The plane of the figure represents a horizontal plane. A thin rigid rod at rest is pivoted without friction about a fixed vertical axis passing through O. Its mass moment of inertia is equal to 0.1 kg∙cm² about O. A point mass of 0.001 kg hits it normally at 200 cm/sat the location shown, and sticks to it. Immediately after the impact, the angular velocity of the rod is ___________ rad/s (in integer).

The correct answer is 10 rad/s.

1. Identify the given parameters from the problem statement and the diagram:
- Mass of the point particle: $m=0.001\text{kg}$
- Velocity of the point particle before impact: $v=200\text{cm/s}$
- Mass moment of inertia of the rod about the pivot O: $I_{\text{rod}}=0.1\text{kg}·{\text{cm}}^2$
- Distance of the impact point from the pivot O (as shown in the diagram): $r=10\text{cm}$

2. Calculate the moment of inertia of the combined system after impact:
Since the particle strikes the rod normally and sticks to it at a distance of $r=10\text{cm}$ from pivot O, the moment of inertia of the combined system ($I_c$) about O is the sum of the moment of inertia of the rod and the point mass:
$I_c=I_{\text{rod}}+m{r}^2$
Substituting the given values:
$I_c=0.1+0.001\times {10}^2$
$I_c=0.1+0.001\times 100$
$I_c=0.1+0.1=0.2\text{kg}·{\text{cm}}^2$

3. Apply the law of conservation of angular momentum about the pivot point O:
Since the rod is pivoted without friction, no external torque acts on the system about the vertical axis passing through O. Thus, the angular momentum before impact equals the angular momentum after impact:
$L_{\text{before}}=L_{\text{after}}$
The angular momentum before impact is due to the point mass:
$L_{\text{before}}=mvr$
The angular momentum after impact is due to the combined system rotating with angular velocity $\omega$:
$L_{\text{after}}=I_c\omega$

4. Solve for the angular velocity ($\omega$):
Equating the angular momentum before and after impact:
$mvr=I_c\omega$
Substituting the numerical values into the equation:
$0.001\times 200\times 10=0.2\times \omega$
$2=0.2\omega$
$\omega =\frac{2}{0.2}=10\text{rad/s}$