Each item restates the prompt and choices, marks the correct letter, and gives a brief justification. Numbers use $g = 9.8~\mathrm{m/s^2}$, $G = 6.67 \times 10^{-11}~\mathrm{N \cdot m^2/kg^2}$.每道题均重申题目和选项，标注正确答案字母，并给出简要说明。数值取 $g = 9.8~\mathrm{m/s^2}$，$G = 6.67 \times 10^{-11}~\mathrm{N \cdot m^2/kg^2}$。
A flywheel of moment of inertia $I = 2.0~\mathrm{kg \cdot m^2}$ spins about its axis at $\omega = 3.0~\mathrm{rad/s}$. Its rotational kinetic energy is一个转动惯量为 $I = 2.0~\mathrm{kg \cdot m^2}$ 的飞轮以 $\omega = 3.0~\mathrm{rad/s}$ 绕轴旋转。其转动动能为
(A) $3~\mathrm{J}$
(B) $6~\mathrm{J}$
(C) $9~\mathrm{J}$
(D) $18~\mathrm{J}$
Answer:答案：(C)
$$K_\text{rot} = \tfrac{1}{2} I \omega^2 = \tfrac{1}{2}(2.0)(3.0)^2 = 9~\mathrm{J}$$
Trap (D) drops the $\tfrac{1}{2}$; (B) forgets to square $\omega$.干扰项 (D) 遗漏了 $\tfrac{1}{2}$；(B) 忘记对 $\omega$ 平方。
Q2EASY6.3 Angular Momentum6.3 角动量No Calculator
A particle of mass $2.0~\mathrm{kg}$ moves at $3.0~\mathrm{m/s}$ in a circle of radius $0.50~\mathrm{m}$. The magnitude of its angular momentum about the center of the circle is一个质量为 $2.0~\mathrm{kg}$ 的质点以 $3.0~\mathrm{m/s}$ 的速度在半径 $0.50~\mathrm{m}$ 的圆形轨道上运动。其相对于圆心的角动量大小为
(A) $1.5~\mathrm{kg \cdot m^2/s}$
(B) $3.0~\mathrm{kg \cdot m^2/s}$
(C) $6.0~\mathrm{kg \cdot m^2/s}$
(D) $12~\mathrm{kg \cdot m^2/s}$
Answer:答案：(B)
For circular motion $\vec v \perp \vec r$, so圆周运动中 $\vec v \perp \vec r$，故
$$L = m v r = 2.0(3.0)(0.50) = 3.0~\mathrm{kg \cdot m^2/s}$$
Q3EASY6.4 Conservation of L6.4 角动量守恒No Calculator
A spinning ice skater pulls her arms in toward her body. With no external torques acting on her, her angular speed一位旋转的花样滑冰运动员将手臂向身体收拢。在没有外力矩作用的情况下，她的角速度
(A)increases增大
(B)decreases减小
(C)stays the same保持不变
(D)depends on initial $\omega$取决于初始角速度 $\omega$
Answer:答案：(A)
$L = I\omega$ is conserved with no external torque. Pulling arms in decreases $I$ (mass closer to the axis), so $\omega$ must increase to compensate. The skater's KE actually goes up, work is done by the arm muscles pulling the mass inward.无外力矩时 $L = I\omega$ 守恒。收拢手臂使 $I$ 减小（质量更靠近轴），故 $\omega$ 必须增大以补偿。运动员的动能实际上增加了，这是手臂肌肉向内拉动质量所做的功。
Q4EASY6.6 Orbital Speed6.6 轨道速度No Calculator
A satellite is in a circular orbit around Earth. As the orbital radius is increased, the satellite's orbital speed一颗卫星围绕地球做圆轨道运动。随着轨道半径增大，卫星的轨道速度
(A)increases增大
(B)decreases减小
(C)stays the same保持不变
(D)depends on satellite mass取决于卫星质量
Answer:答案：(B)
Newton's 2nd law in centripetal form gives向心力形式的牛顿第二定律给出
