Unit 5: Torque & Rotational Dynamics第 5 单元：力矩与转动动力学
Extend Newton's laws to rotating systems. Master rotational kinematics, torque, rotational inertia, equilibrium, and combined translation-rotation problems using calculus-based methods.
把 Newton 定律推广到转动系统。用微积分方法掌握转动运动学（rotational kinematics）、力矩（torque）、转动惯量（rotational inertia）、平衡，以及平动—转动耦合问题。
10–15% Exam Weight考试占分 10–15%~14–20 Class Periods约 14–20 课时6 Topics6 个专题
Topic 5.1专题 5.1
Rotational Kinematics转动运动学
If you have mastered one-dimensional kinematics from Unit 1, you already know the skeleton of rotational kinematics — only the variable names change. Instead of position $x$, velocity $v$, and acceleration $a$, you work with angular displacement $\theta$, angular velocity $\omega$, and angular acceleration $\alpha$. The mathematics is identical; the physics shifts from straight-line motion to rotation about a fixed axis.
Angular displacement measures the angle, in radians, through which a point on a rigid body rotates about a specified axis. A rigid system maintains its shape — different points can move in different directions during rotation, so it cannot be modeled as a single point particle.
Key Insight
If the overall rotation of a system about its own axis is negligible compared to another motion (e.g., Earth spinning on its axis versus orbiting the Sun), the system can be approximated as a single object for that analysis.
关键洞察
如果一个系统绕自身轴的转动相对其他运动可以忽略（例如地球绕自转轴的自转 vs. 绕太阳的公转），就可以在分析中把整个系统近似为一个质点。
Angular Velocity & Acceleration角速度与角加速度
Angular Velocity (Calculus Definition)角速度（微积分定义）
$$\omega = \frac{d\theta}{dt}$$
Angular velocity is the time derivative of angular position.
Angular acceleration is the second derivative of angular position.
角加速度是角位置对时间的二阶导数。
Constant Angular Acceleration Equations恒定角加速度方程组
When $\alpha$ is constant, three rotational kinematic equations mirror their linear counterparts exactly. Replace $x \to \theta$, $v \to \omega$, and $a \to \alpha$:
Graphs of $\theta(t)$, $\omega(t)$, and $\alpha(t)$ relate to each other just like graphs of $x(t)$, $v(t)$, and $a(t)$: the slope of a $\theta$-vs-$t$ graph gives $\omega$, the slope of an $\omega$-vs-$t$ graph gives $\alpha$, and the area under an $\omega$-vs-$t$ graph gives $\Delta\theta$.
Exam Note
The AP exam will test magnitudes of $\theta$, $\omega$, and $\alpha$ using vector conventions, but the directions of these vectors (e.g., via the right-hand rule) are not assessed. Directions are limited to "clockwise" or "counterclockwise" relative to a given axis.
应试提醒
AP 考试会在矢量约定下考查 $\theta$、$\omega$、$\alpha$ 的大小，但不考它们作为矢量的方向（即右手定则给出的轴向）。方向描述仅限于"顺时针"或"逆时针"。
Worked Example — Spinning Wheel例题 —— 转动的轮
A wheel starts from rest and accelerates uniformly at $\alpha = 3.0\;\text{rad/s}^2$. Find $\omega$ and $\Delta\theta$ after $4.0\;\text{s}$.
That is about $3.8$ full revolutions since $24/(2\pi) \approx 3.82$.约合 $3.8$ 圈，因为 $24/(2\pi) \approx 3.82$。
Worked Example — Non-Constant α (Calculus)例题 —— 变 α（微积分）
A disk has angular acceleration $\alpha(t) = 6t - 2\;\text{rad/s}^2$ and initial angular velocity $\omega_0 = 4\;\text{rad/s}$. Find $\omega(t)$ and $\omega$ at $t = 3\;\text{s}$.
