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Rolling & Angular Momentum

Source lecture(s): SC133 Lec 13

Intuition

Two grand finales of rotational mechanics. Rolling marries translation to rotation: a wheel that rolls without slipping is doing both at once, locked together by its rim. Angular momentum is rotation's conserved currency — the reason spinning tops stand up, skaters spin faster with arms pulled in, and planets sweep equal areas. When no external torque acts, \(L\) simply cannot change.

Rolling without slipping

The contact point is momentarily at rest on the ground; the geometry locks

\[v_\text{cm} = \omega R, \qquad a_\text{cm} = \alpha R\]

Kinetic energy splits into two clean pieces:

\[K = \underbrace{\tfrac12 M v_\text{cm}^2}_{\text{translation}} + \underbrace{\tfrac12 I_\text{cm}\omega^2}_{\text{rotation}} = \tfrac12 M v_\text{cm}^2\left(1 + \frac{I_\text{cm}}{MR^2}\right)\]

The great downhill race. From energy conservation on a slope of height \(h\):

\[v = \sqrt{\frac{2gh}{1 + I_\text{cm}/MR^2}} \qquad a = \frac{g\sin\theta}{1 + I_\text{cm}/MR^2}\]

Mass and radius cancel; only the shape factor \(I/MR^2\) matters. Sphere (\(\tfrac25\)) beats disk (\(\tfrac12\)) beats hoop (\(1\)) — every time, and a frictionless sliding block beats them all (no energy diverted into spin). Static friction enables rolling but does no work: the contact point never moves.

Angular momentum

For a particle: \(\vec L = \vec r\times\vec p\). For a rigid body about its axis:

\[L = I\omega \qquad\qquad \boxed{\,\sum\vec\tau = \frac{d\vec L}{dt}\,}\]

Conservation: zero net external torque ⇒ \(\vec L\) constant, even while the body rearranges itself:

\[I_1\omega_1 = I_2\omega_2\]
  • Skater / diver / neutron star: pull mass inward, \(I\) drops, \(\omega\) soars. A collapsing stellar core shrinks \(r\) by \(\sim10^3\), spinning up from once per ~month to hundreds of times per second — pulsars are conservation of \(L\) made audible.
  • Kepler's second law is the same statement for orbits: gravity is central (zero torque about the Sun), so \(L = mvr_\perp\) is constant — equal areas in equal times.
  • Helicopter tail rotors exist because spinning up the main rotor would counter-spin the fuselage.

Worked example: skater's spin

A skater at \(\omega_1 = 2\,\text{rev/s}\) with arms out (\(I_1 = 4\,\text{kg·m}^2\)) pulls in to \(I_2 = 1.6\,\text{kg·m}^2\):

\[\omega_2 = \frac{I_1}{I_2}\omega_1 = 5\,\text{rev/s}\]

Kinetic energy rises from \(\tfrac12 I_1\omega_1^2\) to \(\tfrac{I_1}{I_2}\) times that (here ×2.5) — pulling your arms in against the centrifugal tendency is real work, and that work is where the extra energy comes from.

Common mistakes

  • "Friction does negative work on a rolling wheel." In pure rolling the contact point is instantaneously at rest — static friction does zero work; it merely converts between translation and rotation.
  • Using \(v = \omega R\) when slipping. The lock holds only for rolling without slipping (a skidding, braking wheel violates it).
  • Conserving \(\omega\) instead of \(L\). When \(I\) changes, it is \(I\omega\) that survives.
  • Conserving \(L\) despite external torque — check the pivot: a rod struck at its pivot feels hinge forces but zero torque about the hinge, which is why we conserve \(L\) about that specific point.

Knowledge graph position

Prerequisites: Rotation, Moment of inertia, Torque, Conservation of energy. Leads to: Equilibrium, Kepler's laws.

Quiz

Q1 (computational). A solid sphere rolls from rest down a 1.4 m-high slope. Speed at the bottom?

Answer

\(v = \sqrt{2gh/(1 + 2/5)} = \sqrt{2(9.8)(1.4)/1.4} = \sqrt{19.6} \approx 4.4\,\text{m/s}\) — versus \(5.2\,\text{m/s}\) for a frictionless slider; 2/7 of the energy is tied up in spin.

Q2 (conceptual). Why is it easy to balance on a moving bicycle and hard on a stationary one?

Answer

The spinning wheels carry angular momentum along the axle; tipping the bike demands a torque to reorient \(\vec L\), and the response (precession) steers the front wheel to counteract the fall. Gyroscopic stiffness + steering geometry do the balancing for you.

Q3 (multiple choice). A spinning cloud of gas collapses under gravity by a factor 10 in radius. Its rotation rate grows by roughly: (a) 10 (b) 100 (c) 1000

Answer

(b). \(I \propto r^2\), so \(\omega \propto 1/r^2\) — a factor 100. This is why everything in astronomy (stars, disks, galaxies) ends up spinning fast.