What is Perfect pivot? — Physics

Definition 1.11 University Physics — Year 2 · Chapter 1 — Rigid-Body Mechanics

A solid rotates about a fixed axis $\Delta$ in a perfect pivot (an ideal bearing) when the contact forces of the bearing have zero moment about $\Delta$ ; they then develop no power ( $\vect v_A = \vect 0$ on the axis, $\mathcal M_\Delta = 0$ ). A real bearing adds a friction torque $-\Gamma_f\operatorname{sgn}\omega$ , or a viscous one $-h\omega$ .

Left: the physical pendulum — the weight’s moment about the pivot drives the rotation, the pivot’s force has no moment. Right: the forces on a wheel rolling on a horizontal ground: weight, normal reaction N and tangential friction T at the contact point, whose magnitude is bounded by Coulomb’s law. — Left: the physical pendulum — the weight’s moment about the pivot drives the rotation, the pivot’s force has no moment. Right: the forces on a wheel rolling on a horizontal ground: weight, normal reaction $\vect N$ and tangential friction $\vect T$ at the contact point, whose magnitude is bounded by Coulomb’s law.

Examples

Example 1.13 (The metre rule)

A metre rule pivoted at one end: $J = \tfrac13M\ell^2$ (Huygens from $\tfrac1{12}M\ell^2$ ), $d = \ell/2$ , $\ell_{\text{eq}} = \tfrac23\ell = 0.667\,\mathrm{m}$ and $T = 1.64\,\mathrm{s}$ — a whole-rule pendulum beats slower than a point mass hung at its centre ( $1.42\,\mathrm{s}$ ) and faster than one at its end ( $2.0\,\mathrm{s}$ ).

Example 1.9 (A flywheel)

A steel disk of $50\,\mathrm{kg}$ and radius $30\,\mathrm{cm}$ : $J = \tfrac12 \times 50 \times 0.09 = 2.25\,\mathrm{kg}\,\mathrm{m}^{2}$ ; at $3000\,\mathrm{rpm}$ ( $\omega = 314\,\mathrm{rad}/\mathrm{s}$ ) it stores $\tfrac12J\omega^2 = 111\,\mathrm{kJ}$ , the kinetic energy of a small car at $50\,\mathrm{km}/\mathrm{h}$ . Energy-storage flywheels spin carbon-fibre rotors at $50\,000\,\mathrm{rpm}$ in vacuum on magnetic bearings: the energy grows as $\omega^2$ , the limit is the tensile strength of the rim.

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