转动运动学与动力学

AP 物理 C：力学· AP Physics C: Mechanics CED — Rotation· 14 分钟阅读

1. 什么是转动运动学与动力学？★★☆☆☆⏱ 3 min

转动运动学与动力学研究刚体的定轴转动，描述转动如何发生、什么引发转动速度的变化。本主题是第5单元转动的核心内容，占AP物理C：力学考试总分的14-20%，同时出现在选择题和自由作答部分，通常会和平动结合组成多模块问题。

按照惯例，逆时针转动定义为正，所有计算中的角量都以弧度为单位。转动运动学描述转动量之间的关系，不涉及转动的起因，而转动动力学将这些量和力矩（力的转动类比）联系起来。本主题直接建立平动和转动力学的类比，简化刚体运动的学习。

2. 转动运动学★★☆☆☆⏱ 4 min

转动运动学不涉及起因，仅描述转动运动，和平动的直线运动学直接类比。核心物理量包括：

角位移 $Δ θ$ ：刚体绕转轴转过的角度变化，单位为弧度
角速度 $ω = \frac{d θ}{d t}$ ：角位移的变化率，单位 rad/s
角加速度 $α = \frac{d ω}{d t} = \frac{d ^{2} θ}{d t ^{2}}$ ：角速度的变化率，单位 rad/s²

对于匀角加速度，我们可以直接推导出和匀加速直线运动类比的运动学公式，只需将 $x \to θ$ , $v \to ω$ , $a \to α$ 替换：

\begin{align*}\omega &= \omega_0 + \alpha t \\\theta &= \omega_0 t + \frac{1}{2}\alpha t^2 \\\omega^2 &= \omega_0^2 + 2\alpha \Delta\theta\end{align*}

对于转动刚体上距离转轴 $r$ 处的任意一点，我们可以将角量和线切向量联系起来，所有点都有指向转轴的向心（径向）加速度：

Arc length: $s = r θ$
Tangential speed: $v = r ω$
Tangential acceleration: $a_{t} = r α$
Centripetal acceleration: $a_{c} = r ω^{2}$

📐 例题

搅拌机叶片从静止开始匀加速，1.5秒后转速达到1800 rpm。求(a)叶片的角加速度，(b)加速过程中叶片转过的总转数，(c)距离转轴0.04 m处一点的切向加速度。

1
Convert final angular speed to radians per second:
$\omega_f = 1800 \frac{\text{rev}}{\text{min}} \times \frac{2\pi \text{rad}}{\text{rev}} \times \frac{1 \text{min}}{60 \text{s}} = 60\pi \text{rad/s}$
2
Calculate angular acceleration:
$\alpha = \frac{\omega_f - \omega_0}{t} = \frac{60\pi - 0}{1.5} = 40\pi \approx 125.7 \text{rad/s}^2$
3
Find total angular displacement and convert to rotations:
$\Delta\theta = \omega_0 t + \frac{1}{2}\alpha t^2 = 0 + 0.5(40\pi)(1.5)^2 = 45\pi \text{rad} \implies \frac{45\pi}{2\pi} = 22.5 \text{rotations}$
4
Calculate tangential acceleration:
$a_t = r\alpha = 0.04(40\pi) \approx 5.03 \text{m/s}^2$

3. 力矩与转动惯量★★★☆☆⏱ 3 min

力矩是力的转动类比：就像力引发线运动的变化，力矩是引发转动运动变化的物理量。对于作用点距离转轴 $r$ 的力 $F$ ，力矩的大小为：

\tau = rF\sin\theta = rF_\perp = r_\perp F

Where $θ$ is the angle between the position vector $r$ (from the axis to the point of application) and $F$ . $F_{⊥}$ is the component of force perpendicular to $r$ , and $r_{⊥}$ is the perpendicular lever arm from the axis to the line of action of the force. By convention, counterclockwise torque is positive, and clockwise torque is negative.

Rotational inertia (or moment of inertia) is the rotational analog of mass, describing how much torque is needed to produce a given angular acceleration. For a system of discrete masses, $I = \sum m_{i} r_{i}^{2}$ , where $r_{i}$ is the distance of mass $m_{i}$ from the axis. For a continuous rigid body, $I = \int r^{2} d m$ . The parallel axis theorem lets you calculate $I$ for any axis parallel to the axis through the center of mass:

I = I_{cm} + Md^2

Where $M$ is the total mass of the object, and $d$ is the distance between the two parallel axes.

📐 例题

一块质量 $M = 3 kg$ 、边长 $L = 0.8 m$ 的均匀实心正方形板，绕过中心且平行于一边的转轴的转动惯量为 $I_{c m} = \frac{1}{6} M L^{2}$ 。求绕板边缘转轴的转动惯量。

1
均匀正方形的质心在其几何中心。中心转轴到边缘转轴的距离为：
$d = \frac{L}{2}$
2
Apply the parallel axis theorem:
$I = I_{cm} + Md^2 = \frac{1}{6}ML^2 + M\left(\frac{L}{2}\right)^2$
3
Simplify the expression:
$I = ML^2\left(\frac{1}{6} + \frac{1}{4}\right) = \frac{5}{12}ML^2$
4
Substitute numerical values:
$I = \frac{5}{12}(3 \text{kg})(0.8 \text{m})^2 = 0.8 \text{kg·m}^2$

4. 转动的牛顿第二定律★★★☆☆⏱ 4 min

Newton's second law for rotation connects net torque to angular acceleration, analogous to Newton's second law for translation $\sum F = m a$ . For rotation about a fixed axis, the law states:

\sum \tau_{\text{axis}} = I_{\text{axis}} \alpha

This means the net torque on a rigid body about the rotation axis equals the product of the rotational inertia about that axis and the angular acceleration. For problems involving both translation and rotation (e.g., pulley systems with massive pulleys, rolling motion), this law is used alongside Newton's second law for translation, with the relation $a = r α$ connecting linear acceleration of a point on the rigid body to angular acceleration when there is no slipping.

