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1.1.A |
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AP 物理 C:力学
AP 物理 C:力学是基于微积分的版本:运动学、力与平动动力学、功、能与功率、线动量、力矩与 转动动力学、转动系统的能量与动量、振动。它覆盖的范围比物理 1 小,但深度大得多;微积分不是 点缀——结论正是由它推导出来的,考试也要求你这样使用它。
具体来说,你需要能自如地对变化量积分求总量,并通过求导从位置得到速度、再得到加速度,无需 提示。转动动力学是多数学生耗时最多的部分,因为转动惯量需要以积分形式建立,而不是背下来。
自由作答部分要求先用符号完成推导,最后才出现数字,因此请练习先得到通式、最后再代入数值。 本站笔记按 College Board 的单元编排,每单元一页,每个推导都完整展示,而不是只引用结论。
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1
运动学
1.1
标量与矢量
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来源:美国大学理事会 AP 课程与考试说明
运动学(kinematics)描述运动。两种量:
- 一个标量(scalar)只有大小(大小(magnitude)):距离、速率、质量、时间。
- 一个矢量(vector)有大小和方向:位移、速度、加速度、力。
在这门基于微积分的课程里你把矢量分解成分量(components)并逐分量地把它们相加;一个矢量的大小是 $\sqrt{v_x^2+v_y^2+\cdots}$。物理 C 也用单位矢量(unit vectors)写矢量:$\vec{v}=v_x\hat{i}+v_y\hat{j}$,其中 $\hat{i}$ 和 $\hat{j}$ 是沿 $x$ 和 $y$ 的长度为一的方向——方便,因为微积分然后能分别作用于每个分量。

把一个矢量分解成它的 x 和 y 分量 
A car speedometer: speed is the magnitude of velocity — a scalar rate of motion 探索Explore the dot product
Drag the components of $\vec{A}$ and $\vec{B}$. The dot product $\vec{A}\cdot\vec{B}=A_xB_x+A_yB_y$ reaches zero exactly when the two vectors are perpendicular — the geometry behind $\cos\theta$.
词汇表 训练英文 中文 拼音 Kinematics 运动学 yùn dòng xué scalar 标量 biāo liàng magnitude 大小 dà xiǎo vector 矢量 shǐ liàng components 分量 fèn liàng unit vectors 单位矢量 dān wèi shǐ liàng 1.2
位移、速度与加速度
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Learning Objective Essential Knowledge 1.2.A
Describe a change in an object's position.- 1.2.A.1 When using the object model, the size, shape, and internal configuration are ignored. The object may be treated as a single point with extensive properties such as mass and charge.
- 1.2.A.2 Displacement is the change in an object's position.
- Equation: $\Delta x = x - x_0$
1.2.B
Describe the average velocity and acceleration of an object.- 1.2.B.1 Averages of velocity and acceleration are calculated considering the initial and final states of an object over an interval of time.
- 1.2.B.2 Average velocity is the displacement of an object divided by the interval of time in which that displacement occurs.
- Equation: $\vec{v}_{\text{avg}} = \dfrac{\Delta \vec{x}}{\Delta t}$
- 1.2.B.3 Average acceleration is the change in velocity divided by the interval of time in which that change in velocity occurs.
- Equation: $\vec{a}_{\text{avg}} = \dfrac{\Delta \vec{v}}{\Delta t}$
- 1.2.B.4 An object is accelerating if either the magnitude and/or direction of the object's velocity are changing.
- 1.2.B.5 Calculating average velocity or average acceleration over a very small time interval yields a value that is very close to the instantaneous velocity or instantaneous acceleration.
1.2.C
Describe the instantaneous position, velocity, and acceleration of an object as a function of time.- 1.2.C.1 As the time interval used to calculate the average value of a quantity approaches zero, the average value of that quantity approaches the value of the quantity at that instant, called the instantaneous value.
- 1.2.C.1.i Instantaneous velocity is the rate of change of the object's position, which is equal to the derivative of position with respect to time.
- Equation: $\vec{v} = \dfrac{d\vec{r}}{dt}$
- Equation: $v_x = \dfrac{dx}{dt}$
- 1.2.C.1.ii Instantaneous acceleration is the rate of change of the object's velocity, which is equal to the derivative of velocity with respect to time.
- Equation: $\vec{a} = \dfrac{d\vec{v}}{dt}$
- Equation: $a_x = \dfrac{dv_x}{dt}$
- 1.2.C.1.i Instantaneous velocity is the rate of change of the object's position, which is equal to the derivative of position with respect to time.
- 1.2.C.2 Time-dependent functions and instantaneous values of position, velocity, and acceleration can be determined using differentiation and integration.
来源:美国大学理事会 AP 课程与考试说明
抛体运动:两个独立方向 因为运动能连续地变化,把速率定义为导数(derivatives):
$$v=\frac{dx}{dt},\qquad a=\frac{dv}{dt}=\frac{d^2x}{dt^2}.$$反转这个,积分以恢复速度和位置:$$v(t)=v_0+\int_0^t a\,dt,\qquad x(t)=x_0+\int_0^t v\,dt.$$把平均与瞬时区分开:平均速度(average velocity)是一个区间上的 $\Delta x/\Delta t$,而瞬时速度(instantaneous velocity)是一个时刻的导数——切线的斜率,以及一个速度表所显示的。对于恒定加速度这些给出熟悉的运动学方程($v=v_0+at$、$x=x_0+v_0t+\tfrac12at^2$、$v^2=v_0^2+2a\,\Delta x$)。最重要的恒定 $a$ 情况是自由落体(free fall):在地球表面附近每个物体,重的或轻的,一旦空气阻力可忽略就以 $g=9.8\ \text{m/s}^2$ 向下加速。当 $a$ 或 $v$ 随时间变化时,直接用积分。
Worked example. 一个质点以 $x(t)=2t^3-3t^2$(米)移动。求导得速度和加速度:
$$v=\frac{dx}{dt}=6t^2-6t,\qquad a=\frac{dv}{dt}=12t-6.$$在 $t=2\ \text{s}$,$v=24-12=12\ \text{m/s}$ 而 $a=24-6=18\ \text{m/s}^2$。这个微积分联系——不只是恒定 $a$ 公式——是区别物理 C 的东西。Worked example. 一个质点从原点静止开始,$a(t)=6t$。积分:$v=\int 6t\,dt=3t^2$ 而 $x=\int 3t^2\,dt=t^3$。在 $t=2\ \text{s}$,$v=12\ \text{m/s}$ 而 $x=8\ \text{m}$。

在一个速度-时间图上斜率是加速度而面积是位移 
A high-speed train: kinematics describes displacement, velocity and acceleration over time 词汇表 训练英文 中文 拼音 derivatives 导数 dǎo shù average velocity 平均速度 píng jūn sù dù instantaneous velocity 瞬时速度 shùn shí sù dù free fall 自由落体 zì yóu luò tǐ 1.3
运动的表示
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Learning Objective Essential Knowledge 1.3.A
Describe the position, velocity, and acceleration of an object using representations of that object's motion.- 1.3.A.1 Motion can be represented by motion diagrams, figures, graphs, equations, and narrative descriptions.
- 1.3.A.2 For constant acceleration, three kinematic equations can be used to describe instantaneous linear motion in one dimension:
- Equation: $v_x = v_{x0} + a_x t$
- Equation: $x = x_0 + v_{x0} t + \dfrac{1}{2} a_x t^2$
- Equation: $v_x^2 = v_{x0}^2 + 2a_x \left( x - x_0 \right)$
- Note: The equations above are written to indicate motion in the $x$-direction, but these equations can be used in any single dimension as appropriate.
- 1.3.A.3 Near the surface of Earth, the vertical acceleration caused by the force of gravity is downward, constant, and has a measured value approximately equal to
- Equation: $a_g = g \approx 10 \ \text{m/s}^2$.
- 1.3.A.4 Graphs of position, velocity, and acceleration as functions of time can be used to find the relationships between those quantities.
- 1.3.A.4.i An object's instantaneous velocity is the rate of change of the object's position, which is equal to the slope of a line tangent to a point on a graph of the object's position as a function of time.
- Equation: $v_x = \dfrac{dx}{dt}$
- 1.3.A.4.ii An object's instantaneous acceleration is the rate of change of the object's velocity, which is equal to the slope of a line tangent to a point on a graph of the object's velocity as a function of time.
- Equation: $a_x = \dfrac{dv_x}{dt}$
- 1.3.A.4.iii The displacement of an object during a time interval is equal to the area under the curve of a graph of the object's velocity as a function of time (i.e., the area bounded by the function and the horizontal axis for the appropriate interval).
- Equation: $\Delta x = \displaystyle\int_{t_1}^{t_2} v_x(t) \ dt$
- 1.3.A.4.iv The change in velocity of an object during a time interval is equal to the area under the curve of a graph of the acceleration of the object as a function of time.
- Equation: $\Delta v_x = \displaystyle\int_{t_1}^{t_2} a_x(t) \ dt$
- 1.3.A.4.i An object's instantaneous velocity is the rate of change of the object's position, which is equal to the slope of a line tangent to a point on a graph of the object's position as a function of time.
Boundary statement: AP Physics C: Mechanics and AP Physics C: Electricity and Magnetism expects that for all situations in which a numerical quantity is required for g, the value $g \approx 10 \ \text{m/s}^2$ will be used. However, students will not be penalized for correctly using the more precise commonly accepted values of $g = 9.81 \ \text{m/s}^2$ or $g = 9.8 \ \text{m/s}^2$.
来源:美国大学理事会 AP 课程与考试说明
在描述、图、表和方程之间流畅地移动:

一个位移-时间图:任何时刻的斜率是速度 - 在一个位置-时间图上,斜率($dx/dt$)是速度。
- 在一个速度-时间图上,斜率是加速度而面积($\int v\,dt$)是位移。
- 在一个加速度-时间图上,面积($\int a\,dt$)是速度的变化。
斜率是导数;面积是积分——这两者在这里是逆运算。

位置、速度和加速度由求导联系 探索Explore the velocity–time graph
Change the start velocity and acceleration. The slope of a $v$–$t$ line is the acceleration $a=\tfrac{dv}{dt}$, and the area underneath is the displacement $\Delta x=\int v\,dt$ — slope and area are the graphical faces of the derivative and the integral.
1.4
参考系与相对运动
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Learning Objective Essential Knowledge 1.4.A
Describe the reference frame of a given observer.- 1.4.A.1 The choice of reference frame will determine the direction and magnitude of quantities measured by an observer in that reference frame.
1.4.B
Describe the motion of objects as measured by observers in different inertial reference frames.- 1.4.B.1 Measurements from a given reference frame may be converted to measurements from another reference frame.
- 1.4.B.2 The observed velocity of an object results from the combination of the object's velocity and the velocity of the observer's reference frame.
- 1.4.B.2.i Combining the motion of an object and the motion of an observer in a given reference frame involves the addition or subtraction of vectors.
- 1.4.B.2.ii The acceleration of any object is the same as measured from all inertial reference frames.
Boundary statement: Unless otherwise stated, the frame of reference of any problem may be assumed to be inertial.
来源:美国大学理事会 AP 课程与考试说明
所有运动都相对于一个参考系(reference frame)。通过矢量加法组合速度:$\vec{v}_{A/C}=\vec{v}_{A/B}+\vec{v}_{B/C}$。这处理横穿河流的船和移动车辆上的乘客——相对运动(relative motion)问题。把下标读作一个链:"A 相对于 B"加"B 相对于 C"给出"A 相对于 C"。
Worked example. 一艘在静水里以 $4.0\ \text{m/s}$ 移动的船直向横穿一条以 $3.0\ \text{m/s}$ 流动的河。相对于地面这艘船以 $\sqrt{4.0^2+3.0^2}=5.0\ \text{m/s}$ 移动,朝下游成角。若河宽 $80\ \text{m}$,穿越仍取 $t=\dfrac{80}{4.0}=20\ \text{s}$ ——只有横流分量穿过河;水流只把船带 $60\ \text{m}$ 下游。
词汇表 训练英文 中文 拼音 reference frame 参考系 cān kǎo xì relative motion 相对运动 xiāng duì yùn dòng 1.5
二维或三维运动
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Learning Objective Essential Knowledge 1.5.A
Describe the motion of an object moving in two or three dimensions.- 1.5.A.1 Motion in two or three dimensions can be analyzed using one-dimensional kinematic relationships if the motion is separated into components.
- 1.5.A.2 Velocity and acceleration may be different in each dimension and may be nonuniform.
- 1.5.A.3 Motion in one dimension may be changed without causing a change in a perpendicular dimension.
- 1.5.A.4 Projectile motion is a special case of two-dimensional motion that has zero acceleration in one dimension and constant, nonzero acceleration in the second dimension.
Boundary statement: AP Physics C: Mechanics only expects students to quantitatively analyze the motion of an object in two dimensions. AP Physics C: Electricity and Magnetism expects students to also qualitatively describe the motion of a particle in three dimensions.
来源:美国大学理事会 AP 课程与考试说明
在二维或三维里,位置是一个矢量 $\vec{r}(t)=\langle x(t),y(t),z(t)\rangle$,而速度和加速度是它的连续导数——每个分量独立处理。
Worked example. $\vec{r}(t)=\big(2t^2\big)\hat{i}+\big(4t-t^3\big)\hat{j}$(米)。对每个分量求导:$\vec{v}=4t\,\hat{i}+(4-3t^2)\,\hat{j}$ 而 $\vec{a}=4\,\hat{i}-6t\,\hat{j}$。在 $t=1\ \text{s}$:$\vec{v}=4\hat{i}+1\hat{j}$,所以速率是 $\sqrt{17}\approx4.1\ \text{m/s}$ ——没有新物理,只是每分量一个导数。

一个以一个角度发射的抛体:水平和竖直运动是独立的 对于抛体运动(projectile motion),水平和竖直运动是独立的,只由时间 $t$ 联系:水平地 $a_x=0$(恒定速度)、竖直地 $a_y=-g$。路径是一条抛物线。这个分量方法延伸到任何沿每个轴的加速度已知的二维运动。
Worked example. 一个球以 $20\ \text{m/s}$、水平之上 $30^{\circ}$ 发射($g=9.8\ \text{m/s}^2$)。分量是 $v_{0x}=20\cos30^{\circ}=17.3\ \text{m/s}$ 而 $v_{0y}=20\sin30^{\circ}=10\ \text{m/s}$。竖直运动设定时间:总飞行 $=\dfrac{2v_{0y}}{g}=\dfrac{20}{9.8}=2.0\ \text{s}$、最大高度 $=\dfrac{v_{0y}^2}{2g}=5.1\ \text{m}$,而射程 $=v_{0x}\times2.0=35\ \text{m}$。

