Rolling With Slipping

For a disk or sphere of radius $R$ rolling without slipping, the velocity of the center of mass $v_{cm}$ and the angular velocity $\omega$ must satisfy $v_{cm} = R\omega$. When this equality holds, the contact point has zero velocity relative to the ground, and static friction (up to its maximum value $f_s \leq \mu_s N$) provides any needed force. Rolling with slipping occurs when $v_{cm} \neq R\omega$. If $v_{cm} > R\omega$, the contact point slides forward (as with a skidding ball), and kinetic friction acts backward on the object. If $v_{cm} < R\omega$ (spin exceeds translation, as with a car wheel spinning on ice), the contact point moves backward relative to the ground, and kinetic friction acts forward.

Once slipping occurs, the friction force is kinetic: $f_k = \mu_k N = \mu_k mg$ (on a horizontal surface). This force acts at the contact point and governs both translational and rotational motion independently. For translation: $ma_{cm} = -f_k = -\mu_k mg$, giving $a_{cm} = -\mu_k g$ (deceleration if $v_{cm} > R\omega$). For rotation about the center of mass: $I\alpha = f_k R$, so $\alpha = \frac{\mu_k mg R}{I}$ (angular acceleration). For a solid sphere, $I = \frac{2}{5}mR^2$, so $\alpha = \frac{5\mu_k g}{2R}$. For a solid cylinder, $I = \frac{1}{2}mR^2$, so $\alpha = \frac{2\mu_k g}{R}$.

As time progresses, kinetic friction decelerates $v_{cm}$ and accelerates $\omega$ (or vice versa). The velocities evolve as: $v_{cm}(t) = v_0 - \mu_k g t$ and $\omega(t) = \omega_0 + \alpha t$. Pure rolling is restored at time $t^*$ when $v_{cm}(t^*) = R\omega(t^*)$: $v_0 - \mu_k g t^* = R\left(\omega_0 + \frac{\mu_k g R}{I/m} \cdot \frac{t^*}{R}\right)$. Solving for $t^*$ gives the moment slipping ends. After $t^*$, if no external torque acts, the object continues in pure rolling with constant velocity. Note: energy is lost to heat during slipping due to kinetic friction doing negative net work on the system.

During pure rolling, the total kinetic energy is $KE = \frac{1}{2}mv_{cm}^2 + \frac{1}{2}I\omega^2$. During slipping, energy is dissipated by friction at a rate $P = f_k \cdot v_{slip}$, where $v_{slip} = |v_{cm} - R\omega|$ is the slip speed. The energy lost to heat from time $0$ to $t^*$ is $\Delta E_{lost} = \int_0^{t^*} f_k v_{slip}\, dt$. This is always positive during slipping, meaning the mechanical energy of the system decreases until pure rolling is achieved. This is fundamentally different from pure rolling, where the contact-point friction does no work (since the contact point has zero velocity).

t^* \approx 1.17\text{ s}, \quad v_f \approx 5.71\text{ m/s}

• Rolling Without Slipping — The idealized case that rolling with slipping departs from; the condition $v_{cm} = R\omega$ defines the boundary between the two regimes.
• Static and Kinetic Friction — Rolling with slipping is triggered when the required friction exceeds the static maximum, switching the contact force to kinetic friction — the key physical transition in this phenomenon.
• Torque and Angular Acceleration — The rotational equation $\tau = I\alpha$ describes how the friction force at the contact point creates angular acceleration during slipping, driving the ball toward the pure rolling condition.
• Moment of Inertia — The time required to reach pure rolling depends directly on $I$; objects with larger moments of inertia (hollow cylinders vs. solid spheres) take longer to spin up and transition out of slipping.
• Energy Dissipation by Friction — Unlike pure rolling, slipping converts mechanical energy into thermal energy via kinetic friction work, making energy analysis essential for understanding the dynamics of the slip-to-roll transition.