= Solution
The <moment of inertia> about an axis is $I=\int r_\perp^2\,dm$. The <parallel axis theorem> states $I=I_{\rm CM}+Ma^2$, where a is the perpendicular separation from a parallel axis through the centre of mass, and M is total mass.
Since $\rho=kr^2$, the density coefficient has units $\mathrm{kg\,m^{-5}}$. The <density-power cylindrical moment of inertia> follows from cylindrical shells:
$$
M=\int_0^R2\pi Lkr^3\,dr=\frac{\pi kLR^4}{2},\qquad
I_{\rm CM}=\int_0^R2\pi Lkr^5\,dr=\frac{\pi kLR^6}{3}
=\boxed{\frac23MR^2}.
$$
For an offset pivot, the <physical pendulum> equation is $(I_{\rm CM}+Ma^2)\ddot\theta=-Mga\sin\theta$. At small amplitude,
$$
\boxed{T=2\pi\sqrt{\frac{a^2+2R^2/3}{ga}}.}
$$
For the <minimum-period offset of a physical pendulum>, minimizing $T^2$ is equivalent to minimizing $a+2R^2/(3a)$. Its derivative vanishes at
$$
\boxed{a_{\min}=R\sqrt{2/3},\qquad
T_{\min}=2\pi\sqrt{\frac{2R\sqrt{2/3}}g}.}
$$
The second derivative is positive. The period diverges as $a^{-1/2}$ near zero and grows as $a^{1/2}$ for large offset, giving one minimum. This assumes the stated ideal pivot can pass inside the body; an externally constrained pivot would restrict the allowable a. <University Physics Volume 1 by William Moebs and others>[§15.4, physical-pendulum torque equation and period]{ref}
\Image[/past-exam-of-the-physics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2011/paper-1-ia2011-pendulum-offset.png]
{title=Pendulum pivot offset}
{source=paper-1-ia2011-pendulum-offset.py}
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