Rigid-body kinematics: constraint first
Введено| English | Значение |
|---|---|
| rigid body/ˈrɪdʒɪd ˈbɒdi/ | an idealised body keeping every pair of its points at a fixed distance |
| slider-crank mechanism | the crank, rod and slider that convert rotation into reciprocating motion |
| instant centre | the point of a body's plane motion that has zero velocity at this instant |
| rolling without slipping/ˈrəʊlɪŋ wɪˈðaʊt ˈslɪpɪŋ/ | the contact condition tying wheel to ground: v = omega r, contact point at rest |
Constraint before differentiation
- A rigid body 刚体 keeps every pair of its points at fixed distance. The shaker is driven by a slider-crank mechanism 曲柄滑块机构: crank $r = 50$ mm, rod $l = 200$ mm, $\omega = 30$ rad/s.
- Piston position $x = r\cos\theta + \sqrt{l^2 - r^2\sin^2\theta}$ - state the geometry, then differentiate.
*The geometry does the differentiating
Quarter-turn, exactly
- At $\theta = 90°$: $x = \sqrt{l^2 - r^2} = 0.1936$ m, $\dot{x} = -r\omega = -1.500$ m/s, $\ddot{x} = +r\omega^2\lambda = +11.25$ m/s².
- At dead centre the acceleration is $-56.25$ m/s² - a 5:1 range from the same crank, the number the fatigue file is written from.
Slider-crank r = 50 mm, omega = 30 rad/s: piston speed magnitude at quarter-turn (m/s, three decimals)?
Сложность 2/5 · 140 монет + небольшой бонус за первое завершение сегодня
v = r omega.
Same position with lambda = r/l = 0.25: table acceleration (m/s², two decimals)?
Сложность 2/5 · 140 монет + небольшой бонус за первое завершение сегодня
a = r omega² lambda.
The dead-centre acceleration extreme, with lambda = 0.25 (m/s², two decimals)?
Сложность 2/5 · 140 монет + небольшой бонус за первое завершение сегодня
r omega² (1 + lambda).
Piston position at quarter-turn (m, four decimals)?
Сложность 2/5 · 140 монет + небольшой бонус за первое завершение сегодня
sqrt(l² - r²).
At quarter-turn the connecting rod rotates while it translates.
Сложность 1/5 · 100 монет + небольшой бонус за первое завершение сегодня
Its IC is at infinity: pure translation, 1.500 m/s everywhere.
The instant centre
- At quarter-turn the rod translates: both ends at 1.500 m/s, its instant centre 瞬心 at infinity.
- The wheel rolls at 2.00 m/s with $r = 0.310$ m: $\omega = v/r = 6.452$ rad/s, contact at rest, top at 4.000 m/s - rolling without slipping 纯滚动 ties them.
Which point moves how?
Attach each velocity to its point on the rolling wheel.
A 0.310 m wheel rolling at 2.00 m/s: angular velocity (rad/s, three decimals)?
Сложность 2/5 · 140 монет + небольшой бонус за первое завершение сегодня
omega = v/r.
Same wheel: top point speed (m/s, three decimals)?
Сложность 2/5 · 140 монет + небольшой бонус за первое завершение сегодня
2v.
The instant centre of the rolling wheel is:
Сложность 1/5 · 100 монет + небольшой бонус за первое завершение сегодня
Zero-velocity point - it moves as the wheel rolls.
Three mechanism slips
- rev/s inside $\omega^2 r$. Radians first, always.
- Treating the IC as a pivot. It travels with the configuration.
- Forgetting the top point. Contact sleeps, centre rides, top flies at $2v$.
Recap
- Constraint geometry first, differentiation second.
- Quarter-turn: $v = r\omega = 1.500$ m/s, $a = r\omega^2\lambda = 11.25$ m/s², extremes 5:1.
- Rolling: $v = \omega r$, IC at contact, top at $2v = 4.000$ m/s.