# Spin Shot Pool Physics

I am trying to calculate the total distance travelled by a pool ball in a shot modelled as shown in the picture. I intend to do a path integral once I've figured out the position vector of the ball at all times. For the moment I am assuming an infinite table so as to not worry about cushion interactions and also the initial spin is only top/back spin. I have tried to label as best I can what I think should be happening at each point and in each interval.

To keep things slightly simpler I have assumed a perfect collision with the red ball (cof elasticity = 1) so that I can make the stun vector 90° to the collision vector. Unless there is good reason not to I'm neglecting all interactions between the cloth and the ball other than friction.

I have reasoned that the overall friction will be a resultant of the friction caused by the translation of the ball combined with the friction caused by the relative slippage to due spin. - Is this correct?

My knowledge of physics is rather limited and I've never dealt with spin-translation interactions with loss of energy before. I know of concepts like moment of inertia and that for a solid sphere it is $I = \frac{2}{5}mr^2$

What I don't know is how spin converts to translation and how much energy is lost.

Any and all help is greatly appreciated!

• Please be specific on what you are asking. Can you give me initial condition before they collide ? What do you mean by perfect collision ? – Si Kucing Nov 21 '17 at 12:10
• The condition at $P_1$ is $$\begin{displaymath}\dot{\bold{r}} = \left( \begin{array}{c} u_x\\ u_y\\ 0 \end{array} \right) \end{displaymath}$$ $$\begin{displaymath}\dot{\bold{θ}} = \left( \begin{array}{c} \omega\\ 0\\ 0 \end{array} \right) \end{displaymath}$$ As shown in the diagram. Not sure why this isn't displaying correctly? – Ben Crossley - hobbyist Nov 21 '17 at 12:13
• By perfect collision I just mean coefficient of elasticity is 1. I'll put that in the question. – Ben Crossley - hobbyist Nov 21 '17 at 12:16
• I think the $\omega^*$ should be $-\omega^*$ if you put $x-$axis to the right and $y-$axis goes up. – Si Kucing Nov 21 '17 at 12:26
• Because this based on the actual case, i think this takes a lot of work (particularly $P_1-P_2$ part) and i dont feel like tackle this now. But you should know that there are some good reference like this and this espescially Impact Mechanics by Stronge. I'll post an answer once i have free time. – Si Kucing Nov 21 '17 at 13:05

$P_0-P_1 :$

With your chosen cartesian coordinate, consider the cue ball with radius $r$ start with initial velocity $\vec{v} = (0,v_0,0)$ and initial angular velocity $\dot{\theta} = (\omega^*,0,0)$ (so its back-spin). I assume we know the distance from $P_0$ to $P_1$ before its hit another ball. Call the distance $s$. Because its spin backward, the friction force must act on the ball to the negative direction of its velocity so the total force will be $\vec{F} = (0,-f,N-mg)$, where $f$ is its kinetic friction with magnitude $f=\mu_kN$. By Newton's law in the $y-$axis and torque equation $$-f = m a_y \quad \text{and} \quad -fr = I \alpha$$ At this situation $a_y \neq \alpha r$ because its slipping. With distance $s$ the final velocity will be $$v^2 = v_0^2 +2a_y s = v_o^2 -2\frac{f}{m}s = v_0^2 - 2\mu_kgs, \quad v = \sqrt{v_0^2 - 2\mu_kgs} \qquad (1)$$ To compute the final angular velocity we need the time travel of the ball with distance $s$, that is $$s = v_ot + \frac{1}{2} a_y t^2 = v_0t - \frac{1}{2}\mu_kg t^2 \iff \frac{1}{2}\mu_kg t^2 -v_0t + s = 0$$ and solve for $t$. The angular velocity will be $$\omega = \omega^* + \alpha t = \omega^*- \frac{fr}{I} t = \omega^* - \frac{\mu_k mgr}{I} t \qquad (2)$$ So at $P_1$ the velocity is $\vec{v} = (0,v,0)$ and $\dot{\theta} = (\omega,0,0)$ and it still slipping (otherwise the ball will not move back after hit the another ball as shown by the video).

$P_1 - P_1$ (collision) :

Consider the collision to be elastic (no deformation of the balls) so the kinetic energy conserved $$\frac{1}{2} m v^2 + \frac{1}{2} I \omega^2 = \frac{1}{2}mv'^2 + \frac{1}{2}I \omega'^2 + \frac{1}{2} M V^2 + \frac{1}{2}I_M \omega_M^2\qquad (3)$$
with $v'$ and $\omega'$ are velocity and angular velocity of the cue ball whereas $M,I_M, V, \omega_M$ are the mass, moment of inertia, velocity and angular velocity of the target ball respectively. Now consider the Impulse momentum equation from Newton's law below $$\sum \vec{F}_i = \sum \frac{d\vec{p_i}}{dt} \quad \text{or} \quad \sum \vec{F_1} + \sum \vec{F_2} = \frac{d\vec{p_1}}{dt} + \frac{d\vec{p_2}}{dt}$$ integrate them, we have impulse momentum equation $$(\vec{p_1})_{\text{initial}} + (\vec{p_2})_{\text{initial}} + \int (\sum \vec{F_1} + \sum \vec{F_2}) dt = (\vec{p_1})_{\text{final}} + (\vec{p_2})_{\text{final}}$$

