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Reading that made me realise that in non-inertial frames it is valid to deal with fictitious changes in momentum. If the frame has acceleration a=a(t) over an interval then conservation of momentum works by adding the "fictitious" term mΔv=m∫a.dt.It's not much different from what you did. The acceleration of the belt is ##\frac{dV}{dt}##. In the non-inertial frame it manifests itself as a fictitious force ##ma## acting on the CM in the opposite direction as the acceleration of the belt. The torque equationabout the point of contact on the beltgives ##MaR=(Mk^2+MR^2)\alpha## (note the use of he parallel axis theorem). Let ##a'_{cm}## be the acceleration of the CM in the non-inertial frame. Then ##a'_{cm}=\alpha/R##. Put this back in the torque equation, cancel the masses and get $$a'_{cm}=\frac{R^2}{k^2+R^2}a=\frac{R^2}{k^2+R^2}\left(-\frac{dV}{dt}\right).$$ The negative sign is introduced because ##a'_{cm}## and ##\frac{dV}{dt}## are in opposite directions. In the inertial frame, the acceleration of the CM is $$a_{cm}=a+a'_{cm}=\frac{dV}{dt}-\frac{R^2}{k^2+R^2}\frac{dV}{dt} =\frac{k^2}{k^2+R^2}\frac{dV}{dt} $$The velocity of the CM is$$v_{cm}(t)=\int a_{cm}dt=\frac{k^2}{k^2+R^2}\int \frac{dV}{dt}dt=\frac{k^2}{k^2+R^2}V(t).$$I am not convinced that this is a better way to approach the problem however.

Something similar may be possible for angular momentum, and if such a rule were taken as standard you could avoid the integration step above.