Energy and Momentum of Rotating Systems

6.1

Rotational Kinetic Energy

Syllabus

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 $K_{\text{rot}} = \dfrac{1}{2} I \omega^2$. Equation: $K_{\text{rot}} = \dfrac{1}{2} I \omega^2$ 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.

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 $K_{\text{rot}} = \dfrac{1}{2} I \omega^2$.
- Equation: $K_{\text{rot}} = \dfrac{1}{2} I \omega^2$
- 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.

Source: College Board AP Course and Exam Description

A spinning body has rotational kinetic energy 转动动能

$$K_{\text{rot}}=\tfrac12 I\omega^2,$$

the rotational twin of $\tfrac12mv^2$, with the rotational inertia 转动惯量 $I$ playing the role of mass. A body that both moves and spins carries both terms:

$$K=\tfrac12 mv_{\text{cm}}^2+\tfrac12 I\omega^2.$$

Worked example. A uniform cylinder ($I=\tfrac12mr^2$) rolls at speed $v$. Its kinetic energy is $K=\tfrac12mv^2+\tfrac12\big(\tfrac12mr^2\big)\big(\tfrac{v}{r}\big)^2=\tfrac34mv^2$ – one third of it is rotational. The same ball of energy bookkeeping decides every rolling problem.

Vocabulary Train

English	Chinese	Pinyin
rotational kinetic energy	转动动能	zhuǎn dòng dòng néng
rotational inertia	转动惯量	zhuǎn dòng guàn liàng

Exercise sheet

6.2

Torque and Work

Syllabus

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: $W = \displaystyle\int_{\theta_1}^{\theta_2} \tau \, d\theta$ 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 the torque as a function of angular position.

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: $W = \displaystyle\int_{\theta_1}^{\theta_2} \tau \, d\theta$
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 the torque as a function of angular position.

Source: College Board AP Course and Exam Description

A torque 力矩 acting through an angular displacement 角位移 does work, and power is torque times angular velocity:

$$W=\int_{\theta_1}^{\theta_2}\tau\,d\theta,\qquad P=\tau\omega.$$

These are the rotational forms of $W=\int F\,dx$ and $P=Fv$ – the whole translational energy toolkit carries over with $F\to\tau$, $x\to\theta$, $v\to\omega$.

Worked example. A motor applies a constant $8.0\ \text{N}\cdot\text{m}$ torque to a flywheel for $5.0$ full turns: $W=\tau\,\Delta\theta=8.0(5.0)(2\pi)=250\ \text{J}$, which appears as rotational kinetic energy if friction is negligible.

Explore

Balance torques on a beam

Torque is force times perpendicular distance, $\tau=Fd$. Equal torques on each side keep the beam in rotational equilibrium.

Vocabulary Train

English	Chinese	Pinyin
torque	力矩	lì jǔ
angular displacement	角位移	jiǎo wèi yí

Exercise sheet

6.3

Angular Momentum and Angular Impulse

Syllabus

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 $L = I\omega$. Equation: $L = I\omega$ 6.3.A.2 The angular momentum of an object about a given point is $\vec{L} = \vec{r} \times \vec{p}$. Equation: $\vec{L} = \vec{r} \times \vec{p}$ 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: $\text{angular impulse} = \displaystyle\int \tau \, dt$ 6.3.B.2 Angular impulse has the same direction as the torque imparting it. 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 momenta of the object or rigid system. Equation: $\Delta L = L - L_0$ 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: $\Delta L = \displaystyle\int_{t_1}^{t_2} \tau \, dt$ 6.3.C.2.ii The rotational form of the impulse–momentum theorem is a direct result of Newton's second law of motion for cases in which rotational inertia is constant. Equation: $\tau_{\text{net}} = \dfrac{dL}{dt} = I\dfrac{d\omega}{dt} = I\alpha$ 6.3.C.3 The net torque exerted on an object or rigid system 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 or rigid system 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.

