A thin flat disc of radius $a$ has density (mass per unit area) $\rho(r, \theta)=\rho_{0}(a-r)$ where $(r, \theta)$ are plane polar coordinates on the disc and $\rho_{0}$ is a constant. The disc is free to rotate about a light, thin rod that is rigidly fixed in space, passing through the centre of the disc orthogonal to it. Find the moment of inertia of the disc about the rod.

The section of the disc lying in $r \geqslant \frac{1}{2} a,-\frac{\pi}{13} \leqslant \theta \leqslant \frac{\pi}{13}$ is cut out and removed. Starting from rest, a constant torque $\tau$ is applied to the remaining part of the disc until its angular speed about the axis reaches $\Omega$. Show that this takes a time

$$\frac{3 \pi \rho_{0} a^{5} \Omega}{32 \tau}$$

After this time, no further torque is applied and the partial disc continues to rotate at constant angular speed $\Omega$. Given that the total mass of the partial disc is $k \rho_{0} a^{3}$, where $k$ is a constant that you need not determine, find the position of the centre of mass, and hence its acceleration. From where does the force required to produce this acceleration arise?