A point mass $m = 1$ is attached to the free end of a mass-less rigid rod of length $L = 1$ . The other end of the rod is attached to the origin with a ball joint, allowing the free end to be positioned at an arbitrary point in $3D$ space (constrained by the length of the rod, or course). The gravitational acceleration is $10$ in the negative $z$ direction.

Initially, the rod is rotating in the $xy$ plane with angular speed $\omega = 7$ (radians per second) with respect to the $z$ axis. As the rod moves, what is the minimum $z$ coordinate reached by its free end?

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Bonus:
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Plot the trajectory

The answer is -0.357.

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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 $\boxed {-0.3563}$ .