Unit A.4: Rigid Body Mechanics HL only单元 A.4：刚体力学 HL only
The HL-only rotational super-topic of Theme A "Space, time and motion". Torque and couples, rotational equilibrium, moment of inertia, the angular suvat equations, Newton's second law for rotation, angular momentum and its conservation, rotational kinetic energy, and rolling without slipping. Every rotational quantity is the rotational analogue of a translational one you already met in A.1–A.3; mastering the analogy is most of the work.主题 A"空间、时间与运动"中仅 HL 的转动超主题。力矩与力偶、转动平衡、转动惯量、角量 suvat 方程、转动的牛顿第二定律、角动量及其守恒、转动动能，以及无滑滚动。每一个转动量都是 A.1–A.3 中已学平动量的转动类比；掌握这一类比就完成了大半。
How to use this guide本指南使用说明
A.4 is the "rotational mirror" of the linear mechanics in A.1–A.3. Almost every formula has a translational twin: force becomes torque, mass becomes moment of inertia, linear momentum becomes angular momentum. The marks come from (i) computing torque with the correct perpendicular distance, (ii) quoting the right moment of inertia from the data booklet, and (iii) keeping rotational and translational equations separate but linked by $a = \alpha r$. Learn the analogy table and the rest follows.A.4 是 A.1–A.3 平动力学的"转动镜像"。几乎每个公式都有平动孪生：力变为力矩，质量变为转动惯量，线动量变为角动量。分数来自：(i) 用正确的垂直距离算力矩；(ii) 从数据手册引用正确的转动惯量；(iii) 把转动方程与平动方程分开但用 $a = \alpha r$ 联系起来。背熟类比表，其余自然顺理成章。
Memorise the analogy table: $\tau = I\alpha$ (mirror of $F=ma$), $L = I\omega$ (mirror of $p=mv$), $E_k = \tfrac12 I\omega^2$ (mirror of $\tfrac12 mv^2$), and the angular suvat $\omega = \omega_0 + \alpha t$. Torque is $\tau = Fr\sin\theta$. For data-booklet $I$, recall: rod about centre $\tfrac{1}{12}ML^2$, solid disc $\tfrac12 MR^2$, solid sphere $\tfrac25 MR^2$.
背熟类比表：$\tau = I\alpha$（对应 $F=ma$）、$L = I\omega$（对应 $p=mv$）、$E_k = \tfrac12 I\omega^2$（对应 $\tfrac12 mv^2$），以及角量 suvat $\omega = \omega_0 + \alpha t$。力矩为 $\tau = Fr\sin\theta$。数据手册的 $I$ 要记住：细杆绕中心 $\tfrac{1}{12}ML^2$、实心圆盘 $\tfrac12 MR^2$、实心球 $\tfrac25 MR^2$。
Be able to derive $\tau = I\alpha$ from $\Sigma\tau = \Sigma m r^2 \alpha$, set up rotational equilibrium with the correct pivot choice, and solve coupled rolling problems where translational ($F = ma$) and rotational ($\tau = I\alpha$) equations are linked by the rolling constraint $a = \alpha r$. Explain angular-momentum conservation when no external torque acts, and split a rolling body's energy into $\tfrac12 mv^2 + \tfrac12 I\omega^2$.
能从 $\Sigma\tau = \Sigma m r^2 \alpha$ 推出 $\tau = I\alpha$，用正确的pivot（转轴）建立转动平衡，并求解滚动耦合问题——平动方程（$F = ma$）与转动方程（$\tau = I\alpha$）由滚动约束 $a = \alpha r$ 联系。能在无external torque（外力矩）时解释角动量守恒，并把滚动体能量拆为 $\tfrac12 mv^2 + \tfrac12 I\omega^2$。
Torque (Moment), Couples and Equilibrium力矩（转动效应）、力偶与平衡 A.4 HL
$\tau = Fr\sin\theta$, where $r$ is the distance from pivot to the point of application and $\theta$ the angle between $\vec{F}$ and $\vec{r}$.
$$ \tau = F r \sin\theta = F d, \qquad d = r\sin\theta \text{ (perpendicular / "moment arm")}. $$
Units $\mathrm{N\,m}$. Maximum torque when $\theta = 90^{\circ}$; zero when the force passes through the pivot.
Couple. Two equal, antiparallel forces separated by a perpendicular distance $d$. Net force is zero but net torque is $\tau = Fd$ — pure rotation.
Rotational equilibrium. A rigid body is in equilibrium when BOTH hold:
$$ \Sigma F = 0 \quad\text{and}\quad \Sigma \tau = 0. $$
Taking moments about any point gives the same answer; choose the pivot to eliminate an unknown force.
