Example 2 Given a frustum of a cone with the top base radius $r_{1}=5 \mathrm{~cm}$. The bottom base radius $r_{2}=10 \mathrm{~cm}$, and the slant height $A_{1} A_{2}=20 \mathrm{~cm}$. A rope is pulled from the midpoint of $A_{1} A_{2}$, wraps around the lateral surface of the frustum to a point $A_{2}$ on the bottom base circumference. Find:
(1) What is the minimum length of the rope?
(2) When the rope is at its minimum length, what is the minimum length of the line segment connecting a point on the rope to a point on the top base circumference?


As shown in Figure $17-5$, let the generatrix of the frustum be extended to intersect at point $P$.
As shown in Figure 17-6, the lateral surface of the frustum is unfolded along the generatrix $A_{1} A_{2}$ into a sector ring $A_{1} A^{\prime}{ }_{1} A^{\prime}{ }_{2} A_{2}$.
Then $\frac{P A_{1}}{P A_{1}+20}=\frac{r_{1}}{r_{2}}$, so $P A_{1}=20 \text{ cm}$.
It is easy to know that the central angle of the sector $\theta=\frac{r_{2}-r_{1}}{A_{1} A_{2}} \cdot 360^{\circ}=\frac{10-5}{20} \cdot 360^{\circ}=90^{\circ}$.
(1) Connect $A^{\prime}{ }_{2} M$. If $A^{\prime}{ }_{2} M$ does not intersect the sector $A^{\prime}{ }_{1} P A_{1}$, then $A^{\prime}{ }_{2} M$ is the shortest length of the rope.
In the right triangle $\triangle A^{\prime}{ }_{2} P M$, $P M=P A_{1}+A_{1} M=30, P A^{\prime}{ }_{2}=40$.
So $A^{\prime}{ }_{2} M=\sqrt{30^{2}+40^{2}}=50$.
Draw $P C \perp A^{\prime}{ }_{2} M$, with the foot of the perpendicular at $C$, and let $P C$ intersect the upper base circle at point $D$.
By $P C \cdot A^{\prime}{ }_{2} M=P M \cdot P A^{\prime}{ }_{2}, P C=\frac{P M \cdot P A^{\prime}{ }_{2}}{A^{\prime}{ }_{2} M}=\frac{30 \times 40}{50}=24>20=P A_{1}$,
Therefore, the line segment $A^{\prime}{ }_{2} M$ does not intersect the sector $A^{\prime}{ }_{1} P A_{1}$, so the shortest length of the rope is $50 \text{ cm}$.
(2) Let $P C$ intersect the arc $A^{\prime}{ }_{1} A_{1}$ at $D$. We will prove that $C D$ is the shortest line segment connecting a point on the rope and a point on the upper base circle.
If there is another line segment $C^{\prime} D^{\prime}$, where $C^{\prime}, D^{\prime}$ are on the rope and the upper base circle respectively, connect $P C^{\prime}, P D^{\prime}$.
Thus, $P D^{\prime}+D^{\prime} C^{\prime} \geqslant P C^{\prime} \geqslant P C=P D+D C$.
Since $P D=P D^{\prime}$, it follows that $D^{\prime} C^{\prime} \geqslant D C$.
Therefore, $C D$ is the shortest line segment connecting a point on the rope and a point on the upper base circle.
So, the shortest connecting line segment is $C D=P C-P D=24-20=4(\text{ cm})$.