Q: Maximum velocity of elastic pendulum released from rest Consider the following problem.

Construct a pendulum with a mass $m$ at the end. However, we don't a rigid rod of length $\ell$, we instead use a spring of natural length $\ell$ and spring constant $k$. We lift the mass so that the spring is horizontal and at its natural length (i.e. $\theta = \pi/2$ in the picture). We then release the mass, letting the pendulum swing. How low is the mass at the lowest point of the swing?


What I have so far:
We can find second-order equations of motion, either using Lagrangian mechanics or re-writing $F=ma$ in terms of its angular and radial components. But just like the regular pendulum, these equations are impossible to solve. I need some kind of invariance to solve this problem.
If we use conservation of energy, we get
$$\frac{1}{2}m(\ell + r)^2 \dot\theta^2 + \frac{1}{2}m\dot r^2 + \frac{1}{2}kr^2 - mg(\ell + r)\cos \theta = 0$$
at any time in the swing.
Since the height of the mass is $-(\ell + r)\cos \theta$, at the lowest point of the swing we have
$$ 0 = \frac{d}{dt} ((\ell + r)\cos \theta)$$
or rather,
$$\dot \theta \tan\theta = \frac{\dot r}{\ell + r}$$
And here I'm stuck. The above isn't enough to find the lowest part of the swing, so I need some other invariant. Looking at momentum seems to do no good; the force applied by the spring varies with time. Looking at angular momentum is similarly difficult.
For context: This is problem 76 of chapter 15 in "Physics for Scientists and Engineers" by Jewett and Serway, 8th edition. I looked at this problem some time ago, and it's nagged me ever since.
Also, in general, the elastic pendulum is chaotic, so any solution should make specific use of the given starting conditions.

A: You can construct a free-body diagram for the pendulum bob at the lowest point of its motion and balance out the forces in the vertical direction to obtain yet another equation.
The weight of the bob subtracted from the force on the bob due to the extension of the spring should provide the necessary centripetal force for the bob to sustain its motion.