$\tau = Fr\sin\theta$，其中 $r$ 为转轴到作用点的距离，$\theta$ 为 $\vec{F}$ 与 $\vec{r}$ 间的夹角。
$$ \tau = F r \sin\theta = F d, \qquad d = r\sin\theta \text{（垂直距离 / "力臂"）}. $$
单位 $\mathrm{N\,m}$。$\theta = 90^{\circ}$ 时力矩最大；力的作用线过转轴时为零。
力偶。两个大小相等、方向相反、相距垂直距离 $d$ 的力。合力为零但合力矩为 $\tau = Fd$——纯转动。
转动平衡。刚体平衡时需同时满足：
$$ \Sigma F = 0 \quad\text{且}\quad \Sigma \tau = 0. $$
对任意点取矩结果相同；选转轴时让某个未知力消去。
A uniform plank of weight $200\ \mathrm{N}$ and length $4.0\ \mathrm{m}$ rests on two supports, one at the left end and one $1.0\ \mathrm{m}$ from the right end. A $300\ \mathrm{N}$ box sits $1.0\ \mathrm{m}$ from the left end. Find the force on each support.一块均匀木板，重 $200\ \mathrm{N}$、长 $4.0\ \mathrm{m}$，搁在两个支点上：一个在左端，另一个距右端 $1.0\ \mathrm{m}$。一个 $300\ \mathrm{N}$ 的箱子放在距左端 $1.0\ \mathrm{m}$ 处。求每个支点受力。
Identify. Static rigid body: use $\Sigma F = 0$ and $\Sigma\tau = 0$. Let the left support force be $R_L$ and the right support force $R_R$ (acting at $x = 3.0\ \mathrm{m}$ from the left). Weight acts at the centre, $x = 2.0\ \mathrm{m}$.
识别。静止刚体：用 $\Sigma F = 0$ 与 $\Sigma\tau = 0$。设左支点力为 $R_L$，右支点力为 $R_R$（作用在距左端 $x = 3.0\ \mathrm{m}$ 处）。重力作用在中心 $x = 2.0\ \mathrm{m}$。
Set up. Take moments about the left support (this eliminates $R_L$). Anticlockwise positive.
列式。对左支点取矩（消去 $R_L$）。取逆时针为正。
$$ R_R(3.0) - (300)(1.0) - (200)(2.0) = 0. $$Execute. $3.0\,R_R = 300 + 400 = 700 \Rightarrow R_R = 233\ \mathrm{N}$. Then $\Sigma F = 0$: $R_L + R_R = 300 + 200 = 500 \Rightarrow R_L = 500 - 233 = 267\ \mathrm{N}$.
求解。$3.0\,R_R = 300 + 400 = 700 \Rightarrow R_R = 233\ \mathrm{N}$。再由 $\Sigma F = 0$：$R_L + R_R = 500 \Rightarrow R_L = 267\ \mathrm{N}$。
Evaluate. The left support carries more because the box is nearer it. Check moments about the right support: $R_L(3.0) = (300)(2.0) + (200)(1.0) = 800 \Rightarrow R_L = 267\ \mathrm{N}$. Consistent.
评估。箱子离左支点更近，故左支点承重更多。对右支点验算：$R_L(3.0) = (300)(2.0) + (200)(1.0) = 800 \Rightarrow R_L = 267\ \mathrm{N}$。一致。
Going deeper: why the pivot choice is free深入：为何转轴可任选
If $\Sigma F = 0$, then the total torque is independent of the chosen reference point. Shifting the pivot by $\vec{c}$ changes each torque by $-\vec{c}\times\vec{F}_i$, and the sum of these changes is $-\vec{c}\times\Sigma\vec{F}_i = 0$. So a body in translational equilibrium has the same net torque about every point.
若 $\Sigma F = 0$，则总力矩与所选参考点无关。把转轴平移 $\vec{c}$ 会使每个力矩改变 $-\vec{c}\times\vec{F}_i$，这些改变之和为 $-\vec{c}\times\Sigma\vec{F}_i = 0$。故平动平衡的物体对每一点的合力矩都相同。
Practical upshot: choose the pivot at the line of action of the force you do not know, so its moment is zero and it drops out of your equation. This is the single biggest time-saver in statics questions.
实用结论：把转轴选在你不知道的那个力的作用线上，使其力矩为零、从方程中消去。这是静力学题中最省时的技巧。
Moment of Inertia转动惯量 A.4 HL
$I = \sum m r^2$:
$$ I = \sum_i m_i r_i^2, \qquad \text{units } \mathrm{kg\,m^2}. $$
Here $r_i$ is each mass's perpendicular distance from the rotation axis. Mass far from the axis contributes much more ($r^2$).