Source: College Board AP Course and Exam Description

Angular momentum 角动量 for a particle is

$$\vec{L}=\vec{r}\times\vec{p},\qquad |L|=mvr\sin\theta,$$

so even a particle moving in a straight line has angular momentum about any point not on that line ($L=mv\,d$, with $d$ the perpendicular distance). For a rigid body spinning about a fixed axis, $L=I\omega$. Newton's second law in rotational form is

$$\vec{\tau}_{\text{net}}=\frac{d\vec{L}}{dt}\quad(=I\alpha\ \text{when }I\text{ is constant}),$$

and a net torque acting over time delivers an angular impulse 角冲量 $\int\tau\,dt=\Delta L$ – the rotational impulse–momentum theorem.

Vocabulary Train

English	Chinese	Pinyin
Angular momentum	角动量	jiǎo dòng liàng
angular impulse	角冲量	jiǎo chōng liàng

Exercise sheet

6.4

Conservation of Angular Momentum

Syllabus

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 rotational 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 farther 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.

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 rotational 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 farther 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.

Source: College Board AP Course and Exam Description

Angular momentum: pull in, spin faster

With zero net external torque, total angular momentum is conserved 守恒:

$$L_{\text{before}}=L_{\text{after}},\qquad I_1\omega_1=I_2\omega_2\ \text{(one rigid body reshaping)}.$$

If $I$ shrinks, $\omega$ grows – a spinning skater speeds up pulling her arms in. The rule survives collisions and shape changes, which is what makes it so useful: pick the axis so that every external force (gravity, the pivot force) exerts no torque about it.

Worked example. A skater spins at $2.0\ \text{rev/s}$ with $I_1=4.0\ \text{kg}\cdot\text{m}^2$. Pulling in her arms drops it to $I_2=1.6\ \text{kg}\cdot\text{m}^2$: $\omega_2=\dfrac{4.0}{1.6}(2.0)=5.0\ \text{rev/s}$. Her kinetic energy rises – the extra energy is the work her muscles do pulling her arms inward.

Pulling mass inward lowers I, so ω rises to conserve L = Iω

Angular momentum about the pivot is conserved as the bullet embeds in the rod

Worked example (rotational collision). A $0.020\ \text{kg}$ bullet at $300\ \text{m/s}$ strikes the tip of a uniform rod ($M=1.5\ \text{kg}$, length $l=0.60\ \text{m}$) hanging from a pivot, and embeds. About the pivot, gravity and the pivot force exert no torque during the strike, so $L$ is conserved: $L=mvl=0.020(300)(0.60)=3.6\ \text{kg}\cdot\text{m}^2/\text{s}$. Afterwards $I=\tfrac13Ml^2+ml^2=0.18+0.0072=0.187\ \text{kg}\cdot\text{m}^2$, so $\omega=\dfrac{3.6}{0.187}\approx19\ \text{rad/s}$. (Linear momentum is not conserved here – the pivot pushes on the rod – and kinetic energy certainly is not: check both before claiming them.)

Figure skaters performing on the ice Figure skating and angular momentum: a skater who pulls their arms in during a spin lowers their rotational inertia, so they spin faster

Vocabulary Train

English	Chinese	Pinyin
conserved	守恒	shǒu héng

Exercise sheet

6.5

Rolling

Syllabus

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 following 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 C: Mechanics.

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 following 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 C: Mechanics.

Source: College Board AP Course and Exam Description

Rolling without slipping

Rolling without slipping 无滑滚动 ties translation to rotation: the contact point is momentarily at rest, so

$$\Delta x_{\text{cm}}=r\,\Delta\theta,\qquad v_{\text{cm}}=r\omega,\qquad a_{\text{cm}}=r\alpha.$$

In rolling without slipping the contact point is at rest, so v = rω

Static friction supplies the torque that keeps the spin matched to the motion, but at a point that is not sliding – so for pure rolling, friction does no work, and energy conservation is safe to use. (Rolling friction is beyond the AP course.)

