加速および回転参照系
導入済み| English | 意味 |
|---|---|
| centrifugal force/ˌsentrɪˈfjuːɡl fɔːs/ | 遠心力 |
| Coriolis force/ˌkɒrɪˈəʊliz fɔːs/ | コリオリの力 |
A decision before an answer
- A person standing still on a turning platform needs a real inward force, although their platform-relative acceleration is zero.
- Your goal: Transform acceleration with translating and rotating frame terms.
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.
Ω points +z and v′ points +x. The Coriolis force points:
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
z×x=y; the minus sign in −2mΩ×v′ reverses it.
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.
A mass stays fixed at nonzero radius on a platform with constant rotation. Which statement is correct?
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
v′=0 removes Coriolis; the real centripetal force is inward even though a′=0.
Check cross-product directions
- Take Ω along +z, an anticlockwise platform viewed from above, and a particle at r along +x. Centrifugal force is along +x with magnitude mΩ²r. If the particle moves outward with v′ along +x, then Ω×v′ is +y, so Coriolis force 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.
For m=2 kg, Ω=2 rad/s along +z, r=1 m along +x and outward v′=3 m/s, centrifugal force is +8 N in x and Coriolis force is −24 N in y. Euler force is zero at constant Ω. If the real force is zero, a′=(4,−12) m/s² at this instant. A separate 10 kg passenger in an elevator accelerating upward at 2 m/s² has N=120 N with g=10.
A 3 kg passenger accelerates downward at 2 m/s² with g=10. The scale force is ____ N.
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
N−30=3(−2), so N=24.
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.
The relative velocity v′ belongs in the Coriolis term. Using the full inertial velocity counts rotation twice.
このケースに当てはまる答えはどれか？
並進および回転する座標系項を含む加速度の変換
Coriolis force can change relative velocity direction while doing zero instantaneous work on it.
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
The cross product is perpendicular to v′, so F_C·v′=0.
Keep the distinctions
- Coriolis force 科里奥利力 — The apparent rotating-frame force −2mΩ×v′ caused by relative motion.
- centrifugal force 离心力 — The apparent rotating-frame force −mΩ×(Ω×r), directed away from the rotation axis.
- Transform acceleration with translating and rotating frame terms.
- Determine centrifugal, Coriolis and Euler directions from cross products.
- Separate apparent forces from real interactions and test inertial limits.
Match each term with its precise meaning in this lesson.
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
Keep the distinctions stated in the teaching example.
Put this lesson’s reasoning or event sequence in order.
難易度 3/5 · 200 コイン + 今日初回完了時に小ボーナス付与
The order follows the stated process; check each stage before the next.