Standard rigid bodies (data booklet). Quote, never re-derive in the exam:
- Thin rod about its centre: $I = \tfrac{1}{12} M L^2$.
- Thin rod about one end: $I = \tfrac{1}{3} M L^2$.
- Solid disc / cylinder about its axis: $I = \tfrac{1}{2} M R^2$.
- Solid sphere about a diameter: $I = \tfrac{2}{5} M R^2$.
- Hollow hoop / ring about its axis: $I = M R^2$.
$I = \sum m r^2$：
$$ I = \sum_i m_i r_i^2, \qquad \text{单位 } \mathrm{kg\,m^2}. $$
其中 $r_i$ 为每个质量到转轴的垂直距离。离轴越远贡献越大（$r^2$）。
标准刚体（数据手册）。考试中直接引用，切勿重新推导：
- 细杆绕中心：$I = \tfrac{1}{12} M L^2$。
- 细杆绕一端：$I = \tfrac{1}{3} M L^2$。
- 实心圆盘 / 圆柱绕轴：$I = \tfrac{1}{2} M R^2$。
- 实心球绕直径：$I = \tfrac{2}{5} M R^2$。
- 空心环绕轴：$I = M R^2$。
Two point masses, $2.0\ \mathrm{kg}$ and $3.0\ \mathrm{kg}$, are fixed at the ends of a light (massless) rod of length $1.2\ \mathrm{m}$. Find the moment of inertia about an axis through the $2.0\ \mathrm{kg}$ mass, perpendicular to the rod.两个质点 $2.0\ \mathrm{kg}$ 与 $3.0\ \mathrm{kg}$ 固定在一根长 $1.2\ \mathrm{m}$ 的轻质（无质量）杆两端。求绕过 $2.0\ \mathrm{kg}$ 质点且垂直于杆的轴的转动惯量。
Identify. Point masses, so use $I = \sum m r^2$. Measure each $r$ from the chosen axis.
识别。是质点，用 $I = \sum m r^2$。每个 $r$ 从所选轴量起。
Set up. The $2.0\ \mathrm{kg}$ mass sits on the axis, so $r_1 = 0$. The $3.0\ \mathrm{kg}$ mass is $1.2\ \mathrm{m}$ away, so $r_2 = 1.2\ \mathrm{m}$.
列式。$2.0\ \mathrm{kg}$ 质点在轴上，故 $r_1 = 0$。$3.0\ \mathrm{kg}$ 质点距 $1.2\ \mathrm{m}$，故 $r_2 = 1.2\ \mathrm{m}$。
$$ I = m_1 r_1^2 + m_2 r_2^2 = (2.0)(0)^2 + (3.0)(1.2)^2. $$Execute. $I = 0 + (3.0)(1.44) = 4.32 \approx 4.3\ \mathrm{kg\,m^2}$.
求解。$I = 0 + (3.0)(1.44) = 4.32 \approx 4.3\ \mathrm{kg\,m^2}$。
Evaluate. A mass on the axis contributes nothing. About the rod's centre of mass instead, $I$ would be smaller, confirming that $I$ depends on the axis, not just the body.
评估。轴上的质量贡献为零。若改绕杆的质心，$I$ 会更小，印证 $I$ 取决于转轴，而非仅取决于物体。
Going deeper: $\sum m r^2$ for a continuous rod becomes an integral深入：连续细杆的 $\sum m r^2$ 化为积分
For a continuous body the sum becomes $I = \int r^2 \, dm$. For a uniform rod of length $L$, mass $M$, linear density $\lambda = M/L$, rotated about its centre:
对连续物体，求和化为 $I = \int r^2 \, dm$。对长 $L$、质量 $M$、线密度 $\lambda = M/L$ 的均匀细杆绕中心：
$$ I = \int_{-L/2}^{L/2} x^2 \, \lambda \, dx = \lambda \left[\frac{x^3}{3}\right]_{-L/2}^{L/2} = \frac{M}{L}\cdot\frac{2}{3}\left(\frac{L}{2}\right)^3 = \frac{1}{12} M L^2. $$This reproduces the data-booklet value. The IB exam never asks you to do this integral, but seeing where $\tfrac{1}{12}$ comes from makes the quoted formulae memorable.
