9.3 Rolling Motion

Characteristics and Kinematics of Rolling Motion

Rolling motion combines spinning and moving forward at the same time. A wheel on a car, a bowling ball down a lane, a bicycle tire on pavement: all of these rotate while also traveling along a surface. The physics of rolling motion explains how these two types of movement connect, and it shows up constantly in problems involving energy, forces, and acceleration.

Characteristics of rolling motion

Pure rolling (also called "rolling without slipping") means the object rotates and translates simultaneously with no sliding between the object and the surface. Two things define it:

• The contact point between the object and the surface has zero instantaneous velocity relative to the surface. At any given moment, that bottom point acts as a temporary pivot.
• The rolling constraint ties the object's linear displacement directly to how far it has rotated: $x = R\theta$, where $R$ is the radius and $\theta$ is the angular displacement in radians. Every full rotation moves the center of mass forward by exactly one circumference ($2\pi R$).

If the object skids or slides at all, this constraint breaks and you're no longer dealing with pure rolling.

Linear vs. angular velocities

The center of mass moves forward with linear velocity $v$, while the object spins with angular velocity $\omega$ (measured in radians per second). For rolling without slipping, these are locked together:

$v = R\omega$

You can derive this by differentiating the rolling constraint with respect to time:

$v = \frac{dx}{dt} = R\frac{d\theta}{dt} = R\omega$

This relationship is worth memorizing. It's the single most-used equation in rolling motion problems, and it connects the translational world to the rotational world in one clean step.

Energy and Dynamics of Rolling Motion

Kinetic energy in rolling

A rolling object has both translational and rotational kinetic energy. You need to account for both:

$KE_{total} = KE_{trans} + KE_{rot} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$

Here $m$ is the object's mass and $I$ is its moment of inertia, which depends on how the mass is distributed. Common values you should know:

Since $\omega = \frac{v}{R}$ for rolling without slipping, you can substitute to write everything in terms of $v$:

$KE_{total} = \frac{1}{2}\left(m + \frac{I}{R^2}\right)v^2$

This is useful because it lets you solve energy problems without tracking $\omega$ separately. Notice that objects with more of their mass concentrated at the rim (larger $I$) store more energy in rotation. That's why a hoop rolls slower than a solid sphere released from the same height: more energy goes into spinning and less into translating.

Forces and equations for rolling

Three forces typically act on a rolling object: the normal force ($N$), gravity ($mg$), and friction ($f$).

A key detail: the friction involved in pure rolling is static friction, not kinetic. The contact point isn't sliding, so static friction is what prevents slipping. Its maximum value is $f_s \leq \mu_s N$, but in many problems the actual friction force is well below this limit.

For rolling on a surface, you apply Newton's second law in two forms simultaneously:

1. Translation: $F_{net} = ma$
2. Rotation: $\tau_{net} = I\alpha$

Friction provides the torque that makes the object spin: $\tau = Rf$. And just as $v = R\omega$, the accelerations are linked by:

$a = R\alpha$

To solve a typical rolling problem (say, a ball rolling down a ramp):

1. Draw a free-body diagram showing gravity, normal force, and friction.
2. Write $F_{net} = ma$ along the direction of motion.
3. Write $\tau_{net} = I\alpha$ about the center of mass.
4. Use $a = R\alpha$ to connect the two equations.
5. Solve for the unknowns (often $a$ and $f$).

Once you have the acceleration, the standard kinematic equations apply for both translation and rotation:

• $v = v_0 + at$ and $\omega = \omega_0 + \alpha t$
• $x = x_0 + v_0 t + \frac{1}{2}at^2$ and $\theta = \theta_0 + \omega_0 t + \frac{1}{2}\alpha t^2$

For problems involving height changes (like rolling down a hill), energy conservation is often faster than using forces:

$\Delta KE + \Delta PE = W_{non\text{-}conservative}$

Since static friction does no work in pure rolling (the contact point doesn't move), $W_{non\text{-}conservative} = 0$ in many cases, and you can set the loss in potential energy equal to the gain in total kinetic energy directly.