= Solution
{c}
<Inertia> is resistance to a change of motion. Translational <inertia> is measured by <mass> in <Newton's second law>; rotational <inertia> about a specified fixed axis is measured by the <moment of inertia>
$$
I=\int r_\perp^2\,dm,
$$
where $r_\perp$ is perpendicular <distance> from that axis. For a thin uniform rod, take axes perpendicular to the rod. With <linear mass density> $\lambda=m/l$,
$$
I_{\rm CM}=\frac ml\int_{-l/2}^{l/2}x^2\,dx=\frac{ml^2}{12},\qquad
\boxed{I_{\rm end}=I_{\rm CM}+m\left(\frac l2\right)^2=\frac{ml^2}{3}.}
$$
The second result uses the <parallel axis theorem>; direct <integration> over $0\leq x\leq l$ gives the same value. The axis specification is essential: an ideal thin rod has zero <moment of inertia> about its own longitudinal axis.
During the short <elastic collision>, neglect the <impulse> from gravity. The pivot can exert an external <impulse>, so total linear <momentum> of rod plus particle need not be conserved. Its <torque> about the pivot is zero, so <conservation of angular momentum> about that point applies. Let $w$ be the particle's signed outgoing horizontal <velocity>, and $\omega$ the rod's outgoing <angular velocity>, positive towards the incident motion. Then
$$
mlv_0=mlw+I_{\rm end}\omega,\qquad
\frac12mv_0^2=\frac12mw^2+\frac12I_{\rm end}\omega^2.
$$
Substituting $\omega=3(v_0-w)/l$ into the <kinetic energy> equation gives
$$
v_0^2=w^2+3(v_0-w)^2,
\qquad (v_0-w)(v_0-2w)=0.
$$
Discarding the zero-<impulse> branch $w=v_0$, the actual <collision> gives
$$
\boxed{w=\frac{v_0}{2}\quad\text{forward},\qquad
\omega=\frac{3v_0}{2l}.}
$$
The rod's lower end initially moves forward at $l\omega=3v_0/2$, so the bodies separate. Afterwards the free particle undergoes <projectile motion>. With the pivot as origin and $z$ downward, its <trajectory> is $x=(v_0/2)t$, $z=l+gt^2/2$, where $t=0$ at the <collision>. The rod instead swings about the fixed pivot in a vertical plane as a <physical pendulum>.
The rod's initial rotational <kinetic energy> is $I_{\rm end}\omega^2/2=3mv_0^2/8$. Reaching the horizontal raises its <centre of mass> by $l/2$, increasing <gravitational potential energy> by $mgl/2$. Consequently,
$$
\boxed{v_0^2\geq\frac{4gl}{3}\ \text{to reach the horizontal},\qquad
v_0=\sqrt{\frac{4gl}{3}}\ \text{to just reach it at rest}.}
$$
For the threshold case, measure the rod's <angle> $\theta$ from the downward vertical and reset $t=0$ at its first horizontal <turning point>, so $\theta(0)=\pi/2$ and $\dot\theta(0)=0$. The restoring <torque> about the pivot gives the exact <equation of motion>
$$
\boxed{\ddot\theta+\frac{3g}{2l}\sin\theta=0.}
$$
At the horizontal, $\ddot\theta=-3g/(2l)$, so the rod immediately swings back. <Conservation of mechanical energy> yields
$$
\frac12\frac{ml^2}{3}\dot\theta^2+\frac{mgl}{2}(1-\cos\theta)=\frac{mgl}{2},
\qquad
\dot\theta^2=\frac{3g}{l}\cos\theta.
$$
On the initial descent its explicit time dependence is specified by the <definite integral>
$$
\boxed{t=\sqrt{\frac{l}{3g}}\int_{\theta(t)}^{\pi/2}\frac{d\alpha}{\sqrt{\cos\alpha}}.}
$$
It continues through the vertical to $-\pi/2$, reverses, and repeats without damping. An alternative expression valid throughout the oscillation uses the <Jacobi elliptic sine>. Put $k=1/\sqrt2$, $\omega_r=\sqrt{3g/(2l)}$, and let $K(k)$ denote the <complete elliptic integral of the first kind>. Then
$$
\boxed{\theta(t)=2\arcsin\!\left[k\,\operatorname{sn}\!\left(K(k)-\omega_rt,k\right)\right],\qquad
T_r=4\sqrt{\frac{2l}{3g}}\,K\!\left(\frac1{\sqrt2}\right).}
$$
This is not <simple harmonic motion> at an <amplitude> of $\pi/2$. Its <oscillation period> exceeds its own small-<amplitude> value $2\pi\sqrt{2l/(3g)}$ by the factor $2K(1/\sqrt2)/\pi\simeq1.18034$.
A <simple pendulum> of <length> $l/2$ obeys $\ddot\theta+(2g/l)\sin\theta=0$. At the same <amplitude> $\pi/2$, its <oscillation period> is $T_p=4\sqrt{l/(2g)}K(1/\sqrt2)$, so
$$
\boxed{\frac{T_r}{T_p}=\frac{2}{\sqrt3}>1.}
$$
The rod swings more slowly. Its equivalent <simple pendulum> <length> is $I_{\rm end}/[m(l/2)]=2l/3$, not the <distance> $l/2$ to its <centre of mass>. The same <oscillation period> ratio holds in the common small-<amplitude> limit, but replacing the actual 90-degree motion by that limit would underestimate both <oscillation periods>.
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