Q5MEDIUM6.1 Disk vs. Hoop KE6.1 圆盘与圆环转动动能比较No Calculator
A solid disk and a thin hoop have the same mass $M$ and the same radius $R$, and rotate about their central axes at the same angular speed $\omega$. The ratio of the disk's rotational kinetic energy to the hoop's rotational kinetic energy is一个实心圆盘和一个细圆环质量均为 $M$，半径均为 $R$，以相同的角速度 $\omega$ 绕各自的中心轴旋转。圆盘的转动动能与圆环的转动动能之比为
(A) $1/4$
(B) $1/2$
(C) $1$
(D) $2$
Answer:答案：(B)
$K_\text{rot} \propto I$ at fixed $\omega$. With $I_\text{disk} = \tfrac{1}{2}MR^2$ and $I_\text{hoop} = MR^2$:固定 $\omega$ 时，$K_\text{rot} \propto I$。其中 $I_\text{disk} = \tfrac{1}{2}MR^2$，$I_\text{hoop} = MR^2$：
A constant torque of $5.0~\mathrm{N \cdot m}$ rotates a wheel through an angular displacement of $4.0~\mathrm{rad}$. The work done on the wheel by the torque is一个大小为 $5.0~\mathrm{N \cdot m}$ 的恒定力矩使一个轮子转过 $4.0~\mathrm{rad}$ 的角位移。该力矩对轮子做的功为
A disk of moment of inertia $I$ spins about a vertical axis at angular speed $\omega_0$. A second disk of moment of inertia $2I$, initially at rest, is gently dropped onto the first disk and sticks to it. The final angular speed of the combined disks is一个转动惯量为 $I$ 的圆盘以角速度 $\omega_0$ 绕竖直轴旋转。另一个转动惯量为 $2I$ 的圆盘从静止开始轻轻叠落在第一个圆盘上并与之粘合。两圆盘合并后的最终角速度为
(A) $\omega_0/2$
(B) $\omega_0/3$
(C) $2\omega_0/3$
(D) $\omega_0$
Answer:答案：(B)
Angular momentum about the vertical axis is conserved (no external torque):绕竖直轴的角动量守恒（无外力矩）：
$$I\,\omega_0 = (I + 2I)\,\omega_f \;\Longrightarrow\; \omega_f = \frac{\omega_0}{3}$$
This is the rotational analogue of a perfectly inelastic collision, KE is not conserved.这是完全非弹性碰撞的转动类比，动能不守恒。
Q8MEDIUM6.5 Sphere Rolling Down Incline6.5 实心球沿斜面滚动No Calculator
A uniform solid sphere is released from rest at the top of an incline of vertical height $h$ and rolls without slipping to the bottom. Its speed at the bottom is一个均匀实心球从高度为 $h$ 的斜面顶端由静止开始释放，无滑动地滚到底部。到达底部时的速度为
(A) $\sqrt{2gh}$
(B) $\sqrt{\dfrac{10gh}{7}}$
(C) $\sqrt{\dfrac{4gh}{3}}$
(D) $\sqrt{gh}$
Answer:答案：(B)
For rolling without slipping ($\omega = v/R$) with $I_\text{sphere} = \tfrac{2}{5}MR^2$:无滑动滚动时（$\omega = v/R$），球的转动惯量 $I_\text{sphere} = \tfrac{2}{5}MR^2$：
$$Mgh = \tfrac{1}{2}M v^2 + \tfrac{1}{2}\bigl(\tfrac{2}{5}MR^2\bigr)\!\left(\tfrac{v}{R}\right)^2 = \tfrac{7}{10} M v^2$$
$$v = \sqrt{\frac{10 g h}{7}}$$
Trap (A) is the frictionless slide; (C) is the answer for a disk ($\sqrt{4gh/3}$).干扰项 (A) 为无摩擦滑动的结果；(C) 为圆盘的结果（$\sqrt{4gh/3}$）。
