Rolling motion, sliding, contact point, friction

Question

I have a question about rolling motion. So suppose we throw a bowling ball with no initial angular velocity, only linear velocity $v_0$. At first it will be sliding on the ground. Let the kinetic friction between the bowling ball and the ground be $\mu_k > 0$. The friction force opposes the direction of linear motion which will act as a torque about the bowling ball's axis and increase its angular velocity. Now my confusion stems from how to handle the following cases:

Assume the ball is moving in the $xy$-plane only and to the right (positive $x$-direction). CM = center of mass

$(1)$ As long as the bowling ball is not in pure rolling motion, the contact point $P$ moves relative to the surface, right? Its velocity is the sum of the velocity of the center of mass and the velocity due to its rotation in the opposite direction. $$v_P = v_\mathrm{CM} - R\omega$$ Initially, $v_P = v_\mathrm{CM}$. To increase $P$'s angular velocity, there needs to be a torque, which should be the kinetic friction force. In effect, the contact point still moves more to the right (in the direction of the CM velocity, i.e. $v_P > 0$) and the net friction will be to the left, still, causing an increase in angular velocity. Does the friction only apply to the contact point, thus only increasing its linear velocity opposing the motion of the CM, or does it also act to decrease the bowling ball's linear velocity until the the rolling condition $$v_\mathrm{CM} = R \omega$$ is satisfied? Will there be energy lost or just converted form linear kinetic energy to rotational kinetic energy until they "match" (pure rolling)?

$(2)$ What exactly happens at the transition from sliding to rolling? Why is friction not doing any work at that point, given that we still have static friction? Mathematically, $v_P = v_\mathrm{Cm} - R\omega = 0$ for pure rolling, so that there's "no motion" of the contact point, which doesn't make sense intuitively to me, yet. Couldn't we still say that the contact point "rubs" the surface?

Answer 1

(1) In a horizontal surface, until reaching the situation of no slip, if we disregard air drag, the kinetic friction is the net force on the ball. So the COM decelerates following $F_f = ma$. This force also generates a torque: $\tau = RF_f = I\alpha$. While $F_f$ in general changes with time, at any instant $t$ during the slip deceleration period: $$m \frac{dv}{dt} = -\frac{I}{R}\frac{d \omega}{dt} \implies \frac{dv}{d\omega} = -\frac{I}{mR}$$ The minus sign is there because $v$ decreases as $\omega$ increases. In the case of a massive ball, with $I = \frac{2mR^2}{5}$: $$\frac{dv}{d\omega} = -\frac{2R}{5} \implies v = -\frac{2R}{5}\omega + v_0$$

When the ball is close to the situation of no slip, $\omega \to \frac{v}{R}$

$$v = -\frac{2v}{5} + v_0 \implies v = \frac{5}{7}v_0$$

I suppose that initially the ball has a velocity $v_0$ without rolling, so $\omega = 0 \implies v = v_0$.

(2) Note that rolling without slip is a pure inertial situation. Theoretically, once this situation is reached, friction is no longer necessary. If the ground is suddenly without friction, the ball keeps rotating and moving the same way.