一个以一个角度发射的抛体:速度分量、最大高度和射程 探索Explore projectile motion
Fire the projectile, then vary the angle and speed. The horizontal motion is steady while gravity acts only downward — together they trace a parabola. Find the angle that gives the greatest range (it peaks near $45^\circ$), and try the Moon.
词汇表 训练英文 中文 拼音 projectile motion 抛体运动 pāo tǐ yùn dòng 1.5
考试技巧
- 在相加之前把每个矢量分解成分量——绝不直接相加成角度的大小。
- 在物理 C 上你被期望用微积分:速度是 $\vec v=\tfrac{d\vec r}{dt}$ 而加速度 $\vec a=\tfrac{d\vec v}{dt}$;用积分反转。
- 携带并检查单位,并用符号处理方向(选择一个正轴并坚持它)。
- 对功类型的量用点积,对力矩和角动量用叉积。
- 草绘矢量——一个图捕捉代数隐藏的符号和方向错误。
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2
力与平动动力学
2.1
系统与质心
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Learning Objective Essential Knowledge 2.1.A
Describe the properties and interactions of a system.- 2.1.A.1 System properties are determined by the interactions between objects within the system.
- 2.1.A.2 If the properties or interactions of the constituent objects within a system are not important in modeling the behavior of the macroscopic system, the system can itself be treated as a single object.
- 2.1.A.3 Systems may allow interactions between constituent parts of the system and the environment, which may result in the transfer of energy or mass.
- 2.1.A.4 Individual objects within a chosen system may behave differently from each other as well as from the system as a whole.
- 2.1.A.5 The internal structure of a system affects the analysis of that system.
- 2.1.A.6 As variables external to a system are changed, the system's substructure may change.
2.1.B
Describe the location of a system's center of mass with respect to the system's constituent parts.- 2.1.B.1 For objects or systems with symmetrical mass distributions, the center of mass is located on lines of symmetry.
- 2.1.B.2 The location of a system's center of mass along a given axis can be calculated using the equation
- Equation: $\vec{x}_{\text{cm}} = \dfrac{\sum m_i \vec{x}_i}{\sum m_i}$
- 2.1.B.3 For a nonuniform solid that can be considered as a collection of differential masses, $dm$, the solid's center of mass can be calculated using the equation
- Equation: $\vec{r}_{\text{cm}} = \dfrac{\int \vec{r}\, dm}{\int dm}$
- 2.1.B.3.i The linear mass density of a rod or other linear rigid body is the derivative of the rod's mass with respect to the position of the differential mass element on the rigid body.
- Equation: $\lambda = \dfrac{d}{d\ell} m(\ell)$
- 2.1.B.3.ii If a function of mass density is given for a solid, the total mass can be determined by integrating the mass density over the length (one dimension), area (two dimensions), or volume (three dimensions) of the solid. For example:
- Equation: $M_{\text{total}} = \int \rho(r)\, dV$
- 2.1.B.4 A system can be modeled as a singular object that is located at the system's center of mass.
来源:美国大学理事会 AP 课程与考试说明
一个系统(system)是你分析的物体,当作它质心(center of mass)处的一个点。对于一组质量,$x_{\text{cm}}=\dfrac{\sum m_i x_i}{\sum m_i}$;对于一个连续物体,$x_{\text{cm}}=\dfrac{1}{M}\int x\,dm$。只有外力移动质心。
词汇表 训练英文 中文 拼音 system 系统 xì tǒng center of mass 质心 zhì xīn 2.2
力与受力图
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Learning Objective Essential Knowledge 2.2.A
Describe a force as an interaction between two objects or systems- 2.2.A.1 Forces are vector quantities that describe the interactions between objects or systems.
- 2.2.A.1.i A force exerted on an object or system is always due to the interaction of that object or system with another object or system.
- 2.2.A.1.ii An object or system cannot exert a net force on itself.
- 2.2.A.2 Contact forces describe the interaction of an object or system touching another object or system and are macroscopic effects of interatomic electric forces.
2.2.B
Describe the forces exerted on an object or system using a free-body diagram.- 2.2.B.1 Free-body diagrams are useful tools for visualizing forces being exerted on a single object or system and for determining the equations that represent a physical situation.
- 2.2.B.2 The free-body diagram of an object or system shows each of the forces exerted on the object or system by the environment.
- 2.2.B.3 Forces exerted on an object or system are represented as vectors originating from the representation of the center of mass, such as a dot. A system is treated as though all of its mass is located at the center of mass.
- 2.2.B.4 A coordinate system with one axis parallel to the direction of acceleration of the object or system simplifies the translation from free-body diagram to algebraic representation. For example, in a free-body diagram of an object on an inclined plane, it is useful to set one axis parallel to the surface of the incline.
Boundary statement: AP Physics C: Mechanics and AP Physics C: Electricity and Magnetism only expect students to depict the forces exerted on objects, not the force components on free-body diagrams. On the AP Physics exams, individual forces represented on a free-body diagram must be drawn as individual straight arrows, originating on the dot and pointing in the direction of the force. Individual forces that are in the same direction must be drawn side by side, not overlapping.
来源:美国大学理事会 AP 课程与考试说明
一个力(force)是一个推或拉(一个矢量,以牛顿)。一个受力图(free-body diagram)画一个物体,带它上每个力的一个箭头——重力、法向、张力、摩擦、施加、阻力。先画它;它设置每个动力学方程。

一个受力图显示作用在一个物体上的每个力 探索Balance the forces on a free-body diagram
A free-body diagram shows every force on one object as an arrow. The object accelerates only if the forces don't cancel — the net force sets $a=F/m$.
词汇表 训练英文 中文 拼音 force 力 lì free-body diagram 受力图 shòu lì tú 2.3
牛顿第三定律
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Learning Objective Essential Knowledge 2.3.A
Describe the interaction of two objects or systems using Newton's third law and a representation of paired forces exerted on each object or system.- 2.3.A.1 Newton's third law describes the interaction of two objects or systems in terms of the paired forces that each exerts on the other.
- Equation: $\vec{F}_{\text{A on B}} = -\vec{F}_{\text{B on A}}$
- 2.3.A.2 Interactions between objects within a system (internal forces) do not influence the motion of a system's center of mass.
- 2.3.A.3 Tension is the macroscopic net result of forces that infinitesimal segments of a string, cable, chain, or similar system exert on each other in response to an external force.
- 2.3.A.3.i An ideal string has negligible mass and does not stretch when under tension.
- 2.3.A.3.ii The tension in an ideal string is the same at all points within the string.
- 2.3.A.3.iii In a string with nonnegligible mass, tension may not be the same at all points within the string.
- 2.3.A.3.iv An ideal pulley is a pulley that has negligible mass and rotates about an axle through its center of mass with negligible friction.
来源:美国大学理事会 AP 课程与考试说明
牛顿第三定律(Newton's third law):A 推 B,B 相等且相反地推回。这一对作用在不同的物体上,所以它在一个受力图内从不抵消。

一个牛顿第三定律对:两个不同物体上大小相等、方向相反的力 词汇表 训练英文 中文 拼音 Newton's third law 牛顿第三定律 niú dùn dì sān dìng lǜ 2.4
牛顿第一定律
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Learning Objective Essential Knowledge 2.4.A
Describe the conditions under which a system's velocity remains constant.- 2.4.A.1 The net force on a system is the vector sum of all forces exerted on the system.
- 2.4.A.2 Translational equilibrium is the configuration of forces such that the net force exerted on a system is zero.
- Derived equation: $\sum \vec{F}_i = 0$
- 2.4.A.3 Newton's first law states that if the net force exerted on a system is zero, the velocity of that system will remain constant.
- 2.4.A.4 Forces may be balanced in one dimension but unbalanced in another. The system's velocity will change only in the direction of the unbalanced force.
- 2.4.A.5 An inertial reference frame is one from which an observer would verify Newton's first law of motion.
来源:美国大学理事会 AP 课程与考试说明
牛顿第一定律(惯性(inertia)):以零合力(net force),速度保持恒定——物体处于平动平衡(translational equilibrium)。
词汇表 训练英文 中文 拼音 inertia 惯性 guàn xìng net force 合力 hé lì translational equilibrium 平动平衡 píng dòng píng héng 2.5
牛顿第二定律
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Learning Objective Essential Knowledge 2.5.A
Describe the conditions under which a system's velocity changes.- 2.5.A.1 Unbalanced forces are a configuration of forces such that the net force exerted on a system is not equal to zero.
- 2.5.A.2 Newton's second law of motion states that the acceleration of a system's center of mass has a magnitude proportional to the magnitude of the net force exerted on the system and is in the same direction as that net force.
- Equation: $\vec{a}_{\text{sys}} = \dfrac{\sum \vec{F}}{m_{\text{sys}}} = \dfrac{\vec{F}_{\text{net}}}{m_{\text{sys}}}$
- 2.5.A.3 The velocity of a system's center of mass will only change if a nonzero net external force is exerted on that system.
来源:美国大学理事会 AP 课程与考试说明
一般形式用动量:
$$\sum\vec{F}=\frac{d\vec{p}}{dt}=m\vec{a}\ \ (\text{for constant mass}).$$一次一个轴地应用它。当力取决于速度或位置时,这变成一个要解的微分方程(differential equation)。Worked example. 一个 $2.0\ \text{kg}$ 的木块沿一个 $30^{\circ}$ 的无摩擦斜面(incline)滑下。只有重力的沿坡道分量驱动它,所以 $a=g\sin30^{\circ}=9.8(0.5)=4.9\ \text{m/s}^2$,与质量无关。
Worked example (two bodies). 一个无摩擦桌子上的 $6.0\ \text{kg}$ 小车被一根越过一个轻滑轮的绳子拉向一个悬挂的 $2.0\ \text{kg}$ 质量。把这一对当作一个系统:只有悬挂的重量驱动它,所以 $a=\dfrac{2.0(9.8)}{8.0}=2.45\ \text{m/s}^2$。然后单独隔离小车以求绳子张力(tension):$T=6.0(2.45)\approx15\ \text{N}$。先系统求 $a$、单个物体求内力——那两步是标准的模式。

A rocket launch: net force changes momentum — F_net = ma for translational dynamics 词汇表 训练英文 中文 拼音 differential equation 微分方程 wēi fēn fāng chéng incline 斜面 xié miàn tension 张力 zhāng lì 2.6
万有引力
大纲
Learning Objective Essential Knowledge 2.6.A
Describe the gravitational interaction between two objects or systems with mass.- 2.6.A.1 Newton's law of universal gravitation describes the gravitational force between two objects or systems as directly proportional to each of their masses and inversely proportional to the square of the distance between the systems' centers of mass.
- Equation: $\left| \vec{F}_g \right| = G \dfrac{m_1 m_2}{r^2}$
- 2.6.A.1.i The gravitational force is attractive.
- 2.6.A.1.ii The gravitational force is always exerted along the line connecting the center of mass of the two interacting systems.
- 2.6.A.1.iii The gravitational force on a system can be considered to be exerted on the system's center of mass.
- 2.6.A.2 A field models the effects of a noncontact force exerted on an object at various positions in space.
- 2.6.A.2.i The magnitude of the gravitational field created by a system of mass $M$ at a point in space is equal to the ratio of the gravitational force exerted by the system on a test object of mass $m$ to the mass of the test object.
- Derived equation: $\left| \vec{g} \right| = \dfrac{\left| \vec{F}_g \right|}{m} = G \dfrac{M}{r^2}$
- 2.6.A.2.ii If the gravitational force is the only force exerted on an object, the observed acceleration of the object (in $\text{m/s}^2$) is numerically equal to the magnitude of the gravitational field strength (in $\text{N/kg}$) at that location.
- 2.6.A.2.i The magnitude of the gravitational field created by a system of mass $M$ at a point in space is equal to the ratio of the gravitational force exerted by the system on a test object of mass $m$ to the mass of the test object.
- 2.6.A.3 The gravitational force exerted by an astronomical body on a relatively small nearby object is called weight.
- Derived equation: $\text{Weight} = F_g = mg$
2.6.B
Describe situations in which the gravitational force can be considered constant.- 2.6.B.1 If the gravitational force between two systems' centers of mass has a negligible change as the relative position of the two systems changes, the gravitational force can be considered constant at all points between the initial and final positions of the systems.
- 2.6.B.2 Near the surface of Earth, the strength of the gravitational field is
- Equation: $g \approx 10\ \text{N/kg}$
2.6.C
Describe the conditions under which the magnitude of a system's apparent weight is different from the magnitude of the gravitational force exerted on that system.- 2.6.C.1 The magnitude of the apparent weight of a system is the magnitude of the normal force exerted on the system.
- 2.6.C.2 If the system is accelerating, the apparent weight of the system is not equal to the magnitude of the gravitational force exerted on the system.
- 2.6.C.3 A system appears weightless when there are no forces exerted on the system or when the force of gravity is the only force exerted on the system.
- 2.6.C.4 The equivalence principle states that an observer in a noninertial reference frame is unable to distinguish between an object's apparent weight and the gravitational force exerted on the object by a gravitational field.
2.6.D
Describe inertial and gravitational mass.- 2.6.D.1 Objects have inertial mass, or inertia, a property that determines how much an object's motion resists changes when interacting with another object.
- 2.6.D.2 Gravitational mass is related to the force of attraction between two systems with mass.
- 2.6.D.3 Inertial mass and gravitational mass have been experimentally verified to be equivalent.
2.6.E
Describe the gravitational force exerted on an object by a uniform spherical distribution of mass.- 2.6.E.1 The net gravitational force exerted on an object by a uniform spherical distribution of mass is the sum of the individual forces from small differential masses that comprise the distribution.
- 2.6.E.2 Newton's shell theorem describes the net gravitational force exerted on an object by a uniform spherical shell of mass.
- 2.6.E.2.i The net gravitational force exerted on an object inside a thin spherical shell is zero.
- 2.6.E.2.ii The net gravitational force exerted on an object outside a thin spherical shell can be determined by treating the shell as a single massive object located at the center of the shell.
- 2.6.E.2.iii An object inside a sphere of uniform density experiences a net gravitational force from only a partial mass of the sphere.
- 2.6.E.2.iv The partial mass of a sphere that contributes to the net gravitational force exerted on an object within that sphere is the portion of the sphere's mass located a distance less than or equal to the object's distance from the center of the sphere and can be calculated using the density of the sphere.
- Derived equation: $m_{\text{partial}} = \rho \dfrac{4}{3} \pi \left( r_{\text{partial}} \right)^3$
- 2.6.E.3 The gravitational force exerted on an object within a uniform sphere can be shown to be proportional to the object's distance from the sphere's center.
- Derived equation: $F_{g,\text{partial}} = -k r_{\text{partial}}$
Boundary statement: AP Physics C: Mechanics does not expect students to mathematically prove or derive Newton's shell theorem.
来源:美国大学理事会 AP 课程与考试说明
轨道运动(开普勒第二定律) 在一个表面附近,重量是 $F_g=mg$。一般地,万有引力定律(Newton's law of gravitation):
$$F_g=\frac{Gm_1m_2}{r^2},$$吸引的且平方反比。引力场是 $g=\dfrac{GM}{r^2}$。
两个质量沿连接它们的线以大小相等、方向相反、平方反比的力互相吸引 词汇表 训练英文 中文 拼音 Newton's law of gravitation 万有引力定律 wàn yǒu yǐn lì dìng lǜ 2.7
动摩擦与静摩擦
大纲
Learning Objective Essential Knowledge 2.7.A
Describe kinetic friction between two surfaces.- 2.7.A.1 Kinetic friction occurs when two surfaces in contact move relative to each other.
- 2.7.A.1.i The kinetic friction force is exerted in a direction opposite the motion of each surface relative to the other surface.
- 2.7.A.1.ii The force of friction between two surfaces does not depend on the size of the surface area of contact.
- 2.7.A.2 The magnitude of the kinetic friction force exerted on an object is the product of the normal force the surface exerts on the object and the coefficient of kinetic friction.
- Equation: $\left| \vec{F}_{f,k} \right| = \left| \mu_k \vec{F}_N \right|$
- 2.7.A.2.i The coefficient of kinetic friction depends on the material properties of the surfaces that are in contact.
- 2.7.A.2.ii Normal force is the perpendicular component of the force exerted on an object by the surface with which it is in contact; it is directed away from the surface.
2.7.B
Describe static friction between two surfaces.- 2.7.B.1 Static friction may occur between the contacting surfaces of two objects that are not moving relative to each other.
- 2.7.B.2 Static friction adopts the value and direction required to prevent an object from slipping or sliding on a surface.
- Equation: $\left| \vec{F}_{f,s} \right| \leq \left| \mu_s \vec{F}_n \right|$
- 2.7.B.2.i Slipping and sliding refer to situations in which two surfaces are moving relative to each other.
- 2.7.B.2.ii There exists a maximum value for which static friction will prevent an object from slipping on a given surface.
- Derived equation: $F_{f,s,\text{max}} = \mu_s F_N$
- 2.7.B.3 The coefficient of static friction is typically greater than the coefficient of kinetic friction for a given pair of surfaces.
来源:美国大学理事会 AP 课程与考试说明
摩擦力(friction)沿一个表面反对滑动:滑动时动摩擦(kinetic)$f_k=\mu_k N$,而滑动之前静摩擦(static)$f_s\le\mu_s N$ 到一个最大值。$N$ 是法向力(normal force)。注意不等式:静摩擦只大到它需要的那么大,而 $\mu_s N$ 是它的上限,不是它的值。
Worked example. 同一个木块在同一个 $30^{\circ}$ 的斜面上,现在带 $\mu_k=0.20$。垂直于坡道:$N=mg\cos30^{\circ}$。沿坡道:$ma=mg\sin30^{\circ}-\mu_k mg\cos30^{\circ}$,所以 $a=g(\sin30^{\circ}-0.20\cos30^{\circ})=9.8(0.500-0.173)=3.2\ \text{m/s}^2$ ——质量又约去。
探索Slide a block down a slope with friction
Friction opposes motion up to a maximum $\mu N$. Tilt the slope until gravity's pull along it beats static friction and the block starts to slide.
词汇表 训练英文 中文 拼音 Friction 摩擦力 mó cā lì kinetic 动摩擦 dòng mó cā static 静摩擦 jìng mó cā normal force 法向力 fǎ xiàng lì 2.8
弹簧力
大纲
Learning Objective Essential Knowledge 2.8.A
Describe the force exerted on an object by an ideal spring.- 2.8.A.1 An ideal spring has negligible mass and exerts a force that is proportional to the change in its length as measured from its relaxed length. A nonideal spring either has nonnegligible mass or exerts a force that is not proportional to the change in its length as measured from its relaxed length.
- 2.8.A.2 The magnitude of the force exerted by an ideal spring on an object is given by Hooke's law:
- Equation: $\vec{F}_s = -k \Delta \vec{x}$
- 2.8.A.3 The force exerted on an object by a spring is always directed toward the equilibrium position of the object–spring system.
2.8.B
Describe the equivalent spring constant of a combination of springs exerting forces on an object.- 2.8.B.1 A collection of springs that exert forces on an object may behave as though they were a single spring with an equivalent spring constant $k_{\text{eq}}$.
- 2.8.B.1.i The inverse of the equivalent spring constant of a set of springs in series is equal to the sum of the inverses of the individual spring constants.
- Derived equation: $\dfrac{1}{k_{\text{eq, series}}} = \sum_i \dfrac{1}{k_i} = \dfrac{1}{k_1} + \dfrac{1}{k_2} + \dots$
- 2.8.B.1.ii The equivalent spring constant of a set of springs arranged in series is smaller than the smallest constituent spring constant.
- 2.8.B.1.iii The equivalent spring constant of a set of springs arranged in parallel is the sum of the individual spring constants.
- Derived equation: $k_{\text{eq, parallel}} = \sum_i k_i = k_1 + k_2 + \dots$
- 2.8.B.1.i The inverse of the equivalent spring constant of a set of springs in series is equal to the sum of the inverses of the individual spring constants.
Boundary statement: AP Physics C: Mechanics only expects students to find the effective spring constant of systems of springs that are arranged either in series or in parallel and does not expect students to find the effective spring constant of a system in which springs are arranged in both series and parallel.
来源:美国大学理事会 AP 课程与考试说明
胡克定律与弹性极限 一个理想弹簧遵循胡克定律(Hooke's law)$F_s=-kx$,一个由弹簧劲度系数(spring constant)$k$ 设定的回复力。它储存的能量是 $U=\tfrac12 kx^2$。

