转动系统的能量与动量

AP 物理 1 · 第 6 主题

训练

讲义词汇表

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 $K = \frac{1}{2} I \omega^2$.
- 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),$\tfrac12 mv^2$ 的旋转孪生:

$$K_{\text{rot}}=\tfrac{1}{2}I\omega^2.$$

一个既移动又旋转的物体(像一个滚动的球)有两种平动和转动动能,而它的总量是 $K=\tfrac12 mv^2+\tfrac12 I\omega^2$。

词汇表训练

英文	中文	拼音
rotational kinetic energy	转动动能	zhuǎn dòng dòng néng

练习卷

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: $W = \tau \Delta\theta$
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 torque as a function of angular position.

来源:美国大学理事会 AP 课程与考试说明

一个力矩作用过一个角位移做功,改变转动动能:

$$W=\tau\,\Delta\theta,\qquad P=\tau\omega.$$

这是 $W=Fd$ 和 $P=Fv$ 的旋转形式,而它把动能定理扩展到旋转。

Worked example. 一个马达在一个飞轮转过 $10\ \text{rad}$ 时对它施加一个稳定的 $8.0\ \text{N m}$ 力矩。做的功是 $W=\tau\,\Delta\theta=8.0\times10=80\ \text{J}$,而若飞轮从静止开始这全都变成转动动能。

探索

Balance torques on a seesaw

Torque is force times perpendicular distance, $\tau = Fd$. The beam balances when the torques on each side are equal — move the forces and distances to find the balance point.

练习卷

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 $L = I\omega$.
6.3.A.2 The magnitude of the angular momentum of an object about a given point is $L = rmv \sin\theta$.
- 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: $\text{angular impulse} = \tau \Delta t$
6.3.B.2 Angular impulse has the same direction as the torque exerted on the object or system.
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 angular momenta of the object or rigid system: $\Delta L = L - L_0$
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: $\Delta L = \tau \Delta t$
- 6.3.C.2.ii The rotational form of the impulse–momentum theorem is a direct result of the rotational form of Newton's second law of motion for cases in which rotational inertia is constant: $\tau_{\text{net}} = \dfrac{\Delta L}{\Delta t} = I \dfrac{\Delta\omega}{\Delta t} = I\alpha$
6.3.C.3 The net torque exerted on an object 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 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.

Boundary statement: While AP Physics 1 expects that students can mathematically manipulate the magnitude of angular momentum using one-dimensional vector conventions, the direction of angular momentum and angular impulse is beyond the scope of the course.

来源:美国大学理事会 AP 课程与考试说明

角动量(angular momentum)是线性动量的旋转版本:

$$L=I\omega.$$

一个合力矩随时间作用传递一个改变它的角冲量(angular impulse):$\tau\,\Delta t=\Delta L$ ——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

练习卷

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 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 further 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

若一个系统上的合外力矩是零,它的总角动量守恒(conserved):

$$I_1\omega_1=I_2\omega_2.$$

所以若 $I$ 减小,$\omega$ 增大以保持 $L$ 恒定——这就是为什么一个旋转的滑冰者在收臂时加速。它也适用于旋转系统的碰撞和爆炸。

Worked example. 一个滑冰者以 $2.0\ \text{rev/s}$ 旋转,转动惯量 $I_1=4.0\ \text{kg m}^2$。她收臂,把她的转动惯量降到 $I_2=1.6\ \text{kg m}^2$。没有外力矩,角动量守恒:

$$\omega_2=\frac{I_1}{I_2}\,\omega_1=\frac{4.0}{1.6}\times2.0=5.0\ \text{rev/s}.$$

她的动能实际上上升——额外的能量来自她的肌肉逆着向外的拉力收臂所做的功。

向内拉质量降低 I,所以 ω 上升以守恒 L = Iω

词汇表训练

英文	中文	拼音
conserved	守恒	shǒu héng

练习卷

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 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 1.

Boundary statement: The precise mathematical relationships between linear and angular quantities while a rigid body is rolling while slipping are beyond the scope of AP Physics 1 and 2, and students will not be expected to model those relationships quantitatively. However, students are expected to qualitatively explain the changes to linear and angular quantities while a rigid body is rolling while slipping.

