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# Maximum period of a vertically spinning ball

Problem: A point mass with mass $$m$$, tethered by a string of length $$R$$, is in unforced circular motion in a gravitational field with strength $$g$$. The plane of motion is parallel to the gravitational field lines. The period of motion is $$T$$. What is the largest possible $$T$$? (ignore friction and any heat loss)

This a problem that I made up, but I'm sure it has been posed and answered before. I cannot find any reference. What I want to know is if my assumption for solving this problem is correct.

Assumption: For the path to be circular, the string must be under tension. Minimum tension is when the mass is at the top of the loop. Maximum period corresponds to zero tension at this point.

Details: Under this assumption, the velocity at that point is horizontal with magnitude $$v=\sqrt{g R}$$, since: $$m v^2 / R = m g.$$ Assuming zero potential at the bottom of the loop, the total energy at the top of the loop is $$\frac12 m (\sqrt{g R})^2 + m g ( 2 R)$$ Total energy at height $$h$$, measured from bottom of the loop is $$\frac12 m v_h^2 + m g h,$$ where $$0 \le h \le 2R$$ and $$v_h$$ is the speed at that height. Using the conservation of energy, equaling these and solving for $$v_h$$ $$v_h = \sqrt{5 g R - 2 g h}.$$ The average speed is (EDIT : In view of Floris' answer, this is where things go south. In hind sight, the error is obvious. If the particle stops mid motion for 100 years, the following formula gives the same average speed!) $$v_{\mbox{avg}} = \frac{1}{2R} \int_0^{2R} \sqrt{5 g R - 2 g h} \, dh = \frac{5 \sqrt{5}-1}{6} \sqrt{gR}.$$ The maximum period is $$T_{\mbox{max}}= \frac{2 \pi R}{v_{\mbox{avg}}} = \frac{12 \pi}{5\sqrt{5}-1} \sqrt{\frac{R}{g}}.$$

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