这正是数据手册的值。IB 考试不会要求做此积分，但理解 $\tfrac{1}{12}$ 的来源能帮助记住所引用的公式。
Rotational Kinematics — the Angular suvat转动运动学——角量 suvat A.4 HL
A wheel starts from rest and accelerates uniformly to $12\ \mathrm{rad\,s^{-1}}$ in $4.0\ \mathrm{s}$. Find the angular acceleration and the number of revolutions made.一个轮子从静止开始均匀加速，在 $4.0\ \mathrm{s}$ 内达到 $12\ \mathrm{rad\,s^{-1}}$。求角加速度与转过的圈数。
Identify. Constant $\alpha$, so use angular suvat. Known: $\omega_0 = 0$, $\omega = 12$, $t = 4.0$. Find $\alpha$ and $\theta$.
识别。恒定 $\alpha$，用角量 suvat。已知：$\omega_0 = 0$、$\omega = 12$、$t = 4.0$。求 $\alpha$ 与 $\theta$。
Set up. $\omega = \omega_0 + \alpha t$ gives $\alpha$; $\theta = \tfrac{1}{2}(\omega_0 + \omega)t$ gives the angle.
列式。$\omega = \omega_0 + \alpha t$ 求 $\alpha$；$\theta = \tfrac{1}{2}(\omega_0 + \omega)t$ 求角度。
$$ \alpha = \frac{\omega - \omega_0}{t} = \frac{12 - 0}{4.0} = 3.0\ \mathrm{rad\,s^{-2}}. $$ $$ \theta = \tfrac{1}{2}(0 + 12)(4.0) = 24\ \mathrm{rad}. $$Execute. Revolutions $= \theta / 2\pi = 24 / (2\pi) \approx 3.8\ \text{rev}$.
求解。圈数 $= \theta / 2\pi = 24 / (2\pi) \approx 3.8\ \text{圈}$。
Evaluate. $24\ \mathrm{rad}$ is a little under 4 full turns — sensible for a wheel reaching about 2 turns per second after 4 seconds.
评估。$24\ \mathrm{rad}$ 略少于 4 整圈——对一个 4 秒后约每秒 2 圈的轮子是合理的。
Going deeper: angular suvat is exactly the linear suvat in disguise深入：角量 suvat 实为线量 suvat 的化身
Because $\omega = d\theta/dt$ and $\alpha = d\omega/dt$ are the rotational copies of $v = ds/dt$ and $a = dv/dt$, every derivation from A.1 transfers symbol-for-symbol. Integrating $\alpha = $ const once gives $\omega = \omega_0 + \alpha t$; integrating again gives $\theta = \omega_0 t + \tfrac{1}{2}\alpha t^2$. Eliminating $t$ yields $\omega^2 = \omega_0^2 + 2\alpha\theta$.
由于 $\omega = d\theta/dt$、$\alpha = d\omega/dt$ 是 $v = ds/dt$、$a = dv/dt$ 的转动版本，A.1 的每个推导都可逐符号迁移。对常数 $\alpha$ 积一次得 $\omega = \omega_0 + \alpha t$；再积一次得 $\theta = \omega_0 t + \tfrac{1}{2}\alpha t^2$；消去 $t$ 得 $\omega^2 = \omega_0^2 + 2\alpha\theta$。
The only caveat: these hold only while $\alpha$ is constant. A torque that changes with angle (e.g. a spring-driven wheel) breaks them, just as variable acceleration breaks the linear suvat.
唯一须注意：这些式子仅在 $\alpha$ 恒定时成立。随角度变化的力矩（如弹簧驱动的轮）会使其失效，正如变加速度使线量 suvat 失效一样。
Newton's Second Law for Rotation: $\tau = I\alpha$转动的牛顿第二定律：$\tau = I\alpha$ A.4 HL
$\tau = I\alpha$: net torque equals moment of inertia times angular acceleration.
$$ \Sigma\tau = I\alpha. $$
This is the exact mirror of $\Sigma F = ma$: torque plays the role of force, $I$ the role of mass, $\alpha$ the role of acceleration.
