Rolling Without Slipping

For a disk or sphere of radius $R$ rolling on a flat surface, the no-slip condition requires that the distance traveled by the center of mass equals the arc length swept out by the rolling body. If the object rotates by angle $\theta$, the center moves a distance $s = R\theta$. Differentiating with respect to time gives the key kinematic constraint: $v_{cm} = R\omega$, where $v_{cm}$ is the speed of the center of mass and $\omega$ is the angular speed. A further differentiation gives $a_{cm} = R\alpha$, linking translational acceleration to angular acceleration $\alpha$.

Every point on a rolling body has a velocity that is the vector sum of the translational velocity of the center of mass and the rotational velocity of that point about the center. For a point at position $\vec{r}$ from the center, its velocity is $\vec{v} = \vec{v}_{cm} + \vec{\omega} \times \vec{r}$. At the contact point (bottom of the wheel), the rotational contribution is $-v_{cm}$ (backward), exactly canceling $v_{cm}$ (forward), giving zero net velocity — confirming the no-slip condition. At the top of the wheel, the rotational contribution adds to the translational velocity, giving a speed of $2v_{cm}$. This point — called the instantaneous center of rotation — lies at the contact point itself.

The total kinetic energy of a body rolling without slipping is the sum of its translational and rotational kinetic energies: $$K_{total} = \frac{1}{2}mv_{cm}^2 + \frac{1}{2}I_{cm}\omega^2$$ Substituting $\omega = v_{cm}/R$, this becomes $$K_{total} = \frac{1}{2}mv_{cm}^2\left(1 + \frac{I_{cm}}{mR^2}\right)$$ The factor $\beta = I_{cm}/(mR^2)$ depends only on the geometry of the object: $\beta = 1/2$ for a solid cylinder, $\beta = 2/5$ for a solid sphere, and $\beta = 1$ for a thin-walled hollow cylinder. Objects with larger $\beta$ store more energy in rotation and therefore roll more slowly down an incline.

Consider a body of mass $m$, radius $R$, and moment of inertia $I_{cm} = \beta mR^2$ rolling without slipping down a ramp inclined at angle $\theta$. Using energy conservation (static friction does no work since the contact point has zero velocity): $$mgh = \frac{1}{2}mv_{cm}^2(1 + \beta)$$ Solving for speed at the bottom: $$v_{cm} = \sqrt{\frac{2gh}{1+\beta}}$$ Using Newton's second law along the incline and torque about the center of mass gives the acceleration: $$a_{cm} = \frac{g\sin\theta}{1+\beta}$$ A solid sphere ($\beta = 2/5$) accelerates as $a = \frac{5}{7}g\sin\theta$, faster than a solid cylinder ($a = \frac{2}{3}g\sin\theta$), which in turn beats a hollow cylinder ($a = \frac{1}{2}g\sin\theta$). Mass and radius cancel entirely — only the shape matters.

Rolling without slipping requires a static friction force, not kinetic friction. When an object rolls on a horizontal surface at constant speed, no friction is needed. However, on an incline, static friction acts at the contact point to provide the torque that produces angular acceleration. Using Newton's second law: $mg\sin\theta - f_s = ma_{cm}$, and torque equation: $f_s R = I_{cm}\alpha = I_{cm}(a_{cm}/R)$. Solving gives $f_s = \frac{\beta mg\sin\theta}{1+\beta}$. Rolling without slipping is possible only if $f_s \leq \mu_s mg\cos\theta$, i.e., if the incline is not too steep or the surface not too slippery.

• $v_{cm}$ — Speed of center of mass (m/s): The translational speed of the geometric center of the rolling object.
• $R$ — Radius (m): The radius of the rolling body, measured from the center to the contact surface.
• $\omega$ — Angular speed (rad/s): The rate of rotation of the rolling body about its central axis.

Problem. A solid sphere of mass $m = 2.0\ \text{kg}$ and radius $R = 0.10\ \text{m}$ starts from rest and rolls without slipping down a ramp of height $h = 0.80\ \text{m}$. (a) Find the speed of the center of mass at the bottom of the ramp. (b) Find the translational and rotational kinetic energies at the bottom. (The moment of inertia of a solid sphere is $I_{cm} = \frac{2}{5}mR^2$.)

Answer. The center of mass speed is approximately 3.35 m/s. Translational KE = 11.2 J, Rotational KE = 4.48 J.

• Moment of Inertia — The moment of inertia $I_{cm}$ determines how much energy is stored in rotation and controls the rolling speed down an incline via the factor $\beta = I_{cm}/mR^2$.
• Rotational Kinetic Energy — Rolling motion stores energy in both translation ($\frac{1}{2}mv_{cm}^2$) and rotation ($\frac{1}{2}I_{cm}\omega^2$); understanding rotational KE is essential for energy conservation in rolling problems.
• Torque and Angular Acceleration — Static friction provides the torque that gives the rolling body its angular acceleration; the equation $\tau = I_{cm}\alpha$ is used alongside Newton's second law to find the rolling acceleration.
• Static Friction — Rolling without slipping requires static friction at the contact point to prevent sliding; the maximum static friction determines whether rolling without slipping is possible on a given surface.
• Conservation of Energy — Since static friction does no work on a rolling body, mechanical energy is conserved, allowing gravitational PE to be converted directly into translational and rotational KE.