= Solution
The <moment of inertia> about a specified axis is $I=\int s^2\,dm$, where $s$ is perpendicular distance from the axis. The <parallel axis theorem> states $I=I_{\rm cm}+md^2$ for two parallel axes separated by $d$, one passing through the <centre of mass> of a body of <mass> $m$.
It is easiest to calculate the eccentric disc by subtraction. Let $m_f$ be the <mass> of the unperforated disc and $m_h$ the removed <mass>. Uniform thickness and <density> give $m_h=m_f/4$ and $m_f-m_h=M$, hence $m_f=4M/3$ and $m_h=M/3$. The hole is tangent to the rim, so its centre, and therefore the shaft axis, lies $R/2$ from the original disc centre. About the shaft axis,
$$
I_{\rm perforated}=\frac{4M}{3}\left(\frac{R^2}{2}+\frac{R^2}{4}\right)-\frac12\frac M3\left(\frac R2\right)^2=\frac{23}{24}MR^2.
$$
The shaft contributes $I_{\rm shaft}=M(R/2)^2/2=MR^2/8$. Thus
$$
\boxed{I=I_{\rm perforated}+I_{\rm shaft}=\frac{13}{12}MR^2.}
$$
The motor's <torque> is $P/\omega$ at nonzero <angular velocity>, while the resisting <torque> is $-\gamma\omega$. The <rotational equation of motion> and its equivalent <energy balance> are
$$
I\dot\omega=\frac P\omega-\gamma\omega,\qquad \frac{d}{dt}\left(\frac12I\omega^2\right)=P-\gamma\omega^2.
$$
At the <steady state>, input <power> equals dissipated <power>, giving
$$
\boxed{\omega_0=\sqrt{P/\gamma}.}
$$
For the acceleration, use $u=\omega^2$. Then $\dot u+(2\gamma/I)u=2P/I$ with $u(0)=0$, so
$$
u(t)=\omega_0^2(1-e^{-2\gamma t/I}),\qquad \boxed{t=-\frac I{2\gamma}\ln\left(1-\frac{\omega^2}{\omega_0^2}\right),\quad0\leq\omega<\omega_0.}
$$
The approach to $\omega_0$ takes infinite time. This <constant-power rotational acceleration> idealizes the start: $P/\omega$ diverges at rest, although the <energy balance> has a well-defined limiting solution. A real motor has finite starting <torque>.
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