来源:美国大学理事会 AP 课程与考试说明

纯滚动(rolling without slipping)把平动和转动运动联系起来:接触点暂时静止,所以

$$v=r\omega \qquad\text{and}\qquad a=r\alpha.$$

一个滚动物体的能量在平动和转动之间分割,所以在一个斜面上它比一个无摩擦滑动的物体加速更慢——一些能量进入旋转。

Worked example. 对于一个纯滚动的实心圆盘($I=\tfrac12 mR^2$),它的动能的多少分数是转动的?用 $v=R\omega$,转动部分是 $\tfrac12 I\omega^2=\tfrac12(\tfrac12 mR^2)\omega^2=\tfrac14 mv^2$,而平动部分是 $\tfrac12 mv^2$。所以总量是 $\tfrac34 mv^2$ 而转动份额是 $\tfrac{1/4}{3/4}=\tfrac13$。一个环,它的质量更远,把它一半的能量储存在旋转里并从一个坡道滚下更慢。

在纯滚动里接触点静止,所以 v = rω

词汇表训练

英文	中文	拼音
Rolling without slipping	纯滚动	chún gǔn dòng

练习卷

6.6

环绕卫星的运动

大纲

Learning Objective

Essential Knowledge

6.6.A
Describe the motions of a system consisting of two objects 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 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.3.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.3.ii The escape velocity of a satellite from a central body of mass $M$ can be derived using conservation of energy laws.
  - Equation (derived): $v_{\text{esc}} = \sqrt{\dfrac{2GM}{r}}$

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）

一个在轨道里的卫星(satellite)处于自由落体:重力提供把它的路径弯曲成一个轨道所需的确切向心力。把重力设为等于向心要求,

$$\frac{GMm}{r^2}=\frac{mv^2}{r}\ \Rightarrow\ v=\sqrt{\frac{GM}{r}}.$$

所以一个更大的轨道意味着一个更慢的速率。对于一个圆形轨道,角动量和机械能都恒定;对于一个椭圆轨道,角动量守恒(绕行星没有力矩)而速率变化——最接近时最快。

重力提供保持一个卫星在轨道里的向心力

Worked example. 求一个刚好在地球之上的低轨道里的卫星的速率,半径 $r=6.4\times10^{6}\ \text{m}$,$GM=4.0\times10^{14}\ \text{m}^3/\text{s}^2$:

$$v=\sqrt{\frac{GM}{r}}=\sqrt{\frac{4.0\times10^{14}}{6.4\times10^{6}}}=\sqrt{6.25\times10^{7}}\approx 7.9\times10^{3}\ \text{m/s},$$

约 $7.9\ \text{km/s}$ ——并注意卫星的质量约去,所以所有低轨道共享这个速率。

逃逸速度(escape velocity)是恰好让一个物体永远离开的发射速率:它的总机械能恰好为零,所以它只在无穷远处减速到零。令 $\tfrac12 mv^2 - \dfrac{GMm}{r}=0$(用一般引力势能 $U_g=-GMm/r$)并解出 $v$,

$$v_{\text{esc}}=\sqrt{\frac{2GM}{r}}.$$

它是相同半径处圆轨道速率的 $\sqrt2$ 倍,而且(像轨道速率一样)不取决于逃逸物体的质量。

探索

Compare orbits at different radii

A satellite is in free fall, its gravity supplying the centripetal force. A larger orbit means a slower speed and a longer period — Kepler's third law.

词汇表训练

英文	中文	拼音
satellite	卫星	wèi xīng
Escape velocity	逃逸速度	táo yì sù dù

练习卷

6.6

考试技巧

当没有外力矩作用时守恒角动量 $L=I\omega$:一个更小的 $I$(收臂)给出一个更大的 $\omega$。
一个滚动物体把它的能量在 $\tfrac12 mv^2$ 和 $\tfrac12 I\omega^2$ 之间分割,由 $v=r\omega$ 联系——所以它从一个坡道加速比一个滑动的更慢。
对于一个圆形轨道把重力设为等于向心要求:$v=\sqrt{GM/r}$,所以一个更大的轨道更慢而卫星的质量约去。
收臂提高旋转和动能——额外的能量来自向内拉所做的功;$L$ 不变。
注意哪个旋转量守恒:$L$(没有力矩)与能量(没有摩擦)是不同的条件。

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