胡克定律:伸长与负载成比例,直到比例极限 组合的弹簧作为一个劲度系数为 $k_{\text{eq}}$ 的等效弹簧。并联(并排、共同分担负载)时劲度系数相加,$k_{\text{eq}}=k_1+k_2$ ——比任一个都更硬。串联(首尾相接)时它们按 $\dfrac{1}{k_{\text{eq}}}=\dfrac{1}{k_1}+\dfrac{1}{k_2}$ 组合——比最软的还软。两根从相反两侧拉一个质量的弹簧也作为并联,给出恢复劲度系数 $k_1+k_2$。无论哪种,简谐运动周期都用 $k_{\text{eq}}$:$T=2\pi\sqrt{m/k_{\text{eq}}}$。
探索Stretch a spring (Hooke's law)
A spring's force is proportional to its extension, $F=kx$ (Hooke's law). Pull harder and the extension grows in step — until the spring's limit.
词汇表 训练英文 中文 拼音 Hooke's law 胡克定律 hú kè dìng lǜ spring constant 弹簧劲度系数 tán huáng jìn dù xì shù 2.9
阻力
大纲
Learning Objective Essential Knowledge 2.9.A
Describe the motion of an object subject to a resistive force.- 2.9.A.1 A resistive force is defined as a velocity-dependent force in the opposite direction of an object's velocity, for example:
- Equation: $\vec{F}_r = -k\vec{v}$
- 2.9.A.2 Applying Newton's second law to an object upon which a resistive force is exerted results in a differential equation for velocity.
- 2.9.A.2.i Using the method of separation of variables, the velocity can be determined by integrating over the proper limits of integration.
- 2.9.A.2.ii The acceleration or position of a moving object that is subject to a velocity-dependent force may be determined using initial conditions of the object and methods of calculus, once a function for velocity is determined.
- 2.9.A.2.iii The position, velocity, and acceleration as functions of time of an object under the influence of a resistive force of the form $\vec{F}_r = -k\vec{v}$ are exponential and have asymptotes that are determined by the initial conditions of the object and the forces exerted on the object.
- 2.9.A.3 Terminal velocity is defined as the maximum speed achieved by an object moving under the influence of a constant force and a resistive force that are exerted on the object in opposite directions. The terminal condition is reached when the net force exerted on the object is zero.
来源:美国大学理事会 AP 课程与考试说明
一个阻力(resistive force)(drag)反对通过一个流体的运动并随速率增长,常常建模为 $F=-bv$ 或 $F=-cv^2$。牛顿第二定律然后给出一个微分方程,例如对一个落下的物体 $m\dfrac{dv}{dt}=mg-bv$。随着速率上升,阻力增强直到它平衡驱动力;物体然后停止加速并以一个恒定的终极速度(terminal velocity)移动,通过把合力(和 $dv/dt$)设为零求出。
Worked example. 对于一个落下的物体,$m\dfrac{dv}{dt}=mg-bv$,终极速度是 $\dfrac{dv}{dt}=0$ 的地方:$mg=bv_T$,所以 $v_T=\dfrac{mg}{b}$。以 $m=0.10\ \text{kg}$ 和 $b=0.50\ \text{kg/s}$,$v_T=\dfrac{0.10(9.8)}{0.50}=2.0\ \text{m/s}$。
要得到完整的运动 $v(t)$,分离变量(separate the variables)并从静止积分:
$$\int_0^{v}\frac{dv'}{mg-bv'}=\int_0^{t}\frac{dt'}{m}\;\Rightarrow\;-\frac{1}{b}\ln\!\frac{mg-bv}{mg}=\frac{t}{m},$$它重排为 $v(t)=v_T\left(1-e^{-bt/m}\right)$ ——速率指数地向 $v_T$ 上升。展示这个积分正是 Physics C 自由回答所奖励的微积分技能,而不只是引用最终公式。
一个通过一个流体落下的物体加速到一个终极速度 
Skydivers in freefall: drag grows with speed until it balances weight at terminal velocity 词汇表 训练英文 中文 拼音 resistive force 阻力 zǔ lì terminal velocity 终极速度 zhōng jí sù dù separate the variables 分离变量 fēn lí biàn liàng 2.10
圆周运动
大纲
Learning Objective Essential Knowledge 2.10.A
Describe the motion of an object traveling in a circular path.- 2.10.A.1 Centripetal acceleration is the component of an object's acceleration directed toward the center of the object's circular path.
- 2.10.A.1.i The magnitude of centripetal acceleration for an object moving in a circular path is the ratio of the object's tangential speed squared to the radius of the circular path.
- Equation: $a_c = \dfrac{v^2}{r}$
- 2.10.A.1.ii Centripetal acceleration is directed toward the center of an object's circular path.
- 2.10.A.1.i The magnitude of centripetal acceleration for an object moving in a circular path is the ratio of the object's tangential speed squared to the radius of the circular path.
- 2.10.A.2 Centripetal acceleration can result from a single force, more than one force, or components of forces that are exerted on an object in circular motion.
- 2.10.A.2.i At the top of a vertical, circular loop, an object requires a minimum speed to maintain circular motion. At this point, and with this minimum velocity, the gravitational force is the only force that causes the centripetal acceleration.
- Derived equation: $v = \sqrt{gr}$
- 2.10.A.2.ii Components of the static friction force and the normal force can contribute to the net force producing centripetal acceleration of an object traveling in a circle on a banked surface.
- 2.10.A.2.iii A component of tension contributes to the net force producing centripetal acceleration experienced by a conical pendulum.
- 2.10.A.2.i At the top of a vertical, circular loop, an object requires a minimum speed to maintain circular motion. At this point, and with this minimum velocity, the gravitational force is the only force that causes the centripetal acceleration.
- 2.10.A.3 Tangential acceleration is the rate at which an object's speed changes and is directed tangent to the object's circular path.
- 2.10.A.4 The net acceleration of an object moving in a circle is the vector sum of the centripetal acceleration and tangential acceleration.
- 2.10.A.5 The revolution of an object traveling in a circular path at a constant speed (uniform circular motion) can be described using period and frequency.
- 2.10.A.5.i The time to complete one full circular path, one full rotation, or a full cycle of oscillatory motion is defined as period, $T$.
- 2.10.A.5.ii The rate at which an object is completing revolutions is defined as frequency, $f$.
- Equation: $T = \dfrac{1}{f}$
- 2.10.A.5.iii For an object traveling at a constant speed in a circular path, the period is given by the derived equation
- Derived equation: $T = \dfrac{2\pi r}{v}$
2.10.B
Describe circular orbits using Kepler's third law.- 2.10.B.1 For a satellite in circular orbit around a central body, the satellite's centripetal acceleration is caused only by gravitational attraction. The period and radius of the circular orbit are related to the mass of the central body.
- Derived equation: $T^2 = \dfrac{4\pi^2}{GM} R^3$
Boundary statement: AP Physics C: Mechanics does not expect students to know Kepler's first or second laws of planetary motion.
来源:美国大学理事会 AP 课程与考试说明
匀速圆周运动 匀速圆周运动有一个指向中心的向心加速度(centripetal acceleration)$a_c=\dfrac{v^2}{r}$,需要一个由一个真实力(张力、重力、摩擦、法向)供应的合向内力 $F_c=\dfrac{mv^2}{r}$。没有向外的力。对于竖直圆和倾斜的弯道,分解真实的力以求哪个提供向心要求。
Worked example. 一个 $0.50\ \text{kg}$ 的球在一根 $1.0\ \text{m}$ 的绳上以 $3.0\ \text{m/s}$ 在一个水平圆里被甩。绳子张力供应整个向心力:$T=\dfrac{mv^2}{r}=\dfrac{0.50(3.0)^2}{1.0}=4.5\ \text{N}$。
Worked example (vertical circle). 在一个半径 $r$ 的竖直环的顶部,重力和法向力都指向中心:$N+mg=\dfrac{mv^2}{r}$。顶部最慢可能的速率是轨道什么都不推($N=0$)的地方:$v_{\min}=\sqrt{gr}$。对于 $r=2.5\ \text{m}$:$v_{\min}=\sqrt{9.8(2.5)}=4.9\ \text{m/s}$。再慢一点小车就在顶部之前离开轨道。

速度沿切线指向;向心力指向中心 词汇表 训练英文 中文 拼音 centripetal acceleration 向心加速度 xiàng xīn jiā sù dù 2.10
考试技巧
- 用 $x_{cm}=\tfrac{1}{M}\int x\,dm$(或对点质量 $\tfrac{\sum m_i x_i}{\sum m_i}$)定位质心并对质量元素用 $\lambda,\sigma,\rho$。
- 合外力移动质心,仿佛所有质量都坐在那里:$\vec F_{net}=M\vec a_{cm}$。
- 清晰地定义你的系统——内力抵消,所以只有外力改变它的动量。
- 利用对称来简化一个质心积分。
- 把质心与重心区分开(在一个均匀场里相同)。
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3
功、能与功率
3.1
平动动能
大纲
Learning Objective Essential Knowledge 3.1.A
Describe the translational kinetic energy of an object in terms of the object's mass and velocity.- 3.1.A.1 An object's translational kinetic energy is given by the equation
- Equation: $K = \dfrac{1}{2}mv^2$
- 3.1.A.2 Translational kinetic energy is a scalar quantity.
- 3.1.A.3 Different observers may measure different values of the translational kinetic energy of an object, depending on the observer's frame of reference.
来源:美国大学理事会 AP 课程与考试说明
动能(kinetic energy)是运动的能量,一个以焦耳(joules)(J)测量的标量(scalar):
$$K=\tfrac{1}{2}mv^2.$$它随速率的平方增长——把速率加倍使 $K$ 变为四倍。一个值得知道的微妙之处:动能取决于观察者的参考系(reference frame)。一个在火车上走的乘客在火车里面测量有一个小的 $K$ 而从地面测量有一个巨大的——两个观察者都对,各在他们自己的参考系里。
词汇表 训练英文 中文 拼音 Kinetic energy 动能 dòng néng scalar 标量 biāo liàng joules 焦耳 jiāo ěr reference frame 参考系 cān kǎo xì 3.2
功
大纲
Learning Objective Essential Knowledge 3.2.A
Describe the work done on an object or system by a given force or collection of forces.- 3.2.A.1 Work is the amount of energy transferred into or out of a system by a force exerted on that system over a distance.
- 3.2.A.1.i The work done by a conservative force exerted on a system is path-independent and only depends on the initial and final configurations of that system.
- 3.2.A.1.ii The work done by a conservative force on a system—or the change in the potential energy of the system—will be zero if the system returns to its initial configuration.
- 3.2.A.1.iii Potential energies are associated only with conservative forces.
- 3.2.A.1.iv The work done by a nonconservative force is path-dependent.
- 3.2.A.1.v The most common nonconservative forces are friction and air resistance.
- 3.2.A.2 Work is a scalar quantity that may be positive, negative, or zero.
- 3.2.A.3 The work done on an object by a variable force is calculated using
- Equation: $W = \displaystyle\int_a^b \vec{F}(r) \cdot d\vec{r}$, where the integral is taken over the path from point $a$ to point $b$.
- 3.2.A.3.i The dot product between two vectors, $\vec{A}$ and $\vec{B}$, results in a scalar quantity of magnitude $\vec{A} \cdot \vec{B} = AB\cos\theta$.
- 3.2.A.3.ii Only the component of the force exerted on a system that is parallel to the displacement of the point of application of the force will change the system's total energy.
- 3.2.A.3.iii If the component of the force exerted on a system that is parallel to the displacement is constant, the work done on the system by the force is given by the derived equation $W = F_{\parallel}d = Fd\cos\theta$.
- 3.2.A.3.iv The component of the force exerted on a system perpendicular to the direction of the displacement of the system's center of mass can change the direction of the system's motion without changing the system's kinetic energy.
- 3.2.A.4 The work–energy theorem states that the change in an object's kinetic energy is equal to the sum of the work (net work) being done by all forces exerted on the object.
- Equation: $\Delta K = \displaystyle\sum W_i = \sum F_{\parallel,i}\, d_i$
- 3.2.A.4.i An external force may change the configuration of a system. The component of the external force parallel to the displacement times the displacement of the point of application of the force gives the change in kinetic energy of the system.
- 3.2.A.4.ii If the system's center of mass and the point of application of the force move the same distance when a force is exerted on a system, then the system may be modeled as an object, and only the system's kinetic energy can change.
- 3.2.A.4.iii The energy dissipated by friction is typically equated to the force of friction times the length of the path over which the force is exerted.
- Equation: $\Delta E_{\text{mech}} = F_f d\cos\theta$
- 3.2.A.5 Work is equal to the area under the curve of a graph of $F_{\parallel}$ as a function of displacement.
Boundary statement: AP Physics C: Mechanics only expects students to analyze the transfer of mechanical energy, although students should be aware that mechanical energy may be dissipated in the form of thermal energy or sound.
来源:美国大学理事会 AP 课程与考试说明
功(work)是一个力作用在一个距离上转移进或出一个系统的能量。对于一个变力它是沿路径的一个积分(integral):
$$W=\int_a^b \vec{F}\cdot d\vec{r}=\int F\cos\theta\,dr.$$对于一个恒定力这简化到 $W=Fd\cos\theta$;在一个力-位置图上,功是曲线下的面积。功是一个带符号的标量,而这个符号是物理,不是记账:
- 正 ——力有一个沿运动的分量(它使物体加速)。
- 负 ——力反对运动(摩擦力(friction)和空气阻力做负功)。
- 零 ——力垂直于运动(一个滑动木块上的法向力(normal force)、一次水平移动上的重力、一次圆形摆动里的张力)。

