Each item restates the prompt and choices, marks the correct letter, and gives a brief justification keyed to the AP topic. Numbers use $g = 9.8~\mathrm{m/s^2}$ unless otherwise noted.每道题均重新列出题干和选项，标注正确字母，并给出与 AP 主题对应的简要说明。数值计算使用 $g = 9.8~\mathrm{m/s^2}$，除非另有说明。
One revolution per second corresponds to $2\pi~\mathrm{rad/s}$.每秒一转对应 $2\pi~\mathrm{rad/s}$。
Q2EASY5.2 Linear & Rotational5.2 线量与转动量No Calculator
A point on the rim of a disk of radius $0.50~\mathrm{m}$ moves with the disk at angular speed $\omega = 4.0~\mathrm{rad/s}$. The linear speed of that point is半径为 $0.50~\mathrm{m}$ 的圆盘边缘上一点随圆盘以角速度 $\omega = 4.0~\mathrm{rad/s}$ 转动。该点的线速度为
(A) $0.50~\mathrm{m/s}$
(B) $2.0~\mathrm{m/s}$
(C) $4.0~\mathrm{m/s}$
(D) $8.0~\mathrm{m/s}$
Answer:答案：(B)
$$v = \omega r = 4.0(0.50) = 2.0~\mathrm{m/s}$$
Q3EASY5.3 Torque5.3 力矩No Calculator
A force of $20~\mathrm{N}$ is applied perpendicular to the handle of a wrench at a distance $0.30~\mathrm{m}$ from the axis of rotation. The magnitude of the torque about the axis is一个 $20~\mathrm{N}$ 的力垂直作用于扳手手柄，作用点距转动轴 $0.30~\mathrm{m}$。关于该轴的力矩大小为
(A) $0.6~\mathrm{N \cdot m}$
(B) $6.0~\mathrm{N \cdot m}$
(C) $60~\mathrm{N \cdot m}$
(D) $66~\mathrm{N \cdot m}$
Answer:答案：(B)
$$\tau = r F \sin\theta = 0.30(20)(1) = 6.0~\mathrm{N \cdot m}$$
Perpendicular force gives the maximum torque ($\sin 90^\circ = 1$).垂直力产生最大力矩（$\sin 90^\circ = 1$）。
Q4EASY5.4 Rotational Inertia5.4 转动惯量No Calculator
The moment of inertia of a thin uniform rod of mass $M$ and length $L$ about an axis through its center, perpendicular to the rod, is质量为 $M$、长度为 $L$ 的细均匀杆，绕过其中心且垂直于杆的轴的转动惯量为
(A) $\dfrac{1}{3}ML^2$
(B) $\dfrac{1}{12}ML^2$
(C) $ML^2$
(D) $\dfrac{1}{2}ML^2$
Answer:答案：(B)
Standard table result for a rod about its center. Trap (A) is the rod about its end (also a common reference value, but distinct from the centroidal axis here).标准公式：杆绕中心轴的转动惯量。干扰项 (A) 是杆绕端点的转动惯量（同样是常用参考值，但与质心轴不同）。
A disk of radius $0.20~\mathrm{m}$ rotates about its central axis at constant angular speed $\omega = 10~\mathrm{rad/s}$. The centripetal acceleration of a point on the rim is半径为 $0.20~\mathrm{m}$ 的圆盘绕中心轴以匀角速度 $\omega = 10~\mathrm{rad/s}$ 转动。圆盘边缘上一点的向心加速度为
(A) $2.0~\mathrm{m/s^2}$
(B) $5.0~\mathrm{m/s^2}$
(C) $20~\mathrm{m/s^2}$
(D) $100~\mathrm{m/s^2}$
Answer:答案：(C)
$$a_c = \omega^2 r = (10)^2(0.20) = 20~\mathrm{m/s^2}$$
Trap (A) is $\omega r$ (linear speed, not acceleration); (D) drops the radius.干扰项 (A) 是 $\omega r$（线速度，非加速度）；(D) 漏掉了半径。
Q7MEDIUM5.3 Torque at an Angle5.3 斜向力矩No Calculator
A force of $30~\mathrm{N}$ is applied at the end of a wrench of length $0.40~\mathrm{m}$, but at an angle of $30^\circ$ to the wrench handle. The magnitude of the torque about the axis is一个 $30~\mathrm{N}$ 的力施加在长 $0.40~\mathrm{m}$ 的扳手末端，方向与扳手柄成 $30^\circ$ 角。关于轴的力矩大小为
