Cơ học cổ điển
Đã giới thiệu| English | Ý nghĩa |
|---|---|
| small oscillations/smɔːl ˌɒsɪˈleɪʃnz/ | Motion close to a stable equilibrium, where the restoring force is nearly linear. |
| free-body diagram/friː ˈbɒdi ˈdaɪəɡræm/ | A sketch of one object showing every force acting on it. |
| generalised coordinates/ˈdʒenərəlaɪzd kəʊˈɔːdɪnəts/ | Any independent variables that fix a system’s configuration. |
Constant speed, still accelerating
- A planet on a circular orbit keeps the same speed, yet it accelerates: its velocity turns, so $a = v^2/r$ toward the centre.
- GRE mechanics tests vector forces, which conservation law applies, small oscillations 小振动 and the Lagrangian.
Newton's law is a vector law
- $\sum \mathbf F = m\mathbf a$ component by component. Draw the free-body diagram 受力分析图, choose axes along the constraint, then write components.
- Block on a frictionless incline of angle $\theta$: along the slope $a = g\sin\theta$; perpendicular, $N = mg\cos\theta$.
Which quantity is conserved?
- Momentum: zero net external impulse. Mechanical energy: no work by non-conservative forces.
- A collision can conserve momentum and lose kinetic energy.
2 kg at 3 m/s hits 1 kg at rest and sticks. Momentum: $6 = 3v$, so $v = 2$ m/s. Kinetic energy falls from 9 J to 6 J — 3 J goes to heat and deformation.
Momentum conserved does not mean kinetic energy conserved: check each law against its own condition.
What is conserved?
Decide which conservation law applies to a system.
A 3 kg cart at 4 m/s hits a 1 kg cart at rest; they stick. How much kinetic energy (J) is lost?
Độ khó 2/5 · 140 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
Momentum is conserved: $3 \times 4 = 4v$, so $v = 3$ m/s. Kinetic energy falls from $\tfrac12 \cdot 3 \cdot 4^2 = 24$ J to $\tfrac12 \cdot 4 \cdot 3^2 = 18$ J, so 6 J is lost.
Momentum is conserved in an inelastic collision even though kinetic energy is not.
Độ khó 1/5 · 100 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
True. With no external impulse, total momentum is always conserved; in an inelastic collision some kinetic energy becomes heat, sound or deformation.
Small oscillations
- Near a stable minimum of $U(x)$, $U \approx U_0 + \tfrac12 U''(x_0)(x - x_0)^2$, so $\omega = \sqrt{U''(x_0)/m}$.
- Spring: $\omega = \sqrt{k/m}$; simple pendulum: $\omega = \sqrt{g/\ell}$. Doubling the mass on a spring divides $\omega$ by $\sqrt2$, not 2.
A 1 kg particle moves in $U(x) = x^4 - 2x^2$ (SI units). Find $\omega$ for small oscillations about $x = 1$, to two decimal places.
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$U'' = 12x^2 - 4 = 8$; $\omega = \sqrt 8$.
The mass on a spring is quadrupled. The oscillation period
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The period is $T = 2\pi\sqrt{m/k}$, so it scales as $\sqrt{m}$: four times the mass doubles the period.
The Lagrangian
- $L = T - V$ in generalised coordinates 广义坐标; $\dfrac{d}{dt}\dfrac{\partial L}{\partial \dot q} = \dfrac{\partial L}{\partial q}$.
- Spring: $L = \tfrac12 m\dot x^2 - \tfrac12 kx^2$ gives $m\ddot x + kx = 0$. If $L$ does not contain $q$, then $\partial L/\partial \dot q$ is conserved.
If $L$ does not depend on $q$, the generalised ____ $\partial L/\partial\dot q$ is conserved.
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Euler–Lagrange gives d/dt of it equal to 0.
Recap
- Turning velocity means acceleration.
- Momentum needs no external impulse; energy needs no non-conservative work.
- $\omega = \sqrt{U''/m}$ near a minimum; Lagrange: $L = T - V$.
- small oscillations: Motion close to a stable equilibrium, where the restoring force is nearly linear.
- free-body diagram: A sketch of one object showing every force acting on it.
- generalised coordinates: Any independent variables that fix a system’s configuration.