Rigid-body kinetics: the last rows of the file
도입됨| English | 의미 |
|---|---|
| mass moment of inertia | the spin analogue of mass, the integral of r squared dm - half m r squared for a disk |
| torque/tɔːk/ | the moment of a force about an axis, meeting inertia through sum tau = I alpha |
| angular acceleration/ˈæŋɡjʊlə əkˌseləˈreɪʃn/ | the rate of change of angular velocity, in radians per second squared |
| tipping | the toppling mode that starts when the resultant force crosses the base edge |
| angular momentum/ˈæŋɡjʊlə məʊˈmentəm/ | the spin storage I omega, conserved when no external torque acts |
Spin stores energy
- The flywheel: 8.0 kg disk at $r = 0.150$ m has $I = \frac{1}{2}mr^2 = 0.0900$ kg·m² - the mass moment of inertia 转动惯量 about its axle.
- Torque 转矩 4.5 N·m gives angular acceleration 角加速度 $\alpha = \tau/I = 50.0$ rad/s²: 225 rad (35.81 rev) to $\omega = 150$ rad/s in 3.00 s, storing 1012.5 J.
*The work tau·theta and the energy half-I-omega-squared agree to the joule
An 8.0 kg disk of radius 0.150 m about its axle: I (kg·m², four decimals)?
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Half m r².
With 4.5 N·m applied: alpha (rad/s², one decimal)?
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tau / I.
Energy stored at 150 rad/s (J, one decimal)?
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Half I omega².
The moment of inertia of a body is a single number, independent of axis choice.
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Axle and edge differ by m r² - always name the axis.
Measure I with a stopwatch
- Drive off: the flywheel coasts 150 → 0 rad/s in 40.0 s, so $\alpha = -3.75$ rad/s² and bearing friction is $\tau_f = 0.3375$ N·m.
- Cross-check: $\tau_f \theta = 0.3375 \times 3000 = 1012.5$ J - the same energy the spin-up stored.
Coasting 150 to 0 rad/s in 40.0 s: bearing friction torque (N·m, four decimals)?
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tau_f = I |alpha|.
Rolling and tipping
- A disk rolling down 15°: $a = \frac{2}{3}g\sin\beta = 1.693$ m/s², energy split 2:1 - 12.70 J carried, 6.35 J spun at $v = 2.253$ m/s over 1.50 m (total 19.04 J = $mgh$).
- The mount on a rough floor: slides at $\mu_s g = 3.434$, tipping 倾覆 at $gb/h = 6.540$ m/s² - it slides first, and no stiffening changes the order.
Which tool answers it?
Route each rigid-body question to its instrument.
A disk rolling down 15 degrees: acceleration (m/s², three decimals)?
난이도 2/5 · 140코인 + 오늘 첫 완료 시 작은 보너스 추가
Two thirds g sin beta.
That 5.0 kg disk after 1.50 m of rolling carries 19.04 J total. The split is:
난이도 1/5 · 100코인 + 오늘 첫 완료 시 작은 보너스 추가
Two thirds translation, one third rotation - the disk geometry taxes every ramp.
The mount, half-width 0.10 m and CG at 0.15 m: tipping threshold (m/s², three decimals)?
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g b/h.
Three rigid-body slips
- Quoting I without its axis. Axle and edge differ by $mr^2$.
- Full width in b/h. Half-width over centre-of-mass height.
- Rolling energy without the factor. Translation plus rotation: $(1 + I/mr^2)$.
Recap
- $\sum\tau = I\alpha$, disk $I = \frac{1}{2}mr^2$: 0.0900, 50.0 rad/s², 1012.5 J both ways.
- Rolling disk $a = \frac{2}{3}g\sin\beta = 1.693$, split 12.70 + 6.35 = 19.04 J.
- Angular momentum 角动量 $L = I\omega$ is conserved when no external torque acts.
- The mount slides at 3.434 before it tips at 6.540 m/s².