只有沿位移的力分量做功 动能定理(work–energy theorem)收集每个力的贡献:合功等于动能的变化,
$$W_{\text{net}}=\sum W_i=\Delta K.$$Worked example. 一个变力 $F(x)=3x^2\ \text{N}$(沿运动)从 $x=0$ 到 $x=2\ \text{m}$ 作用:$W=\displaystyle\int_0^2 3x^2\,dx=\big[x^3\big]_0^2=8\ \text{J}$。单独作用在一个从静止开始的物体上,它会把动能提高到恰好 $8\ \text{J}$。
词汇表 训练英文 中文 拼音 Work 功 gōng integral 积分 jī fēn friction 摩擦力 mó cā lì normal force 法向力 fǎ xiàng lì work–energy theorem 动能定理 dòng néng dìng lǐ 3.3
势能
大纲
Learning Objective Essential Knowledge 3.3.A
Describe the potential energy of a system.- 3.3.A.1 A system composed of two or more objects has potential energy if the objects within that system only interact with each other through conservative forces.
- 3.3.A.2 Potential energy is a scalar quantity associated with the position of objects within a system.
- 3.3.A.3 The definition of zero potential energy for a given system is a decision made by the observer considering the situation to simplify or otherwise assist in analysis.
- 3.3.A.4 The relationship between conservative forces exerted on a system and the system's potential energy is
- Equation: $\Delta U = -\displaystyle\int_a^b \vec{F}_{cf}(r) \cdot d\vec{r}$
- 3.3.A.5 The conservative forces exerted on a system in a single dimension can be determined using the slope of the system's potential energy with respect to position in that dimension; these forces point in the direction of decreasing potential energy.
- Equation: $F_x = -\dfrac{dU(x)}{dx}$
- 3.3.A.6 Graphs of a system's potential energy as a function of its position can be useful in determining physical properties of that system.
- 3.3.A.6.i Stable equilibrium is a location at which a small displacement in an object's position results in a force exerted on the object opposite to the direction of the small displacement, accelerating the object back toward the equilibrium position.
- 3.3.A.6.ii Unstable equilibrium is a location at which a small displacement in an object's position results in a force exerted on the object in the same direction as the small displacement, accelerating the object away from the equilibrium position.
- 3.3.A.6.iii In a given dimension, stable equilibrium positions exist at locations where the potential energy as a function of position in that dimension has a local minimum.
- 3.3.A.6.iv In a given dimension, unstable equilibrium positions occur at locations where the potential energy as a function of position in that dimension has a local maximum.
- 3.3.A.7 The potential energy of common physical systems can be described using the physical properties of that system.
- 3.3.A.7.i The elastic potential energy of an ideal spring is given by the following equation, where $\Delta x$ is the distance the spring has been stretched or compressed from its equilibrium length.
- Equation: $U_s = \dfrac{1}{2}k(\Delta x)^2$
- 3.3.A.7.ii The general form for the gravitational potential energy of a system consisting of two approximately spherical distributions of mass (e.g., moons, planets, or stars) is given by the equation
- Equation: $U_g = -G\dfrac{m_1 m_2}{r}$
- 3.3.A.7.iii Because the gravitational field near the surface of a planet is nearly constant, the change in gravitational potential energy in a system consisting of an object with mass $m$ and a planet with gravitational field of magnitude $g$ when the object is near the surface of the planet may be approximated by the equation
- Equation: $\Delta U_g = mg\Delta y$
- 3.3.A.7.i The elastic potential energy of an ideal spring is given by the following equation, where $\Delta x$ is the distance the spring has been stretched or compressed from its equilibrium length.
- 3.3.A.8 The total potential energy of a system containing more than two objects is the sum of the potential energy of each pair of objects within the system.
来源:美国大学理事会 AP 课程与考试说明
势能(potential energy)是一个系统通过它各部分的位置储存的能量——它只对保守力(conservative forces)存在,它们的功与路径无关。它通过功定义:
$$\Delta U=-\int_a^b\vec{F}\cdot d\vec{r},$$而你自由地选择 $U=0$ 的地方——只有 $U$ 的变化重要,所以挑选使问题最简单的零。标准的结果:
- 一个表面附近的重力:$\Delta U_g=mg\,\Delta y$。
- 一般的重力:$U_g=-\dfrac{Gm_1m_2}{r}$(在无限间距为零)。
- 弹簧:$U_s=\tfrac12k(\Delta x)^2$,$\Delta x$ 从自然长度测量。
对于几个物体的系统,把每对的势能相加。把定义反过来,一个保守力是它势能的负导数(derivative):
$$F_x=-\frac{dU}{dx}.$$力在 $U(x)$ 曲线上指向"下坡"。平衡坐在斜率为零的地方:一个最小值是一个稳定平衡(stable equilibrium)(被移开,力推回),一个最大值是一个不稳定平衡(unstable equilibrium)(被移开,力推开)。

在一条势能曲线上力指向下坡而 E = U 标记转折点 Worked example. 给定 $U(x)=2x^3-6x$(焦耳),力是 $F=-\dfrac{dU}{dx}=6-6x^2$。平衡坐在 $F=0$:$x=\pm1$。因为 $\dfrac{d^2U}{dx^2}=12x$ 在 $x=+1$ 为正(一个最小值——稳定)而在 $x=-1$ 为负(一个最大值——不稳定),这两点表现相反。
探索Store elastic potential energy in a spring
Stretching a spring stores elastic potential energy $\tfrac12 kx^2$ — the area under the force-extension line. Release it and that energy becomes kinetic.
词汇表 训练英文 中文 拼音 Potential energy 势能 shì néng conservative forces 保守力 bǎo shǒu lì derivative 导数 dǎo shù stable equilibrium 稳定平衡 wěn dìng píng héng unstable equilibrium 不稳定平衡 bù wěn dìng píng héng 3.4
能量守恒
大纲
Learning Objective Essential Knowledge 3.4.A
Describe the energies present in a system.- 3.4.A.1 A system composed of only a single object can only have kinetic energy.
- 3.4.A.2 A system that contains objects that interact via conservative forces or that can change its shape reversibly may have both kinetic and potential energies.
3.4.B
Describe the behavior of a system using conservation of mechanical energy principles.- 3.4.B.1 Mechanical energy is the sum of a system's kinetic and potential energies.
- 3.4.B.2 Any change to a type of energy within a system must be balanced by an equivalent change of other types of energies within the system or by a transfer of energy between the system and its surroundings.
- 3.4.B.3 A system may be selected so that the total energy of that system is constant.
- 3.4.B.4 If the total energy of a system changes, that change will be equivalent to the energy transferred into or out of the system.
3.4.C
Describe how the selection of a system determines whether the energy of that system changes.- 3.4.C.1 Energy is conserved in all interactions.
- 3.4.C.2 If the work done on a selected system is zero and there are no nonconservative interactions within the system, the total mechanical energy of the system is constant.
- 3.4.C.3 If the work done on a selected system is nonzero, energy is transferred between the system and the environment.
Boundary statement: AP Physics C: Mechanics expects students to know that mechanical energy can be dissipated as thermal energy or sound by nonconservative forces.
来源:美国大学理事会 AP 课程与考试说明
能量守恒:动能↔势能 能量在所有相互作用里守恒——问题只是它去哪里。机械能(mechanical energy)是和 $E=K+U$。若一个系统上的外功是零而它里面没有任何东西通过非保守力(nonconservative forces)作用,那么
$$K_1+U_1=K_2+U_2.$$当摩擦或阻力作用时,它们把机械能转换成热能(thermal energy)($\Delta E_{\text{mech}}=-F_fd$),而这个平衡必须包括那一项。若外功确实被做,系统的总能量恰好变化转移的能量:$W_{\text{ext}}=\Delta E_{\text{sys}}$。选择系统就是选择记账——一个单一物体只能有动能;包括地球或弹簧,系统也能储存势能。

一个摆把重力势能换成动能并换回 一条势能图是一幅完整的运动图:高度 $E$ 处的水平线是总能量、间隙 $E-U(x)$ 是每个 $x$ 处的动能,而交叉 $E=U$ 是转折点(turning points),物体在那里暂时停止并反转。
Worked example. 一个 $2.0\ \text{kg}$ 的木块从高度 $1.5\ \text{m}$ 从静止沿一个坡道滑下,以 $4.0\ \text{m/s}$ 到达底部。能量核算:$mgh=29.4\ \text{J}$ 可得;$\tfrac12mv^2=16\ \text{J}$ 作为动能到达;所以摩擦沿途把 $29.4-16=13\ \text{J}$ 转换成热能。

A roller coaster trades energy back and forth: highest (most PE) at the top, fastest (most KE) at the bottom 探索Watch energy convert as an object falls
With no friction, mechanical energy is conserved: as an object falls, gravitational potential energy turns into kinetic energy while the total stays fixed.
词汇表 训练英文 中文 拼音 Mechanical energy 机械能 jī xiè néng nonconservative forces 非保守力 fēi bǎo shǒu lì thermal energy 热能 rè néng turning points 转折点 zhuǎn zhé diǎn 3.5
功率
大纲
Learning Objective Essential Knowledge 3.5.A
Describe the transfer of energy into, out of, or within a system in terms of power.- 3.5.A.1 Power is the rate at which energy changes with respect to time, either by transfer into or out of a system or by conversion from one type to another within a system.
- 3.5.A.2 Average power is the amount of energy being transferred or converted, divided by the time it took for that transfer or conversion to occur.
- Equation: $P_{\text{avg}} = \dfrac{\Delta E}{\Delta t}$
- 3.5.A.3 Because work is the change in energy of an object or system due to a force, average power is the total work done, divided by the time during which that work was done.
- Equation: $P_{\text{avg}} = \dfrac{W}{\Delta t}$
- 3.5.A.4 The instantaneous power delivered to an object by a force is given by the equation
- Equation: $P_{\text{inst}} = \dfrac{dW}{dt}$
- 3.5.A.5 The instantaneous power delivered to an object by the component of a constant force parallel to the object's velocity can be described with the derived equation
- Equation: $P_{\text{inst}} = F_{\parallel}v = Fv\cos\theta$
来源:美国大学理事会 AP 课程与考试说明
功率(power)是能量转移的速率,以瓦特(watts)(W):
$$P_{\text{avg}}=\frac{\Delta E}{\Delta t}=\frac{W}{\Delta t},\qquad P_{\text{inst}}=\frac{dW}{dt}=\vec{F}\cdot\vec{v}.$$它测量功被做得多快,不是多少。点积重要:只有沿速度的力分量传递功率。

功率是功-时间图的斜率:同样的功在更少时间里意味着更多功率 Worked example. 一个木块从一个无摩擦的 $3.0\ \text{m}$ 高坡道的顶部从静止释放,以 $v=\sqrt{2gh}=7.7\ \text{m/s}$ 到达底部(从 $mgh=\tfrac12mv^2$)。一个然后以稳定的 $7.7\ \text{m/s}$ 逆着一个 $20\ \text{N}$ 阻力驱动它的马达传递 $P=Fv=20(7.7)\approx150\ \text{W}$。
Worked example. 一辆 $1200\ \text{kg}$ 的车以稳定的 $15\ \text{m/s}$ 爬一个每 $20\ \text{m}$ 道路上升 $1.0\ \text{m}$ 的山。引擎必须供应重力的功率消耗:$P=mg\,v\sin\theta=1200(9.8)(15)\big(\tfrac{1}{20}\big)\approx8.8\ \text{kW}$ ——在加空气阻力之前。
Exam skill. 能量 FRQ 奖励核算句子:命名你的系统、陈述哪些力对它做功,并在代入数字之前写平衡($W_{\text{ext}}=\Delta K+\Delta U+\Delta E_{\text{thermal}}$)。"摩擦存在,所以机械能不守恒"是一个评分的陈述。
词汇表 训练英文 中文 拼音 Power 功率 gōng lǜ watts 瓦特 wǎ tè 3.5
考试技巧
- 用动能定理 $W_{net}=\Delta KE$ 并对一个变力把功计算为 $W=\int \vec F\cdot d\vec r$。
- 从一个力-位置图把功读作曲线下的面积。
- 功率是 $P=\tfrac{dW}{dt}=\vec F\cdot\vec v$;注意瞬时 vs 平均。
- 把力分成保守的(定义一个势能)和非保守的(耗散能量)。
- 当力变化或路径复杂时选择能量方法而不是运动学。
- 3.1.A.1 An object's translational kinetic energy is given by the equation
-
4
线动量
4.1
线动量
大纲
Learning Objective Essential Knowledge 4.1.A
Describe the linear momentum of an object or system.- 4.1.A.1 Linear momentum is defined by the equation $\vec{p} = m\vec{v}$.
- 4.1.A.2 Momentum is a vector quantity and has the same direction as the velocity.
- 4.1.A.3 Momentum can be used to analyze collisions and explosions.
- 4.1.A.3.i A collision is a model for an interaction where the forces exerted between the involved objects in the system are much larger than the net external force exerted on those objects during the interaction.
- 4.1.A.3.ii As only the initial and final states of a collision are analyzed, the object model may be used to analyze collisions.
- 4.1.A.3.iii An explosion is a model for an interaction in which forces internal to the system move objects within that system apart.
来源:美国大学理事会 AP 课程与考试说明
动量(linear momentum)是质量乘速度:
$$\vec{p}=m\vec{v}.$$它是一个沿速度指向的矢量(vector),而它是分析碰撞(collisions)和爆炸(explosions)的那个量。一组物体能被当作一个以它质心(center of mass)速度移动的系统:
$$\vec{v}_{\text{cm}}=\frac{\sum m_i\vec{v}_i}{\sum m_i},$$所以系统的总动量是它的总质量乘 $\vec{v}_{\text{cm}}$ ——对任何数量的部分只有一个物体量的记账。
词汇表 训练英文 中文 拼音 Linear momentum 动量 dòng liàng vector 矢量 shǐ liàng collisions 碰撞 pèng zhuàng explosions 爆炸 bào zhà center of mass 质心 zhì xīn 4.2
动量变化与冲量
大纲
Learning Objective Essential Knowledge 4.2.A
Describe the impulse delivered to an object or system.- 4.2.A.1 The rate of change of a system's momentum is equal to the net external force exerted on that system.
- Equation: $\vec{F}_{\text{net}} = \dfrac{d\vec{p}}{dt}$
- 4.2.A.2 Impulse is defined as the integral of a force exerted on an object or system over a time interval.
- Equation: $\vec{J} = \displaystyle\int_{t_1}^{t_2} \vec{F}_{\text{net}}(t)\,dt$
- 4.2.A.3 Impulse is a vector quantity and has the same direction as the net force exerted on the system.
- 4.2.A.4 The impulse delivered to a system by a net external force is equal to the area under the curve of a graph of the net external force exerted on the system as a function of time.
- 4.2.A.5 The net external force exerted on a system is equal to the slope of a graph of the momentum of the system as a function of time.
4.2.B
Describe the relationship between the impulse exerted on an object or system and the change in momentum of the object or system.- 4.2.B.1 Change in momentum is the difference between a system's final momentum and its initial momentum.
- Equation: $\Delta\vec{p} = \vec{p} - \vec{p}_0$
- 4.2.B.2 The impulse–momentum theorem relates the impulse delivered to an object and the object's change in momentum.
- 4.2.B.2.i The impulse exerted on an object is equal to the object's change in momentum.
- Equation: $\vec{J} = \displaystyle\int_{t_1}^{t_2} \vec{F}_{\text{net}}(t)\,dt = \Delta\vec{p}$
- 4.2.B.2.ii Newton's second law of motion is a direct result of the impulse–momentum theorem applied to systems with constant mass.
- Equation: $\vec{F}_{\text{net}} = \dfrac{d\vec{p}}{dt} = m\dfrac{d\vec{v}}{dt} = m\vec{a}$
- 4.2.B.2.iii The impulse–momentum theorem also describes the behavior of a system in which the velocity is constant but the mass changes with respect to time.
- Equation: $\vec{F}_{\text{net}} = \dfrac{d\vec{p}}{dt} = \dfrac{dm}{dt}\vec{v}$
- 4.2.B.2.i The impulse exerted on an object is equal to the object's change in momentum.
来源:美国大学理事会 AP 课程与考试说明
牛顿第二定律实际上是一个关于动量的陈述:
$$\vec{F}_{\text{net}}=\frac{d\vec{p}}{dt},$$当质量恒定时它简化到 $m\vec{a}$ ——而当它不恒定时处理一个变化的质量(特例 $\vec{F}=\dfrac{dm}{dt}\vec{v}$ 只在速度恒定时成立,例如一条堆到秤上的链,或落到匀速传送带上的沙;火箭不适用,因为它在加速)。一个力传递的冲量(impulse)是它的时间积分,而冲量-动量定理(impulse–momentum theorem)说它等于动量的变化:
$$\vec{J}=\int_{t_1}^{t_2}\vec{F}_{\text{net}}\,dt=\Delta\vec{p}.$$冲量是一个沿合力的矢量。从图两种方式读它:
- 在一个力-时间图上,冲量是曲线下的面积。
- 在一个动量-时间图上,合力是每个时刻的斜率(slope)。

