So far we've studied pure rotation (an object spins in place) or pure translation (an object moves in a straight line). But in reality, wheels and spheres spin and move simultaneously. This combination is called rolling motion.

Definition: Rolling Without Slipping

Rolling without slipping occurs when:

The No-Slip Condition

Consider a disk of radius \( R \) rolling without slipping. If:

Then:

\[ v_{cm} = \omega R \]

Proof: At the contact point, velocity must be zero:

\[ v_{\text{contact}} = v_{cm} - \omega R = 0 \implies v_{cm} = \omega R \]

Example: A ball of radius 0.1 m rolls without slipping on the ground. Its center moves at 2 m/s. What is its angular velocity?

\[ \omega = \frac{v_{cm}}{R} = \frac{2}{0.1} = 20 \text{ rad/s} \]

Velocity Components

At any instant, every point on a rolling object has two velocity components:

  1. Translational velocity: velocity of the center of mass (\( v_{cm} \))
  2. Rotational velocity: velocity relative to the center (\( \omega r' \), where \( r' \) is distance from axis)

Velocity Map

Different points on a disk:

Energy of Rolling Motion

In rolling, total kinetic energy has two parts:

\[ K_{\text{total}} = K_{\text{translational}} + K_{\text{rotational}} \]

\[ K_{\text{total}} = \frac{1}{2}m v_{cm}^2 + \frac{1}{2}I\omega^2 \]

Using the No-Slip Condition

From \( v_{cm} = \omega R \):

\[ \omega = \frac{v_{cm}}{R} \]

\[ K_{\text{total}} = \frac{1}{2}m v_{cm}^2 + \frac{1}{2}I\left(\frac{v_{cm}}{R}\right)^2 \]

\[ K_{\text{total}} = \frac{1}{2}m v_{cm}^2 + \frac{I}{2R^2} v_{cm}^2 \]

\[ K_{\text{total}} = \frac{1}{2}\left(m + \frac{I}{R^2}\right) v_{cm}^2 \]

Shape Factor

We can write this more compactly:

\[ K_{\text{total}} = \frac{1}{2}m v_{cm}^2 \left(1 + \frac{I}{mR^2}\right) \]

Note: The value \( \frac{I}{mR^2} \) depends on object shape:

Numerical Example: Three Rolling Objects

Three objects (solid cylinder, sphere, hoop) with mass m = 2 kg and radius R = 0.1 m roll without slipping. Each has center-of-mass velocity v_cm = 3 m/s.

Solid cylinder: \[ I = \frac{1}{2}mR^2 = \frac{1}{2} \times 2 \times (0.1)^2 = 0.01 \text{ kg·m}^2 \]

\[ K = \frac{1}{2} \times 2 \times 3^2 + \frac{1}{2} \times 0.01 \times \left(\frac{3}{0.1}\right)^2 = 9 + 4.5 = 13.5 \text{ J} \]

Sphere: \[ I = \frac{2}{5}mR^2 = \frac{2}{5} \times 2 \times (0.1)^2 = 0.008 \text{ kg·m}^2 \]

\[ K = \frac{1}{2} \times 2 \times 3^2 + \frac{1}{2} \times 0.008 \times 30^2 = 9 + 3.6 = 12.6 \text{ J} \]

Hoop: \[ I = mR^2 = 2 \times (0.1)^2 = 0.02 \text{ kg·m}^2 \]

\[ K = \frac{1}{2} \times 2 \times 3^2 + \frac{1}{2} \times 0.02 \times 30^2 = 9 + 9 = 18 \text{ J} \]

Result: Sphere has minimum energy (most efficient); hoop has maximum.

Rolling Down an Inclined Plane

When an object is released on an incline, two things happen:

  1. Translation: object moves down
  2. Rotation: object spins

Deriving the Acceleration

Using energy conservation or force equations:

\[ a_{cm} = \frac{g \sin\theta}{1 + \frac{I}{mR^2}} \]

Comparison: If the object slides without rolling (\( I = 0 \)): \[ a_{\text{sliding}} = g\sin\theta \]

Note: A rolling object accelerates slower because energy is shared between translation and rotation.

What You Should Know

Preview of §11.2

Now that we understand rolling motion, what causes it? The answer: torque. Let's examine the forces at work and see how they produce rotation.

📚 See also: Halliday Vol 1, Ch 11, §11.1 — Rolling without slipping. 🔗 Reference: §10.4 (Rotational kinetic energy) — which we used for total energy here.

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