= Solution
{c}
The free rod has the same <mass> $m$ and length $l$ as in part (a). There is no pivot. Put the rod's initial <center of mass> at $y=0$ and the impact end at $y=l/2$. At impact the combined <center of mass> is
$$
y_C=\frac{m(0)+m(l/2)}{2m}=\frac l4.
$$
<Conservation of momentum> gives the combined <center of mass> <velocity> $V_C=v_0/2$. Evaluate <angular momentum> about this combined <center of mass>; the incident ball's perpendicular lever arm is $l/4$, so its initial magnitude is $mv_0l/4$.
After the <perfectly inelastic collision>, the <parallel axis theorem> gives the combined <moment of inertia>
$$
I_C=\left(\frac{ml^2}{12}+m\frac{l^2}{16}\right)+m\frac{l^2}{16}
=\frac{5ml^2}{24}.
$$
With negligible external angular <impulse>, <conservation of angular momentum> therefore gives
$$
\boxed{|\omega|=\frac{mv_0l/4}{5ml^2/24}=\frac{6v_0}{5l}.}
$$
Its sense is that of the incident ball's <angular momentum>, clockwise in the source diagram. As a consistency check, the final <kinetic energy> is
$$
K_f=\frac12(2m)\left(\frac{v_0}{2}\right)^2
+\frac12\frac{5ml^2}{24}\left(\frac{6v_0}{5l}\right)^2
=\frac25mv_0^2.
$$
Thus $mv_0^2/10$, or one fifth of the initial <kinetic energy>, is dissipated in sticking. Treating the rod's original centre as a fixed pivot would give an incorrect <angular velocity>.
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