The analogy table.
| Translation | Rotation |
|---|---|
| $x$ (displacement) | $\theta$ (angle) |
| $v$ (velocity) | $\omega$ (angular velocity) |
| $a$ (acceleration) | $\alpha$ (angular acceleration) |
| $m$ (mass) | $I$ (moment of inertia) |
| $F = ma$ | $\tau = I\alpha$ |
| $p = mv$ | $L = I\omega$ |
| $E_k = \tfrac12 mv^2$ | $E_k = \tfrac12 I\omega^2$ |
$\tau = I\alpha$：合力矩等于转动惯量乘以角加速度。
$$ \Sigma\tau = I\alpha. $$
这是 $\Sigma F = ma$ 的精确镜像：力矩对应力，$I$ 对应质量，$\alpha$ 对应加速度。
类比表。
| 平动 | 转动 |
|---|---|
| $x$（位移） | $\theta$（角度） |
| $v$（速度） | $\omega$（角速度） |
| $a$（加速度） | $\alpha$（角加速度） |
| $m$（质量） | $I$（转动惯量） |
| $F = ma$ | $\tau = I\alpha$ |
| $p = mv$ | $L = I\omega$ |
| $E_k = \tfrac12 mv^2$ | $E_k = \tfrac12 I\omega^2$ |
A solid disc of mass $4.0\ \mathrm{kg}$ and radius $0.30\ \mathrm{m}$ is free to rotate about its central axis. A constant tangential force of $6.0\ \mathrm{N}$ is applied at the rim. Find the angular acceleration.一个质量 $4.0\ \mathrm{kg}$、半径 $0.30\ \mathrm{m}$ 的实心圆盘可绕中心轴自由转动。在轮缘施加恒定切向力 $6.0\ \mathrm{N}$。求角加速度。
Identify. Rotational dynamics: $\tau = I\alpha$. Need $\tau$ from the force and $I$ from the data booklet.
识别。转动动力学：$\tau = I\alpha$。由力求 $\tau$，由数据手册求 $I$。
Set up. Tangential force at the rim ($\theta = 90^{\circ}$): $\tau = Fr = (6.0)(0.30) = 1.8\ \mathrm{N\,m}$. Solid disc: $I = \tfrac{1}{2}MR^2 = \tfrac{1}{2}(4.0)(0.30)^2 = 0.18\ \mathrm{kg\,m^2}$.
列式。轮缘切向力（$\theta = 90^{\circ}$）：$\tau = Fr = (6.0)(0.30) = 1.8\ \mathrm{N\,m}$。实心圆盘：$I = \tfrac{1}{2}MR^2 = \tfrac{1}{2}(4.0)(0.30)^2 = 0.18\ \mathrm{kg\,m^2}$。
$$ \alpha = \frac{\tau}{I} = \frac{1.8}{0.18} = 10\ \mathrm{rad\,s^{-2}}. $$Evaluate. Doubling the radius would change both $\tau$ (up) and $I$ (up by $R^2$), so $\alpha$ would actually fall — a useful sanity-check on rotational intuition.
评估。半径加倍会同时改变 $\tau$（增大）与 $I$（按 $R^2$ 增大），故 $\alpha$ 实际会下降——对转动直觉是有用的检验。
Going deeper: deriving $\tau = I\alpha$ from $F = ma$深入：由 $F = ma$ 推出 $\tau = I\alpha$
Treat the body as a set of particles. Particle $i$ at radius $r_i$ feels a tangential force $F_i = m_i a_{t,i} = m_i r_i \alpha$ (all share the same $\alpha$ in a rigid body). Its torque is $\tau_i = F_i r_i = m_i r_i^2 \alpha$. Summing:
把物体看作一组粒子。半径 $r_i$ 处的粒子 $i$ 受切向力 $F_i = m_i a_{t,i} = m_i r_i \alpha$（刚体中各粒子 $\alpha$ 相同）。其力矩为 $\tau_i = F_i r_i = m_i r_i^2 \alpha$。求和：
$$ \Sigma\tau = \left(\sum_i m_i r_i^2\right)\alpha = I\alpha. $$The bracket is exactly $I = \sum m r^2$. So moment of inertia is what naturally appears when you add up the torques needed to give every particle the same angular acceleration.
括号正是 $I = \sum m r^2$。故转动惯量正是把"使每个粒子获得相同角加速度所需的力矩"加起来时自然出现的量。
Angular Momentum and Its Conservation角动量及其守恒 A.4 HL
$L = I\omega$ — the rotational analogue of $p = mv$.
$$ L = I\omega, \qquad \text{units } \mathrm{kg\,m^2\,s^{-1}}. $$
Torque–angular-momentum link. $\Sigma\tau = dL/dt$ (mirror of $\Sigma F = dp/dt$).
Conservation. If no external torque acts, $L$ is conserved:
$$ I_1\omega_1 = I_2\omega_2. $$
Reducing $I$ (pulling mass inward) speeds up the spin; increasing $I$ slows it. This is the spinning-skater effect.
