Rolling Dynamics assumed rolling without slipping from the start. A bowling ball doesn't begin that way: it lands on the lane sliding, with almost no spin. Kinetic friction then works on it until it rolls.
Why it works
While the ball slides, kinetic friction $\mu_kMg$ acts backward at the contact point. It slows the center, $v = v_0 - \mu_kgt$, and its torque $\mu_kMgR$ spins the ball up: for a solid ball ($I = \tfrac{2}{5}MR^2$), $R\omega = \tfrac{5}{2}\mu_kgt$. Sliding stops when $v = R\omega$: $$v_0 - \mu_kgt = \tfrac{5}{2}\mu_kgt \;\Rightarrow\; t = \dfrac{2v_0}{7\mu_kg}, \quad v = \tfrac{5}{7}v_0.$$ A slicker route: friction acts at the contact point and gravity balances the normal force, so angular momentum about any point on the floor stays constant: $Mv_0R = MvR + I\omega$ gives the same $\tfrac{5}{7}v_0$ in one line, without $\mu_k$.
Key idea
An object launched sliding with no spin skids until $v = R\omega$, then rolls at $v = \dfrac{v_0}{1 + I/(MR^2)}$: $\tfrac{5}{7}v_0$ for a solid ball, $\tfrac{2}{3}v_0$ for a disk, $\tfrac{1}{2}v_0$ for a hoop. During the skid, kinetic friction turns kinetic energy into heat; once it rolls, that stops.
Worked example: A bowling ball
A ball lands sliding at $7\ \text{m/s}$ with no spin; $\mu_k = 0.20$. How long does it skid, how fast does it then roll, and how far does it skid?
- $t = \dfrac{2(7)}{7(0.20)(9.8)} = 1.02\ \text{s}$.
- $v = \tfrac{5}{7}(7) = 5.0\ \text{m/s}$.
- Distance at constant deceleration $\mu_kg = 1.96\ \text{m/s}^2$: $d = 7(1.02) - \tfrac{1}{2}(1.96)(1.02)^2 = 6.12\ \text{m}$.
Answer: $1.02\ \text{s}$; $5.0\ \text{m/s}$; $6.12\ \text{m}$.
Worked example: Energy lost to friction
What fraction of the ball's starting kinetic energy is lost while it skids?
- Start: $\tfrac{1}{2}Mv_0^2$. End: $\tfrac{1}{2}Mv^2(1 + \tfrac{2}{5}) = \tfrac{7}{10}M\left(\tfrac{5}{7}v_0\right)^2 = \tfrac{5}{14}Mv_0^2$.
- Lost: $\tfrac{1}{2} - \tfrac{5}{14} = \tfrac{2}{14} = \tfrac{1}{7}$ of $Mv_0^2$, which is $\tfrac{2}{7}$ of the starting energy.
Answer: $\tfrac{2}{7}$, about 29%.
Spin at launch changes the outcome, and the angular momentum shortcut still works. If the ball starts with angular velocity $\omega_0$ (positive in the rolling direction), then $Mv_0R + I\omega_0 = (MR^2 + I)\dfrac{v}{R}$ gives $v = \dfrac{v_0 + cR\omega_0}{1 + c}$ with $c = I/(MR^2)$. Backspin ($\omega_0 < 0$) makes the final speed smaller; enough topspin can make the ball speed up as friction grips.
Common mistake
Assuming a thrown bowling ball rolls without slipping from the moment it lands, so it keeps its full speed and $v = R\omega$ at once. A ball can't spin up without a torque, and the only torque comes from kinetic friction while it slides. Check: at landing $R\omega = 0$ but $v = v_0$, so $v \ne R\omega$: it must skid first, and the final speed is less than $v_0$.