What is Moment about a point; about an axis? — Physics

Definition 15.1 University Physics — Year 1 · 第 15 章 — Angular Momentum

The moment (or torque) about a point $O$ of a force $\vect F$ applied at $M$ is the vector

\vect{\mathcal M}_O(\vect F) = \vect{OM}\wedge\vect F \qquad (\mathrm{N}\,\mathrm{m}),

perpendicular to the plane of $\vect{OM}$ and $\vect F$ , of norm $OM\cdot F\sin\alpha = F\,d$ , where $d$ — the lever arm — is the distance from $O$ to the line of action of $\vect F$ . The moment about an oriented axis $\Delta$ (unit vector $\vect e_\Delta$ ) through $O$ is the scalar $\mathcal M_\Delta = \vect{\mathcal M}_O\cdot\vect e_\Delta$ , independent of the choice of $O$ on $\Delta$ ; it vanishes if $\vect F$ is parallel to $\Delta$ or meets it. Two opposite forces on different lines of action form a couple, whose moment $\vect F\wedge\vect{AB}$ is the same about every point.

The moment of F about O: the cross product OM F, of norm F\,d with d the distance from O to the line of action (the lever arm), directed perpendicular to the figure — here toward the reader. — The moment of $\vect F$ about $O$ : the cross product $\vect{OM}\wedge\vect F$ , of norm $F\,d$ with $d$ the distance from $O$ to the line of action (the lever arm), directed perpendicular to the figure — here toward the reader.

示例

Example 15.2 (The door)

A $20\,\mathrm{N}$ push perpendicular to a door at $0.80\,\mathrm{m}$ from the hinges has moment $16\,\mathrm{N}\,\mathrm{m}$ about the hinge axis; the same push at $0.10\,\mathrm{m}$ , $2\,\mathrm{N}\,\mathrm{m}$ ; along the door (line of action through the hinges), zero. The hinge’s own reaction, meeting the axis, has no moment about it: that is what makes the axis the natural place to take moments.

Example 15.6 (The pendulum by moments)

Simple pendulum, axis $\Delta$ through the pivot perpendicular to the plane of swing: $L_\Delta = m\ell^2\dot\theta$ ; the tension meets the axis (no moment), the weight’s moment is $-mg\ell\sin\theta$ (lever arm $\ell\sin\theta$ , restoring). The theorem gives $m\ell^2\ddot\theta = -mg\ell\sin\theta$ , the equation of Example 12.12 without ever writing the tension — the chief advantage of moments: forces through the axis disappear.

Example 15.10 (The skater)

Arms out, a skater’s body has $J \approx 4\,\mathrm{kg}\,\mathrm{m}^{2}$ plus two $3\,\mathrm{kg}$ arms at $0.8\,\mathrm{m}$ : $J_1 = 4 + 2 \times 3 \times 0.64 = 7.8\,\mathrm{kg}\,\mathrm{m}^{2}$ . Arms in ( $0.2\,\mathrm{m}$ ): $J_2 = 4.2\,\mathrm{kg}\,\mathrm{m}^{2}$ . The ice’s reaction and the weight have no moment about the vertical axis: $J\omega$ is conserved and the spin rate rises by $J_1/J_2 = 1.85$ — from $2$ to $3.7$ turns per second — while the kinetic energy rises by the same factor, paid by the muscles pulling the arms in.

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