$L = I\omega$——$p = mv$ 的转动类比。
$$ L = I\omega, \qquad \text{单位 } \mathrm{kg\,m^2\,s^{-1}}. $$
力矩—角动量关系。$\Sigma\tau = dL/dt$（$\Sigma F = dp/dt$ 的镜像）。
守恒。无外力矩时，$L$ 守恒：
$$ I_1\omega_1 = I_2\omega_2. $$
减小 $I$（把质量收向轴）使转速加快；增大 $I$ 则减慢。这就是旋转滑冰者效应。
A skater spins at $2.0\ \mathrm{rad\,s^{-1}}$ with arms out, moment of inertia $4.5\ \mathrm{kg\,m^2}$. She pulls her arms in, reducing her moment of inertia to $1.5\ \mathrm{kg\,m^2}$. Find her new angular velocity.一名滑冰者张开双臂以 $2.0\ \mathrm{rad\,s^{-1}}$ 旋转，转动惯量 $4.5\ \mathrm{kg\,m^2}$。她收回双臂，使转动惯量减至 $1.5\ \mathrm{kg\,m^2}$。求新角速度。
Identify. No external torque about the spin axis (friction at the skate is negligible), so angular momentum is conserved: $I_1\omega_1 = I_2\omega_2$.
识别。绕旋转轴无外力矩（冰刀摩擦可忽略），故角动量守恒：$I_1\omega_1 = I_2\omega_2$。
Set up. $I_1 = 4.5$, $\omega_1 = 2.0$, $I_2 = 1.5$. Solve for $\omega_2$.
列式。$I_1 = 4.5$、$\omega_1 = 2.0$、$I_2 = 1.5$。求 $\omega_2$。
$$ \omega_2 = \frac{I_1\omega_1}{I_2} = \frac{(4.5)(2.0)}{1.5} = \frac{9.0}{1.5} = 6.0\ \mathrm{rad\,s^{-1}}. $$Evaluate. Cutting $I$ to a third triples $\omega$. Note the kinetic energy rises ($\tfrac12 I\omega^2$ goes from $9.0$ to $27\ \mathrm{J}$) — the skater's muscles do work pulling the arms in; energy is not conserved here, but angular momentum is.
评估。$I$ 减至三分之一使 $\omega$ 增至 3 倍。注意动能上升（$\tfrac12 I\omega^2$ 从 $9.0$ 增至 $27\ \mathrm{J}$）——滑冰者的肌肉在收臂时做功；此处能量不守恒，但角动量守恒。
Going deeper: where does the extra kinetic energy come from?深入：多出的动能从何而来？
Angular momentum is conserved because no external torque acts, but kinetic energy is NOT, because the skater does internal work. As she pulls her arms inward against the outward (centrifugal, in the rotating frame) tendency, her muscles transfer chemical energy into rotational kinetic energy. With $L = I\omega$ fixed, $E_k = \tfrac12 I\omega^2 = \tfrac{L^2}{2I}$, so halving $I$ doubles $E_k$.
角动量守恒是因为无外力矩，但动能不守恒，因为滑冰者做了内部功。她克服向外（在旋转参考系中为离心）趋势把手臂收向身体时，肌肉把化学能转化为转动动能。$L = I\omega$ 固定时，$E_k = \tfrac12 I\omega^2 = \tfrac{L^2}{2I}$，故 $I$ 减半使 $E_k$ 加倍。
Reverse the move (arms out) and the muscles do negative work: $E_k$ falls and the spin slows. The exam loves the contrast "$L$ conserved but $E_k$ not" — state it explicitly for the mark.
反向操作（张臂）时肌肉做负功：$E_k$ 下降、转速减慢。考试很喜欢"$L$ 守恒而 $E_k$ 不守恒"的对比——明确写出以得分。
Rotational Kinetic Energy and Rolling Without Slipping转动动能与无滑滚动 A.4 HL
$E_k = \tfrac12 I\omega^2$ — the rotational analogue of $\tfrac12 mv^2$.
$$ E_{k,\text{rot}} = \tfrac{1}{2} I\omega^2. $$
Rolling without slipping. The contact point is instantaneously at rest, which forces the constraint
$$ v = \omega R \qquad\text{and}\qquad a = \alpha R. $$
A rolling body carries BOTH translational and rotational KE:
$$ E_{k,\text{total}} = \tfrac{1}{2} m v^2 + \tfrac{1}{2} I\omega^2. $$
Down a slope. Energy conservation: $mgh = \tfrac12 mv^2 + \tfrac12 I\omega^2$. Bodies with smaller $I/mR^2$ (e.g. a solid sphere, $\tfrac25$) reach the bottom faster than those with larger (a hoop, $1$).