Adjust initial angular velocity and constant angular acceleration. This graph is θ versus time, so it becomes a parabola whenever α is not zero — it is not showing an object flying upward in space.调整初始角速度与恒定角加速度。注意这是 θ–时间 图像，所以只要 α 不为零它就是抛物线——并不代表物体在空中飞起来。
θ at t=5t=5 时的 θ
50.0
rad
ω at t=5t=5 时的 ω
15.0
rad/s
Revolutions圈数
7.96
at t=5 st=5 s 时
A disk rotates with $\omega(t) = 4t - t^2$ (rad/s). At what time does the angular acceleration equal zero?圆盘以 $\omega(t) = 4t - t^2$（rad/s）转动。何时角加速度为零？
Take the derivative: $\alpha = d\omega/dt = 4 - 2t$. Setting this equal to zero gives $t = 2$ s.求导：$\alpha = d\omega/dt = 4 - 2t$；令其为零得 $t = 2$ s。
Topic 5.2专题 5.2
Connecting Linear & Rotational Motion平动与转动的联系
Every point on a rotating rigid body simultaneously undergoes rotational motion (described by $\theta$, $\omega$, $\alpha$) and linear motion (described by $s$, $v$, $a_T$). The bridge between these descriptions is the distance $r$ from the point to the axis of rotation.
Key Insight
All points on a rigid body share the same $\omega$ and $\alpha$. However, points farther from the axis have greater linear speed $v = r\omega$ and tangential acceleration $a_T = r\alpha$. This is why the outer edge of a record moves faster even though every point completes a revolution in the same time.
The tangential acceleration $a_T = r\alpha$ accounts only for changes in the speed along the circular path. The centripetal (radial) acceleration $a_c = v^2/r = r\omega^2$ is always present whenever the point moves in a circle and points inward toward the axis.
The outer point moves 3× faster linearly,外侧点平动速率为内侧点的 3 倍，
This is consistent with the ratio $r_2/r_1 = 3$.与半径比 $r_2/r_1 = 3$ 一致。
Exam Tip — Rolling Without Slipping
For an object that rolls without slipping, the contact point has zero velocity relative to the ground. This gives the constraint $v_{\text{cm}} = R\omega$ and $a_{\text{cm}} = R\alpha$, which is essential for combining translational and rotational equations of motion.
A wheel of radius $0.20$ m has a constant angular acceleration of $5.0$ rad/s². What is the tangential acceleration of a point on the rim?半径 $0.20$ m 的轮以恒定角加速度 $5.0$ rad/s² 转动。轮缘上一点的切向加速度是多少？
Torque is to rotation what force is to translation. A force pushes an object across the floor; a torque makes it spin. Torque depends on three factors: the magnitude of the force, the distance from the axis of rotation to the point where the force is applied, and the angle between the force vector and the position vector.
Equivalently: $\tau = Fd$, where $d = r\sin\theta$ is the lever arm (moment arm).
等价地：$\tau = Fd$，其中 $d = r\sin\theta$ 是力臂（lever arm / moment arm）。
Only the force component perpendicular to $\vec{r}$ produces torque. The parallel component pushes along $\vec{r}$ without causing rotation. The lever arm $d$ is the perpendicular distance from the axis to the line of action of the force.
只有与 $\vec{r}$ 垂直的力分量才产生力矩；沿 $\vec{r}$ 方向的分量只是把物体推沿 $\vec{r}$，不会引起转动。力臂$d$ 是转轴到力的作用线（line of action）的垂直距离。
Physical Intuition
This is why door handles are placed far from the hinges — maximizing $r$ maximizes torque for a given applied force. A force applied at the hinge ($r = 0$) produces zero torque regardless of its magnitude.
When analyzing torque, draw a force diagram (extended free-body diagram) that shows not just the magnitude and direction of each force, but also where each force is applied on the extended body relative to the pivot. This spatial information is essential for calculating torques.
The cross product $\vec{A} \times \vec{B}$ yields a vector perpendicular to both $\vec{A}$ and $\vec{B}$, with magnitude $AB\sin\theta$. The direction follows the right-hand rule: curl your fingers from $\vec{A}$ toward $\vec{B}$, and your thumb points in the direction of $\vec{A} \times \vec{B}$.
