1. The problem statement, all variables and given/known data A solid ball of mass M and radius R and moment of inertia I is placed onto a table with an initial velocity v_0 to the right and angular momentum w_0 anticlockwise (i.e the ball has backspin). Due to friction, the angular velocity and linear velocity changes as the ball both rolls and slips along the table. Find the velocity v_0 (in terms of the other variables) such that the ball will roll some distance and then stops completely. 2. Relevant equations I. v=v_0+at, where a=-f/m II. w=w_0+(alpha)*t, where alpha=torque/I=-fR/I 3. The attempt at a solution Due to friction, the velocity and angular velocity (in the anticlockwise direction) should both decrease. The velocity is zero when t=mv_0/f. The angular velocity is zero when t=Iw_0/(fR). My confusion is the necessary condition for the ball to stop completely (meaning no linear motion, and no rotation). Naively, at some point in time, t, both the velocity and angular velocity are zero, so we should equate the two expressions above and solve for v_0=Iw_0/(mR). However I have an uneasy feeling about this, because I seem to get the same answer by assuming that the ball will stop when the no-slip condition is satisfied, i.e v=Rw. Using this equation we get v_0+at=R(w_0+(alpha)*t). Solving for t, we find t=(v_0-Rw_0)/(alpha*R-a)=(v_0-Rw_0)/(-f(R^2)/I+f/m). The velocity is zero when t=mv_0/f, so equating these two expressions, we have: (v_0-Rw_0)/(-f(R^2)/I+f/m)=mv_0/f (v_0-Rw_0)/(1/m-(R^2)/I)=mv_0 I(v_0-Rw_0)/(I-(R^2)m)=v_0 I(v_0-Rw_0)=v_0(I-(R^2)m) Iv_0-IRw_0=v_0I-v_0(R^2)m IRw_0=v_0(R^2)m v_0=Iw_0/(mR) So I get the same answer by both methods, which might seem encouraging, but I'm not sure about my reasoning for the conditions under which the ball stops completely.