Modelling a backspin pool shot I'm trying to derive equations for $\theta(t)$ and $r(t)$ when modelling a backspin shot in pool.
I am of the understanding that the rotational deceleration is $\ddot \theta(t)=\dfrac{m\mu_sgR}{I}$ where $\mu_s$ is the coefficient of sliding friction.
However, the values that I have for the real world gives me a crazy deceleration.


*

*mass of whiteball $96$g = $0.096$kg

*radius of whiteball $\frac{15}{16}$in = $2.38$cm = $0.0238$m

*coefficient of sliding resistance = $0.3$

*$g = 9.8$m/s²

*Moment of inertia $I$ = $\frac{2}{5}mR^2$ = $0.0000218$ kg m²


Subbing all of this in gives $\ddot \theta(t)=\dfrac{0.096 \times 0.3 \times 9.8 \times 0.0238}{0.0000218}\approx 300$
Why am I getting an angular acceleration of $300$ rad / second?
 A: I think the critical part is to understand the way the rotation of the ball is connected to its translation, which isn't trivial (at least not for me).
Of course there might be a strange result at the end since your model assumes constant frictional force and rotational parameters can be hard to estimate.
From what you worked out (seems to be plausible regarding $L=I\cdot \dot{\theta}, ~ \dot{L}=M, ~ M=F \cdot r$ ) you can try to find the above mentioned connection between translation and rotation to estimate how this result affects the outcome, i.e. how long it takes the ball to reach the turning point. 
A simple approach is to think of the ball as a point mass for a moment and to use $$m\cdot a = F = m\cdot g\cdot \mu_s.$$
However, i don't think that this calculation is accurate, because the force acts on the ball by slowing down his forward motion and affecting his rotation. The latter seems to be somehow unregarded because this actually assumes the force to act on the center of mass, which implies no effect on it's rotation. Nevertheless you get
$$ v(t) = v_0 - a \cdot t = v_0 - m  g \mu_s \cdot t $$
$$ \dot{\theta}(t) = \omega_0 - \ddot{\theta}\cdot t = \omega_0 - \frac{m \mu_s g R}{I}\cdot t$$
By muliplying the last equation with $R$  and using $I = \frac25 m R^2$ you  obtain the rotational speed: $$v_r (t) =\omega_0 R - \frac{5 \mu_s g}{2} \cdot t$$
Maybe you can use these equations to estimate if your result is possible.
Good luck! 
Paul