(A) $3.0~\mathrm{N \cdot m}$
(B) $6.0~\mathrm{N \cdot m}$
(C) $10.4~\mathrm{N \cdot m}$
(D) $12~\mathrm{N \cdot m}$
Answer:答案：(B)
$$\tau = r F \sin\theta = 0.40(30)\sin 30^\circ = 12(0.5) = 6.0~\mathrm{N \cdot m}$$
Only the component perpendicular to the handle delivers torque. Trap (C) uses $\cos 30^\circ$; (D) drops the angle.只有垂直于手柄的分量产生力矩。干扰项 (C) 使用了 $\cos 30^\circ$；(D) 忽略了角度。
A solid uniform disk of mass $M$ and radius $R$ rotates about an axis perpendicular to the disk through a point on its edge. Its moment of inertia about that axis is质量为 $M$、半径为 $R$ 的实心均匀圆盘，绕过其边缘上一点且垂直于圆盘的轴转动。关于该轴的转动惯量为
(A) $\dfrac{1}{2}MR^2$
(B) $MR^2$
(C) $\dfrac{3}{2}MR^2$
(D) $2MR^2$
Answer:答案：(C)
Parallel-axis theorem with $d = R$:以 $d = R$ 应用平行轴定理：
A uniform horizontal beam of mass $10~\mathrm{kg}$ and length $4.0~\mathrm{m}$ is hinged at the wall at one end. The far end is supported by a vertical cable. The tension in the cable is一根质量为 $10~\mathrm{kg}$、长 $4.0~\mathrm{m}$ 的均匀水平梁，一端用铰链固定在墙上，另一端由竖直绳索支撑。绳索中的张力为
(A) $25~\mathrm{N}$
(B) $49~\mathrm{N}$
(C) $98~\mathrm{N}$
(D) $196~\mathrm{N}$
Answer:答案：(B)
Take torques about the hinge; weight acts at the center (lever arm $L/2$):以铰链为支点取力矩；重力作用于中心（力臂 $L/2$）：
$$T \cdot L = M g \cdot \frac{L}{2} \;\Longrightarrow\; T = \frac{M g}{2} = \frac{10(9.8)}{2} = 49~\mathrm{N}$$
Trap (C) reports $Mg$ (the cable carrying the full weight, would be true only if the beam were massless and pivoted at the load).干扰项 (C) 给出 $Mg$（绳索承担全部重量，仅当梁无质量且以载荷为支点时成立）。
Q10MEDIUM5.6 Newton's 2nd Law (Rotation)5.6 转动牛顿第二定律No Calculator
A net torque of $6.0~\mathrm{N \cdot m}$ acts on a wheel with moment of inertia $I = 2.0~\mathrm{kg \cdot m^2}$. The wheel's angular acceleration is一个合力矩 $6.0~\mathrm{N \cdot m}$ 作用在转动惯量 $I = 2.0~\mathrm{kg \cdot m^2}$ 的轮子上。该轮子的角加速度为
A wheel starts from rest and undergoes constant angular acceleration $\alpha = 2.0~\mathrm{rad/s^2}$. Its angular displacement after $5.0~\mathrm{s}$ is一个轮子从静止开始，以匀角加速度 $\alpha = 2.0~\mathrm{rad/s^2}$ 转动。$5.0~\mathrm{s}$ 后的角位移为
Trap (A) is $\alpha t$ (final $\omega$, not $\theta$); (D) drops the $\tfrac{1}{2}$.干扰项 (A) 是 $\alpha t$（末角速度，非角位移）；(D) 漏掉了 $\tfrac{1}{2}$。
Q12MEDIUM5.4 Comparing Moments of Inertia5.4 转动惯量比较No Calculator
A thin hoop, a solid disk, and a solid sphere all have the same mass $M$ and the same radius $R$, and rotate about axes through their centers. Ranked from greatest to least moment of inertia, they are一个细圆环、一个实心圆盘和一个实心球体，质量均为 $M$，半径均为 $R$，均绕通过各自中心的轴转动。按转动惯量从大到小排列，顺序为
The more concentrated the mass near the axis, the smaller $I$. The hoop has all its mass at radius $R$; the sphere distributes mass throughout the volume, much of it near the axis.质量越集中于轴附近，$I$ 越小。圆环的质量全部在半径 $R$ 处；球体的质量分布在整个体积内，大部分靠近轴。
