转动动能

AP 物理1· 第6单元：转动运动，学习目标6.E.2· 12 分钟阅读

1. 转动动能的定义与推导★★☆☆☆⏱ 3 min

任何刚体都可以建模为大量微小质点的集合，当刚体转动时，每个质点都以不同的切向速度运动。转动刚体的总动能等于所有这些质点各自的平动动能之和。

🔬 推导

目标:

推导转动动能的公式

起点:

刚体内所有质点的动能之和

1
对于距离转动轴为 $r_{i}$ 的每个质点 $m_{i}$ ，其切向速度满足 $v_{i} = r_{i} ω$
2
$K_{total} = \sum \frac{1}{2} m_i v_i^2 = \sum \frac{1}{2} m_i (r_i \omega)^2$
3
将对刚体内所有质点均为常数的 $\frac{1}{2} ω^{2}$ 提取到求和符号外
4
$K_{total} = \frac{1}{2} \omega^2 \sum m_i r_i^2$
5
求和项 $\sum m_{i} r_{i}^{2}$ 就是转动惯量 $I$ 的定义式

结果:

转动动能的表达式为 $K_{r} = \frac{1}{2} I ω^{2}$

📘 定义

转动动能

$K_{r} = \frac{1}{2} I ω^{2}$

刚体因绕固定轴或质心转动而具有的动能

例:

A solid disk with $I = 2 kg \cdot m^{2}$ rotating at 3 rad/s has $K_{r} = 9 J$

📐 例题

一根长2 m、质量0.5 kg的细杆以4 rad/s的角速度绕自身中心转动，计算其转动动能。细杆绕中心的转动惯量为 $I = \frac{1}{12} M L^{2}$ 。

1
首先计算该细杆的转动惯量
2
$I = \frac{1}{12} \times 0.5 \text{ kg} \times (2 \text{ m})^2 = \frac{1}{12} \times 0.5 \times 4 = 0.167 \text{ kg} \cdot \text{m}^2$
3
将转动惯量I和角速度代入转动动能公式
4
$K_r = \frac{1}{2} \times 0.167 \text{ kg} \cdot \text{m}^2 \times (4 \text{ rad/s})^2 = 0.5 \times 0.167 \times 16 = 1.33 \text{ J}$

2. 滚动运动的总动能★★★☆☆⏱ 3 min

对于在表面上滚动的物体，其运动是质心平动与绕质心转动的组合，总动能为这两个独立分量的和。

K_{total} = K_{translational} + K_{rotational} = \frac{1}{2} M v_{cm}^2 + \frac{1}{2} I_{cm} \omega^2

📐 例题

一个质量3 kg、半径0.2 m的实心球体以2 m/s的质心速度做无滑滚动，计算其总动能。实心球的转动惯量为 $I_{s p h er e} = \frac{2}{5} M R^{2}$ 。

1
首先利用无滑条件求出角速度
2
$\omega = \frac{v_{cm}}{R} = \frac{2 \text{ m/s}}{0.2 \text{ m}} = 10 \text{ rad/s}$
3
计算平动动能
4
$K_t = \frac{1}{2} M v_{cm}^2 = 0.5 \times 3 \text{ kg} \times (2 \text{ m/s})^2 = 6 \text{ J}$
5
计算转动惯量与转动动能
6
$I = \frac{2}{5} \times 3 \text{ kg} \times (0.2 \text{ m})^2 = 0.048 \text{ kg} \cdot \text{m}^2, \quad K_r = 0.5 \times 0.048 \times 10^2 = 2.4 \text{ J}$
7
将两个分量相加得到总动能
8
$K_{total} = 6 \text{ J} + 2.4 \text{ J} = 8.4 \text{ J}$

3. 转动系统的能量守恒★★★★☆⏱ 4 min

机械能守恒定律对转动系统的适用方式与平动系统完全相同，只要你在总能量求和中加入转动动能项即可。无滑滚动中的静摩擦力做功为零，因此不会从系统中消耗机械能。

📐 例题

一个质量为M、半径为R的实心圆柱体从静止开始沿高度为3 m的斜面做无滑滚动下滑，利用能量守恒求它到达斜面底部时的速度。实心圆柱的转动惯量为 $I_{cy l in d er} = \frac{1}{2} M R^{2}$ 。