Sign Convention
By convention, counterclockwise torques are positive and clockwise torques are negative (when viewed from the standard direction). Be consistent within each problem.
正负号约定
按惯例，从标准方向看去时，逆时针为正、顺时针为负。一道题里前后保持一致即可。
Torque Explorer力矩探究器
Adjust the force magnitude, distance from pivot, and signed angle measured from the rod. The diagram, lever arm, and formula τ = rF sinθ now use the same convention.调整力的大小、到支点的距离，以及从杆量起的带符号角。图示、力臂与公式 τ = rF sinθ 采用同一套约定。
Torque τ (signed)力矩 τ（带符号）
20.0
N·m
Lever Arm |d|力臂 |d|
0.40
m
F⊥ ComponentF⊥ 分量
50.0
N
Worked Example — Torque on a Bolt例题 —— 螺栓受到的力矩
A mechanic applies F = 80 N at the end of a 0.25 m wrench at 60° from the wrench handle. Find the torque.
技师在 0.25 m 长扳手的末端施加 F = 80 N 的力， 与扳手柄方向夹角 60°。求力矩。
If the force were perpendicular ($\theta = 90°$), the torque would hit its maximum of 20 N·m.若力垂直于扳手（$\theta = 90°$），力矩将达到最大值 20 N·m。
A uniform rod of length $L$ is hinged at one end. A force $F$ is applied perpendicularly at a distance $L/4$ from the hinge. What is the torque about the hinge?长度为 $L$ 的均匀杆一端铰接（hinged），在距铰链 $L/4$ 处施加一个垂直于杆的力 $F$。绕铰链的力矩是多少？
$FL$
$FL/2$
$FL/8$
$FL/4$
Correct! Since the force is perpendicular, $\tau = rF = (L/4)(F) = FL/4$.正确！由于力垂直于杆，$\tau = rF = (L/4)(F) = FL/4$。
The force is perpendicular to $\vec{r}$, so $\sin\theta = 1$ and $\tau = rF = (L/4)(F) = FL/4$.力垂直于 $\vec{r}$，故 $\sin\theta = 1$，$\tau = rF = (L/4)(F) = FL/4$。
Topic 5.4专题 5.4
Rotational Inertia转动惯量
Rotational inertia (moment of inertia) is the rotational analog of mass. While mass measures resistance to changes in linear motion, rotational inertia $I$ measures resistance to changes in rotational motion. Crucially, $I$ depends not only on total mass but on how that mass is distributed relative to the axis of rotation.
转动惯量（rotational inertia / moment of inertia）是平动中质量的转动对应物。质量度量物体抵抗平动状态改变的能力，而转动惯量 $I$ 度量它抵抗转动状态改变的能力。关键在于：$I$ 不仅取决于总质量，还取决于质量相对转轴如何分布。
Point Mass质点
$$I = mr^2$$
$r$ = perpendicular distance from the axis
$r$ = 到转轴的垂直距离
System of Point Masses质点组
$$I_{\text{tot}} = \sum_i m_i r_i^2$$
Continuous Body (Integral Form)连续体（积分形式）
$$I = \int r^2\,dm$$
Key Insight
Mass far from the axis contributes much more to $I$ than the same mass close to the axis (because of the $r^2$ factor). A hollow hoop ($I = MR^2$) has greater rotational inertia than a solid disk ($I = \frac{1}{2}MR^2$) of the same mass and radius — all the hoop's mass sits at the maximum distance $R$.
$I$ is minimized when the axis passes through the center of mass.
当转轴通过质心时 $I$ 取最小值。
Exam Note — Derivations
The AP exam expects you to derive $I$ using calculus for: (1) thin rods of uniform or non-uniform density about an arbitrary perpendicular axis, and (2) cylindrical shells, disks, or bodies composed of coaxial rings/shells about a central axis. You should also understand qualitatively why mass farther from the axis increases $I$.