Q13MEDIUM5.6 Disk Angular Acceleration5.6 圆盘角加速度No Calculator
A net torque $\tau$ acts on a uniform solid disk of mass $M$ and radius $R$ about its central axis. The angular acceleration of the disk is合力矩 $\tau$ 作用于质量为 $M$、半径为 $R$ 的均匀实心圆盘的中心轴上。圆盘的角加速度为
Trap (A) uses $I = MR^2$ (hoop); (C) doubles the half-factor in the wrong place.干扰项 (A) 使用了 $I = MR^2$（圆环的公式）；(C) 在错误位置引入了额外的 2 倍因子。
Q14HARD5.6 Atwood with Massive Pulley5.6 有质量滑轮的阿特伍德机Calculator
Two masses, $m_1 = 4.0~\mathrm{kg}$ and $m_2 = 2.0~\mathrm{kg}$, are connected by a massless string passing over a uniform-disk pulley of mass $M = 4.0~\mathrm{kg}$ and radius $R$. The string does not slip on the pulley. Take $g = 10~\mathrm{m/s^2}$. The acceleration of the masses is closest to两个质量 $m_1 = 4.0~\mathrm{kg}$ 和 $m_2 = 2.0~\mathrm{kg}$ 通过无质量绳索跨过质量为 $M = 4.0~\mathrm{kg}$、半径为 $R$ 的均匀圆盘滑轮相连。绳索在滑轮上不打滑。取 $g = 10~\mathrm{m/s^2}$。两质量的加速度最接近
(A) $2.0~\mathrm{m/s^2}$
(B) $2.5~\mathrm{m/s^2}$
(C) $3.3~\mathrm{m/s^2}$
(D) $5.0~\mathrm{m/s^2}$
Answer:答案：(B)
No-slip $\Rightarrow a = R\alpha$. The pulley adds an effective "mass" $I/R^2 = M/2$ to the system inertia:不打滑条件 $\Rightarrow a = R\alpha$。滑轮为系统等效惯性增加了 $I/R^2 = M/2$：
A uniform ladder of mass $M$ leans at angle $\theta$ above the horizontal against a frictionless vertical wall. The coefficient of static friction between the ladder and the floor is $\mu$. The minimum value of $\theta$ for which the ladder does not slip is given by质量为 $M$ 的均匀梯子以与水平面成 $\theta$ 角靠在无摩擦的竖直墙壁上。梯子与地面之间的静摩擦系数为 $\mu$。梯子不滑动的最小 $\theta$ 值满足
(A) $\tan\theta = \mu$
(B) $\tan\theta = 2\mu$
(C) $\tan\theta = \dfrac{1}{2\mu}$
(D) $\tan\theta = \dfrac{1}{\mu}$
Answer:答案：(C)
Wall is frictionless, so $N_w$ is horizontal. Floor: $N_f = Mg$, friction $f \le \mu N_f$ (horizontal).
Torques about the floor contact, length $L$, weight at $L/2$:墙壁无摩擦，故 $N_w$ 水平。地面：$N_f = Mg$，摩擦力 $f \le \mu N_f$（水平）。以地面接触点为支点取力矩，杆长 $L$，重力作用于 $L/2$ 处：
$$N_w L\sin\theta = M g\,\tfrac{L}{2}\cos\theta \;\Longrightarrow\; N_w = \frac{Mg}{2\tan\theta}$$
Horizontal balance: $f = N_w$. At the slip threshold $f = \mu N_f = \mu M g$:水平方向平衡：$f = N_w$。在临界滑动时 $f = \mu N_f = \mu M g$：
$$\mu M g = \frac{M g}{2\tan\theta} \;\Longrightarrow\; \tan\theta = \frac{1}{2\mu}$$
Steeper than $\tan\theta = \mu$ (which would apply for a simple block on an incline).比 $\tan\theta = \mu$（适用于斜面上简单物块的情形）更陡。
A point mass $m$ is rigidly attached to one end of a uniform rod of mass $M$ and length $L$. The system rotates about an axis through the rod's free end (i.e. the end opposite the point mass), perpendicular to the rod. Its moment of inertia about that axis is质量为 $m$ 的质点固定在质量为 $M$、长度为 $L$ 的均匀杆的一端。该系统绕过杆的自由端（即与质点相对的一端）且垂直于杆的轴转动。关于该轴的转动惯量为
