A right circular cone has base radius 1 cm and slant height 3 cm is given.  $P$ is a point on the circumference of the base and the shortest path from $P$ around the cone and back to $P$ is drawn (see diagram).  What is the minimum distance from the vertex $V$ to this path?

[asy]
import graph;

unitsize(1 cm);

filldraw(shift(-0.15,0.37)*rotate(17)*yscale(0.3)*xscale(1.41)*(Circle((0,0),1)),gray(0.9),nullpen);
draw(yscale(0.3)*(arc((0,0),1.5,0,180)),dashed);
draw(yscale(0.3)*(arc((0,0),1.5,180,360)));
draw((1.5,0)--(0,4)--(-1.5,0));
draw((0,0)--(1.5,0),Arrows);
draw(((1.5,0) + (0.3,0.1))--((0,4) + (0.3,0.1)),Arrows);
draw(shift(-0.15,0.37)*rotate(17)*yscale(0.3)*xscale(1.41)*(arc((0,0),1,0,180)),dashed);
draw(shift(-0.15,0.37)*rotate(17)*yscale(0.3)*xscale(1.41)*(arc((0,0),1,180,360)));

label("$V$", (0,4), N);
label("1 cm", (0.75,-0.5), N);
label("$P$", (-1.5,0), SW);
label("3 cm", (1.7,2));
[/asy]
1. **Unfold the Cone:**
   To find the shortest path from point \( P \) around the cone and back to \( P \), we can unfold the cone into a plane. When the cone is unfolded, it forms a sector of a circle. The radius of this sector is the slant height of the cone, which is 3 cm.

2. **Calculate the Angle of the Sector:**
   The circumference of the base of the cone is \( 2\pi \times 1 = 2\pi \) cm. The length of the arc of the sector is also \( 2\pi \) cm. The radius of the sector is 3 cm. The angle \( \theta \) of the sector can be found using the formula for the arc length:
   \[
   \text{Arc length} = \theta \times \text{radius}
   \]
   \[
   2\pi = \theta \times 3
   \]
   \[
   \theta = \frac{2\pi}{3}
   \]

3. **Determine the Shortest Path:**
   The shortest path from \( P \) around the cone and back to \( P \) is a straight line in the unfolded sector. This line will be a chord of the sector. The chord subtends an angle of \( \frac{2\pi}{3} \) at the center of the sector.

4. **Calculate the Length of the Chord:**
   The length of the chord \( P_sP_e \) can be calculated using the formula for the chord length in a circle:
   \[
   \text{Chord length} = 2 \times \text{radius} \times \sin\left(\frac{\theta}{2}\right)
   \]
   \[
   P_sP_e = 2 \times 3 \times \sin\left(\frac{\frac{2\pi}{3}}{2}\right)
   \]
   \[
   P_sP_e = 6 \times \sin\left(\frac{\pi}{3}\right)
   \]
   \[
   \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}
   \]
   \[
   P_sP_e = 6 \times \frac{\sqrt{3}}{2} = 3\sqrt{3}
   \]

5. **Find the Midpoint of the Chord:**
   The midpoint \( L \) of the chord \( P_sP_e \) is the point on the chord that is closest to the vertex \( V \) of the cone. The distance from the center \( K \) of the sector to the midpoint \( L \) of the chord is:
   \[
   KL = \text{radius} \times \cos\left(\frac{\theta}{2}\right)
   \]
   \[
   KL = 3 \times \cos\left(\frac{\pi}{3}\right)
   \]
   \[
   \cos\left(\frac{\pi}{3}\right) = \frac{1}{2}
   \]
   \[
   KL = 3 \times \frac{1}{2} = \frac{3}{2}
   \]

6. **Calculate the Minimum Distance from \( V \) to the Path:**
   The minimum distance from the vertex \( V \) to the path is the distance from \( V \) to the midpoint \( L \) of the chord. Since \( KL = \frac{3}{2} \) and the radius of the sector is 3 cm, the distance from \( V \) to \( L \) is:
   \[
   VL = \sqrt{3^2 - \left(\frac{3}{2}\right)^2}
   \]
   \[
   VL = \sqrt{9 - \frac{9}{4}}
   \]
   \[
   VL = \sqrt{\frac{36}{4} - \frac{9}{4}}
   \]
   \[
   VL = \sqrt{\frac{27}{4}}
   \]
   \[
   VL = \frac{3\sqrt{3}}{2}
   \]

The final answer is \(\boxed{\frac{3\sqrt{3}}{2}}\)