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CM.5 · Accelerating and rotating reference frames

GRE · GRE Subject Test · GRE 物理 · 知识点 27

训练
27.1

加速与旋转参考系

A person standing still on a turning platform needs a real inward force, although their platform-relative acceleration is zero.

Prerequisites: 26.

  • Transform acceleration with translating and rotating frame 旋转参考系 terms
  • Determine centrifugal, Coriolis and Euler directions from cross products
  • Separate apparent forces from real interactions and test inertial limits
词汇 训练
English 中文 拼音
rotating frame 旋转参考系 xuán zhuǎn cān kǎo xì
27.2

Translate the frame

In a translating frame with origin acceleration A, Newton’s equation becomes m a′=F_real−mA. An upward-accelerating elevator therefore has N−mg=ma in the ground frame, or N−mg−ma=0 for its stationary passenger in the elevator frame. These are the same prediction. The additional term is an apparent force caused by the chosen accelerating coordinates; it is not a new contact with another body. Uniform translation with A=0 changes velocity but needs no apparent force.

27.3

Differentiate rotating axes

For a rotating basis, differentiating a vector adds Ω×that vector. Applying this twice gives a=a_origin+a′+2Ω×v′+Ω×(Ω×r)+Ωdot×r. Here r and v′ are measured relative to the moving origin in its rotating axes; all vectors in a calculation must be expressed in the same basis at the same instant. Move the last three rotational terms to the force side to obtain Coriolis −2mΩ×v′, centrifugal −mΩ×(Ω×r), and Euler −mΩdot×r. A constant rotation removes the Euler term, not the Coriolis term.

27.4

Check cross-product directions

Take Ω along +z, an anticlockwise platform viewed from above, and a particle at r along +x. Centrifugal force 离心力 is along +x with magnitude mΩ²r. If the particle moves outward with v′ along +x, then Ω×v′ is +y, so Coriolis force 科里奥利力 points −y. Reversing the relative velocity reverses Coriolis; keeping v′=0 makes it vanish. Coriolis is perpendicular to v′ and does no instantaneous work on that relative motion. Centrifugal force is outward from the rotation axis, not necessarily outward from the chosen origin in an arbitrary three-dimensional position.

词汇 训练
English 中文 拼音
Coriolis force/ˌkɒrɪˈəʊliz fɔːs/ 科里奥利力 kē lǐ ào lì lì
centrifugal force/ˌsentrɪˈfjuːɡl fɔːs/ 离心力 lí xīn lì
27.5

Balance real and apparent forces

A body at rest on the platform has a′=v′=0. With constant rotation and a fixed origin, its real force must be mΩ×(Ω×r), inward, cancelling the outward apparent term in the rotating equation. If rotation changes, a tangential real force must also balance the Euler term. Always check Ω→0 and A→0: ordinary inertial Newtonian motion must return. Do not mix the real inward centripetal requirement with an added outward real reaction on the same body; interaction partners belong in separate free-body diagrams.

27.6

Worked method

In a rotating frame, Coriolis force uses the relative velocity.

$$\mathbf F_C=-2m\boldsymbol\Omega\times\mathbf v',\qquad \mathbf F_{cf}=-m\boldsymbol\Omega\times(\boldsymbol\Omega\times\mathbf r).$$
For m = 2 kg, Omega = 2 rad/s along z, r = 1 m along x and v' = 3 m/s along x,
$$F_{cf}=m\Omega^2r=(2\,\mathrm{kg})(2\,\mathrm{rad/s})^2(1\,\mathrm m)=8\,\mathrm N\quad(+x).$$
$$F_C=2m\Omega v'=2(2\,\mathrm{kg})(2\,\mathrm{rad/s})(3\,\mathrm{m/s})=24\,\mathrm N\quad(-y).$$
Euler force is zero only because Omega is constant. These are coordinate terms, not extra physical contacts.

Accelerating and rotating reference frames: GRE original diagram
Accelerating and rotating reference frames: original GRE teaching diagram.
27.7

Check conditions and vocabulary

The relative velocity v′ belongs in the Coriolis term. Using the full inertial velocity counts rotation twice.

Coriolis force: The apparent rotating-frame force −2mΩ×v′ caused by relative motion.

centrifugal force: The apparent rotating-frame force −mΩ×(Ω×r), directed away from the rotation axis.

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