冲量是力-时间曲线下的面积,等于平均力乘接触时间 把同样的 $\Delta p$ 分散到一个更长的时间降低力——那就是为什么气囊、溃缩区和软着陆工作,以及为什么你着陆时弯膝。
Worked example. 一个时变力 $F(t)=10t\ \text{N}$ 作用在一个 $2.0\ \text{kg}$ 的物体上 $2.0\ \text{s}$。冲量是 $J=\displaystyle\int_0^2 10t\,dt=\big[5t^2\big]_0^2=20\ \text{N}\cdot\text{s}$,所以速率变化 $\Delta v=J/m=10\ \text{m/s}$。
词汇表 训练英文 中文 拼音 impulse 冲量 chōng liàng impulse–momentum theorem 冲量-动量定理 chōng liàng - dòng liàng dìng lǐ slope 斜率 xié lǜ 4.3
线动量守恒
大纲
Learning Objective Essential Knowledge 4.3.A
Describe the behavior of a system using conservation of linear momentum.- 4.3.A.1 A collection of objects with individual momenta can be described as one system with one center-of-mass velocity.
- 4.3.A.1.i For a collection of objects, the velocity of a system's center of mass can be calculated using the equation
- Equation: $\vec{v}_{\text{cm}} = \dfrac{\sum \vec{p}_i}{\sum m_i} = \dfrac{\sum (m_i \vec{v}_i)}{\sum m_i}$
- 4.3.A.1.ii The velocity of a system's center of mass is constant in the absence of a net external force.
- 4.3.A.1.i For a collection of objects, the velocity of a system's center of mass can be calculated using the equation
- 4.3.A.2 The total momentum of a system is the sum of the momenta of the system's constituent parts.
- 4.3.A.3 In the absence of net external forces, any change to the momentum of an object within a system must be balanced by an equivalent and opposite change of momentum elsewhere within the system. Any change to the momentum of a system is due to a transfer of momentum between the system and its surroundings.
- 4.3.A.3.i The impulse exerted by one object on a second object is equal and opposite to the impulse exerted by the second object on the first. This is a direct result of Newton's third law.
- 4.3.A.3.ii A system may be selected so that the total momentum of that system is constant.
- 4.3.A.3.iii If the total momentum of a system changes, that change will be equivalent to the impulse exerted on the system.
- Equation: $\vec{J} = \Delta\vec{p}$
- 4.3.A.4 Correct application of conservation of momentum can be used to determine the velocity of a system immediately before and immediately after collisions or explosions.
Boundary statement: AP Physics C: Mechanics only expects students to quantitatively analyze collisions and interactions in one or two dimensions. Three-dimensional collisions may be analyzed qualitatively.
4.3.B
Describe how the selection of a system determines whether the momentum of that system changes.- 4.3.B.1 Momentum is conserved in all interactions.
- 4.3.B.2 If the net external force on the selected system is zero, the total momentum of the system is constant.
- 4.3.B.3 If the net external force on the selected system is nonzero, momentum is transferred between the system and the environment.
来源:美国大学理事会 AP 课程与考试说明
内力成牛顿第三定律对,所以它们在任何系统内抵消:它们能在部分之间移动动量但从不改变总量。动量只通过一个合外力进入或离开一个系统($\vec{J}=\Delta\vec{p}$)。所以,以零合外力,总动量守恒(conserved):
$$\sum\vec{p}_{\text{before}}=\sum\vec{p}_{\text{after}}.$$动量在每个碰撞里守恒,无论多剧烈——选择系统足够大,使碰撞力是内部的。沿每个轴分别应用守恒;AP 要求一维或二维的定量工作。

一个正面碰撞:之前的总动量等于之后的总动量 Worked example (explosion). 一个静止的 $6.0\ \text{kg}$ 弹壳分裂成一个以 $9.0\ \text{m/s}$ 向东移动的 $2.0\ \text{kg}$ 片和一个 $4.0\ \text{kg}$ 片。总动量保持零,所以重的片以 $v=\dfrac{2.0(9.0)}{4.0}=4.5\ \text{m/s}$ 向西移动。动能来自储存的(化学或弹簧)能量——动量守恒不要求动能守恒。
探索Collide two carts and conserve momentum
In any collision the total momentum $\sum mv$ before equals the total after. Set the masses and speeds and check the momentum bookkeeping.
词汇表 训练英文 中文 拼音 conserved 守恒 shǒu héng 4.4
弹性碰撞与非弹性碰撞
大纲
Learning Objective Essential Knowledge 4.4.A
Describe whether an interaction between objects is elastic or inelastic.- 4.4.A.1 An elastic collision between objects is one in which the initial kinetic energy of the system is equal to the final kinetic energy of the system.
- 4.4.A.2 In an elastic collision, the final kinetic energies of each of the objects within the system may be different from their initial kinetic energies.
- 4.4.A.3 An inelastic collision between objects is one in which the total kinetic energy of the system decreases.
- 4.4.A.4 In an inelastic collision, some of the initial kinetic energy is not restored to kinetic energy but is transformed by nonconservative forces into other forms of energy.
- 4.4.A.5 In a perfectly inelastic collision, the objects stick together and move with the same velocity after the collision.
来源:美国大学理事会 AP 课程与考试说明
碰撞中的动量守恒 所有碰撞都守恒动量;它们在动能(kinetic energy)发生什么上不同:
类型 动量 动能 弹性碰撞(elastic collision) 守恒 总量守恒(各自份额可能变化) 非弹性碰撞(inelastic collision) 守恒 减少——一些变成热、声、形变(deformation) 完全非弹性碰撞(perfectly inelastic collision) 守恒 最大可能的损失——物体粘住并共享一个速度 策略:总是先写动量守恒;只在问题说"弹性"时加动能方程。两个有用的弹性事实:一维弹性碰撞里相等的质量简单地交换速度,而在质心参考系里每个物体只反转它的速度。
Worked example. 一辆 $1000\ \text{kg}$ 以 $20\ \text{m/s}$ 的车撞一辆静止的 $1500\ \text{kg}$ 的车而它们锁在一起——完全非弹性。动量:$v=\dfrac{1000(20)}{2500}=8.0\ \text{m/s}$。动能从 $2.0\times10^5\ \text{J}$ 落到 $\tfrac12(2500)(8.0)^2=8.0\times10^4\ \text{J}$:约 $60\%$ 损失,即使动量恰好守恒。
Worked example (2D). 一个向东以 $4.0\ \text{m/s}$ 移动的冰球撞一个静止的相同冰球;掠射撞击后,一个以 $2.0\ \text{m/s}$、东偏北 $60^\circ$ 移动。守恒每个轴:东-西,$m(4.0)=m(2.0)\cos60^\circ+mv_x$,所以 $v_x=3.0\ \text{m/s}$;南-北,$0=m(2.0)\sin60^\circ-mv_y$,所以 $v_y=1.7\ \text{m/s}$。第二个冰球以 $\sqrt{3.0^2+1.7^2}=3.5\ \text{m/s}$、东偏南约 $30^\circ$ 移动。

一个掠射碰撞,沿两个垂直坐标轴分解 Exam skill. 在 FRQ 上,用条件、不是口号来论证:"碰撞期间两冰球系统上的合外力是零,所以它的总动量恒定。"若被问碰撞是否弹性,计算之前和之后的动能并比较——绝不假设。
探索Compare elastic and inelastic collisions
Momentum is always conserved, but kinetic energy is only conserved in an elastic collision. In an inelastic one the carts stick and some energy becomes heat.
词汇表 训练英文 中文 拼音 kinetic energy 动能 dòng néng elastic collision 弹性碰撞 tán xìng pèng zhuàng inelastic collision 非弹性碰撞 fēi tán xìng pèng zhuàng deformation 形变 xíng biàn perfectly inelastic collision 完全非弹性碰撞 wán quán fēi tán xìng pèng zhuàng 4.4
考试技巧
- 冲量等于动量变化:$\vec J=\int \vec F\,dt=\Delta\vec p$,而它是一个力-时间图下的面积。
- 每当合外力是零时动量守恒——总是碰撞的首选。
- 把弹性(动能守恒)与非弹性(动能减少)碰撞区分开;在完全非弹性碰撞中物体粘住并以一个速度运动。
- 在二维里对每个分量(x 和 y)分别应用守恒。
- 与质心联系:总动量 $=M\vec v_{cm}$。
-
5
力矩与转动动力学
5.1
转动运动学
大纲
Learning Objective Essential Knowledge 5.1.A
Describe the rotation of a system with respect to time using angular displacement, angular velocity, and angular acceleration.- 5.1.A.1 Angular displacement is the measurement of the angle, in radians, through which a point on a rigid system rotates about a specified axis.
- Equation: $\Delta\theta = \theta - \theta_0$
- 5.1.A.1.i A rigid system is one that holds its shape but in which different points on the system move in different directions during rotation. A rigid system cannot be modeled as an object.
- 5.1.A.1.ii One direction of angular displacement about an axis of rotation—clockwise or counterclockwise—is typically indicated as mathematically positive, with the other direction becoming mathematically negative.
- 5.1.A.1.iii If the rotation of a system about an axis may be well described using the motion of the system's center of mass, the system may be treated as a single object. For example, the rotation of Earth about its axis may be considered negligible when considering the revolution of Earth about the center of mass of the Earth–Sun system.
- 5.1.A.2 Angular velocity is the rate at which angular position changes with respect to time.
- Equation: $\omega = \dfrac{d\theta}{dt}$
- 5.1.A.3 Angular acceleration is the rate at which angular velocity changes with respect to time.
- Equation: $\alpha = \dfrac{d\omega}{dt}$
- 5.1.A.4 Angular displacement, angular velocity, and angular acceleration around one axis are analogous to linear displacement, velocity, and acceleration in one dimension and demonstrate the same mathematical relationships.
- 5.1.A.4.i For constant angular acceleration, the mathematical relationships between angular displacement, angular velocity, and angular acceleration can be described with the following equations:
- $\omega = \omega_0 + \alpha t$
- $\theta = \theta_0 + \omega_0 t + \dfrac{1}{2}\alpha t^2$
- $\omega^2 = \omega_0^2 + 2\alpha(\theta - \theta_0)$
- 5.1.A.4.ii Graphs of angular displacement, angular velocity, and angular acceleration as functions of time can be used to find the relationships between those quantities.
- 5.1.A.4.i For constant angular acceleration, the mathematical relationships between angular displacement, angular velocity, and angular acceleration can be described with the following equations:
Boundary statement: AP Physics C: Mechanics expects students to be able to mathematically manipulate the magnitudes of angular displacement, angular velocity, and angular acceleration using vector conventions. However, the directions of said vectors will not be assessed on the exam.
Descriptions of the directions of rotational kinematics quantities for a point or rigid body are limited to clockwise and counterclockwise with respect to a given axis of rotation.
来源:美国大学理事会 AP 课程与考试说明
旋转用角量映照线性运动,定义为导数:
$$\omega=\frac{d\theta}{dt},\qquad \alpha=\frac{d\omega}{dt}=\frac{d^2\theta}{dt^2}.$$这里 $\theta$ 是以弧度(radians)的角位移(angular displacement)、$\omega$ 是角速度(angular velocity),而 $\alpha$ 是角加速度(angular acceleration)。(AP 用它们的大小和符号;三维矢量方向不被评估。)对于恒定 $\alpha$,运动学方程是重新标注字母的线性方程:
$$\omega=\omega_0+\alpha t,\qquad \Delta\theta=\omega_0t+\tfrac12\alpha t^2,\qquad \omega^2=\omega_0^2+2\alpha\,\Delta\theta.$$
一弧度是弧长等于半径的角 Worked example. 一个风扇叶片以 $\alpha=-6.0\ \text{rad/s}^2$ 从 $30\ \text{rad/s}$ 减慢到静止:它取 $t=5.0\ \text{s}$ 并转过 $\Delta\theta=\dfrac{0-30^2}{2(-6.0)}=75\ \text{rad}$ ——约 $12$ 转。
词汇表 训练英文 中文 拼音 angular displacement 角位移 jiǎo wèi yí radians 弧度 hú dù angular velocity 角速度 jiǎo sù dù angular acceleration 角加速度 jiǎo jiā sù dù 5.2
联系线运动与转动
大纲
Learning Objective Essential Knowledge 5.2.A
Describe the linear motion of a point on a rotating rigid system that corresponds to the rotational motion of that point, and vice versa.- 5.2.A.1 For a point at a distance $r$ from a fixed axis of rotation, the linear distance $s$ traveled by the point as the system rotates through an angle $\Delta\theta$ is given by the equation $\Delta s = r\Delta\theta$.
- 5.2.A.2 Derived relationships of linear velocity and of the tangential component of acceleration to their respective angular quantities are given by the following equations:
- $s = r\theta$
- $v = r\omega$
- $a_T = r\alpha$
- 5.2.A.3 For a rigid system, all points within that system have the same angular velocity and angular acceleration.
Boundary statement: AP Physics C: Mechanics expects students to be able to mathematically manipulate the magnitudes of angular displacement, angular velocity, and angular acceleration using vector conventions. However, the directions of the vectors will not be assessed on the exam.
Descriptions of the directions of rotational kinematics quantities for a point or rigid body are limited to clockwise and counterclockwise with respect to a given axis of rotation.
来源:美国大学理事会 AP 课程与考试说明
一个离轴半径 $r$ 的点沿它的弧以以下移动
$$s=r\theta,\qquad v=r\omega,\qquad a_t=r\alpha,$$所以更远的点移动更快。一个旋转的点一般一次有两个加速度分量:切向加速度(tangential acceleration)$a_t=r\alpha$(沿弧加速)和向心加速度(centripetal acceleration)$a_c=\dfrac{v^2}{r}=r\omega^2$(转动,朝向轴)。它们垂直,所以 $a=\sqrt{a_t^2+a_c^2}$。