Worked Example — Parallel Axis Theorem (Rod About End)例题 —— 平行轴定理（细杆绕端点）
Find $I$ of a uniform thin rod of mass $M$, length $L$, about one of its ends.
求质量 $M$、长度 $L$ 的均匀细杆绕一端的转动惯量 $I$。
We know $I_\text{cm} = \tfrac{1}{12}ML^2$ for the axis through the center, and the end is a distance $d = L/2$ from the center.已知绕中点轴的 $I_\text{cm} = \tfrac{1}{12}ML^2$，且端点到中点距离 $d = L/2$。
(c) Physical pendulum — rod pivoted a distance $\ell$ from its center(c) 复摆 —— 杆绕距中点 $\ell$ 处的支点
This $I$ feeds directly into the physical-pendulum period $T = 2\pi\sqrt{I/(Mg\ell)}$.这个 $I$ 可直接代入复摆（physical pendulum）周期 $T = 2\pi\sqrt{I/(Mg\ell)}$。
$$I = \tfrac{1}{12}ML^2 + M\ell^2$$
With $\ell = L/2$ (pivot at end) this collapses back to $\tfrac{1}{3}ML^2$ as above.当 $\ell = L/2$（支点在端点）时，退化为上面的 $\tfrac{1}{3}ML^2$。
Rule of Thumb
The parallel-axis theorem only works between an axis through the center of mass and a parallel axis offset by distance $d$. You can't use it to jump between two non-cm axes directly — go through the cm axis as a "hub."
Worked Example — Non-Uniform Rod: $I$ by Integration例题 —— 非均匀杆：用积分求 $I$
A thin rod of length $L$ and total mass $M$ has linear mass density
$$\lambda(x)=\lambda_0\frac{x}{L}$$
where $x$ is measured from one end. Find its rotational inertia about the end where $x=0$.
Two objects — a solid disk and a thin hoop — have the same mass $M$ and radius $R$. Which has greater rotational inertia about the central axis, and by what factor?实心圆盘与薄圆环具有相同的质量 $M$ 与半径 $R$。绕中心轴时哪个转动惯量更大？相差几倍？
The disk; factor of 2圆盘；2 倍
The hoop; factor of 1.5圆环；1.5 倍
The hoop; factor of 2圆环；2 倍
They are equal两者相等
Correct! $I_{\text{hoop}} = MR^2$ and $I_{\text{disk}} = \frac{1}{2}MR^2$. The hoop's rotational inertia is exactly twice the disk's because all its mass sits at the maximum radius.正确！$I_{\text{hoop}} = MR^2$，$I_{\text{disk}} = \frac{1}{2}MR^2$。圆环转动惯量恰好是圆盘的 2 倍——它的全部质量都分布在最大半径处。
Just as translational equilibrium requires zero net force and constant velocity, rotational equilibrium requires zero net torque and constant angular velocity. These two kinds of equilibrium are independent — an object can be in rotational equilibrium without being in translational equilibrium, and vice versa.
Newton's first law in rotational form: a system maintains constant angular velocity if and only if the net torque on it is zero. The converse is equally important — if torques are not balanced, $\omega$ must be changing.
这是 Newton 第一定律的转动版本：系统保持角速度不变当且仅当合力矩为零。其逆命题同样重要——若力矩不平衡，$\omega$ 必定在变。
Strategy — Static Equilibrium
For objects at rest, enforce both conditions: $\sum \vec{F} = 0$ (translational) and $\sum \tau = 0$ (rotational). The choice of pivot is arbitrary — choose the pivot to eliminate an unknown force, simplifying the algebra.
