Solving the Hinged Stick Problem for Time-Based Angular Acceleration/Speed

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The discussion centers on the hinged meter stick demonstration, illustrating how its free end accelerates faster than gravity, allowing a ball to land in a cup despite the cup's initial higher elevation. The user seeks to find an analytical solution for angular speed and acceleration as functions of time, starting from the derived equation w = sqrt(3g[1-cos(theta)]/L). However, substituting w = d(theta)/dt leads to a complex integral, specifically the integral of [1-cos(theta)]^(-1/2)d(theta), which appears to be unsolvable with standard integral tables. The transformation 1-cos(theta) = (sin(theta/2))^2 is mentioned as a potential simplification. The discussion highlights the challenges in deriving time-based angular motion equations for this physical scenario.
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A popular demonstration related to torque and angular acceleration involves a hinged meter stick, where one end is free to rotate while the other end serves as a rotation axis. Towards the end of the board, you might find a ball sitting on a golf tee and at the edge, there is usually a cup nailed to the free end. As the stick falls, its free end accelerates faster than gravity. Eventually the ball lands in the cup, even though the cup is initially at higher elevation.
Assume that the stick is initially vertical and θ represents the vertical angle between the normal and the current position of the stick. The stick has mass M and length L.

Question: I can solve for the angular acceleration and angular speed as a function of angle. Is there an analytical solution to the angular speed or angular acceleration as a function of time?
 
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w = d(theta)/dt. So you got the relation between the two. That is enough.
 
It is not quite that simple. After substituting w = d(theta)/dt, you get an integral that is not listed in any integral table that I have found:

Start with w = sqrt(3g[1-cos(theta)]/L) from energy considerations

After the substitution, you get the integral of [1-cos(theta)]^(-1/2)d(theta). Is this solvable?
 
1-cos theta =(sin theta/2)^2
 
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