Where index $i=1,2$ represent out system of two ball, and the sum above represent all force terms that act on each balls. Those force are the internal forces : contact force and frictions, and the external forces : weight, normal forces and frictions from the table. The internal forces are comes in pairs so if we add the forces $\sum \vec{F_1} + \sum \vec{F_2}$, those will cancels each other. The normal force and the weight will cancel each other. So we just have frictions from the table that contribute, and in fact in the small interval (as they collide) the product of theese constant force and the time interval is so small so they can be neglected, $$\int (\sum \vec{F_1} + \sum \vec{F_2}) dt = \int_{0}^{\delta} (\vec{f_1} +\vec{f_2} ) dt = (\vec{f_1} +\vec{f_2}) \delta \approx 0$$ Therefore the momentum is conserved. So $$(\vec{p_1})_{\text{initial}} + (\vec{p_2})_{\text{initial}} = (\vec{p_1})_{\text{final}} + (\vec{p_2})_{\text{final}} \qquad (4)$$ We know that the target ball is not move at the beginning so $(\vec{p_2})_{\text{initial}} = \vec{0}$. Also the direction of the velocity of the target ball will be perpendicular to the tag line so $\vec{p_2})_{\text{final}} = M\vec{V} = (-MV \cos \alpha, MV \sin \alpha,0)$. And also $(\vec{p_1})_{\text{initial}} = (0,mv,0)$. We dont know the direction of the cue ball after collision so lets just say $(\vec{p_1})_{\text{final}} = (mv'_x,mv_y',0)$. Therefore $x$-component and the $y-$component of the momentum eq. $(4)$ $$0 = mv_x'- MV \cos \alpha, \qquad mv = mv_y' + MV \sin \alpha \qquad (5)$$ rewrite and squared both sides $$(mv_x')^2 = (MV \cos \alpha)^2, \qquad (mv_y')^2 = (mv - MV \sin \alpha)^2$$ and then add them, you'll have $$m^2v'^2 = m^2v^2 + M^2 V^2 - 2mM vV \sin \alpha \qquad (6)$$ Now we have to make other assumption that the angular velocity of the ball does not affected by the collision so in the equation $(3)$, $\omega'=\omega$ and $\omega_M = 0$ (actually i think this is the case in pool because the surface of the balls is very smooth). Otherwise it will be very difficult to solve because now the angular velocity not vanish due to the horizontal friction and vertical friction (i'm sorry i dont have the picture) and we have to analyze them using another equation that is angular impulse-momentum equation derived from integrating torque equation. By this assumption $(3)$ become $$mv^2 = mv'^2 + MV^2$$ Combine this and $(6)$ we have $$0 = V^2 (1+ \frac{M}{m}) - 2vV \sin \alpha \implies V = \frac{2v \sin \alpha}{1+ \frac{M}{m}}$$ And therefore by $(5)$ you'll have $v' =\sqrt{ v_x'^2+v_y'^2}$ also.

$P_1 - P_2$ (after collision) :

We have to analize two ball after the collision. Lets consider the easier first that is the target ball. Right after the collision, the target ball moving with velocity $V$ in the direction of the line perpedicular to the tag line. Call this line as $y'-$axis and the tag line $x'-$axis. Because of this, the target ball must slip for a moment and we have to find when this slip is over. By Newton's Law and torque eq. (along $y'$-axis) and the fact that $f = \mu_kMg$ $$-f =-\mu_k Mg= Ma, \qquad -fr=-\mu_kMg r=I_M \alpha$$ So the velocity and the angular velocity of this target ball will be $$V_M = V + at = V-\mu_k g t, \quad \omega_M = \omega_M0 + \alpha t = 0 - \frac{\mu_kMgr}{I_M} t$$ The minus sign in the $\omega_M$ just means that the roatation is top spin. The non-slip condition happen (at some time $t^*$) if the magnitude of $V_M$ and $\omega_M r$ are equall. So set $|V_M| = |\omega_M|r$ $$V - \mu_k g t^* = \frac{\mu_kMgr}{I_M} t^* \implies t^* = \frac{V}{\mu_k g(1+ \frac{rM}{I_M})}$$ Put this back to the $V_M = V + at^* = V-\mu_k g t^*$ we have the final velocity of the target ball until non-slip condition $$V_M = \frac{V}{\frac{I_M}{Mr}+1}$$ And we're done with the target ball.

For the cue ball after collision, the cue ball has condition $\vec{v'} = (v_x',v_y',0)$ with back spin $\dot{\theta}= (\omega,0,0)$ as before collision. Because of the back spin, there is a friction force $\vec{f_1} = (0,-f_1,0)$ and because of the movement in the $+x$-direction (by $v_x'$) we have the second friction force $\vec{f_2}= (-f_2,0,0)$ with their resultant $\vec{f}=\vec{f_1}+\vec{f_2}$ satisfy $|\vec{f}| = \mu_kmg$. And to this point i simply have a doubt about how to proceed to analized this. I'll update it another time if i know.