Speed at Natural Roll

A cue ball leaves the tip sliding. The cloth rubs its underside, which does two things at once: it slows the ball down, and it spins the ball up. Sooner or later the surface speed catches the travel speed, the sliding stops, and the ball rolls. The question is what speed it has when it gets there — and the answer is the most elegant thing on this whole page.

A ball struck with no spin reaches natural roll having lost exactly two sevenths of its speed. Not roughly. Exactly, every single time.

Not "about 5/7 on typical cloth". Exactly 5/7, on new worsted and on a threadbare bar table, on a soft stroke and a break shot, in Winnipeg and on the Moon. The coefficient of friction does not appear in the equation. Neither does gravity. Neither does the mass of the ball. $\mu$ decides how FAR the ball slides and how LONG it takes. It has no say whatever in how much speed is lost. That is a genuinely surprising result and it deserves a moment before we prove it.

The proof is one line, and it is a beautiful use of a trick every mechanics course teaches and almost none applies to anything: choose your axis where the forces are. Friction acts at the single point where the ball touches the cloth. About that contact point, friction exerts NO TORQUE — its moment arm is zero. So the angular momentum of the ball about the contact point is conserved for the entire slide, however long it lasts and whatever $\mu$ happens to be.

At the start, that angular momentum is $mv_0R + I\omega_0$, the orbital part plus the spin part, with $I = \tfrac{2}{5}mR^2$. At the end the ball is rolling, so $\omega_f = v_f/R$, and the angular momentum is $mv_fR + \tfrac{2}{5}mR^2(v_f/R) = \tfrac{7}{5}mv_fR$. Set them equal and cancel $mR$:

\[v_0 + \tfrac{2}{5}R\omega_0 = \tfrac{7}{5}v_f \quad\Longrightarrow\quad v_f = \tfrac{5}{7}v_0 + \tfrac{2}{7}R\omega_0\]

Put $\omega_0 = 0$ and there it is: $v_f = \tfrac{5}{7}v_0$. The sevenths come from the $1 + \tfrac{2}{5} = \tfrac{7}{5}$, which comes from the moment of inertia of a solid sphere, which comes from geometry. That is all that is in it.

Read the general form as a weighted average and it explains the three families of shot at once. The final speed is five parts starting speed to two parts starting spin, and the spin ratio $R\omega_0/v_0$ from the tip-offset page tells you which side of the balance you are on:

• Ratio below 1 — the ball slows down reaching roll. A centre-ball hit is ratio 0 and loses 2/7.
• Ratio exactly 1 — nothing changes. Struck at $b = 0.4R$, the ball is already rolling and $v_f = v_0$.
• Ratio above 1 — the ball SPEEDS UP. Topspin is being converted into travel. This is why a follow shot runs on, and why a hard follow can arrive at the object ball faster than it left the tip.
• Ratio below $-\tfrac{5}{2}$ — $v_f$ comes out negative. The ball stops, reverses, and rolls back toward you. That is draw, and this equation predicts it with no extra physics bolted on. The threshold, $R\omega_0 = -\tfrac{5}{2}v_0$, is well past the miscue limit, which is why a full-table draw needs the object ball close: the cue ball must still have backspin left when it arrives.

Two practical consequences follow immediately. First, a stun shot has a shelf life. The cue ball leaves the tip with no spin and is rolling a metre or so later, so the 90° rule applies only within that distance of the object ball. Second, speed control on a soft shot is not what you think. A gently struck centre-ball cue ball has already given up 2/7 of its speed before it has done anything useful. Players who hit soft and low compound the loss; players who hit soft with a touch of follow avoid it entirely. Nobody teaches it in those terms, but every good player has found it by feel.

Coriolis had all of this in 1835. It is worth sitting with how much of the modern instructional canon — natural roll, the tangent line, the 30° rule — was worked out in one book by a mechanician who is remembered for something else entirely.