转动运动学

AP物理1· AP物理1 课程与考试说明（CED）—— 转动运动· 14 分钟阅读

1. 核心转动运动学物理量★★☆☆☆⏱ 3 min

转动运动学描述转动刚体的位置、速度和加速度，不涉及引发转动的力矩或作用力（该部分分析属于转动动力学内容）。该知识点约占AP物理1考试总分的4-6%，会出现在选择题和简答题中。

📘 定义

标准转动符号体系

Greek letters for rotational quantities

按照AP考试惯例，我们使用希腊字母以避免与直线运动变量混淆： $θ$ （西塔）= 角位置， $ω$ （欧米伽）= 角速度， $α$ （阿尔法）= 角加速度。所有标准公式中的角度都必须使用弧度为单位，不能使用角度。

AP物理1的标准符号约定为：逆时针转动为正，顺时针转动为负。角位移定义为 $Δ θ = θ_{f} - θ_{i}$ ，即一段时间间隔内角位置的变化量。

✓ 快速检测

测试你对基础约定的理解：

A wheel rotates 3 full turns clockwise. What is its angular displacement in radians?
- $6 π$ rad
- $- 6 π$ rad
- $3$ rad
- $- 3$ rad
显示答案
$-6\pi$ rad —
Correct! Clockwise rotation is negative per standard convention, and 3 full turns = $3 \times 2 π = 6 π$ radians.

2. 恒角加速度运动学★★★☆☆⏱ 4 min

当角加速度 $α$ 为常数（即转动均匀加速或减速）时，我们使用三个与一维匀加速直线运动学完全对应的运动学方程：

\begin{align} \omega_f &= \omega_i + \alpha t \\ \Delta \theta &= \omega_i t + \frac{1}{2} \alpha t^2 \\ \omega_f^2 &= \omega_i^2 + 2 \alpha \Delta \theta \end{align}

📐 例题

一台搅拌机的叶片从120 rad/s开始均匀减速，15秒后完全停止。求叶片在停止过程中的总角位移是多少？

1
1. 列出已知变量： $ω_{i} = 120$ rad/s， $ω_{f} = 0$ rad/s， $t = 15$ s， $Δ θ = ?$
2
1. 利用平均角速度关系，无需先计算 $α$ ：
3
$\Delta \theta = \frac{1}{2}(\omega_i + \omega_f)t$
4
1. 代入数值：
5
$\Delta \theta = 0.5 (120 + 0)(15) = 900 \text{ radians}$
6
1. 先计算 $α$ 进行验证： $α = \frac{ω _{f} - ω _{i}}{t} = \frac{- 120}{15} = - 8$ rad/s²。代入第二个运动学方程： $Δ θ = (120) (15) + 0.5 (- 8) (15)^{2} = 900$ 弧度，结果一致。

3. 切向物理量与转动物理量的关联★★★☆☆⏱ 4 min

对于绕固定轴转动的刚体，刚体内所有点的角速度 $ω$ 和角加速度 $α$ 都相同，与该点到转轴的距离无关。但每个点的直线（切向）速度和加速度都不同，因为每个点沿圆周路径运动的半径等于它到转动轴的距离 $r$ 。

根据弧度的定义，弧长（沿圆周路径运动的直线距离）为 $s = r θ$ 。由于 $r$ 是常数，该比例关系同样适用于变化率：切向速度 $v_{t} = r ω$ ，切向加速度 $a_{t} = r α$ 。

请勿将切向加速度与向心（径向）加速度混淆：向心加速度指向圆周路径的中心，使质点保持圆周运动，大小为 $a_{c} = r ω^{2}$ 。切向加速度沿路径切线方向，仅当角加速度不为零时存在。总直线加速度是这两个互相垂直的加速度的矢量和。

📐 例题

半径为2.5 m的旋转木马从静止开始以恒定角加速度0.20 rad/s²运动8.0秒。求坐在木马边缘的儿童在 $t = 8.0$ s时的总线加速度大小是多少？

1
1. 利用恒定角加速度公式求 $t = 8.0$ s时的末角速度 $ω_{f}$ ：
2
$\omega_f = \omega_i + \alpha t = 0 + (0.20)(8.0) = 1.6 \text{ rad/s}$
3
1. 计算切向加速度：
4
$a_t = r\alpha = (2.5)(0.20) = 0.50 \text{ m/s²}$
5
1. 计算向心加速度：
6
$a_c = r\omega^2 = (2.5)(1.6)^2 = 6.4 \text{ m/s²}$
7
1. 求总加速度大小（由于 $a_{t}$ 和 $a_{c}$ 相互垂直）：
8
$a_{total} = \sqrt{a_t^2 + a_c^2} = \sqrt{0.50^2 + 6.4^2} \approx 6.4 \text{ m/s²}$

4. 无滑动滚动的运动学关系★★★★☆⏱ 3 min

转动运动学在AP考试中的常见应用场景是无滑动滚动，即车轮、轮胎、球体等物体沿表面滚动时不发生打滑的情况。当物体做无滑动滚动时，物体质心的直线运动距离恰好等于轮胎与表面接触部分的弧长，由此可得到核心关系：

