Angular momentum and rolling
Angular momentum L = I omega survives anything except an external torque — and rolling ties spin to travel with v = omega R.
Angular momentum
A spinning body carries angular momentum L = I omega, the rotational twin of p = mv, and it changes only under a net external torque: tau = dL/dt. So whenever outside torques vanish or cancel — good bearings, no air, nothing pushing off-axis — L stays fixed no matter how the body reconfigures itself.
L = I\omega, \quad \tau = \frac{dL}{dt}Rolling without slipping
A wheel rolling without slipping pins its spin to its travel: the contact point is momentarily at rest, so each radian of rotation advances the axle exactly R meters — v = omega R. A roller therefore carries both translational energy (1/2)mv^2 and rotational energy (1/2)I omega^2: for a rolling disk a third of the total is spin, for a hoop half, which is why a hoop loses a downhill race to a disk.
v = \omega R, \quad KE = \tfrac{1}{2}mv^2 + \tfrac{1}{2}I\omega^2Worked example
A skater spins with arms out at I = 4.0 kg m^2 and omega = 2.0 rad/s, so L = 8.0 kg m^2/s. She pulls her arms in, dropping to I = 1.6 kg m^2; conservation forces omega = 8.0/1.6 = 5.0 rad/s — two and a half times faster. Her kinetic energy climbs from (1/2)(4.0)(2.0)^2 = 8.0 J to (1/2)(1.6)(5.0)^2 = 20 J. No law is broken: pulling her arms in against the outward pull takes muscular work, and that work shows up as spin.
\omega_2 = \frac{(4.0)(2.0)}{1.6} = 5.0\ \text{rad/s}Flashcard
Angular momentum of a spinning body
Check your understanding
A diver leaves the board rotating at 2.0 rad/s with moment of inertia 12 kg·m^2 about her spin axis. She tucks, reducing her moment of inertia to 4.0 kg·m^2. What is her new angular speed?