Two satellites orbit the same planet in circular orbits. The orbital radius of satellite 2 is twice that of satellite 1. The ratio $T_2/T_1$ of their orbital periods is两颗卫星在同一行星的圆形轨道上运行。卫星 2 的轨道半径是卫星 1 的两倍。它们的轨道周期之比 $T_2/T_1$ 为
A constant net torque of $4.0~\mathrm{N \cdot m}$ acts on a wheel of moment of inertia $2.0~\mathrm{kg \cdot m^2}$, initially at rest, for $3.0~\mathrm{s}$. The wheel's final angular speed is一个大小为 $4.0~\mathrm{N \cdot m}$ 的恒定合力矩作用在转动惯量为 $2.0~\mathrm{kg \cdot m^2}$ 的轮子上，持续时间 $3.0~\mathrm{s}$，轮子初始静止。轮子的最终角速度为
(A) $2.0~\mathrm{rad/s}$
(B) $4.0~\mathrm{rad/s}$
(C) $6.0~\mathrm{rad/s}$
(D) $12~\mathrm{rad/s}$
Answer:答案：(C)
Angular impulse $= \Delta L$:角冲量 $= \Delta L$：
$$\tau\,\Delta t = I\,\omega \;\Longrightarrow\; \omega = \frac{4.0(3.0)}{2.0} = 6.0~\mathrm{rad/s}$$
Q11MEDIUM6.5 Rolling KE Fraction6.5 滚动动能占比No Calculator
A uniform solid disk rolls without slipping along a horizontal surface with center-of-mass speed $v$. The fraction of its total kinetic energy that is rotational is一个均匀实心圆盘以质心速度 $v$ 在水平面上无滑动地滚动。其总动能中转动动能所占的比例为
(A) $1/4$
(B) $1/3$
(C) $1/2$
(D) $2/3$
Answer:答案：(B)
With $I = \tfrac{1}{2}MR^2$ and $\omega = v/R$:利用 $I = \tfrac{1}{2}MR^2$，$\omega = v/R$：
A uniform rod of mass $M$ and length $L$, pivoted at one end and initially at rest, is struck and embedded by a bullet of mass $m$ moving with speed $v$ at the rod's free end (perpendicular to the rod). The angular speed of the rod-plus-bullet system just after the collision is一根质量为 $M$、长度为 $L$ 的均匀细杆，一端铰支，初始静止，被质量为 $m$、速度为 $v$ 的子弹垂直射入其自由端并嵌入其中。碰撞后，杆与子弹组合体的角速度为
(A) $\dfrac{mv}{ML/3}$
(B) $\dfrac{mv}{L\bigl(M/3 + m\bigr)}$
(C) $\dfrac{mv}{L\bigl(M + m\bigr)/3}$
(D) $\dfrac{mv}{ML}$
Answer:答案：(B)
Conserve angular momentum about the pivot. Bullet's $L_\text{before} = m v L$ (perpendicular at distance $L$). After collision, $I_\text{tot} = \tfrac{1}{3}ML^2 + mL^2 = L^2(M/3 + m)$.对铰支点应用角动量守恒。子弹的初始角动量 $L_\text{before} = m v L$（在距铰支点 $L$ 处垂直射入）。碰撞后，$I_\text{tot} = \tfrac{1}{3}ML^2 + mL^2 = L^2(M/3 + m)$。
$$m v L = L^2\!\left(\tfrac{M}{3} + m\right)\,\omega \;\Longrightarrow\; \omega = \frac{m v}{L\bigl(M/3 + m\bigr)}$$
Q13MEDIUM6.1 Work to Stop Wheel6.1 使轮子停转所做的功No Calculator
A wheel of moment of inertia $0.50~\mathrm{kg \cdot m^2}$ spins at $4.0~\mathrm{rad/s}$. The magnitude of the work that must be done on the wheel to bring it to rest is一个转动惯量为 $0.50~\mathrm{kg \cdot m^2}$ 的轮子以 $4.0~\mathrm{rad/s}$ 旋转。使轮子停止旋转所需做的功的大小为
