A soap bubble of radius $a(t)$ is attached to the end of a long, narrow straw of internal radius $\epsilon$ and length $L$, the other end of which is open to the atmosphere. The pressure difference between the inside and outside of the bubble is $2 \gamma / a$, where $\gamma$ is the surface tension of the soap bubble. At time $t=0, a=a_{0}$ and the air in the straw is at rest. Assume that the flow of air through the straw is irrotational and consider the pressure drop along the straw to show that subsequently

$$a^{3} \ddot{a}+2 a^{2} \dot{a}^{2}=-\frac{\gamma \epsilon^{2}}{2 \rho L},$$

where $\rho$ is the density of air.

By multiplying the equation by $2 a \dot{a}$ and integrating, or otherwise, determine an implicit equation for $a(t)$ and show that the bubble disappears in a time

$$t=\frac{\pi}{2} \frac{a_{0}^{2}}{\epsilon}\left(\frac{\rho L}{2 \gamma}\right)^{1 / 2}$$

[Hint: The substitution $a=a_{0} \sin \theta$ can be used.]