=0\\
I=\frac{mR^2}{2h}(\frac{gt^2}{2}-h)
$$