随着半径转过一个角,一个点以速率 v 沿一条弧移动 词汇表 训练英文 中文 拼音 tangential acceleration 切向加速度 qiè xiàng jiā sù dù centripetal acceleration 向心加速度 xiàng xīn jiā sù dù 5.3
力矩
大纲
Learning Objective Essential Knowledge 5.3.A
Identify the torques exerted on a rigid system.- 5.3.A.1 Torque results only from the force component perpendicular to the position vector from the axis of rotation to the point of application of the force.
- 5.3.A.2 The lever arm is the perpendicular distance from the axis of rotation to the line of action of the exerted force.
5.3.B
Describe the torques exerted on a rigid system.- 5.3.B.1 Torques can be described using force diagrams.
- 5.3.B.1.i Force diagrams are similar to free-body diagrams and are used to analyze the torques exerted on a rigid system.
- 5.3.B.1.ii Similar to free-body diagrams, force diagrams represent the relative magnitude and direction of the forces exerted on a rigid system. Force diagrams also depict the location at which those forces are exerted relative to the axis of rotation.
- 5.3.B.2 The torque exerted on a rigid system about a chosen pivot point by a given force is described by $\vec{\tau} = \vec{r} \times \vec{F}$.
- 5.3.B.2.i The cross-product between two vectors, $\vec{A}$ and $\vec{B}$, results in a vector quantity of magnitude $\vec{A} \times \vec{B} = AB\sin\theta$.
- 5.3.B.2.ii The direction of the vector resulting from the cross-product of vectors $\vec{A}$ and $\vec{B}$ is perpendicular to both vectors $\vec{A}$ and $\vec{B}$ and therefore is normal to the plane defined by vectors $\vec{A}$ and $\vec{B}$.
- 5.3.B.2.iii The direction of the vector resulting from the cross-product of vectors $\vec{A}$ and $\vec{B}$ can be qualitatively determined by applying the appropriate right-hand rule.
来源:美国大学理事会 AP 课程与考试说明
力矩的原理 力矩(torque)是一个力的转动效应——旋转的原因,正如力是平动的原因。它是一个矢量积(vector product):
$$\vec{\tau}=\vec{r}\times\vec{F},\qquad \tau=rF\sin\theta=F\,r_\perp,$$其中 $r_\perp$,力臂(moment arm),是从轴到力作用线的垂直距离。更多的力、施加得离轴更远、更垂直:更多力矩。一个直指(或离开)轴的力有零力矩。在一个平面里,给逆时针和顺时针力矩相反的符号并相加。
作为一个矢量积,$\vec\tau=\vec r\times\vec F$ 指向 $\vec r$ 和 $\vec F$ 所在平面的垂直方向,其方向由右手定则(right-hand rule)确定:把右手手指从 $\vec r$ 弯向 $\vec F$,拇指就指向 $\vec\tau$ ——逆时针转动时指出纸面,顺时针时指入纸面。同样的规则给出角动量 $\vec L=\vec r\times\vec p$ 的方向。

一个力的力矩取决于从轴的垂直距离 
A gear train: torques and rotational inertia couple through the gear ratio 探索Balance torques on a beam
Torque is force times perpendicular distance, $\tau=Fd$. The beam is in rotational equilibrium when the torques on each side are equal.
词汇表 训练英文 中文 拼音 Torque 力矩 lì jǔ vector product 矢量积 shǐ liàng jī moment arm 力臂 lì bì right-hand rule 右手定则 yòu shǒu dìng zé Rotational inertia 转动惯量 zhuǎn dòng guàn liàng 5.4
转动惯量
大纲
Learning Objective Essential Knowledge 5.4.A
Describe the rotational inertia of a rigid system relative to a given axis of rotation.- 5.4.A.1 Rotational inertia measures a rigid system's resistance to changes in rotation and is related to the mass of the system and the distribution of that mass relative to the axis of rotation.
- 5.4.A.2 The rotational inertia of an object rotating a perpendicular distance $r$ from an axis is described by the equation $I = mr^2$.
- 5.4.A.3 The total rotational inertia of a collection of objects about an axis is the sum of the rotational inertias of each object about that axis.
- Equation: $I_{\text{tot}} = \sum I_i = \sum m_i r_i^2$
- 5.4.A.4 For a solid that can be considered as a collection of differential masses, $dm$, the solid's rotational inertia can be calculated using the equation $I = \int r^2\, dm$, where $r$ is the perpendicular distance from $dm$ to the axis of rotation.
5.4.B
Describe the rotational inertia of a rigid system rotating about an axis that does not pass through the system's center of mass.- 5.4.B.1 A rigid system's rotational inertia in a given plane is at a minimum when the rotational axis passes through the system's center of mass.
- 5.4.B.2 The parallel axis theorem uses the following equation to relate the rotational inertia of a rigid system about any axis that is parallel to an axis through its center of mass:
- Equation: $I' = I_{\text{cm}} + Md^2$
Boundary statement: AP Physics C: Mechanics only expects students to use calculus in the derivations of the rotational inertia of thin rods of uniform or nonuniform density about an arbitrary axis perpendicular to the rod, as well as derivations of the rotational inertia of a thin cylindrical shell, disk, or rigid bodies that can be considered to be made up of coaxial rings or shells about an axis that passes through their centers (e.g., annular rings).
Students should have a qualitative understanding of the factors that affect rotational inertia; for example, how rotational inertia is greater when mass is farther from the axis of rotation, which is why a hoop has more rotational inertia than a solid puck of the same mass and radius.
来源:美国大学理事会 AP 课程与考试说明
转动惯量(rotational inertia)$I$ 测量对角加速度的抵抗——质量的旋转类比,但它取决于质量坐在哪里。对于点质量,$I=\sum m_ir_i^2$;对于一个连续物体,
$$I=\int r^2\,dm.$$离轴远的质量主导,因为 $r^2$。
Worked example (rod). 一根均匀的杆(质量 $M$、长度 $L$)绕它的中心:以线密度(linear density)$\lambda=M/L$,$dm=\lambda\,dx$,所以 $I=\displaystyle\int_{-L/2}^{L/2}x^2\,\frac{M}{L}\,dx=\frac{1}{12}ML^2$。AP 也期望非均匀杆:若在 $[0,L]$ 上 $\lambda(x)=cx$,先求 $M=\int_0^L cx\,dx=\tfrac12cL^2$,然后绕轻的一端 $I=\int_0^L x^2(cx)\,dx=\tfrac14cL^4=\tfrac12ML^2$。
Worked example (disk from rings). 一个实心圆盘是一嵌套的环:一个半径 $r$、宽度 $dr$ 的环有 $dm=\dfrac{M}{\pi R^2}\,2\pi r\,dr$,所以
$$I=\int_0^R r^2\,dm=\frac{2M}{R^2}\int_0^R r^3\,dr=\tfrac12MR^2.$$同样的同轴壳方法处理圆柱壳和环形环。
平行轴定理(parallel-axis theorem)把任何已知的 $I$ 移到一个相距 $d$ 的平行轴:
$$I'=I_{\text{cm}}+Md^2.$$对于绕一端的杆:$I=\tfrac1{12}ML^2+M\big(\tfrac{L}{2}\big)^2=\tfrac13ML^2$。
词汇表 训练英文 中文 拼音 linear density 线密度 xiàn mì dù parallel-axis theorem 平行轴定理 píng xíng zhóu dìng lǐ 5.5
转动平衡与转动形式的牛顿第一定律
大纲
Learning Objective Essential Knowledge 5.5.A
Describe the conditions under which a system's angular velocity remains constant.- 5.5.A.1 A system may exhibit rotational equilibrium (constant angular velocity) without being in translational equilibrium, and vice versa.
- 5.5.A.1.i Free-body and force diagrams describe the nature of the forces and torques exerted on an object or rigid system.
- 5.5.A.1.ii Rotational equilibrium is a configuration of torques such that the net torque exerted on the system is zero.
- Equation: $\sum \tau_i = 0$
- 5.5.A.1.iii The rotational analog of Newton's first law is that a system will have a constant angular velocity only if the net torque exerted on the system is zero.
- 5.5.A.2 A rotational corollary to Newton's second law states that if the torques exerted on a rigid system are not balanced, the system's angular velocity must be changing.
Boundary statement: AP Physics C: Mechanics does not expect students to simultaneously analyze rotation in multiple planes.
来源:美国大学理事会 AP 课程与考试说明
牛顿第一定律有一个旋转形式:以零合力矩,角速度恒定——一个处于转动平衡(rotational equilibrium)的物体要么不旋转要么稳定地旋转。完全的静平衡(static equilibrium)需要 $\sum F=0$ 和 $\sum\tau=0$(绕任何轴)。梁和梯子的标准做法:把轴放在一个未知力作用的点,使那个力从力矩方程里掉出来。

在平衡时绕轴的顺时针和逆时针力矩相等 Worked example. 一根均匀的 $20\ \text{kg}$ 梁,$4.0\ \text{m}$ 长,在它左端被枢转并由它右端的一根竖直绳保持水平。一个 $40\ \text{kg}$ 的孩子坐在离枢轴 $1.0\ \text{m}$ 处。绕枢轴的力矩:$T(4.0)=20g(2.0)+40g(1.0)$,所以 $T=\dfrac{392+392}{4.0}=196\ \text{N}$。然后竖直力平衡给出枢轴向上的推:$F_p=(60)(9.8)-196=392\ \text{N}$。把枢轴选为轴完全从力矩方程里移除了 $F_p$。

A beam balance: rotational equilibrium when clockwise and anticlockwise torques match 探索Find the balance point
For rotational equilibrium the total clockwise torque equals the total anticlockwise torque. Move the forces and distances until the beam balances.
词汇表 训练英文 中文 拼音 rotational equilibrium 转动平衡 zhuǎn dòng píng héng static equilibrium 静平衡 jìng píng héng 5.6
转动形式的牛顿第二定律
大纲
Learning Objective Essential Knowledge 5.6.A
Describe the conditions under which a system's angular velocity changes.- 5.6.A.1 Angular velocity changes when the net torque exerted on the object or system is not equal to zero.
- 5.6.A.2 The rate at which the angular velocity of a rigid system changes is directly proportional to the net torque exerted on the rigid system and is in the same direction. The angular acceleration of the rigid system is inversely proportional to the rotational inertia of the rigid system.
- Equation: $\alpha_{\text{sys}} = \dfrac{\Sigma\tau}{I_{\text{sys}}} = \dfrac{\tau_{\text{net}}}{I_{\text{sys}}}$
- 5.6.A.3 To fully describe a rotating rigid system, linear and rotational analyses may need to be performed independently.
来源:美国大学理事会 AP 课程与考试说明
一个合力矩产生与转动惯量成比例的角加速度:
$$\alpha=\frac{\sum\tau}{I}\qquad\Big(\text{more generally }\sum\vec{\tau}=\frac{d\vec{L}}{dt}\Big).$$像平动动力学一样解旋转动力学,$\tau\leftrightarrow F$、$I\leftrightarrow m$、$\alpha\leftrightarrow a$ ——受力图、方程、求解。