$F_1$ acts at the left end $(x = 0)$ and $F_2$ acts at $x = 2.0\;\text{m}$.$F_1$ 作用于左端（$x = 0$），$F_2$ 作用于 $x = 2.0\;\text{m}$ 处。
Weight Mg acts at center (x = 1.5 m).重力 Mg 作用于中点（$x = 1.5\;\text{m}$）。
Step 1第 1 步
Torques about the left end (eliminates $F_1$):对左端取力矩（消去 $F_1$）：
$$\Sigma\tau = 0$$
$$F_2(2.0) - Mg(1.5) = 0$$
$$F_2 = Mg(1.5)/2.0$$
$$F_2 = (12)(9.8)(0.75) = 88.2\;\text{N}$$
Step 2第 2 步
Vertical force balance:竖直方向力平衡：
$$F_1 + F_2 = Mg$$
$$F_1 = 117.6 - 88.2 = 29.4\;\text{N}$$
Worked Example — Ladder Against a Wall例题 —— 倚墙的梯子
A uniform ladder of mass M = 15 kg and length L = 4.0 m leans against a frictionless wall at angle 60° from the floor. Find the normal force from the wall and the friction from the floor.
质量 M = 15 kg、长度 L = 4.0 m 的均匀梯子 以与地面成 60° 的角倚靠在无摩擦墙上。 求墙的法向力与地面的摩擦力。
Step 1第 1 步
Identify forces明确受力
$N_w$ is the normal force from the wall (horizontal, at the top).$N_w$：墙的法向力（水平、作用于顶端）。
$N_f$ is the normal force from the floor (vertical, at the bottom).$N_f$：地面的法向力（竖直、作用于底端）。
$f$ is the friction force from the floor (horizontal, at the bottom).$f$：地面摩擦力（水平、作用于底端）。
$Mg$ acts at the center of the ladder, a distance $L/2$ from the bottom.$Mg$ 作用在梯子中点，距底端 $L/2$。
Step 2第 2 步
Torques about the bottom (eliminates $N_f$ and $f$):对底端取力矩（消去 $N_f$ 与 $f$）：
Exam Tip — Choosing the Pivot
In equilibrium problems, you can compute torques about any point. Strategically choosing the pivot where an unknown force acts eliminates that unknown (since $r = 0$ for that force), reducing the number of unknowns in your torque equation.
A meter stick has a $2.0$ kg mass at the $80$ cm mark and is balanced on a fulcrum at the $60$ cm mark. If the uniform meter stick has mass $m$, which equation correctly determines $m$? (Torques about the fulcrum; stick's weight acts at 50 cm.)米尺在 80 cm 处挂有 2.0 kg 的物体，支点（fulcrum）位于 60 cm 处时平衡。设均匀米尺质量为 $m$，下列哪个方程能正确求出 $m$？（对支点取力矩；米尺重力作用于 50 cm 处。）
$mg(60) = 2g(80)$
$mg(10) = 2g(20)$
$mg(50) = 2g(80)$
$mg(10) = 2g(80)$
Correct! About the fulcrum at 60 cm: the stick's center of mass at 50 cm is 10 cm left, and the 2.0 kg mass at 80 cm is 20 cm right. Balancing: $mg(10) = 2g(20)$.正确！以 60 cm 处的支点为基准：米尺质心在 50 cm，距支点 10 cm（左侧）；2.0 kg 物体在 80 cm，距支点 20 cm（右侧）。平衡式：$mg(10) = 2g(20)$。
Measure distances from the fulcrum. Center of mass is 10 cm left; hanging mass is 20 cm right. The equation is $mg(10) = 2g(20)$.距离要从支点量起。质心在左 10 cm，挂物在右 20 cm。方程为 $mg(10) = 2g(20)$。
Topic 5.6专题 5.6
Newton's Second Law — Rotational FormNewton 第二定律 —— 转动形式
The crown jewel of rotational dynamics: just as $\vec{F}_{\text{net}} = m\vec{a}$ governs translation, the net torque on a rigid body determines its angular acceleration, scaled by its rotational inertia:
Newton's Second Law — RotationalNewton 第二定律 —— 转动形式
$$\tau_{\text{net}} = I\alpha$$
Equivalently: $\alpha = \tau_{\text{net}} / I$
等价地：$\alpha = \tau_{\text{net}} / I$
For a given net torque, a body with larger rotational inertia has smaller angular acceleration — it is "harder to spin up."
