Prerequisite materialTorque and the Equilibrium of Rigid BodiesCircular Motion and Centrifugal Force

Rotation of Rigid Bodies and Angular Momentum

A uniform solid sphere and a thin hoop with the same mass and radius are released simultaneously from rest on slopes with the same angle of inclination.

球が坂に沿って1 m進んだ瞬間、輪は何m進んでいるか?

The equations explain the phenomenon

  1. 1The rotational equation follows from the equations of motion of the particles.dLdt=⁠N……
  2. 2The component of a rigid body's angular momentum along its rotation axis is calculated.Lz=⁠I ω⁠,I = Σ mi ρi²……
  3. 3The rotational equation is compared with the equation of motion.I dωdt=⁠Nz……
  4. 4Rolling without slipping is described by 2 equations of motion.M a=⁠M g sinθ− f⁠,I α=⁠f R……
  5. The condition for rolling without slipping is that the contact-point velocity is 0 → the center-of-mass speed satisfies v = Rω → differentiation with respect to time gives a = Rα.
  6. 5The condition for rolling without slipping eliminates the frictional force from the equations.a=⁠g sinθ1 + I/MR²……
  7. 6The condition for rolling without slipping is checked.μs≧k tanθ1 + k……
  8. 7The distances traveled by the sphere and hoop are compared at the same time.xhoopxsphere=⁠ahoopasphere=⁠(1 + 25)/⁠(1 + 1)=⁠0.70……
  9. Angular momentum now explains other phenomena.
  10. 8Energy is used to compare the speeds after descending through the same height.M g h=⁠½ M v²+⁠½ I ω²=⁠½ M v² (1 + IMR²)……
  11. 9Conservation of angular momentum determines how fast a skater spins after pulling their arms inward.I₁ ω₁=⁠I₂ ω₂……
  12. 10Why a tilted top's axis rotatesΩ≒⁠M g d/⁠(I ω)……
  13. 11The result is reconsidered as a conservation law.N = 0⇒⁠L = constant……