\Delta x_{cm} = r \Delta \theta \implies v_{cm} = r \omega \implies a_{cm} = r \alpha

该关系仅适用于无滑动滚动场景；如果物体发生打滑（例如汽车轮胎在冰面上空转），则该关系不成立。

📐 例题

外半径为0.35 m的自行车轮胎沿山坡做无滑动滚动。自行车从静止开始以恒定加速度1.5 m/s²运动6.0秒，求这段时间内轮胎一共完成了多少整圈转动？

1
1. 利用无滑动滚动关系求角加速度：
2
$\alpha = \frac{a_{cm}}{r} = \frac{1.5}{0.35} \approx 4.29 \text{ rad/s²}$
3
1. 计算总角位移（已知 $ω_{i} = 0$ ）：
4
$\Delta \theta = \omega_i t + \frac{1}{2}\alpha t^2 = 0 + 0.5(4.29)(6.0)^2 = 77.2 \text{ radians}$
5
1. 将弧度转换为圈数（1圈 = $2 π$ 弧度）：
6
$N = \frac{77.2}{2\pi} \approx 12.3, \text{ so } 12 \text{ complete revolutions}$
7
1. 用直线运动验证：总运动距离 $Δ x = 0.5 a_{c m} t^{2} = 27$ m，轮胎周长 $2 π r \approx 2.2$ m，因此 $N = 27/2.2 \approx 12.3$ ，与计算结果一致。

5. 常见陷阱

错误做法:

Leaving angular quantities in degrees or rpm when plugging into kinematic equations

原因:

Problems often give initial values in rpm or degrees as a trap; all rotational kinematic formulas require radians to work correctly

正确做法:

Convert all angular speeds to rad/s and angles to radians immediately after listing known variables at the start of every problem

错误做法:

Mixing up tangential acceleration $a_{t} = r α$ and centripetal acceleration $a_{c} = r ω^{2}$ when asked for total acceleration

原因:

Students confuse the acceleration that changes rotation speed with the acceleration required to keep the point moving in a circle

正确做法:

Label both accelerations explicitly: tangential comes from $α$ , centripetal comes from $ω$ , add them as perpendicular vectors for total acceleration

错误做法:

Applying constant angular acceleration kinematic equations to problems where $α$ is not constant

原因:

The equations follow the same structure as linear kinematics, so students assume they work for all rotation problems

正确做法:

Before using the kinematic equations, confirm the problem explicitly states angular acceleration is constant (described as uniform acceleration/deceleration in AP problems)

错误做法:

Using the rolling without slipping relation $v_{c m} = r ω$ for a sliding/slipping object

原因:

Students treat rolling without slipping as a universal rule for all rotating objects that move linearly, but it is a special case

正确做法:

Only use this relationship if the problem explicitly states the object rolls without slipping, or no slipping can be inferred from context (e.g., a car driving normally on pavement)

错误做法:

Taking clockwise rotation as positive, violating the standard convention

原因:

Many problems show clockwise rotation, so students default to the wrong sign, leading to incorrect displacement or acceleration values

正确做法:

Always stick to counterclockwise = positive at the start of the problem, assign signs to all variables based on this rule before substituting into equations

6. 速查表

Category	Formula	Notes
Angular Displacement	$Δ θ = θ_{f} - θ_{i}$	Counterclockwise = positive, clockwise = negative
Average angular velocity	$\overset{ω}{ˉ} = Δ θ /Δ t$	Units: radians per second (rad/s)
Average angular acceleration	$\overset{α}{ˉ} = Δ ω /Δ t$	Units: radians per second squared (rad/s²)
Constant $α$ kinematics (1)	$ω_{f} = ω_{i} + α t$	Only for constant angular acceleration
Constant $α$ kinematics (2)	$Δ θ = ω_{i} t + \frac{1}{2} α t^{2}$	Only for constant angular acceleration
Constant $α$ kinematics (3)	$ω_{f}^{2} = ω_{i}^{2} + 2 α Δ θ$	Only for constant angular acceleration
Arc length (linear distance)	$s = r θ$	Requires $θ$ in radians
Tangential speed	$v_{t} = r ω$	Requires $ω$ in rad/s, $r$ = distance from axis
Tangential acceleration	$a_{t} = r α$	Requires $α$ in rad/s²
Centripetal acceleration	$a_{c} = r ω^{2} = v_{t}^{2} / r$	Points toward axis of rotation
Rolling without slipping	$v_{c m} = r ω, a_{c m} = r α$	Only applies to objects rolling with no slipping

深入阅读

syllabus官方AP物理1课程与考试说明第6单元官方参考资料

下一步

Rotational kinematics is the foundation for all rotational motion topics that follow in AP Physics 1. Next, you will use the angular acceleration and kinematic relationships you learned here to connect rotation to torque and rotational inertia in rotational dynamics. Without a solid mastery of converting between linear and rotational quantities, you will not be able to solve for angular acceleration from net torque, or calculate the motion of rolling objects, which are common high-weight FRQ topics. This topic also connects to energy and momentum, where you will calculate rotational kinetic energy and angular momentum for rotating rigid bodies, both of which require correct use of angular velocity from kinematics.