3D Pendulum

Let at time $t$ , the angle that the rod makes with vertical be $α$ . Then the angular velocity of the point mass is given by $\omega=\dfrac{k}{\sin^2 α}$ , where $k$ is a constant (since the line of centrifugal force intersects the $z$ -axis and the line of weight force is parallel to that axis, the moments of both about that axis are zero).

Let $β=\dfrac{dα}{dt}$ . Then $β\dfrac{dβ}{dα}=\dfrac{k^2\cos α}{\sin^3 α}-g\sin α$

$\implies β^2=2g\cos α-\dfrac{k^2}{\sin^2 α}$ . Here I assumed that the rod was horizontal initially, so that $k=7$ .

When $α$ is minimum, $β=0\implies 2g\sin^2 α=49\cos α\implies \cos α=\dfrac{\sqrt {4+(\frac{49}{2g})^2}-\frac{49}{2g}}{2}\approx 0.35633646$ .

Hence the minimum $z$ coordinate is $\boxed {-0.3563}$ .

Let the pendulum be parameterized in spherical coordinates such that:

$A = \theta$ $B = \phi$

Coordinates of the bob at an arbitrary instant are:

$x = \sin{A} \cos{B}$ $y = \sin{A} \sin{B}$ $z = \cos{A}$

The kinetic energy and potential of the rod is:

$\mathcal{T} = \frac{1}{2}\left(\dot{x}^2+\dot{y}^2+\dot{z}^2\right)$ $\mathcal{V} = 10\cos{A}$

Using the Lagrangian approach or the energy conservation principle, the equations of motion can be derived which are (working left out):

$\ddot{A} = (\sin{A}\cos{A}) \dot{B}^2 + 10\sin{A}$ $\ddot{B} = -2\dot{B}\dot{A} \cot{A}$

Initial conditions are: $A(0) = \frac{\pi}{2}$ $B(0) = 0$ $\dot{A}(0) = 0$ $\dot{B}(0) = 7$

Numerical integration does the rest beyond this point. However, I am presenting an analytical approach similar to that of @Alak Bhattacharya as follows. Consider the second equation of motion

$\ddot{B} = -2\dot{B}\dot{A} \cot{A}$ $\implies \ddot{B} = \frac{d\dot{B}}{dt}= \frac{d\dot{B}}{dA} \frac{dA}{dt}= -2\dot{B}\dot{A} \cot{A}$ $\implies \frac{d\dot{B}}{dA} = -2\dot{B} \cot{A}$

Separating variables and integrating gives:

$\ln{\dot{B}} = -2\ln\left(\sin{A}\right) + \ln{K}$

Where $K$ is an integration constant. SImplifying gives:

$\dot{B} = \frac{K}{\sin^2{A}}$

Applying initial condition yields that $K=7$ . Thus:

$\dot{B} = \frac{7}{\sin^2{A}}$

Using the above result in the first equation of motion:

$\ddot{A} = (\sin{A}\cos{A}) \dot{B}^2 + 10\sin{A}$

$\implies \dot{A}\frac{d \dot{A}}{dA} = \sin{A}\cos{A} \left(\frac{49}{\sin^4{A}}\right) + 10\sin{A}$ $\implies \dot{A} \ d\dot{A} = \left(\left(\frac{49\cos{A}}{\sin^3{A}}\right) + 10\sin{A}\right)dA$

Integrating both sides again:

$\dot{A}^2 = -\frac{49}{\sin^2{A}} - 20\cos{A} + C$

Applying initial conditions again:

$\dot{A}^2 = -\frac{49}{\sin^2{A}} - 20\cos{A} + 49$

Finally, consider:

$z = \cos{A}$

For $z$ to be minimum:

$\dot{z} = -\sin{A} \ \dot{A} = 0$

$\implies \dot{A} = 0$ $\implies -\frac{49}{\sin^2{A}} - 20\cos{A} + 49 = 0$

This leads to a quadratic equation in $\cos{A}$ . Solving and realising that $A> \pi/2$ since the bob moves downward results in:

$\cos{A} \approx -0.3563$

Hence the minimum Z-coordinate is:

$z_{min} = -0.3563$

Thanks to @Alak Bhattacharya and @Karan Chatrath for the detailed solutions. I will simply add some plots of the trajectory in the $xz$ plane. The plots are for $10$ seconds, $30$ seconds, and $100$ seconds. The trajectory is essentially confined to a "band" on the surface of a sphere.