Q14HARD6.4 Person on Rotating Platform6.4 人站在旋转台上Calculator
A horizontal turntable of moment of inertia $I_p = 100~\mathrm{kg \cdot m^2}$ rotates freely about a vertical axis at $\omega_0 = 1.0~\mathrm{rad/s}$. A person of mass $50~\mathrm{kg}$ stands at rest on the turntable at $r = 1.0~\mathrm{m}$ from the axis. The person then walks to the axis (treat the person as a point mass throughout). The new angular speed of the turntable is一个转动惯量为 $I_p = 100~\mathrm{kg \cdot m^2}$ 的水平转台以 $\omega_0 = 1.0~\mathrm{rad/s}$ 绕竖直轴自由旋转。一个质量为 $50~\mathrm{kg}$ 的人静止站在距轴 $r = 1.0~\mathrm{m}$ 处。该人随后走向轴心（全程将人视为质点）。转台的新角速度为
(A) $0.50~\mathrm{rad/s}$
(B) $1.0~\mathrm{rad/s}$
(C) $1.5~\mathrm{rad/s}$
(D) $2.0~\mathrm{rad/s}$
Answer:答案：(C)
Angular momentum about the axis is conserved. With person on the platform, person and platform share $\omega$.绕轴的角动量守恒。人站在转台上时，人与转台共享同一 $\omega$。
$$\bigl(I_p + m r^2\bigr)\,\omega_0 = I_p\,\omega_f$$
A satellite is placed in a circular orbit such that its orbital period equals one Earth day. Take $GM_\mathrm{E} = 4.0 \times 10^{14}~\mathrm{m^3/s^2}$ and $T_\mathrm{day} = 8.64 \times 10^4~\mathrm{s}$. The orbital radius (measured from Earth's center) is closest to一颗卫星被置于圆形轨道上，使其轨道周期等于地球自转一天。取 $GM_\mathrm{E} = 4.0 \times 10^{14}~\mathrm{m^3/s^2}$，$T_\mathrm{day} = 8.64 \times 10^4~\mathrm{s}$。轨道半径（从地心算起）最接近
Trap (D) is roughly the lunar orbital radius; (B) is the altitude above Earth's surface ($r - R_\mathrm{E}$).干扰项 (D) 约为月球轨道半径；(B) 为高于地球表面的轨道高度（$r - R_\mathrm{E}$）。
Q16HARD6.5 Race Down the Incline6.5 斜面竞速No Calculator
A solid sphere, a uniform solid disk, and a thin hoop, all of mass $M$ and radius $R$, are released simultaneously from rest at the top of the same incline and roll without slipping to the bottom. Which arrives first?一个实心球、一个均匀实心圆盘和一个细圆环，质量均为 $M$，半径均为 $R$，同时从同一斜面顶端由静止释放，无滑动地滚到底部。哪个最先到达？
(A)Sphere实心球
(B)Disk圆盘
(C)Hoop圆环
(D)All arrive together三者同时到达
Answer:答案：(A)
For rolling-without-slipping, energy conservation gives无滑动滚动时，能量守恒给出
$$v_\text{bottom} = \sqrt{\frac{2 g h}{1 + I/(MR^2)}}$$
Smaller $I/(MR^2)$ $\Rightarrow$ larger $v$. With $\tfrac{2}{5}, \tfrac{1}{2}, 1$ for sphere/disk/hoop, the sphere wins. Mass and radius drop out, only the shape's $I/(MR^2)$ ratio matters.$I/(MR^2)$ 越小，$v$ 越大。球、圆盘、圆环对应的值分别为 $\tfrac{2}{5}$、$\tfrac{1}{2}$、$1$，故实心球最先到达。质量和半径消去，只有形状的 $I/(MR^2)$ 比值起作用。
Q17HARD6.3 Direction of Angular Momentum6.3 角动量的方向No Calculator
