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力矩与转动动力学

AP 物理 C:力学 · 第 5 主题

训练
讲义 词汇表
5.1

转动运动学

大纲
Learning ObjectiveEssential 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.

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)、$\omega$角速度(angular velocity),而 $\alpha$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
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$$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 ObjectiveEssential 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
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 ObjectiveEssential 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$$\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
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
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
练习卷
5.4

转动惯量

大纲
Learning ObjectiveEssential 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]$$\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 ObjectiveEssential 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$$\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
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
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 ObjectiveEssential 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
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}$$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 $R=0.20\ \text{m}$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}$$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。
  • $v=r\omega$$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。

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