1
写出初始与末态的总机械能，初始动能为零
2
$E_{initial} = U_g = Mgh, \quad E_{final} = K_t + K_r = \frac{1}{2}Mv^2 + \frac{1}{2}I\omega^2$
3
代入 $I = \frac{1}{2} M R^{2}$ 和无滑条件 $ω = v / R$
4
$E_{final} = \frac{1}{2}Mv^2 + \frac{1}{2} \times (\frac{1}{2}MR^2) \times (\frac{v}{R})^2 = \frac{1}{2}Mv^2 + \frac{1}{4}Mv^2 = \frac{3}{4}Mv^2$
5
令初始能量与末态能量相等，两边消去质量M
6
$Mgh = \frac{3}{4}Mv^2 \implies v = \sqrt{\frac{4gh}{3}}$
7
代入g = 9.8 m/s²和h = 3 m
8
$v = \sqrt{\frac{4 \times 9.8 \times 3}{3}} = \sqrt{39.2} \approx 6.26 \text{ m/s}$

✓ 快速检测

测试你对能量守恒规则的理解：

Which of the following has the highest total kinetic energy when rolling without slipping at the same center of mass speed?
- Solid sphere of mass M
- Hollow sphere of mass M
- Solid cylinder of mass M
- All have equal total KE
显示答案
Hollow sphere of mass M —
The hollow sphere has the largest moment of inertia, so it stores more energy as rotational KE at the same speed.

4. 常见陷阱

错误做法:

Using only $K = \frac{1}{2} m v^{2}$ to calculate total KE of a rolling rigid body

原因:

This ignores the rotational motion of the object around its center of mass, which can make up more than 50% of total kinetic energy for some shapes

正确做法:

Always sum both translational $\frac{1}{2} m v_{c m}^{2}$ and rotational $\frac{1}{2} I ω^{2}$ components for rolling objects

错误做法:

Applying $v_{c m} = R ω$ to objects that are slipping

原因:

The no-slip condition only holds when static friction prevents relative motion between the object and the surface at the contact point

正确做法:

Only use $v_{c m} = R ω$ when the problem explicitly states rolling without slipping

错误做法:

Using total mass $m$ instead of moment of inertia $I$ in the rotational KE formula

原因:

Rotational inertia depends on mass distribution relative to the rotation axis, not just total mass of the object

正确做法:

Use the standard moment of inertia formula for the object's shape and given rotation axis

错误做法:

Including work done by static friction in rolling without slipping energy calculations

原因:

The point of instantaneous contact between the rolling object and surface has zero displacement, so static friction does zero net work

正确做法:

Treat rolling without slipping as a conservative system if no other non-conservative forces are acting

错误做法:

Using g = 9.8 m/s² when the problem explicitly specifies g = 10 m/s² for simplicity

原因:

AP Physics 1 problems often use g = 10 to reduce calculation load, and you will lose points if you use the wrong value

正确做法:

Always use the exact value of gravitational acceleration stated in the problem prompt

5. 速查表

Quantity	Formula	Applicable Condition
Translational KE	$K_{t} = \frac{1}{2} M v_{c m}^{2}$	Center of mass motion of any rigid body
Rotational KE	$K_{r} = \frac{1}{2} I ω^{2}$	Rotation around fixed axis or center of mass
Total Rolling KE	$K_{t o t a l} = \frac{1}{2} M v_{c m}^{2} + \frac{1}{2} I ω^{2}$	Combined translation and rotation motion
No-slip Relation	$v_{c m} = R ω$	Rolling without slipping, R = object radius
Gravitational PE	$U_{g} = m g h$	Uniform gravity, h = height of center of mass

下一步

Mastering rotational kinetic energy is critical for tackling the full suite of AP Physics 1 rotational motion questions, from multiple choice trick questions to multi-part FRQs that combine energy, forces, and torque. This concept also forms the foundation for understanding angular momentum conservation, one of the highest-weighted topics on the AP Physics 1 exam. After completing this module, you will be able to solve complex problems that mix linear and rotational motion without needing to use messy rotational kinematic equations, saving valuable time during your exam. Practice applying these rules to ramp problems, pulley systems with hanging masses, and rotating collision scenarios to solidify your mastery.