$E_k = \tfrac12 I\omega^2$——$\tfrac12 mv^2$ 的转动类比。
$$ E_{k,\text{rot}} = \tfrac{1}{2} I\omega^2. $$
无滑滚动。接触点瞬时静止，给出约束
$$ v = \omega R \qquad\text{且}\qquad a = \alpha R. $$
滚动体同时具有平动与转动动能：
$$ E_{k,\text{total}} = \tfrac{1}{2} m v^2 + \tfrac{1}{2} I\omega^2. $$
下斜坡。能量守恒：$mgh = \tfrac12 mv^2 + \tfrac12 I\omega^2$。$I/mR^2$ 越小（如实心球 $\tfrac25$）越快到底，越大（圆环 $1$）越慢。
A solid sphere is released from rest and rolls without slipping down a slope, descending a vertical height of $1.4\ \mathrm{m}$. Find its translational speed at the bottom (take $g = 9.81\ \mathrm{m\,s^{-2}}$).一个实心球从静止释放，无滑滚下斜坡，竖直下降 $1.4\ \mathrm{m}$。求其到达底部的平动速度（取 $g = 9.81\ \mathrm{m\,s^{-2}}$）。
Identify. Rolling without slipping, no slipping losses, so use energy conservation with both KE terms. Solid sphere: $I = \tfrac25 mR^2$.
识别。无滑滚动、无打滑损耗，故用含两项动能的能量守恒。实心球：$I = \tfrac25 mR^2$。
Set up. $mgh = \tfrac12 mv^2 + \tfrac12 I\omega^2$. Substitute $I = \tfrac25 mR^2$ and the constraint $\omega = v/R$:
列式。$mgh = \tfrac12 mv^2 + \tfrac12 I\omega^2$。代入 $I = \tfrac25 mR^2$ 与约束 $\omega = v/R$：
$$ mgh = \tfrac12 mv^2 + \tfrac12\left(\tfrac25 mR^2\right)\frac{v^2}{R^2} = \tfrac12 mv^2 + \tfrac15 mv^2 = \tfrac{7}{10} mv^2. $$Execute. $m$ cancels: $gh = \tfrac{7}{10}v^2 \Rightarrow v = \sqrt{\tfrac{10}{7}gh} = \sqrt{\tfrac{10}{7}(9.81)(1.4)} \approx 4.4\ \mathrm{m\,s^{-1}}$.
求解。$m$ 约去：$gh = \tfrac{7}{10}v^2 \Rightarrow v = \sqrt{\tfrac{10}{7}gh} = \sqrt{\tfrac{10}{7}(9.81)(1.4)} \approx 4.4\ \mathrm{m\,s^{-1}}$。
Evaluate. A sliding (frictionless) block would reach $v = \sqrt{2gh} \approx 5.2\ \mathrm{m\,s^{-1}}$. The sphere is slower because some energy goes into spin, not translation.
评估。滑动（无摩擦）滑块会达到 $v = \sqrt{2gh} \approx 5.2\ \mathrm{m\,s^{-1}}$。球较慢，因部分能量转入旋转而非平动。
Going deeper: why the mass and radius cancel深入：为何质量与半径都约去
Writing $I = k m R^2$ (with $k = \tfrac25$ for a sphere, $\tfrac12$ for a disc, $1$ for a hoop) and using $\omega = v/R$, energy conservation gives $mgh = \tfrac12 mv^2(1 + k)$. Both $m$ and $R$ cancel, leaving
写 $I = k m R^2$（球 $k = \tfrac25$、盘 $\tfrac12$、环 $1$），用 $\omega = v/R$，能量守恒给出 $mgh = \tfrac12 mv^2(1 + k)$。$m$ 与 $R$ 都约去，得
$$ v = \sqrt{\frac{2gh}{1 + k}}. $$So the final speed depends only on $k$ — the mass distribution — not on how big or heavy the body is. This is why a solid sphere ($k = \tfrac25$) always beats a hoop ($k = 1$) down the same slope, regardless of their masses or radii.
故末速度仅取决于 $k$——质量分布——与物体大小或轻重无关。这就是为何实心球（$k = \tfrac25$）在同一斜坡上总是胜过圆环（$k = 1$），无论它们的质量或半径如何。
Exam Strategy and Common Pitfalls考试策略与常见陷阱
- Whenever you are stuck, write the linear equation and swap each symbol. $F\!\to\!\tau$, $m\!\to\!I$, $a\!\to\!\alpha$, $v\!\to\!\omega$, $p\!\to\!L$. The rotational law is then immediate.