Rod about an end (parallel-axis from center): $I_\text{rod, end} = \tfrac{1}{3}ML^2$. Point mass at distance $L$: $mL^2$.杆绕端点的转动惯量（由质心轴用平行轴定理得到）：$I_\text{rod, end} = \tfrac{1}{3}ML^2$。距轴 $L$ 处质点：$mL^2$。
Trap (B) uses the centroidal $\tfrac{1}{12}ML^2$ (wrong axis); (C) halves the point-mass distance.干扰项 (B) 使用质心处的 $\tfrac{1}{12}ML^2$（轴选错）；(C) 将质点距离减半。
Q17HARD5.3 Vector Torque5.3 矢量力矩No Calculator
A force $\vec F = 3\,\hat\jmath~\mathrm{N}$ is applied at the position $\vec r = 2\,\hat\imath~\mathrm{m}$ relative to a fixed pivot. The torque about that pivot is一个力 $\vec F = 3\,\hat\jmath~\mathrm{N}$ 施加在相对于固定转轴位置为 $\vec r = 2\,\hat\imath~\mathrm{m}$ 处。关于该转轴的力矩为
(A) $6\,\hat\imath~\mathrm{N \cdot m}$
(B) $6\,\hat\jmath~\mathrm{N \cdot m}$
(C) $6\,\hat k~\mathrm{N \cdot m}$
(D) $-6\,\hat k~\mathrm{N \cdot m}$
Answer:答案：(C)
$$\vec\tau = \vec r \times \vec F = (2\,\hat\imath) \times (3\,\hat\jmath) = 6\,(\hat\imath \times \hat\jmath) = 6\,\hat k~\mathrm{N \cdot m}$$
Right-hand rule: fingers from $+\hat\imath$ to $+\hat\jmath$ curl in the $+\hat k$ sense. Trap (D) reverses the cross-product order.右手定则：手指从 $+\hat\imath$ 转向 $+\hat\jmath$ 卷曲方向为 $+\hat k$。干扰项 (D) 颠倒了叉积的顺序。
Q18HARD5.6 Yo-Yo on Fixed String5.6 固定绳上的悠悠球No Calculator
A uniform solid disk of mass $M$ and radius $R$ has a light cord wrapped around it; the upper end of the cord is fixed. The disk is released from rest and falls vertically as the cord unwinds. The downward acceleration of the disk's center is质量为 $M$、半径为 $R$ 的均匀实心圆盘上绕有一轻绳，绳的上端固定。圆盘从静止开始释放，随绳展开竖直下落。圆盘中心的向下加速度为
(A) $\dfrac{g}{2}$
(B) $\dfrac{2g}{3}$
(C) $\dfrac{g}{3}$
(D) $g$
Answer:答案：(B)
Translation: $Mg - T = M a$. Rotation about the disk's center: $TR = I\alpha = \tfrac{1}{2}MR^2 \alpha$ with $a = R\alpha$ (no slip), so $T = \tfrac{1}{2}M a$. Substitute:平动方程：$Mg - T = M a$。绕圆盘中心的转动方程：$TR = I\alpha = \tfrac{1}{2}MR^2 \alpha$，结合不打滑条件 $a = R\alpha$，得 $T = \tfrac{1}{2}M a$。代入：
$$M g - \tfrac{1}{2}M a = M a \;\Longrightarrow\; a = \tfrac{2g}{3}$$
The cord-tension term steals $\tfrac{1}{3}$ of $g$ to spin up the disk.绳中张力项"抢走"了 $\tfrac{1}{3}$ 的 $g$ 用于使圆盘转动加速。
Each FRQ walks every part in the canonical setup → execute → evaluate structure. Numbers use $g = 9.8~\mathrm{m/s^2}$.每道自由作答题按"建模→执行→评估"的标准结构逐步讲解。数值计算使用 $g = 9.8~\mathrm{m/s^2}$。
Pulley: tension $T$ at rim (tangential, producing torque about the axle), axle reaction at the center, gravity $M g$ at center (no torque).滑轮：边缘处张力 $T$（切向，对轴产生力矩），轴心处轴承反力，中心处重力 $M g$（无力矩）。
(b)Newton's 2nd law.牛顿第二定律。
$$m g - T = m a \qquad T R = I \alpha = \tfrac{1}{2} M R^2 \alpha$$
No-slip $a = R\alpha$ collapses the rotational equation to $T = \tfrac{1}{2} M a$. Substitute:不打滑条件 $a = R\alpha$ 将转动方程化简为 $T = \tfrac{1}{2} M a$。代入：
$$m g - \tfrac{1}{2} M a = m a \;\Longrightarrow\; a = \frac{m g}{m + M/2} = \frac{1.0(9.8)}{1.0 + 1.0} = 4.9~\mathrm{m/s^2}$$
