Let's say you hit a cue ball with a pool stick that causes an impulse. If the pool stick hits the ball above the ball's center of mass, is the direction of friction different than if the pool stick hits the ball below the center of mass? What determines the direction? Assume the ball is rolling with slipping.
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2 Answers
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I believe that the direction of the friction would actually be different depending where you hit below or above the center of mass. If you hit the ball below the centre the ball still rolls forward but at first the spin is in the opposite direction but changes direction due to the friction and overall forward inertia of the ball. The friction would have to be in the opposite direction of the spin because the direction of the spin would determine the angle at which the ball is interacting with surface. This is primarily from my experience with actually playing billiards. I'm from a completely mathematics background therefore somebody with a physics background could probably answer this better.
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This turned out to be more complicated than I expected, here is my attempt at making sense of the problem using some physics and I think that Ryan M has the right idea. The force of friction will always oppose motion since it dissipates energy, if the ball is slipping there will be friction which opposes this motion, and if it is rolling there will be friction which opposes that too.
To tackle this question,we have to take into account the effect of friction on the translational component ($\dot x$) and the rotational component ($\dot \theta$) of the billiard ball. The harder you hit the ball at the angle that causes the slipping, the higher the translational component (it would be purely rolling motion for a small initial $\dot \theta$). Indeed, $\dot x $ will not be independent of $\dot\theta$ , so $\dot x (\dot \theta)$ and the EOMs for these degrees of freedom are in fact coupled. A useful relation that will help me prove my conclusion (that the direction of friction against the rotational component changes with a reversal of the spin) is that as the ball spins faster, there will be a larger translational component in the motion (due to slipping), so $$\frac{\partial\dot x}{\partial\dot\theta} > 0$$
and the Lagrangian (the kinetic energy for this problem) is: $$\mathcal{L}(\theta,\dot\theta,x,\dot x)=\frac{1}{2}m(\dot x(\dot\theta))^2+\frac{I \dot \theta^2}{2}$$
the EOMs are given by: $$\frac{d}{dt}\frac{\partial\mathcal{L}}{\partial\dot q} - \frac{\partial\mathcal{L}}{\partial q} = -\frac{\partial\mathcal{R}}{\partial \dot q}$$
Where q is a degree of freedom of the system and the function $\mathcal{R}$ accounts for the dissipation of energy due to friction. When we take the derivative with respect to $\dot \theta$ we will have to use the chain rule on $\dot x (\dot \theta)$, but since I have assumed the slope of that function is positive for this scenario, as long as the friction opposes the translational motion (which it does) and as long as $\frac{\partial\mathcal{R}}{\partial \dot q}$ is a negative constant ($\mathcal{R}$ first order in $\dot x$ and $\dot \theta$), it will also oppose the rotational motion. This post might also help/give you a different perspective.
