# Conceptual rolling

If you could roll a ball down a slope with friction, we would all know which direction the ball is rolling. However, say we have a frictionless slope and we roll the ball in the opposite direction and place it on an incline. Would the ball roll down the slope rotating in the opposite direction? My intuition says that it would, because friction is the force that causes rotation, but it doesn't seem to make sense that a ball could roll the wrong way down a slope .

• Would the ball just be under slip condition? There is no force that is acting parallel to the center that causes it to rotate. Gravity acts along COM and hence no torque associated with it. I would assume that the ball would just slip and do not rotate Commented Jun 28, 2015 at 0:07
• I meant that you roll it first then submit it to the slip conditions down the slope Commented Jun 28, 2015 at 0:08
• If i understand correctly, you are trying to say that there is an initial angular acceleration associated with the ball? If that is the case, then the equation of motion will be coupled while writing Newton's EOM with $\alpha$ = $r.a$ where alpha is the angular acceleration. I think simple energy arguments will say that it will continue in the same condition forever. Commented Jun 28, 2015 at 0:12
• If you are assuming zero friction, then the rotation and translation movements of the ball are completely independent (and the word "roll" is inappropriate). The rotation will just continue with the same angular velocity as it had initially (which you did not specify, and could be anything), and the sliding up/down the slope will go on as usual (though faster than in the friction case, since there is no "inertia" associated with spinning up the ball). Commented Jun 28, 2015 at 5:01

The direction of the motion at any time $t$ is the direction of the velocity vector $\textbf{v}(t)$ as derived by solving the equations of motion; likewise $\omega(t)$ gives you back the direction of rotation according to the right hand rule.
is not entirely correct. Any force with non-zero torque generates rotation according to $\textbf{T}=\dot{\textbf{L}}$ and in the case at hand gravity may do according to the axes of inertia and other symmetries. I do not fully understand your scenario but anyway you should solve together $$\textbf{F}=\dot{\textbf{p}},\qquad \textbf{T}=\dot{\textbf{L}}$$ to derive $\textbf{v}(t)$ and $\omega(t)$ and this will give you the answer.