$$T = \tfrac{1}{2} M a = \tfrac{1}{2}(2.0)(4.9) = 4.9~\mathrm{N}$$
Horizontal beam $M = 20~\mathrm{kg}$, $L = 4.0~\mathrm{m}$, hinged at wall; cable from far end at $30^\circ$ above horizontal to wall; load $m = 10~\mathrm{kg}$ at far end.水平梁 $M = 20~\mathrm{kg}$，$L = 4.0~\mathrm{m}$，铰接在墙壁；绳索从远端以与水平面成 $30^\circ$ 角连至墙壁；远端悬挂载荷 $m = 10~\mathrm{kg}$。
(a)FBD. Beam weight $M g$ at center, load $m g$ at far end (down), cable tension $T$ at far end at $30^\circ$ above horizontal, hinge reaction components $H_x$ (horizontal) and $H_y$ (vertical) at the wall end.受力分析图。梁重 $M g$ 作用于中心，载荷 $m g$ 作用于远端（向下），绳张力 $T$ 在远端与水平方向成 $30^\circ$ 角，铰链反力分量 $H_x$（水平）和 $H_y$（竖直）在墙端。
(b)Torque about the hinge. Only the vertical component of $T$ (i.e. $T\sin 30^\circ$) does work against gravity:以铰链为支点取力矩。只有 $T$ 的竖直分量（即 $T\sin 30^\circ$）与重力做功：
$$T \sin 30^\circ \cdot L = M g \cdot \tfrac{L}{2} + m g \cdot L$$
Rod $M = 0.40~\mathrm{kg}$, $L = 1.20~\mathrm{m}$ with point mass $m = 0.10~\mathrm{kg}$ at one end; pivot at the opposite (light) end. Released from horizontal.杆 $M = 0.40~\mathrm{kg}$，$L = 1.20~\mathrm{m}$，一端附有质点 $m = 0.10~\mathrm{kg}$；枢轴在另一端（无质点端）。从水平位置释放。
(a)Moment of inertia about the pivot. Rod about end: $\tfrac{1}{3}ML^2$. Point mass at distance $L$: $mL^2$.关于枢轴的转动惯量。杆绕端点：$\tfrac{1}{3}ML^2$。距轴 $L$ 处的质点：$mL^2$。
FRQ 4HARD5.6 Rolling Cylinder on Incline5.6 斜面上滚动的圆柱体Calculator
Uniform solid cylinder ($M$, $R$) rolls without slipping down an incline of angle $\theta$.均匀实心圆柱体（$M$，$R$）在倾角为 $\theta$ 的斜面上无滑动滚动。
(a)FBD. Weight $Mg$ vertical (components $Mg\sin\theta$ down-slope and $Mg\cos\theta$ into-slope), normal $N$ out of the slope, static friction $f_s$ up the slope (opposes the contact point's tendency to slip backward as the cylinder accelerates down).受力分析图。重力 $Mg$ 竖直（分量：$Mg\sin\theta$ 沿斜面向下，$Mg\cos\theta$ 垂直斜面向内），法向力 $N$ 垂直斜面向外，静摩擦力 $f_s$ 沿斜面向上（阻止圆柱向下加速时接触点向后滑动的趋势）。
(b)Newton's laws.牛顿定律。
$$M g\sin\theta - f_s = M a \qquad f_s R = I_\text{cm}\,\alpha = \tfrac{1}{2}MR^2 \alpha$$
$$M g\sin\theta - \tfrac{1}{2}M a = M a \;\Longrightarrow\; a = \tfrac{2}{3}\,g\sin\theta$$
$$f_s = \tfrac{1}{2}M a = \tfrac{1}{3}M g\sin\theta$$
For comparison, a frictionless slide has $a = g\sin\theta$, rolling costs $\tfrac{1}{3}$ of that, used to spin up the cylinder.作为对比，无摩擦滑动时 $a = g\sin\theta$，滚动时损失其 $\tfrac{1}{3}$ 用于使圆柱旋转加速。
(d)Minimum $\mu_s$ for rolling without slipping. The static-friction constraint $f_s \le \mu_s N$ becomes无滑动滚动所需最小 $\mu_s$。静摩擦约束 $f_s \le \mu_s N$ 变为
$$\tfrac{1}{3} M g\sin\theta \le \mu_s\,(M g\cos\theta) \;\Longrightarrow\; \mu_s \ge \tfrac{1}{3}\tan\theta$$
Steep inclines or low-$\mu_s$ surfaces eventually drive the cylinder to slip.斜面越陡或 $\mu_s$ 越小，最终会导致圆柱打滑。