Solution (source code)

= Solution

The area element in <plane polar coordinates> is $r\,dr\,d\theta$. Thus the axial <moment of inertia> is
$$
I_0=\int_0^{2\pi}\int_0^a r^2\rho_0(a-r)r\,dr\,d\theta
=2\pi\rho_0\left[\frac{ar^4}{4}-\frac{r^5}{5}\right]_0^a
=\boxed{\frac{\pi\rho_0a^5}{10}}.
$$
For the removed material, the angular width is $2\pi/13$, and
$$
\int_{a/2}^a r^3(a-r)\,dr=\frac{13a^5}{320}.
$$
Its <moment of inertia> is therefore $I_{\rm cut}=\pi\rho_0a^5/160$, leaving
$$
I=I_0-I_{\rm cut}=\frac{3\pi\rho_0a^5}{32}.
$$
The <torque> equation $I\dot\Omega=\tau$ gives constant <angular acceleration>, so the time to reach the prescribed <angular speed> from rest is
$$
\boxed{t=\frac{I\Omega}{\tau}=\frac{3\pi\rho_0a^5\Omega}{32\tau}.}
$$

To find the <centre of mass after removing material>, let $\mathbf e_1$ point along the bisector of the missing sector in the disc. The original disc has zero first <mass> moment about its centre. The missing sector has zero transverse first moment by reflection symmetry, while its component along $\mathbf e_1$ is
$$
Q_{\rm cut}=\rho_0\int_{a/2}^a r^2(a-r)\,dr\int_{-\pi/13}^{\pi/13}\cos\theta\,d\theta
=\frac{11\rho_0a^4}{96}\sin\frac\pi{13},
$$
because the radial integral is $11a^4/192$. The remaining first moment is $-Q_{\rm cut}\mathbf e_1$. Dividing by the given <mass> $k\rho_0a^3$ gives the body-fixed <centre of mass>
$$
\boxed{\mathbf R=-\frac{11a}{96k}\sin\frac\pi{13}\,\mathbf e_1.}
$$
It lies away from the missing sector. In the inertial frame, let $t_*$ be the moment the applied <torque> stops and let $\psi=\psi_*+\Omega(t-t_*)$ give the orientation of that sector's bisector. Then, with $d=11a\sin(\pi/13)/(96k)$,
$$
\mathbf R(t)=-d(\cos\psi,\sin\psi,0),\qquad\boxed{\ddot{\mathbf R}=-\Omega^2\mathbf R}.
$$
The <centre of mass> has inward <centripetal acceleration> of magnitude $\Omega^2d$. \b[The required real <force> is the constraint reaction exerted by the fixed rod and its support]. Its horizontal resultant is $k\rho_0a^3\ddot{\mathbf R}$; a zero applied axial <torque> does not imply a zero resultant <force>.