The disc is rigidly attached, so its rotation axis is perpendicular to its face. Its moment of inertia about its center is : with surface mass density , an annulus of radius contributes , whose integral is . The parallel axis theorem adds at the pivot. Adding the rod's moment of inertia gives
The rod and disc contribute restoring torques and , respectively. Using i in the rotational equation of motion givesOnly the disc's center of mass enters its gravitational torque; its finite radius enters its moment of inertia. The formula is also regular at , although the source happens to exclude that value.
The small-angle approximation reduces ii to , whereThe coefficient is positive, so the solution is simple harmonic motion, . Here is the angular frequency; the ordinary frequency and period are
For a fixed attachment ratio , making the disc much heavier than the rod givesIf the disc is also small relative to the attachment distance, , its moment of inertia about its own center is negligible compared with . ThusThis is a simple pendulum with a point mass at the disc's center and effective length ; it is length only if . Even before the small-angle approximation, its equation tends to .
If is allowed to tend to zero together with the other parameters, the bare conditions and do not alone justify that reduction. The needed dominance conditions are , , and , so that the rod contributes negligible gravitational torque and moment of inertia, and the disc behaves as a point mass. They follow automatically in the combined limit with fixed .
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