A particle moves in a circular path in the $xy$-plane, traveling counter-clockwise as viewed from the $+z$ axis. Its angular momentum about the center of the circle points in the一个质点在 $xy$ 平面内沿圆形轨道运动，从 $+z$ 轴方向俯视为逆时针。其相对于圆心的角动量指向
(A)$+x$ direction$+x$ 方向
(B)$-x$ direction$-x$ 方向
(C)$+z$ direction$+z$ 方向
(D)$-z$ direction$-z$ 方向
Answer:答案：(C)
$\vec L = \vec r \times \vec p$. Right-hand rule with $\vec r$ in the $xy$-plane and $\vec p$ tangent to the circle (also in the $xy$-plane) gives $\vec L$ perpendicular to the plane, along $+\hat k$ for counter-clockwise motion viewed from $+z$ (the "spin direction of the angular-velocity vector"). Trap (D) reverses the sense.$\vec L = \vec r \times \vec p$。$\vec r$ 在 $xy$ 平面内，$\vec p$ 与圆相切（也在 $xy$ 平面内），右手定则给出 $\vec L$ 垂直于该平面，从 $+z$ 方向观察逆时针运动时沿 $+\hat k$ 方向（即角速度矢量的方向）。干扰项 (D) 方向相反。
Q18HARD6.6 Total Orbital Energy6.6 轨道总机械能No Calculator
A satellite of mass $m$ orbits a planet of mass $M$ in a circular orbit of radius $r$. The total mechanical energy of the satellite is质量为 $m$ 的卫星在质量为 $M$ 的行星的半径为 $r$ 的圆形轨道上运行。卫星的总机械能为
(A) $-\dfrac{GMm}{r}$
(B) $-\dfrac{GMm}{2r}$
(C) $+\dfrac{GMm}{2r}$
(D) $+\dfrac{GMm}{r}$
Answer:答案：(B)
For a circular orbit, $v^2 = GM/r$, so $K = \tfrac{1}{2}m v^2 = \tfrac{GMm}{2r}$. With $U = -GMm/r$:圆轨道上 $v^2 = GM/r$，故 $K = \tfrac{1}{2}m v^2 = \tfrac{GMm}{2r}$。其中 $U = -GMm/r$：
$$E = K + U = \frac{G M m}{2r} - \frac{G M m}{r} = -\frac{G M m}{2r}$$
The negative sign signals a bound state. Note $E = -K$ and $U = 2E$, the "virial" relations.负号表明这是束缚态。注意 $E = -K$，$U = 2E$，即"维里定理"关系。
(d)Fraction of $K_i$ lost: $(1.62 - 0.81)/1.62 = 0.50$, exactly half. The missing $0.81~\mathrm{J}$ is dissipated as heat / sound / surface deformation during the brief slip between the two disks as they reach a common $\omega$. This is the rotational analogue of a perfectly inelastic linear collision (KE not conserved).$K_i$ 的损失比例：$(1.62 - 0.81)/1.62 = 0.50$，恰好一半。缺失的 $0.81~\mathrm{J}$ 在两圆盘达到共同 $\omega$ 过程中通过短暂滑动以热能、声音和表面形变的形式耗散。这是完全非弹性线性碰撞的转动类比（动能不守恒）。
FRQ 2MEDIUM6.5 Sphere Rolling Down Incline6.5 实心球沿斜面滚动Calculator
$M = 0.50~\mathrm{kg}$, $R = 0.10~\mathrm{m}$ solid sphere, released from rest at vertical height $h = 0.80~\mathrm{m}$ on an incline of $\theta = 30^\circ$, rolls without slipping.$M = 0.50~\mathrm{kg}$，$R = 0.10~\mathrm{m}$ 的实心球，从倾角 $\theta = 30^\circ$、竖直高度 $h = 0.80~\mathrm{m}$ 处由静止释放，无滑动地滚动。