一个有质量的滑轮:两个张力不同,因为滑轮需要合力矩 Worked example (massive pulley). 质量 $m_1=4.0\ \text{kg}$ 和 $m_2=2.0\ \text{kg}$ 从一根越过一个转动惯量 $I=0.50\ \text{kg}\cdot\text{m}^2$、半径 $R=0.20\ \text{m}$ 的滑轮(pulley)的绳悬挂。三个牛顿定律方程,每个物体一个:
$$m_1g-T_1=m_1a,\qquad T_2-m_2g=m_2a,\qquad (T_1-T_2)R=I\alpha=\frac{Ia}{R}.$$把它们相加(以 $I/R^2=12.5\ \text{kg}$):$a=\dfrac{(m_1-m_2)g}{m_1+m_2+I/R^2}=\dfrac{19.6}{18.5}=1.1\ \text{m/s}^2$。然后 $T_1=m_1(g-a)=35\ \text{N}$ 而 $T_2=m_2(g+a)=22\ \text{N}$。张力必须不同——否则没有什么会使滑轮旋转。(若一个问题说滑轮是轻的,那么 $I\approx0$ 而张力又变得相等。)
Exam skill. 旋转 FRQ 给设置评分:每个质量和滑轮分别的受力图、陈述绳约束 $a=R\alpha$,以及符号惯例一致。单个最常见的错误是假设一根越过一个有质量滑轮的绳始终一个张力。
词汇表 训练英文 中文 拼音 pulley 滑轮 huá lún 5.6
考试技巧
- 使用旋转类比:$\theta,\omega=\tfrac{d\theta}{dt},\alpha=\tfrac{d\omega}{dt}$,以恒定 $\alpha$ 方程映照线性运动学。
- 用 $v=r\omega$ 和 $a_t=r\alpha$ 在线性和角之间转换(加向心 $a_c=\tfrac{v^2}{r}=\omega^2 r$)。
- 始终保持弧度并固定一个正的旋转方向。
- 通过 $\tau=I\alpha$ 把力矩与角加速度关联。
- 画旋转轴——力臂是到它的垂直距离。
- 5.1.A.1 Angular displacement is the measurement of the angle, in radians, through which a point on a rigid system rotates about a specified axis.
-
6
转动系统的能量与动量
6.1
转动动能
大纲
Learning Objective Essential Knowledge 6.1.A
Describe the rotational kinetic energy of a rigid system in terms of the rotational inertia and angular velocity of that rigid system.- 6.1.A.1 The rotational kinetic energy of an object or rigid system is related to the rotational inertia and angular velocity of the rigid system and is given by the equation $K_{\text{rot}} = \dfrac{1}{2} I \omega^2$.
- Equation: $K_{\text{rot}} = \dfrac{1}{2} I \omega^2$
- 6.1.A.1.i The rotational inertia of an object about a fixed axis can be used to show that the rotational kinetic energy of that object is equivalent to its translational kinetic energy, which is its total kinetic energy.
- 6.1.A.1.ii The total kinetic energy of a rigid system is the sum of its rotational kinetic energy due to its rotation about its center of mass and the translational kinetic energy due to the linear motion of its center of mass.
- 6.1.A.2 A rigid system can have rotational kinetic energy while its center of mass is at rest due to the individual points within the rigid system having linear speed and, therefore, kinetic energy.
- 6.1.A.3 Rotational kinetic energy is a scalar quantity.
来源:美国大学理事会 AP 课程与考试说明
一个旋转的物体有转动动能(rotational kinetic energy)
$$K_{\text{rot}}=\tfrac12 I\omega^2,$$$\tfrac12mv^2$ 的旋转孪生,转动惯量(rotational inertia)$I$ 扮演质量的角色。一个既移动又旋转的物体携带两个项:
$$K=\tfrac12 mv_{\text{cm}}^2+\tfrac12 I\omega^2.$$Worked example. 一个均匀圆柱($I=\tfrac12mr^2$)以速率 $v$ 滚动。它的动能是 $K=\tfrac12mv^2+\tfrac12\big(\tfrac12mr^2\big)\big(\tfrac{v}{r}\big)^2=\tfrac34mv^2$ ——它的三分之一是转动的。同样的能量核算球决定每个滚动问题。
词汇表 训练英文 中文 拼音 rotational kinetic energy 转动动能 zhuǎn dòng dòng néng rotational inertia 转动惯量 zhuǎn dòng guàn liàng 6.2
力矩与功
大纲
Learning Objective Essential Knowledge 6.2.A
Describe the work done on a rigid system by a given torque or collection of torques.- 6.2.A.1 A torque can transfer energy into or out of an object or rigid system if the torque is exerted over an angular displacement.
- 6.2.A.2 The amount of work done on a rigid system by a torque is related to the magnitude of that torque and the angular displacement through which the rigid system rotates during the interval in which that torque is exerted.
- Equation: $W = \displaystyle\int_{\theta_1}^{\theta_2} \tau \, d\theta$
- 6.2.A.3 Work done on a rigid system by a given torque can be found from the area under the curve of a graph of the torque as a function of angular position.
来源:美国大学理事会 AP 课程与考试说明
一个力矩(torque)作用过一个角位移(angular displacement)做功,而功率是力矩乘角速度:
$$W=\int_{\theta_1}^{\theta_2}\tau\,d\theta,\qquad P=\tau\omega.$$这些是 $W=\int F\,dx$ 和 $P=Fv$ 的旋转形式——整个平动能量工具箱以 $F\to\tau$、$x\to\theta$、$v\to\omega$ 转移过来。
Worked example. 一个马达对一个飞轮施加一个恒定的 $8.0\ \text{N}\cdot\text{m}$ 力矩达 $5.0$ 整转:$W=\tau\,\Delta\theta=8.0(5.0)(2\pi)=250\ \text{J}$,若摩擦可忽略它作为转动动能出现。
探索Balance torques on a beam
Torque is force times perpendicular distance, $\tau=Fd$. Equal torques on each side keep the beam in rotational equilibrium.
词汇表 训练英文 中文 拼音 torque 力矩 lì jǔ angular displacement 角位移 jiǎo wèi yí 6.3
角动量与角冲量
大纲
Learning Objective Essential Knowledge 6.3.A
Describe the angular momentum of an object or rigid system.- 6.3.A.1 The magnitude of the angular momentum of a rigid system about a specific axis can be described with the equation $L = I\omega$.
- Equation: $L = I\omega$
- 6.3.A.2 The angular momentum of an object about a given point is $\vec{L} = \vec{r} \times \vec{p}$.
- Equation: $\vec{L} = \vec{r} \times \vec{p}$
- 6.3.A.2.i The selection of the axis about which an object is considered to rotate influences the determination of the angular momentum of that object.
- 6.3.A.2.ii The measured angular momentum of an object traveling in a straight line depends on the distance between the reference point and the object, the mass of the object, the speed of the object, and the angle between the radial distance and the velocity of the object.
6.3.B
Describe the angular impulse delivered to an object or rigid system by a torque.- 6.3.B.1 Angular impulse is defined as the product of the torque exerted on an object or rigid system and the time interval during which the torque is exerted.
- Equation: $\text{angular impulse} = \displaystyle\int \tau \, dt$
- 6.3.B.2 Angular impulse has the same direction as the torque imparting it.
- 6.3.B.3 The angular impulse delivered to an object or rigid system by a torque can be found from the area under the curve of a graph of the torque as a function of time.
6.3.C
Relate the change in angular momentum of an object or rigid system to the angular impulse given to that object or rigid system.- 6.3.C.1 The magnitude of the change in angular momentum can be described by comparing the magnitudes of the final and initial momenta of the object or rigid system.
- Equation: $\Delta L = L - L_0$
- 6.3.C.2 A rotational form of the impulse–momentum theorem relates the angular impulse delivered to an object or rigid system and the change in angular momentum of that object or rigid system.
- 6.3.C.2.i The angular impulse exerted on an object or rigid system is equal to the change in angular momentum of that object or rigid system.
- Equation: $\Delta L = \displaystyle\int_{t_1}^{t_2} \tau \, dt$
- 6.3.C.2.ii The rotational form of the impulse–momentum theorem is a direct result of Newton's second law of motion for cases in which rotational inertia is constant.
- Equation: $\tau_{\text{net}} = \dfrac{dL}{dt} = I\dfrac{d\omega}{dt} = I\alpha$
- 6.3.C.2.i The angular impulse exerted on an object or rigid system is equal to the change in angular momentum of that object or rigid system.
- 6.3.C.3 The net torque exerted on an object or rigid system is equal to the slope of the graph of the angular momentum of an object as a function of time.
- 6.3.C.4 The angular impulse delivered to an object or rigid system is equal to the area under the curve of a graph of the net external torque exerted on an object as a function of time.
来源:美国大学理事会 AP 课程与考试说明
一个粒子的角动量(angular momentum)是
$$\vec{L}=\vec{r}\times\vec{p},\qquad |L|=mvr\sin\theta,$$所以即使一个沿直线移动的粒子也有绕不在那条线上的任何点的角动量($L=mv\,d$,$d$ 是垂直距离)。对于一个绕一个固定轴旋转的刚体,$L=I\omega$。旋转形式的牛顿第二定律是
$$\vec{\tau}_{\text{net}}=\frac{d\vec{L}}{dt}\quad(=I\alpha\ \text{when }I\text{ is constant}),$$而一个合力矩随时间作用传递一个角冲量(angular impulse)$\int\tau\,dt=\Delta L$ ——旋转冲量-动量定理。
词汇表 训练英文 中文 拼音 Angular momentum 角动量 jiǎo dòng liàng angular impulse 角冲量 jiǎo chōng liàng 6.4
角动量守恒
大纲
Learning Objective Essential Knowledge 6.4.A
Describe the behavior of a system using conservation of angular momentum.- 6.4.A.1 The total angular momentum of a system about a rotational axis is the sum of the angular momenta of the system's constituent parts about that rotational axis.
- 6.4.A.2 Any change to a system's angular momentum must be due to an interaction between the system and its surroundings.
- 6.4.A.2.i The angular impulse exerted by one object or system on a second object or system is equal and opposite to the angular impulse exerted by the second object or system on the first. This is a direct result of Newton's third law.
- 6.4.A.2.ii A system may be selected so that the total angular momentum of that system is constant.
- 6.4.A.2.iii The angular speed of a nonrigid system may change without the angular momentum of the system changing if the system changes shape by moving mass closer to or farther from the rotational axis.
- 6.4.A.2.iv If the total angular momentum of a system changes, that change will be equivalent to the angular impulse exerted on the system.
6.4.B
Describe how the selection of a system determines whether the angular momentum of that system changes.- 6.4.B.1 Angular momentum is conserved in all interactions.
- 6.4.B.2 If the net external torque exerted on a selected object or rigid system is zero, the total angular momentum of that system is constant.
- 6.4.B.3 If the net external torque exerted on a selected object or rigid system is nonzero, angular momentum is transferred between the system and the environment.
来源:美国大学理事会 AP 课程与考试说明
角动量:收拢加快旋转 以零合外力矩,总角动量守恒(conserved):
$$L_{\text{before}}=L_{\text{after}},\qquad I_1\omega_1=I_2\omega_2\ \text{(one rigid body reshaping)}.$$若 $I$ 缩小,$\omega$ 增长——一个旋转的滑冰者收臂时加速。这个规则度过碰撞和形状变化,这就是使它如此有用的东西:挑选轴,使每个外力(重力、枢轴力)绕它不施加力矩。
Worked example. 一个滑冰者以 $2.0\ \text{rev/s}$、$I_1=4.0\ \text{kg}\cdot\text{m}^2$ 旋转。收臂把它降到 $I_2=1.6\ \text{kg}\cdot\text{m}^2$:$\omega_2=\dfrac{4.0}{1.6}(2.0)=5.0\ \text{rev/s}$。她的动能上升——额外的能量是她的肌肉向内收臂所做的功。

向内拉质量降低 I,所以 ω 上升以守恒 L = Iω 
随着子弹嵌入杆,绕枢轴的角动量守恒 Worked example (rotational collision). 一颗 $0.020\ \text{kg}$ 以 $300\ \text{m/s}$ 的子弹撞一根从一个枢轴悬挂的均匀杆($M=1.5\ \text{kg}$、长度 $l=0.60\ \text{m}$)的尖端,并嵌入。绕枢轴,重力和枢轴力在撞击期间不施加力矩,所以 $L$ 守恒:$L=mvl=0.020(300)(0.60)=3.6\ \text{kg}\cdot\text{m}^2/\text{s}$。之后 $I=\tfrac13Ml^2+ml^2=0.18+0.0072=0.187\ \text{kg}\cdot\text{m}^2$,所以 $\omega=\dfrac{3.6}{0.187}\approx19\ \text{rad/s}$。(线性动量在这里不守恒——枢轴推杆——而动能肯定不守恒:在声称它们之前检查两者。)
词汇表 训练英文 中文 拼音 conserved 守恒 shǒu héng 6.5
滚动
大纲
Learning Objective Essential Knowledge 6.5.A
Describe the kinetic energy of a system that has translational and rotational motion.- 6.5.A.1 The total kinetic energy of a system is the sum of the system's translational and rotational kinetic energies.
- Equation: $K_{\text{tot}} = K_{\text{trans}} + K_{\text{rot}}$
6.5.B
Describe the motion of a system that is rolling without slipping.- 6.5.B.1 While rolling without slipping, the translational motion of a system's center of mass is related to the rotational motion of the system itself with the following equations:
- Equation: $\Delta x_{\text{cm}} = r\Delta\theta$
- Equation: $v_{\text{cm}} = r\omega$
- Equation: $a_{\text{cm}} = r\alpha$
- 6.5.B.2 For ideal cases, rolling without slipping implies that the frictional force does not dissipate any energy from the rolling system.
6.5.C
Describe the motion of a system that is rolling while slipping.- 6.5.C.1 When slipping, the motion of a system's center of mass and the system's rotational motion cannot be directly related.
- 6.5.C.2 When a rotating system is slipping relative to another surface, the point of application of the force of kinetic friction exerted on the system moves with respect to the surface, so the force of kinetic friction will dissipate energy from the system.
Boundary statement: Rolling friction is beyond the scope of AP Physics C: Mechanics.
来源:美国大学理事会 AP 课程与考试说明
纯滚动(无滑动) 无滑滚动(rolling without slipping)把平动与旋转绑定:接触点暂时静止,所以
$$\Delta x_{\text{cm}}=r\,\Delta\theta,\qquad v_{\text{cm}}=r\omega,\qquad a_{\text{cm}}=r\alpha.$$
在无滑滚动里接触点静止,所以 v = rω 静摩擦供应保持旋转与运动匹配的力矩,但在一个不滑动的点——所以对于纯滚动,摩擦不做功,而能量守恒安全可用。(滚动摩擦超出 AP 课程。)
让形状沿一个斜面赛跑显示能量分割。以 $I=\beta mr^2$,能量守恒给出
$$a_{\text{cm}}=\frac{g\sin\theta}{1+\beta}:$$一个球($\beta=\tfrac25$)击败一个圆盘(disk)($\beta=\tfrac12$),它击败一个圆环(hoop)($\beta=1$)——质量和半径完全约去。圆环更多的能量被锁在旋转里,所以它的中心移动更慢。

滚动赛跑:I/mr² 最小的形状最先到达底部 Worked example. 一个实心球从静止沿一个 $30^\circ$ 的斜面滚下:$a=\dfrac{g\sin30^\circ}{1+\tfrac25}=\dfrac{9.8(0.50)}{1.4}=3.5\ \text{m/s}^2$,而一个无摩擦滑块是 $4.9\ \text{m/s}^2$ ——滚动物体总是输给滑动物体的赛跑。
词汇表 训练英文 中文 拼音 Rolling without slipping 无滑滚动 wú huá gǔn dòng disk 圆盘 yuán pán hoop 圆环 yuán huán 6.6
轨道卫星的运动
大纲
Learning Objective Essential Knowledge 6.6.A
Describe the motions of a system consisting of two objects or systems interacting only via gravitational forces.- 6.6.A.1 In a system consisting only of a massive central object and an orbiting satellite with mass that is negligible in comparison to the central object's mass, the motion of the central object itself is negligible.
- 6.6.A.2 The motion of satellites in orbits is constrained by conservation laws.
- 6.6.A.2.i In circular orbits, the system's total mechanical energy, the system's gravitational potential energy, and the satellite's angular momentum and kinetic energy are constant.
- 6.6.A.2.ii In elliptical orbits, the system's total mechanical energy and the satellite's angular momentum are constant, but the system's gravitational potential energy and the satellite's kinetic energy can each change.
- 6.6.A.2.iii The gravitational potential energy of a system consisting of a satellite and a massive central object is defined to be zero when the satellite is an infinite distance from the central object.
- Equation: $U_g = -G\dfrac{m_1 m_2}{r}$
- 6.6.A.3 The total energy of a system consisting of a satellite orbiting a central object in a circular path can be written in terms of the gravitational potential energy of that system or the kinetic energy of the satellite.
- Equation: $K = -\dfrac{1}{2}U$
- Equation: $E_{total} = \dfrac{1}{2}U = -\dfrac{GMm}{2r}$
- 6.6.A.4 The escape velocity of a satellite is the satellite's velocity such that the mechanical energy of the satellite–central-object system is equal to zero.
- 6.6.A.4.i When the only force exerted on a satellite is gravity from a central object, a satellite that reaches escape velocity will move away from the central body until its speed reaches zero at an infinite distance from the central body.
- 6.6.A.4.ii The escape velocity of a satellite from a central body of mass $M$ can be derived using conservation of energy laws.
- Equation: $v_{\text{esc}} = \sqrt{\dfrac{2GM}{r}}$
来源:美国大学理事会 AP 课程与考试说明
轨道运动(开普勒第二定律) 重力为一个圆形轨道里的卫星(satellite)供应向心力(centripetal force):
$$\frac{GMm}{r^2}=\frac{mv^2}{r}\quad\Rightarrow\quad v=\sqrt{\frac{GM}{r}}$$——更大的轨道更慢。以引力势能(gravitational potential energy)$U_g=-\dfrac{GMm}{r}$,一个圆形轨道遵循 $K=-\tfrac12U$,所以总机械能(total mechanical energy)是
$$E=K+U=\frac{U}{2}=-\frac{GMm}{2r},$$负的,因为卫星被束缚。要从半径 $r$ 逃逸,总能量必须达到零,给出逃逸速度(escape velocity)
$$v_{\text{esc}}=\sqrt{\frac{2GM}{r}}.$$
重力提供保持一个卫星在轨道里的向心力 在一个椭圆轨道(elliptical orbit)里,$E$ 和 $L$ 固定:重力指向焦点,所以它绕它不施加力矩。守恒的 $L$ 意味着卫星在相等的时间里扫过相等的面积——开普勒第二定律(Kepler's second law)——因此在最接近时移动最快、在远点最慢。能量守恒连接两端的速率。
Worked example. 对于一个绕地球($GM=4.0\times10^{14}\ \text{m}^3/\text{s}^2$)在 $r=7.0\times10^{6}\ \text{m}$ 的卫星:轨道速率 $v=\sqrt{GM/r}=7.6\times10^{3}\ \text{m/s}$,而从那个半径逃逸需要 $v_{\text{esc}}=\sqrt{2GM/r}=1.1\times10^{4}\ \text{m/s}$ ——恰好是圆形速率的 $\sqrt2$ 倍。
Exam skill. 轨道 FRQ 是伪装的能量-和-角动量问题:在两个感兴趣的点写 $E=\tfrac12mv^2-\dfrac{GMm}{r}$ 和 $L=mvr\sin\theta$ 并解这一对——绝不假设圆的轨道公式适用于一个椭圆。
探索Compare orbits at different radii
An orbiting satellite is in free fall, gravity supplying the centripetal force. A larger orbit means a slower speed and longer period (Kepler's third law).
词汇表 训练英文 中文 拼音 centripetal force 向心力 xiàng xīn lì satellite 卫星 wèi xīng gravitational potential energy 引力势能 yǐn lì shì néng total mechanical energy 总机械能 zǒng jī xiè néng escape velocity 逃逸速度 táo yì sù dù elliptical orbit 椭圆轨道 tuǒ yuán guǐ dào Kepler's second law 开普勒第二定律 kāi pǔ lēi dì èr dìng lǜ 6.6
考试技巧
- 计算转动惯量 $I=\int r^2\,dm$ 并用平行轴定理 $I=I_{cm}+Md^2$ 移轴。
- 转动动能是 $\tfrac12 I\omega^2$;一个滚动物体既有平动又有转动 KE。
- 当合外力矩是零时守恒角动量 $L=I\omega$(一个收臂的旋转滑冰者)。
- 对无滑滚动问题用能量守恒($v=r\omega$ 把两个运动绑定)。
- 知道标准 $I$ 值(圆环、圆盘、杆、球)以及轴在哪里。
- 6.1.A.1 The rotational kinetic energy of an object or rigid system is related to the rotational inertia and angular velocity of the rigid system and is given by the equation $K_{\text{rot}} = \dfrac{1}{2} I \omega^2$.
-
7
振动
7.1
简谐运动的定义
大纲
Learning Objective Essential Knowledge 7.1.A
Describe simple harmonic motion.- 7.1.A.1 Simple harmonic motion is a special case of periodic motion.
- 7.1.A.2 SHM results when the magnitude of the restoring force exerted on an object is proportional to that object's displacement from its equilibrium position.
- Derived equation: $ma_x = -k\Delta x$
- 7.1.A.2.i A restoring force is a force that is exerted in a direction opposite to the object's displacement from an equilibrium position.
- 7.1.A.2.ii An equilibrium position is a location at which the net force exerted on an object or system is zero.
来源:美国大学理事会 AP 课程与考试说明
简谐运动 简谐运动(simple harmonic motion)(SHM)是周期运动(periodic motion)的一个特殊情况。它每当两件事为真时出现:有一个合力为零的平衡位置(equilibrium)位置,而移开物体产生一个回复力(restoring force)——一个指回向平衡的力——它的大小与位移成比例:
$$F=-k\,\Delta x.$$牛顿第二定律然后给出定义性的微分方程(differential equation):
$$\frac{d^2x}{dt^2}=-\frac{k}{m}\,x=-\omega^2 x,\qquad \omega=\sqrt{\frac{k}{m}}.$$它的解是正弦的,$x(t)=A\cos(\omega t+\phi)$。你不需要证明这个解——你需要认出这个方程:任何运动遵循"$\ddot{x}=-\omega^2x$"的系统都是一个简谐振子,无论它由什么制成。(一个挂在一个竖直弹簧上的质量以相同的方式工作:重力只移动平衡点;绕它的振动不变。)

