Get it to roll without slipping down an incline starting from rest and have it report its speed at the bottom of the incline. Energy conservation says that for a vertical drop ##h## the potential energy is converted into kinetic energy of the center of mass and rotational energy about the center of mass,$$mgh=\frac{1}{2}mV_{\text{cm}}^2+\frac{1}{2}I_{\text{cm}}\omega^2.$$As you noted, you can write the moment of inertia about the CoM as ##I_{\text{cm}}=qmR^2## where ##q## is the constant fraction that you want to determine. If this thing rolls without slipping, ##\omega =\dfrac{V_{\text{cm}}}{R}## in which case the energy conservation equation becomes $$mgh=\frac{1}{2}mV_{\text{cm}}^2+\frac{1}{2}(qmR^2)\left( \frac{V_{\text{cm}}}{R}\right)^2.$$After the obvious cancellations you get $$gh=\frac{1}{2}V_{\text{cm}}^2+\frac{1}{2}qV_{\text{cm}}^2\implies q=\frac{2gh}{V_{\text{cm}}^2}-1.$$So all you have to do is measure the vertical height by which it drops and have it tell you how fast it is moving after it drops by that height and plug in. Its mass or radius don't matter with this technique. This is counterintuitive when what you are trying to determine is its moment of inertia but there you have it. Just make sure there is no slipping.
