Slide Time Before Natural Roll

Same derivation as the slide-distance page, read at the moment the two speeds meet instead of integrated to a distance. The ball's centre decelerates at $\mu g$ while its surface spins up at $\tfrac{5}{2}\mu g$, the gap closes at $\tfrac{7}{2}\mu g$, and sliding ends after $t = 2v_0/7\mu g$.

The striking thing is that this is linear in $v_0$ while the distance is quadratic. Hit the ball twice as hard and it slides for twice as long, but travels four times as far during that slide. The two facts sit together comfortably once you see why: it is covering the longer time at a higher average speed.

Actual durations are shorter than most people guess. On cloth with $\mu = 0.2$, a 3 m/s cue ball — a firm shot — slides for about 0.44 s. A 1 m/s roll-in slides for 0.15 s. The whole sliding phase of a break shot, at maybe 8 m/s, lasts a bit over a second. These are times you can just about perceive if you know to look, which is why some players can tell you by eye whether a shot arrived stunning or rolling.

The two pages check each other, and that is the main reason both are here. The deceleration is constant, so the average speed across the slide is the plain mean of the two ends, $(v_0 + \tfrac{5}{7}v_0)/2 = \tfrac{6}{7}v_0$. Multiply by the time:

\[\tfrac{6}{7}v_0 \times \frac{2v_0}{7\mu g} = \frac{12v_0^2}{49\mu g}\]

which is the slide distance exactly. Two formulas derived separately that agree to the last digit is not a coincidence; it is the same derivation read twice, and if either had a slip in it they would not meet.

Where the time form is more useful than the distance form is anywhere the ball is not travelling in a straight line. A cue ball struck with draw and sidespin curves as it slides — the sideways friction component bends its path, which is what a masse shot is doing in an extreme form — and for a curving path a time is meaningful where a straight-line distance is not. It is also the right quantity if you want to know whether the cue ball is still sliding when it reaches a ball a known time away, which is the honest way to ask whether the 90° rule applies.

The same caution about $\mu$ applies here as on the distance page, with one addition: because $t$ is linear in $v_0$, an error in $\mu$ shows up proportionally in the time but only as a square root when you back $\mu$ out of a measured distance. If you are calibrating your own cloth, a measured DISTANCE is the more forgiving measurement and a measured TIME is the more sensitive one. Use distance to find $\mu$; use time to check it.