拘束条件、変分方程式および循環座標
導入済み| English | 意味 |
|---|---|
| holonomic constraint/ˌhɒləˈnɒmɪk kənˈstreɪnt/ | 完全拘束 |
| cyclic coordinate/ˈsaɪklɪk kəʊˈɔːdɪnət/ | 循環座標 |
A decision before an answer
- A pendulum tension changes throughout a swing, but its angular equation can be derived without solving for tension.
- Your goal: Derive Euler–Lagrange motion after imposing a holonomic constraint.
Choose independent coordinates
- A constraint removes an independent coordinate before you form the kinetic energy. For a fixed-length pendulum choose θ from the downward vertical: x=l sinθ and y=−l cosθ. Differentiating both coordinates gives v²=l²θdot², not l² sin²θ θdot².
- Thus T=ml²θdot²/2 and V=mgl(1−cosθ), with zero potential at the bottom. The fixed length is a holonomic constraint: it is a relation among coordinates and possibly time. A rolling velocity constraint needs its own analysis; do not automatically treat every constraint as a coordinate substitution.
For a pendulum angle θ measured from downward vertical, which kinetic energy uses both Cartesian velocity components?
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
Adding xdot² and ydot² gives l²θdot².
Derive the equation
- For ideal constraints and the usual conservative system, the stationary-action equation is d/dt(∂L/∂qdot)−∂L/∂q=0 with L=T−V. The variations vanish at the two time endpoints. A pendulum gives ∂L/∂θdot=ml²θdot and ∂L/∂θ=−mgl sinθ, so ml²θddot+mgl sinθ=0.
- The momentum derivative acts on every time-dependent factor: for a varying radius, d(mr²θdot)/dt contains 2mr rdot θdot. The variational method does not mean mechanical energy is conserved in a time-dependent system.
In L=m(rdot²+r²θdot²)/2−V(r), which statement follows from the cyclic angle?
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
∂L/∂θ=0 gives d(mr²θdot)/dt=0; radius and angular speed may change.
Find cyclic momentum
- For planar central motion, L=m(rdot²+r²θdot²)/2−V(r). The angle θ is cyclic because L has no explicit θ dependence, even though L depends on θdot. Its canonical momentum pθ=mr²θdot is constant. The radial equation is m rddot=mrθdot²−dV/dr.
- The first term is part of the coordinate acceleration, not a new outward real force in an inertial frame. Conservation of pθ makes angular speed increase as r decreases; constant angular speed is a different, externally driven situation.
With l=2 m and g=10 m/s², θddot=−5 sinθ. At θ=0.10 rad the exact acceleration is −0.4992 rad/s², close to −0.500 from linearisation. The small-angle period is 2π/sqrt(5)=2.810 s. For a central-force orbit, shrinking r from 3 m to 1.5 m while pθ is conserved multiplies θdot by 4, not 2.
For V=9q²/2 and local inertia M=1, small-oscillation angular frequency is ____ rad/s.
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″=9 and sqrt(9/1)=3.
Linearise with conditions
- Near a stable equilibrium q0, expand V to second order and use a constant local inertia M: V≈V(q0)+V″(q0)(q−q0)²/2. Then the small-displacement frequency is sqrt(V″/M). For a pendulum sinθ≈θ in radians, so ω=sqrt(g/l); the approximation requires small amplitude.
- A negative V″ gives instability, not an oscillation with an imaginary measurable frequency. Check the full nonlinear equation first, then state the approximation and compare units: V″/M must have units of inverse time squared.
A cyclic coordinate is absent from L itself; its velocity need not be absent. Do not discard the time derivative of a changing r² factor.
このケースに当てはまる答えはどれか？
ホロノミカル拘束条件を適用した後のオイラー・ラグランジュ運動方程式を導出する
The small-angle pendulum approximation is unchanged if the angle is inserted numerically in degrees.
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
sinθ≈θ requires radians; a degree value needs conversion.
Keep the distinctions
- holonomic constraint 完整约束 — A constraint expressible as a relation among coordinates and possibly time.
- cyclic coordinate 循环坐标 — A coordinate absent explicitly from the Lagrangian, with conserved canonical momentum under the Euler–Lagrange equation.
- Derive Euler–Lagrange motion after imposing a holonomic constraint.
- Use a cyclic coordinate to identify conserved canonical momentum.
- Linearise a stable equilibrium and check the resulting frequency.
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.