# Velocity and Angular Velocity Equations for Cue-Ball Interactions

I'm trying to create an eight-ball pool simulation in Python, and I'm trying to simulate the interaction between a cue stick and a cue ball. Suppose I thrust a cue stick with mass $$m_s$$ at a velocity $$v_s$$ to hit a cue ball at a position $$(a, b)$$ from its center in the $$xz$$ plane. I also know the ball's mass, $$m_b$$, and its radius, $$R$$. I would like to determine the velocity components $$\{v_x, v_y, v_z\}$$ and angular velocity components $$\{\omega_x, \omega_y, \omega_z\}$$ after the cue stick hits the cue ball.

Here is a picture describing the interaction between the cue stick and the ball in the above example:

I have come up with some equations using the help of this article, but my results seem incorrect after comparing them with the results of another pool simulator. I would like to confirm whether the equations (highlighted in bold at the bottom) are valid, and if not, adjust them such that they are valid.

By the conservation of linear momentum: $$m_sv_{s_i} = m_sv_{s_f} + m_bv_b \tag{1}\label{1}$$

Assuming an elastic collision, by the conservation of energy (the kinetic energy of the cue is converted into the final kinetic energy of the cue, the kinetic energy of the cue ball, and the rotational kinetic energy of the cue ball): $$\frac{1}{2}m_s(v_{s_i})^2 = \frac{1}{2}m_s(v_{s_f})^2 + \frac{1}{2}m_b(v_b)^2 + \frac{1}{2}(\frac{2}{5}m_bR^2)(\omega_x)^2 + \frac{1}{2}(\frac{2}{5}m_bR^2)(\omega_z)^2 \tag{2}\label{2}$$

Since the linear impulse between the tip and the ball is equal to the change in momentum of both the stick and the ball: $$Ft = m_s(v_{s_i} - v_{s_f}) = m_bv_b \tag{3}\label{3}$$

Since the angular impulse is equal to the change in angular momentum: $$bFt = (\frac{2}{5}m_bR^2)(\omega_x) \tag{4}\label{4}$$ $$aFt = (\frac{2}{5}m_bR^2)(\omega_z) \tag{5}\label{5}$$

Then, using $$(1)$$, the final stick speed is: $$v_{s_i} - \frac{m_b}{m_s}v_b = v_{s_f} \tag{6}\label{6}$$

Then, substituting $$(3)$$ into $$(4)$$ and $$(3)$$ into $$(5)$$, $$bm_bv_b = (\frac{2}{5}m_bR^2)(\omega_x) \\ \frac{5}{2}\frac{v_b}{R^2}b = \omega_x \tag{7}\label{7}$$ and similarly:

$$\frac{5}{2}\frac{v_b}{R^2}a = \omega_z \tag{8}\label{8}$$

Substituting $$(6)$$, $$(7)$$, and $$(8)$$ into (2) yields: $$v_b = \frac{2v_s}{1 + \frac{m_b}{m_s} + \frac{5}{2}(\frac{a+b}{R})^2} \tag{9}\label{9}$$

Since $$v_b\sin(\theta) = v_x$$ and $$v_b\cos(\theta) = v_y$$, we have the velocity components $$\{v_b\sin(\theta), v_b\cos(\theta), 0\}$$ and the angular velocity components $$\{\frac{5}{2}\frac{v_b}{R^2}b, 0, \frac{5}{2}\frac{v_b}{R^2}a\}$$, where $$v_b$$ is obtained from equation $$(9)$$.

EDIT:

Dug a little deeper and found what I was looking for in a paper titled Pool Physics Simulation by Event Prediction I. In this article, the authors consider the cue angled to the horizontal $$xy$$ plane. Assuming the time duration of the collision between the cue stick and the ball is small, the force exerted on the ball can be "treated as a perfectly elastic impulse" and thus the velocity of the ball can be expressed as: $$v= (0, -\frac{F}{m}\cos(\theta), -\frac{F}{m}\sin(\theta))$$

The authors claim that "the magnitude of the force $$F$$ in terms of the impact parameters is derived simultaneously solving the equations for the conservation of linear momentum and conservation of energy before and after the cue impact to obtain:" $$F=\frac{2mV_0}{1+\frac{m}{M}+\frac{5}{2R^2}(a^2 + b^2\cos^2(\theta) + c^2\sin^2(\theta) -2bc\cos(\theta)\sin(\theta))}$$

where $$m$$ is the ball mass and $$M$$ is the cue mass and $$c = |\sqrt{R^2 - a^2 - b^2}|$$.