在同样净力矩下，转动惯量越大的物体角加速度越小——它"越难被转起来"。
Key Insight — Combined Problems
Many real problems involve both translation and rotation — a disk rolling down a ramp, a pulley with a hanging mass. Write separate equations for translational ($F_{\text{net}} = ma$) and rotational ($\tau_{\text{net}} = I\alpha$) motion, then connect them via $a = r\alpha$.
Compare: a sliding block (no rotation) would have $a = g\sin 30° = 4.9\;\text{m/s}^2$. Rolling is slower because energy goes into rotation.对比：纯下滑（不转动）的滑块加速度 $a = g\sin 30° = 4.9\;\text{m/s}^2$。滚动较慢，是因为部分能量分配给了转动。
Rolling Race Simulator滚动竞速模拟器
Watch four shapes race down the same incline! The shape with the smallest $I/(mR^2)$ wins. Press Start to race, or adjust the angle and race again.观察四种形状沿同一斜面竞速！$I/(mR^2)$ 最小者胜出。点击"开始"启动，或调整角度后重新比赛。
Solid Sphere实心球
—
m/s²
Solid Cylinder实心圆柱
—
m/s²
Hollow Sphere空心球
—
m/s²
Hoop圆环
—
m/s²
Common Mistake
In rolling problems, the friction force does not do work (the contact point has zero velocity). Do not subtract friction work when using energy methods for rolling without slipping. However, friction is the force that provides the torque causing rotation.
A net torque of $12$ N·m is applied to a solid cylinder of mass $8.0$ kg and radius $0.50$ m about its central axis. What is its angular acceleration?对质量 $8.0$ kg、半径 $0.50$ m 的实心圆柱绕其中心轴施加 $12$ N·m 的净力矩。角加速度是多少？
How Unit 5 Appears on the AP Exam第 5 单元在 AP 考试中的形式
MC
Multiple Choice — Common StylesMultiple Choice —— 常考题型
Conceptual traps: Distinguish rotational from translational quantities. All points share the same $\omega$ and $\alpha$, but have different $v$ and $a_T$ depending on radius.
Functional dependence: "If the force doubles, what happens to the torque?" Use proportional reasoning from $\tau = rF\sin\theta$.
函数关系："力翻倍，力矩怎么变？"由 $\tau = rF\sin\theta$ 用比例推理即可。
Compare shapes: Which object rolls faster down an incline? Use $a = g\sin\phi/(1 + I/mR^2)$.
形状比较：哪种物体沿斜面滚得更快？用 $a = g\sin\phi/(1 + I/mR^2)$。
FR
Free Response — Common StylesFree Response —— 常考题型
Mathematical routines: Derive $I$ via integration. Define $dm$, establish limits, and show every step. Partial credit for correct setup even if the final answer is wrong.
Combined translation-rotation: Atwood machines with massive pulleys, rolling objects on inclines. Always write separate $F = ma$ and $\tau = I\alpha$ equations, then connect with $a = r\alpha$.
QQT questions: Predict how changing one variable affects another (e.g., "if mass is moved farther from the axis, how does $\alpha$ change?"), then justify with equations.
Top Mistakes That Lose Points1. Using degrees instead of radians — all rotational equations require radians.2. Confusing $\omega$ and $v$ — angular velocity is the same for all points; tangential speed $v = r\omega$ depends on $r$.3. Wrong axis for $I$ — a rod about its center ($ML^2/12$) ≠ a rod about its end ($ML^2/3$).4. Forgetting the constraint $a = r\alpha$ in combined translation-rotation problems.5. Choosing a bad pivot in equilibrium — pick the pivot where an unknown force acts to eliminate it.6. Sign errors in torque — assign a positive direction at the start and apply consistently.7. Claiming friction does work in rolling without slipping — the contact point has zero velocity.
