质点运动学：一个时钟，两条坐标轴
引入| English | 释义 | 拼音 |
|---|---|---|
| kinematics/ˌkɪnɪˈmætɪks/ | 运动学 | yùn dòng xué |
| particle/ˈpɑːtɪkl/ | 质点 | zhì diǎn |
| projectile motion/prəˈdʒektaɪl ˈməʊʃn/ | 抛体运动 | pāo tǐ yùn dòng |
| trajectory/trəˈdʒektəri/ | 轨迹 | guǐ jì |
| polar coordinates/ˈpəʊlə kəʊˈɔːdɪnəts/ | 极坐标 | jí zuò biāo |
| angular velocity/ˈæŋɡjʊlə vəˈlɒsɪti/ | 角速度 | jiǎo sù dù |
Reading the trace
- Before any force is known, certification needs position, velocity and acceleration from the motion itself. Kinematics 运动学 is reading the trace of a particle 质点 - a body whose size stops mattering.
- The drop test: released from 1.80 m, the till lands at $\sqrt{2gh} = 5.943$ m/s after 0.6058 s.
Dropped from 1.80 m from rest: landing speed (m/s, three decimals)?
难度 2/5 · 今日首次完成可获 140 金币及少量额外奖励
v = sqrt(2gh) = sqrt(2 × 9.81 × 1.80).
Same drop: flight time (s, three decimals)?
难度 2/5 · 今日首次完成可获 140 金币及少量额外奖励
t = sqrt(2h/g).
One clock, two axes
- Projectile motion 抛体运动: the till leaves a 0.95 m counter at 1.5 m/s horizontal, $a_x = 0$, $a_y = -g$, and both components share one clock $t$.
- The trajectory 轨迹 is a parabola; it lands 0.6601 m out at 4.570 m/s, 70.8° below horizontal - $v_y = gt = 4.317$ m/s at $t = 0.4401$ s.
*Displacement reads the endpoints; the components share one clock
Which description, which quantity?
Route each motion reading to the component that owns it.
Leaving a 0.95 m counter at 1.5 m/s horizontal: horizontal distance to landing (m, three decimals)?
难度 2/5 · 今日首次完成可获 140 金币及少量额外奖励
t = sqrt(2 × 0.95/9.81) = 0.4401 s; x = 1.5 × 0.4401.
Same slide-off: landing speed (m/s, three decimals)?
难度 2/5 · 今日首次完成可获 140 金币及少量额外奖励
v = sqrt(1.5^2 + 4.317^2).
The horizontal and vertical components of the slide-off:
难度 1/5 · 今日首次完成可获 100 金币及少量额外奖励
One t governs both axes; the vertical equation sets it.
Polar on the turntable
- On the turntable the reading is polar coordinates 极坐标: inward $a_r = \omega^2 r = 2.400$ and tangential $a_\theta = r\dot\omega = 1.800$ m/s².
- Total $\sqrt{2.400^2 + 1.800^2} = 3.000$ m/s² - a 3-4-5 triangle drawn by the physics at angular velocity 角速度 $\omega = 2.00$ and $\dot\omega = 3.00$ rad/s².
A point moving in a circle at constant speed has zero acceleration.
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Direction changes: a_r = omega squared r points inward.
Turntable r = 0.600 m at omega = 2.00 rad/s: inward acceleration (m/s², three decimals)?
难度 2/5 · 今日首次完成可获 140 金币及少量额外奖励
a_r = omega² r.
Same turntable speeding up at 3.00 rad/s²: total acceleration magnitude (m/s², three decimals)?
难度 2/5 · 今日首次完成可获 140 金币及少量额外奖励
sqrt(2.400² + 1.800²) - the 3-4-5 triangle.
Three kinematics slips
- Two clocks for one fall. There is one $t$; the horizontal ride ends exactly when the vertical one does.
- Revolutions inside $\omega^2 r$. Convert rev/s to rad/s first.
- Constant speed means no acceleration. False: on a curve, direction changes are acceleration too.
Recap
- Differentiate the trace; one clock for all components.
- Projectile: $a_y = -g$, $a_x = 0$; landing 4.570 m/s at 70.8°.
- Polar: $a_r = \omega^2 r$ inward, $a_\theta = r\dot\omega$; 2.400 + 1.800 = 3.000 m/s².