How did the authors derive this equation for force?

Furthermore, this pool simulation claims that the numerator should be $$2MV_0$$. Which one is correct?

• Are you assuming the cueball can slip on the surface or does it have to roll? This will change the effect of the force.
– LPZ
Commented Jun 6, 2022 at 9:45
• @lpz Thanks for the reply! May I ask why this assumption changes the effect of the force? The authors say that "when a ball is struck by the cue, it begins its motion by sliding across the surface of the table; after some time, the interplay between the table friction and the ball's linear and angular velocities cause the ball to begin rolling." Commented Jun 6, 2022 at 9:53
• It's just that friction changes the modelling. The collision will impart a given angular momentum and linear momentum which will result in a certain angular velocity and linear velocity. However, these two speeds may not be compatible with the no slip condition in general, hence the question of friction which will enforce the compatibility after dissipation. In particular in the infinite friction scenario, ie only rolling, this will impose a relation between the imparted impulse and its point of application. I guess you must have assumed no friction, decoupling the two motions
– LPZ
Commented Jun 6, 2022 at 15:35
• For example, a classic exercise is to calculate the height of the edge of a pool table so that the balls bounce off it without slipping after collision.
– LPZ
Commented Jun 6, 2022 at 15:38

How did the authors derive this equation for force?

with those equations you can obtain the "force" F

\begin{align*} &M\,(\vec{v}_c-\vec v_0)=-F\,\hat e_F\\ &m\,\vec{v}_s=F\,\hat e_F\\ &I\,\vec{\omega}=\vec{r}\times\,\left(F\,\hat e_F\right) \end{align*} and the kinetic energy \begin{align*} &2\,T=M\,\left(\vec{v}_c\cdot\vec{v}_c-\vec{v}_0\cdot\vec{v}_0\right)+m\,\vec{v}_s \cdot\vec{v}_s+I\,\vec{\omega}\cdot\vec{\omega}=0 \end{align*}

where \begin{align*} &\hat{e}_F=-\begin{bmatrix} 0 \\ \cos(\theta) \\ \sin(\theta)\\ \end{bmatrix}\quad , \vec{r}= \begin{bmatrix} a \\ b \\ c \\ \end{bmatrix}\quad , \vec{v}_0= v_0\,\hat{e}_F \end{align*}

you have ten scalar equations for the ten unknowns

\begin{align*} &\vec{v}_c\quad ,\vec{v}_s\quad ,\vec{\omega}\quad ,F\\ \end{align*}

$$\Rightarrow$$

\begin{align*} &F=\frac{2\,m\,v_0}{1+\frac{m}{M}+\frac{m}{I}\,\left(a^2+b^2\,\sin^2(\theta)+ c^2\,\cos^2(\theta)-2\,b\,c\sin(\theta)\,\cos(\theta)\right)}\\ &I=\frac{2}{5}\,m\,R^2 \end{align*}

Notice

that the linear momentum $$~p=M\,\left(\vec{v}_c-\vec{v}_0\right)+m\,\vec{v}_s~$$ is equal zero (conserved ) and the force is $$~\int F\,dt~$$

• $$\vec v_0~$$ initial velocity cue
• $$\vec v_c~$$ velocity cue after collision
• $$\vec v_s~$$ velocity ball after collision
• $$\vec\omega~$$ angular velocity ball after collision
• $$~m~$$ ball mass
• $$~M~$$ cue mass

The result in the document that you gave is:

$$F=\frac{2\,m\,v_0}{1+\frac{m}{M}+\frac{m}{I}\,\left(a^2+\color{red}{b^2\,\cos^2(\theta)+ c^2\,\sin^2(\theta)}-2\,b\,c\sin(\theta)\,\cos(\theta)\right)}$$

you obtain this result with

\begin{align*} &\hat{e}_F=-\begin{bmatrix} 0 \\ \sin(\theta) \\ \cos(\theta)\\ \end{bmatrix} \end{align*}