# Direction of friction for object that rolls with slipping

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.

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.