📐 例题

一个实心圆盘形质量滑轮，质量 $M = 2 kg$ ，半径 $R = 0.1 m$ ，绕中心固定转轴转动。缠绕在滑轮上的绳子悬挂着一个质量 $m = 1 kg$ 的物块。绳子在滑轮上不打滑，求滑轮的角加速度。

1
Write Newton's second law for the hanging block ( $T$ = tension, $a$ = linear acceleration):
$mg - T = ma$
2
Write Newton's second law for the pulley. Rotational inertia of a solid disk is $I = \frac{1}{2} M R^{2}$ :
$TR = I\alpha$
3
Relate linear and angular acceleration for no slipping:
$a = R\alpha$
4
Substitute $I$ and $a = R α$ into the torque equation:
$TR = \left(\frac{1}{2}MR^2\right)\alpha \implies T = \frac{1}{2}Ma$
5
Substitute $T$ into the block equation and solve for $a$ :
$mg - \frac{1}{2}Ma = ma \implies a = \frac{mg}{m + M/2} = \frac{(1)(9.8)}{1 + 1} = 4.9 \text{m/s}^2$
6
Solve for angular acceleration:
$\alpha = \frac{a}{R} = \frac{4.9}{0.1} = 49 \text{rad/s}^2$

5. 常见陷阱

错误做法:

计算切向加速度、弧长或任何导出转动量时，使用度或转而非弧度。

原因:

很多题目给出的角速度单位是转每分钟，因此学生忘记转换为弧度，这是所有标准转动公式要求的单位。

正确做法:

解题开始时就将所有角单位转换为弧度或弧度每秒。

错误做法:

忘记转动物体上角加速度非零的点仍然存在向心加速度。

原因:

学生只关注角加速度带来的切向加速度，忘记任何做圆周运动的点都需要向心加速度来维持其运动轨迹。

正确做法:

When asked for total acceleration, always calculate both $a_{t} = r α$ and $a_{c} = r ω^{2}$ , then combine them with the Pythagorean theorem.

错误做法:

计算均匀刚体绕端轴转动的重力力矩时，使用 $r = L$ （全长）而非 $r = L /2$ 。

原因:

学生错误认为重力作用在物体端点，而非质心。

正确做法:

对于任意均匀刚体，重力始终作用在质心，因此使用转轴到质心的距离计算重力力矩。

错误做法:

Forgetting to add the $M d^{2}$ term when using the parallel axis theorem, or using $d$ as distance from the nearest end of the object.

原因:

Students memorize the formula but misinterpret what $d$ measures.

正确做法:

Explicitly locate the center of mass, measure $d$ from the center of mass to the new axis, then add $M d^{2}$ to $I_{c m}$ .

错误做法:

Neglecting the torque from tension in problems with massive pulleys, treating tension as the same on both sides of the pulley.

原因:

Students are used to massless pulleys where tension is equal, but this does not hold for massive pulleys with angular acceleration.

正确做法:

For massive pulleys, write separate torque equations that account for different tensions on each side of the pulley.

6. 速查表

分类	公式	备注
角-线关系	$s = r θ$ , $v = r ω$ , $a_{t} = r α$ , $a_{c} = r ω^{2}$	$θ, ω, α$ must be in radians
Constant $α$ Kinematics	$ω = ω_{0} + α t$ ; $θ = ω_{0} t + \frac{1}{2} α t^{2}$ ; $ω^{2} = ω_{0}^{2} + 2 α Δ θ$	Direct analog to linear constant acceleration kinematics
Torque Magnitude	$τ = r F sin θ = r F_{⊥} = r_{⊥} F$	Counterclockwise torque = positive by convention
Discrete Rotational Inertia	$I = \sum m_{i} r_{i}^{2}$	$r_{i}$ = distance from rotation axis
Parallel Axis Theorem	$I = I_{c m} + M d^{2}$	$d$ = distance between parallel axes
Newton's 2nd Law (Rotation)	$\sum τ_{axis} = I_{axis} α$	Use $I$ for the actual rotation axis
Common $I_{c m}$ Values	Rod (cm): $\frac{1}{12} M L^{2}$ ; Solid disk: $\frac{1}{2} M R^{2}$ ; Solid sphere: $\frac{2}{5} M R^{2}$ ; Thin hoop: $M R^{2}$	All for axis through center of mass
Total Acceleration	$a_{total} = a_{t}^{2} + a_{c}^{2}$	$a_{t}$ tangential, $a_{c}$ radial; components are perpendicular

真题中的出现

AI 根据考纲规律估算的考点位置,请对照官方真题核实准确性。仅作复习重点参考。

2023 · MCQ
匀角加速度问题
2022 · FRQ
质量滑轮系统问题

深入阅读

unit_overview第5单元转动概述本主题的父单元概述

下一步

转动运动学和动力学是整个第5单元转动的基础主题。接下来你会将这些核心关系应用到转动动能、角动量和无滑滚动运动中，这些是本单元剩余的内容。如果不掌握力矩、转动惯量、角加速度之间的关系，以及定轴转动的运动学规律，你将无法解决涉及滚动运动或守恒角动量的多模块问题，这些都是AP考试中分值很高的常见自由作答问题。