§10.5 — Rolling Motion

Billiard balls on a table, bicycles on a street, car wheels — they all roll. Rolling is a perfect blend:

• Linear motion: center of mass moves forward
• Rotational motion: object spins around itself

If an object rolls without slipping (what we call pure rolling), there's a simple geometric link between the two motions.

No-Slip Condition: Linear-Angular Relation

If a wheel rolls without slipping:

• The contact point with the ground is instantaneously at rest
• The center moves a distance equal to the arc length traveled

The relationship:

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

where:

• \( v_{\text{cm}} \) = velocity of center of mass
• \( R \) = radius
• \( \omega \) = angular velocity

Geometric Proof

If a wheel completes one full rotation (\( \theta = 2\pi \) rad):

• Contact point travels the circumference (\( 2\pi R \))
• Center of mass also travels circumference (\( 2\pi R \))

\[ s = R\theta \implies v_{\text{cm}} = R\omega \]

Example

A basketball (radius 12 cm) rolls such that its center moves at 2 m/s. What is its angular velocity?

\[ \omega = \frac{v_{\text{cm}}}{R} = \frac{2}{0.12} \approx 16.7 \text{ rad/s} \]

Revolutions per second:

\[ f = \frac{\omega}{2\pi} = \frac{16.7}{6.28} \approx 2.7 \text{ Hz} \]

The ball spins roughly 2.7 times per second!

Energy in Rolling

A rolling object has two types of kinetic energy:

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

Substitute \( \omega = v_{\text{cm}}/R \):

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

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

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

Effective Rolling Mass

For convenience, define:

\[ m_{\text{eff}} = m + \frac{I}{R^2} \]

Then:

\[ K_{\text{total}} = \frac{1}{2}m_{\text{eff}} v_{\text{cm}}^2 \]

The rolling object acts like a linear object with greater mass!

Numerical Example

Solid cylinder: mass 3 kg, radius 0.2 m:

• \( I = \frac{1}{2}MR^2 = \frac{1}{2} \times 3 \times (0.2)^2 = 0.06 \) kg·m²
• \( m_{\text{eff}} = 3 + \frac{0.06}{(0.2)^2} = 3 + 1.5 = 4.5 \) kg

The rolling object is 1.5 times heavier (in energy terms).

If \( v_{\text{cm}} = 4 \) m/s:

\[ K_{\text{total}} = \frac{1}{2} \times 4.5 \times 16 = 36 \text{ J} \]

Breakdown:

• \( K_{\text{trans}} = \frac{1}{2} \times 3 \times 16 = 24 \) J
• \( K_{\text{rot}} = \frac{1}{2} \times 0.06 \times (20)^2 = 12 \) J
• Total = 36 J ✓

Incline Problems

Case 1: Sliding Down (No Friction)

An object sliding down a frictionless incline:

\[ a = g\sin\theta \]

(Just ordinary incline kinematics)

Case 2: Rolling Down (No Slip)

If an object rolls without slipping:

Linear (Newton II): \[ mg\sin\theta - f = ma_{\text{cm}} \]

Rotational (Newton II): \[ fR = I\alpha \]

No-slip condition: \[ a_{\text{cm}} = R\alpha \]

Solving for \( a_{\text{cm}} \):

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

Comparison

Different objects on the same incline (angle \( \theta \)):

Shape	\( I \)	\( a_{\text{cm}} \)
Sliding (no friction)	—	\( g\sin\theta \approx 0.5g \)
Sphere	\( \frac{2}{5}MR^2 \)	\( \frac{5g\sin\theta}{7} \approx 0.36g \)
Cylinder	\( \frac{1}{2}MR^2 \)	\( \frac{2g\sin\theta}{3} \approx 0.33g \)
Hoop	\( MR^2 \)	\( \frac{g\sin\theta}{2} \approx 0.25g \)

If a sphere and hoop start at the top together, the sphere reaches the bottom first! It wastes less energy on rotation.

Braking and Friction

When rolling friction (like brake pads) stops a wheel:

• Linear friction slows translational motion
• Rotational friction reduces spin

If friction is too strong, the wheel can't simultaneously slow both motions — slipping begins.

Example: Billiard ball hit with backspin:

• Ball moves forward
• But spins backward (reverse spin)
• Table friction reduces this slip until motion stops
• When reverse spin reaches zero, ball enters pure rolling
• If friction persists, ball stops completely

Achieving Pure Rolling

In practice, for a ball to start rolling without slipping:

1. Start with slip: ball has velocity but little spin
2. Ground friction: accelerates it forward and adds spin
3. Equilibrium: when \( v = R\omega \) is satisfied, pure rolling begins

This friction is necessary to eliminate slip!

What You Should Know

• No-slip condition: \( v_{\text{cm}} = R\omega \)
• Total energy: \( K = \frac{1}{2}m_{\text{eff}} v_{\text{cm}}^2 \)
• Incline acceleration: \( a_{\text{cm}} = \frac{g\sin\theta}{1 + I/(mR^2)} \)
• Sphere is fastest (least rotational energy)
• Friction is necessary to prevent slipping

Preview of §10.6

Now that we understand how objects roll, a deeper question: what happens when a spinning object gets unexpected rotation (when its axis isn't fixed)? How does a gyroscope maintain balance? This is gyroscopic motion and precession.

📚 See also: Halliday Vol 1, Ch 10, §10.5 — Rolling motion. 🔗 Reference: §10.4 (Rotational energy) — to understand energy distribution.

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