Worked Example — Atwood Machine with Massive Pulley (FRQ Style)例题 —— 带质量滑轮的 Atwood 机（FRQ 风格）
Two blocks ($m_1 = 4.0\;\text{kg}$, $m_2 = 6.0\;\text{kg}$) are connected by a massless string over a solid disk pulley ($M = 2.0\;\text{kg}$, $R = 0.10\;\text{m}$). The string does not slip on the pulley. Find the system's acceleration and the angular acceleration of the pulley.
Worked Example — Rolling Without Slipping Down an Incline (FRQ Style)例题 —— 斜面上无滑滚动（FRQ 风格）
A solid cylinder of mass $m$ and radius $R$ starts from rest at the top of a rough incline (angle $\phi$, height $h$). It rolls without slipping to the bottom. Find (a) the acceleration, (b) the friction force, and (c) the speed at the bottom.
If $I = 0$ (sliding block): $a = g\sin\phi$ and $v = \sqrt{2gh}$. Both are larger than our answers — rotation absorbs energy. ✓若 $I = 0$（纯下滑滑块）：$a = g\sin\phi$、$v = \sqrt{2gh}$，都比我们的结果大——转动吸收了一部分能量。✓
Friction must satisfy $f_s \leq \mu_s N = \mu_s mg\cos\phi$. If $\mu_s < \frac{\tan\phi}{3}$, the cylinder slips. ✓摩擦力须满足 $f_s \leq \mu_s N = \mu_s mg\cos\phi$。若 $\mu_s < \frac{\tan\phi}{3}$，圆柱会打滑。✓
Matches the general formula $a = g\sin\phi/(1 + I/(mR^2))$ with $I/(mR^2) = 1/2$: $a = g\sin\phi/1.5 = 2g\sin\phi/3$. ✓与通用公式 $a = g\sin\phi/(1 + I/(mR^2))$ 在 $I/(mR^2) = 1/2$ 时的结果吻合：$a = g\sin\phi/1.5 = 2g\sin\phi/3$。✓
2. A torque $\tau$ gives a disk angular acceleration $\alpha$. If both the torque and the rotational inertia are doubled, $\alpha$ becomes:力矩 $\tau$ 使圆盘的角加速度为 $\alpha$。若力矩和转动惯量都加倍，新的 $\alpha$ 是：
$\alpha = \tau/I$. Doubling both numerator and denominator: $\alpha' = 2\tau/(2I) = \alpha$.$\alpha = \tau/I$。分子分母都加倍：$\alpha' = 2\tau/(2I) = \alpha$。
3. A force $\vec{F}$ is applied at position $\vec{r}$ from the pivot. If $\vec{F} \parallel \vec{r}$, the resulting torque is:力 $\vec{F}$ 作用于支点出发的位置矢量 $\vec{r}$ 处。若 $\vec{F} \parallel \vec{r}$，所产生的力矩为：
$rF$
$rF/2$
$rF\cos\theta$
Zero零
Correct! When $\vec{F} \parallel \vec{r}$, the angle is $0°$ or $180°$, so $\sin\theta = 0$ and $\tau = 0$.正确！$\vec{F} \parallel \vec{r}$ 时夹角为 0° 或 180°，$\sin\theta = 0$，故 $\tau = 0$。
$\tau = rF\sin\theta$. When force is parallel to position vector, $\sin\theta = 0$, so torque is zero.$\tau = rF\sin\theta$。当力与位置矢量平行时 $\sin\theta = 0$，力矩为零。
4. A thin rod ($I_{\text{cm}} = \frac{1}{12}ML^2$) is rotated about an axis $L/4$ from its center. By the parallel axis theorem, $I' = $?细杆（$I_{\text{cm}} = \frac{1}{12}ML^2$）绕距其中点 $L/4$ 的轴转动。由平行轴定理，$I' = $？
5. A wheel starts at rest and reaches $\omega = 20$ rad/s after $5.0$ s of constant $\alpha$. How many revolutions does it complete?轮从静止开始，在恒定 $\alpha$ 下经 5.0 s 达到 $\omega = 20$ rad/s。期间转了多少圈？