在 SHM 里加速度总是指回向平衡,与位移相反 词汇表 训练英文 中文 拼音 Simple harmonic motion 简谐运动 jiǎn xié yùn dòng periodic motion 周期运动 zhōu qī yùn dòng equilibrium 平衡位置 píng héng wèi zhì restoring force 回复力 huí fù lì differential equation 微分方程 wēi fēn fāng chéng period 周期 zhōu qī 7.2
简谐运动的频率与周期
大纲
Learning Objective Essential Knowledge 7.2.A
Describe the frequency and period of an object exhibiting SHM.- 7.2.A.1 The period of SHM is related to the angular frequency, $\omega$, of the object's motion by the following equation:
- Equation: $T = \dfrac{2\pi}{\omega} = \dfrac{1}{f}$
- 7.2.A.1.i The period of an object–ideal-spring oscillator is given by the equation
- Equation: $T_s = 2\pi\sqrt{\dfrac{m}{k}}.$
- 7.2.A.1.ii The period of a simple pendulum displaced by a small angle is given by the equation
- Equation: $T_p = 2\pi\sqrt{\dfrac{l}{g}}.$
来源:美国大学理事会 AP 课程与考试说明
角频率(angular frequency)$\omega$ 设定周期(period)和频率(frequency):
$$T=\frac{2\pi}{\omega}=\frac{1}{f}.$$对于两个标准系统:
$$T_{\text{spring}}=2\pi\sqrt{\frac{m}{k}},\qquad T_{\text{pendulum}}=2\pi\sqrt{\frac{\ell}{g}}.$$两个经典的概念陷阱:SHM 的周期从不取决于振幅(amplitude),而每个系统忽略一个明显的变量——摆的周期不涉及它的质量,而弹簧的周期不涉及 $g$(一个弹簧-木块振子在轨道里保持完美的时间;一个摆钟不)。
探索Time a pendulum's swing
A pendulum's period depends on its length and gravity, not its mass or (small) amplitude: $T=2\pi\sqrt{L/g}$. Lengthen it and each swing takes longer.
词汇表 训练英文 中文 拼音 angular frequency 角频率 jiǎo pín lǜ frequency 频率 pín lǜ amplitude 振幅 zhèn fú 7.3
简谐运动的表示与分析
大纲
Learning Objective Essential Knowledge 7.3.A
Describe the displacement, velocity, and acceleration of an object exhibiting SHM.- 7.3.A.1 For an object exhibiting SHM, the displacement of that object measured from its equilibrium position can be represented by the equations $x = A\cos(2\pi ft)$ or $x = A\sin(2\pi ft)$.
- 7.3.A.1.i Minima, maxima, and zeros of displacement, velocity, and acceleration are features of harmonic motion.
- 7.3.A.1.ii Recognizing the positions or times at which the displacement, velocity, and acceleration for SHM have extrema or zeros can help in qualitatively describing the behavior of the motion.
- 7.3.A.2 The position as a function of time for an object exhibiting SHM is a solution of the second-order differential equation derived from the application of Newton's second law.
- Derived equation: $\dfrac{d^2 x}{dt^2} = -\omega^2 x$
- 7.3.A.3 Characteristics of SHM, such as velocity and acceleration, can be determined by or derived from the equation $x = A\cos(\omega t + \phi).$
- 7.3.A.3.i The acceleration of an object exhibiting SHM is related to the object's angular frequency and position.
- Derived equation: $a = -\omega^2 x$
- 7.3.A.3.ii It can be shown that the maximum velocity and acceleration of an object exhibiting SHM are related to the angular frequency of the object's motion.
- Derived equations: $v_{\max} = A\omega$
- $a_{\max} = A\omega^2$
- 7.3.A.3.i The acceleration of an object exhibiting SHM is related to the object's angular frequency and position.
- 7.3.A.4 In the presence of a sinusoidal external force, a system may exhibit resonance.
- 7.3.A.4.i Resonance occurs when an external force is exerted at the natural frequency of an oscillating system.
- 7.3.A.4.ii Resonance increases the amplitude of oscillating motion.
- 7.3.A.4.iii The natural frequency of a system is the frequency at which the system will oscillate when it is displaced from its equilibrium position.
- 7.3.A.5 Changing the amplitude of a system exhibiting SHM will not change its period.
- 7.3.A.6 Properties of SHM can be determined and analyzed using graphical representations.
Boundary statement: AP Physics C: Mechanics only expects students to know the solution to the second-order differential equation that describes SHM, as well as be able to identify SHM. AP Physics C: Mechanics does not expect students to mathematically prove that the solution is correct.
来源:美国大学理事会 AP 课程与考试说明
把 $x(t)=A\cos(\omega t+\phi)$ 求导两次:
$$v=-A\omega\sin(\omega t+\phi),\qquad a=-A\omega^2\cos(\omega t+\phi)=-\omega^2x.$$所以在摆动的两端($x=\pm A$)速率是零而加速度最大;通过平衡($x=0$)加速度是零而速率最大,$v_{\max}=A\omega$。振幅 $A$ 和相位常数(phase constant)$\phi$ 来自初始条件:物体从哪里开始以及它移动多快。消去 $t$(或用能量,下面)给出速率作为位置的一个函数:
$$v=\pm\,\omega\sqrt{A^2-x^2}.$$
位移在简谐运动里随时间正弦地变化 Worked example. 一个 $k=200\ \text{N/m}$ 弹簧上的 $0.50\ \text{kg}$ 质量:$\omega=\sqrt{k/m}=20\ \text{rad/s}$、$T=2\pi/\omega=0.31\ \text{s}$。以 $A=0.10\ \text{m}$:在平衡 $v_{\max}=A\omega=2.0\ \text{m/s}$,而在 $x=0.050\ \text{m}$ 速率是 $v=\omega\sqrt{A^2-x^2}=20\sqrt{0.10^2-0.050^2}=1.7\ \text{m/s}$。
词汇表 训练英文 中文 拼音 phase constant 相位常数 xiàng wèi cháng shù 7.4
简谐振子的能量
大纲
Learning Objective Essential Knowledge 7.4.A
Describe the mechanical energy of a system exhibiting SHM.- 7.4.A.1 The total energy of a system exhibiting SHM is the sum of the system's kinetic and potential energies.
- Relevant equation: $E_{\text{total}} = U + K$
- 7.4.A.2 Conservation of energy indicates that the total energy of a system exhibiting SHM is constant.
- 7.4.A.3 The kinetic energy of a system exhibiting SHM is at a maximum when the system's potential energy is at a minimum.
- 7.4.A.4 The potential energy of a system exhibiting SHM is at a maximum when the system's kinetic energy is at a minimum.
- 7.4.A.4.i The minimum kinetic energy of a system exhibiting SHM is zero.
- 7.4.A.4.ii Changing the amplitude of a system exhibiting SHM will change the maximum potential energy of the system and, therefore, the total energy of the system.
- Relevant equation for a spring–object system: $E_{\text{total}} = \dfrac{1}{2}kA^2$
来源:美国大学理事会 AP 课程与考试说明
振子的总能量是和 $E=U+K$,而没有摩擦它恒定——能量只是在弹簧的势能和质量的动能之间来回交换:
$$E=\tfrac12kA^2=\tfrac12kx^2+\tfrac12mv^2.$$在两端全是势能(那里 $K=0$),在平衡全是动能(那里 $U$ 最小)。因为 $E\propto A^2$,把振幅加倍使能量变为四倍。

动能和势能在一个循环上交换而总能量保持恒定 Worked example. 能量在哪里均匀分割?令 $\tfrac12kx^2=\tfrac12E=\tfrac14kA^2$,所以 $x=A/\sqrt2\approx0.71A$ ——比到中间更接近端。对于上面的系统($A=0.10\ \text{m}$):$x=0.071\ \text{m}$。
不受打扰时,一个系统以它自己的固有频率(natural frequency)振荡(由 $k$ 和 $m$ 决定:$f_0=\frac{1}{2\pi}\sqrt{k/m}$)。用一个周期性的力在那个相同频率驱动它,振幅急剧增大——共振(resonance)。每一次恰好定时的推动加入能量的速度快于阻尼移除的速度,所以当驱动频率匹配 $f_0$ 时驱动振幅尖锐地达到峰值(荡秋千按节奏推、桥被风摇晃)。
探索Trade kinetic and potential energy in SHM
In simple harmonic motion, energy sloshes between kinetic (fastest at the centre) and potential (greatest at the extremes) while the total stays constant.
词汇表 训练英文 中文 拼音 natural frequency 固有频率 gù yǒu pín lǜ resonance 共振 gòng zhèn 7.5
单摆与物理摆
大纲
Learning Objective Essential Knowledge 7.5.A
Describe the properties of a physical pendulum.- 7.5.A.1 A physical pendulum is a rigid body that undergoes oscillation about a fixed axis.
- 7.5.A.2 For small amplitudes of motion, the period of a physical pendulum is derived from the application of Newton's second law in rotational form.
- Relevant equation: $T_{\text{phys}} = 2\pi\sqrt{\dfrac{I}{mgd}}$
- 7.5.A.2.i When displaced from equilibrium, the gravitational force exerted on a physical pendulum's center of mass provides a restoring torque.
- Derived equation: $\tau = -mgd\sin\theta$
- 7.5.A.2.ii For small amplitudes of motion, the small-angle approximation can be applied to the restoring torque.
- Derived equation: $\sin\theta \approx \theta$
- $\tau = -mgd\theta = I\alpha$
- 7.5.A.2.iii The small-angle approximation and Newton's second law in rotational form yield a second-order differential equation that describes SHM:
- Equation: $\dfrac{d^2\theta}{dt^2} = -\omega^2\theta$
- 7.5.A.3 A simple pendulum is a special case of physical pendulums in which the hanging object can be modeled as a point mass at a distance, $l$, from the pivot point.
- Relevant equation: $T_p = 2\pi\sqrt{\dfrac{\ell}{g}}$
- 7.5.A.4 A torsion pendulum is a case of SHM where the restoring torque is proportional to the angular displacement of a rotating system. For example, a horizontal disk that is suspended from a wire attached to its center of mass may undergo rotational oscillations about the wire in the horizontal plane.
- Derived equation: $I\alpha = -k\Delta\theta$
来源:美国大学理事会 AP 课程与考试说明
能量守恒:动能↔势能 一个物理摆(physical pendulum)是任何绕一个固定枢轴摆动的刚体。把它移开一个角 $\theta$,而重力,作用在距枢轴一个距离 $d$ 的质心处,供应一个回复力矩
$$\tau=-mgd\sin\theta.$$对于小角度,应用小角度近似(small-angle approximation)$\sin\theta\approx\theta$ 和旋转形式的牛顿第二定律($\tau=I\alpha$):
$$\frac{d^2\theta}{dt^2}=-\frac{mgd}{I}\,\theta=-\omega^2\theta\quad\Rightarrow\quad T_{\text{phys}}=2\pi\sqrt{\frac{I}{mgd}}.$$这是与之前相同的"$\ddot\theta=-\omega^2\theta$"模式——认出它就是这个推导。一个单摆(simple pendulum)是一个轻绳上的点质量的特殊情况:$I=m\ell^2$ 和 $d=\ell$ 给出 $T=2\pi\sqrt{\ell/g}$。

作用在质心处的重力提供一个物理摆的回复力矩 Worked example. 一根均匀的杆(质量 $M$、长度 $L$)从一端摆动:$I=\tfrac13ML^2$、$d=\tfrac{L}{2}$,所以
$$T=2\pi\sqrt{\frac{\tfrac13ML^2}{Mg\,\tfrac{L}{2}}}=2\pi\sqrt{\frac{2L}{3g}}$$——比一个长度 $L$ 的单摆更短,因为杆的质量坐得离枢轴更近。
一个扭摆(torsion pendulum)——一个从一根扭转的金属丝悬挂的圆盘——又是 SHM 一次:金属丝的回复力矩与扭转角成比例,$I\alpha=-\kappa\,\Delta\theta$,给出 $T=2\pi\sqrt{I/\kappa}$。
Exam skill. 每个摆 FRQ 想要相同的三个步骤:写绕枢轴的回复力矩、应用小角度近似,并把结果匹配到 $\ddot\theta=-\omega^2\theta$ 以读出 $\omega$。明确地陈述小角度步骤——它是一个评分点,而它是为什么大振幅摆动不是简谐的。
词汇表 训练英文 中文 拼音 physical pendulum 物理摆 wù lǐ bǎi small-angle approximation 小角度近似 xiǎo jiǎo dù jìn sì simple pendulum 单摆 dān bǎi torsion pendulum 扭摆 niǔ bǎi 7.5
考试技巧
- 从一个线性回复力 $F=-kx$ 辨别 SHM,它给出 $\tfrac{d^2x}{dt^2}=-\omega^2 x$,$\omega=\sqrt{k/m}$。
- 写解 $x=A\cos(\omega t+\phi)$ 并通过求导得到 $v,a$;周期 $T=\tfrac{2\pi}{\omega}$ 与振幅无关。
- 能量在 $\tfrac12 kx^2$ 和 $\tfrac12 mv^2$ 之间交换,总量 $\tfrac12 kA^2$。
- 对于一个摆,用小角度近似 $\sin\theta\approx\theta$ 达到 SHM。
- 把相位匹配到起点:从最大位移静止释放用余弦。