(a)Energy conservation with $I = \tfrac{2}{5}MR^2$ and $\omega = v/R$:利用 $I = \tfrac{2}{5}MR^2$，$\omega = v/R$ 进行能量守恒：
$$M g h = \tfrac{1}{2} M v^2 + \tfrac{1}{2}\bigl(\tfrac{2}{5}MR^2\bigr)\!\left(\tfrac{v}{R}\right)^2 = \tfrac{7}{10} M v^2$$
(c)Newton's 2nd law. Translation: $M g\sin\theta - f_s = M a$. Rotation about CM: $f_s R = I\alpha = \tfrac{2}{5}MR^2\alpha$, and $a = R\alpha$ gives $f_s = \tfrac{2}{5}M a$. Substitute:牛顿第二定律。平动：$M g\sin\theta - f_s = M a$。绕质心转动：$f_s R = I\alpha = \tfrac{2}{5}MR^2\alpha$，$a = R\alpha$ 给出 $f_s = \tfrac{2}{5}M a$。代入：
$$M g\sin\theta - \tfrac{2}{5} M a = M a \;\Longrightarrow\; a = \tfrac{5}{7}\,g\sin\theta$$
(d)Static-friction force is $f_s = \tfrac{2}{5}M a = \tfrac{2}{7} M g\sin\theta$. Constraint $f_s \le \mu_s\,M g\cos\theta$:静摩擦力 $f_s = \tfrac{2}{5}M a = \tfrac{2}{7} M g\sin\theta$。约束条件 $f_s \le \mu_s\,M g\cos\theta$：
Energy after the collision is $K_\text{rot} = \tfrac{1}{2}I\omega^2 \approx 42.9~\mathrm{J}$. To swing up by angle $\theta$ from vertical, the CM rises by $x_\text{cm}(1 - \cos\theta)$. Setting $K_\text{rot} = (M+m) g\,x_\text{cm}(1 - \cos\theta)$:碰撞后的转动动能为 $K_\text{rot} = \tfrac{1}{2}I\omega^2 \approx 42.9~\mathrm{J}$。摆动至偏离竖直方向 $\theta$ 时，质心上升 $x_\text{cm}(1 - \cos\theta)$。令 $K_\text{rot} = (M+m) g\,x_\text{cm}(1 - \cos\theta)$：
⚠ This implies $\cos\theta \approx -19.3$, which is impossible. The post-collision rotational KE ($42.9~\mathrm{J}$) exceeds the maximum gravitational PE barrier ($2(M+m)g\,x_\text{cm} \approx 4.23~\mathrm{J}$) by nearly 10×, so the system passes over the top and continues to rotate. There is no momentary stop with the given parameters; the rod-plus-bullet executes full rotations. For a finite swing-up angle, $v_0$ would need to be roughly $10\times$ smaller (e.g. $\sim 20~\mathrm{m/s}$ gives $\theta \approx 37^\circ$).⚠ 这意味着 $\cos\theta \approx -19.3$，这在物理上不可能。碰撞后的转动动能（$42.9~\mathrm{J}$）是最大重力势能壁垒（$2(M+m)g\,x_\text{cm} \approx 4.23~\mathrm{J}$）的近 10 倍，因此系统越过最高点并继续旋转。在给定参数下不存在瞬间停止，杆与子弹组合体做完整圆周运动。若要得到有限摆角，$v_0$ 需约减小 10 倍（例如 $\sim 20~\mathrm{m/s}$ 时 $\theta \approx 37^\circ$）。
Even though angular momentum is conserved, most of the bullet's KE is dissipated into the rod's deformation as the bullet embeds.尽管角动量守恒，子弹大部分动能在嵌入过程中以杆的形变方式耗散。
Positive, raising the orbit costs energy (the satellite ends with more total mechanical energy, even though it moves slower; the PE rise outweighs the KE drop).结果为正，说明升高轨道需要输入能量（卫星的总机械能增大，尽管速度降低，势能的增加超过了动能的减少）。