- 卡住时先写线量方程再逐符号替换。$F\!\to\!\tau$、$m\!\to\!I$、$a\!\to\!\alpha$、$v\!\to\!\omega$、$p\!\to\!L$。转动定律即得。
- Keep angles in radians. $s = r\theta$, $v = r\omega$, $a_t = r\alpha$ all require radians, not degrees.
- 角度用弧度。$s = r\theta$、$v = r\omega$、$a_t = r\alpha$ 都需弧度，不是度。
- $\tau = Fr\sin\theta$ uses the angle between force and arm. Mixing up $\sin$ and $\cos$ is the single most common torque error.
- $\tau = Fr\sin\theta$ 用力与力臂的夹角。混淆 $\sin$ 与 $\cos$ 是最常见的力矩错误。
- For equilibrium, choose the pivot on an unknown force's line of action. That force drops out, leaving one equation in one unknown.
- 求平衡时把转轴选在某未知力的作用线上。该力消去，只剩一个未知量的方程。
- Angular momentum is conserved when no external torque acts. State this explicitly before writing $I_1\omega_1 = I_2\omega_2$.
- 无外力矩时角动量守恒。写 $I_1\omega_1 = I_2\omega_2$ 前先明确说明这一点。
- Energy is NOT generally conserved when $I$ changes. A skater pulling arms in gains $E_k$; a clay-on-turntable collision loses it. Say which.
- $I$ 改变时能量通常不守恒。滑冰者收臂使 $E_k$ 增加；黏土落转盘的碰撞使其损失。要说明哪种。
- Always include both KE terms: $\tfrac12 mv^2 + \tfrac12 I\omega^2$. Then apply the constraint $v = \omega R$ to eliminate $\omega$.
- 务必包含两项动能 $\tfrac12 mv^2 + \tfrac12 I\omega^2$。再用约束 $v = \omega R$ 消去 $\omega$。
- Quote $I$ from the data booklet; never re-derive. State the axis explicitly so the marker sees you chose the right one.
- 从数据手册引用 $I$，切勿重新推导。明确写出转轴，让阅卷者看到你选对了。
Flashcards闪卡
Unit A.4 Practice Quiz单元 A.4 练习测验
Readiness Checklist备考清单
Tick each item when you can do it cold, without notes, on your first attempt.
每一条都要"裸做"做对（不看笔记、一次过）才打勾。
- Compute torque with $\tau = Fr\sin\theta$, using the correct perpendicular distance用 $\tau = Fr\sin\theta$ 并取正确垂直距离计算力矩
- Recognise a couple and state its torque $Fd$ with zero net force识别力偶并写出其力矩 $Fd$（合力为零）
- Apply $\Sigma F = 0$ and $\Sigma\tau = 0$, choosing a pivot to eliminate an unknown force用 $\Sigma F = 0$ 与 $\Sigma\tau = 0$，选转轴消去未知力
- Compute $I = \sum m r^2$ for point masses about a stated axis对给定轴用 $I = \sum m r^2$ 计算质点系的转动惯量
- Quote the data-booklet $I$ for a rod, disc, sphere or hoop, stating the axis引用数据手册中杆、盘、球、环的 $I$，并注明转轴
- Use the angular suvat equations for constant-$\alpha$ motion对恒定 $\alpha$ 运动使用角量 suvat 方程
- Convert between linear and angular quantities via $s = r\theta$, $v = r\omega$, $a = r\alpha$用 $s = r\theta$、$v = r\omega$、$a = r\alpha$ 在线量与角量间换算
- Apply $\tau = I\alpha$ and the full rotation-translation analogy table应用 $\tau = I\alpha$ 与完整的转动—平动类比表
- Use $L = I\omega$ and conserve $L$ when no external torque acts使用 $L = I\omega$，无外力矩时让 $L$ 守恒
- Explain the spinning-skater effect and why $E_k$ is not conserved while $L$ is解释旋转滑冰者效应以及为何 $L$ 守恒而 $E_k$ 不守恒
- Compute rotational kinetic energy $E_k = \tfrac12 I\omega^2$计算转动动能 $E_k = \tfrac12 I\omega^2$
- Solve a rolling-without-slipping problem with $\tfrac12 mv^2 + \tfrac12 I\omega^2$ and $v = \omega R$用 $\tfrac12 mv^2 + \tfrac12 I\omega^2$ 与 $v = \omega R$ 求解无滑滚动问题
IB Paper-Style PracticeIB 试卷风格练习
A.4 Practice and Solutions are on the roadmap, to ship under Practice Questions/Unit_A4_*.html with the bilingual built-in pattern.
A.4 配套 Practice 与 Solutions 在排期，上线后位于 Practice Questions/Unit_A4_*.html。