Racing shapes down an incline shows the energy split. With $I=\beta mr^2$, energy conservation gives

$$a_{\text{cm}}=\frac{g\sin\theta}{1+\beta}:$$

a sphere ($\beta=\tfrac25$) beats a disk 圆盘 ($\beta=\tfrac12$), which beats a hoop 圆环 ($\beta=1$) – mass and radius cancel completely. More of the hoop's energy is locked in rotation, so its center moves slower.

Rolling race: the shape with the smallest I/mr² reaches the bottom first

Worked example. A solid sphere rolls from rest down a $30^\circ$ incline: $a=\dfrac{g\sin30^\circ}{1+\tfrac25}=\dfrac{9.8(0.50)}{1.4}=3.5\ \text{m/s}^2$, versus $4.9\ \text{m/s}^2$ for a frictionless slider – rolling objects always lose the race against sliding ones.

Vocabulary Train

English	Chinese	Pinyin
Rolling without slipping	无滑滚动	wú huá gǔn dòng
disk	圆盘	yuán pán
hoop	圆环	yuán huán

Exercise sheet

6.6

Motion of Orbiting Satellites

Syllabus

Learning Objective	Essential Knowledge
6.6.A Describe the motions of a system consisting of two objects or systems 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 total energy of a system consisting of a satellite orbiting a central object in a circular path can be written in terms of the gravitational potential energy of that system or the kinetic energy of the satellite. Equation: $K = -\dfrac{1}{2}U$ Equation: $E_{total} = \dfrac{1}{2}U = -\dfrac{GMm}{2r}$ 6.6.A.4 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.4.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.4.ii The escape velocity of a satellite from a central body of mass $M$ can be derived using conservation of energy laws. Equation: $v_{\text{esc}} = \sqrt{\dfrac{2GM}{r}}$

Learning Objective

Essential Knowledge

6.6.A
Describe the motions of a system consisting of two objects or systems 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 total energy of a system consisting of a satellite orbiting a central object in a circular path can be written in terms of the gravitational potential energy of that system or the kinetic energy of the satellite.
- Equation: $K = -\dfrac{1}{2}U$
- Equation: $E_{total} = \dfrac{1}{2}U = -\dfrac{GMm}{2r}$
6.6.A.4 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.4.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.4.ii The escape velocity of a satellite from a central body of mass $M$ can be derived using conservation of energy laws.
  - Equation: $v_{\text{esc}} = \sqrt{\dfrac{2GM}{r}}$

Source: College Board AP Course and Exam Description

Orbital motion (Kepler's 2nd law)

Gravity supplies the centripetal force 向心力 for a satellite 卫星 in a circular orbit:

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

– larger orbits are slower. With gravitational potential energy 引力势能 $U_g=-\dfrac{GMm}{r}$, a circular orbit obeys $K=-\tfrac12U$, so the total mechanical energy 总机械能 is

$$E=K+U=\frac{U}{2}=-\frac{GMm}{2r},$$

negative because the satellite is bound. To escape from radius $r$, total energy must reach zero, giving the escape velocity 逃逸速度

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

Gravity provides the centripetal force that keeps a satellite in orbit

In an elliptical orbit 椭圆轨道, $E$ and $L$ are fixed: gravity points at the focus, so it exerts no torque about it. Conserved $L$ means the satellite sweeps equal areas in equal times – Kepler's second law 开普勒第二定律 – and therefore moves fastest at closest approach, slowest at the far point. Energy conservation connects speeds at the two ends.

Worked example. For a satellite at $r=7.0\times10^{6}\ \text{m}$ around Earth ($GM=4.0\times10^{14}\ \text{m}^3/\text{s}^2$): orbital speed $v=\sqrt{GM/r}=7.6\times10^{3}\ \text{m/s}$, while escaping from that radius needs $v_{\text{esc}}=\sqrt{2GM/r}=1.1\times10^{4}\ \text{m/s}$ – exactly $\sqrt2$ times the circular speed.

Exam skill. Orbit FRQs are energy-and-angular-momentum problems in disguise: write $E=\tfrac12mv^2-\dfrac{GMm}{r}$ and $L=mvr\sin\theta$ at the two points of interest and solve the pair – never assume the orbit formulae for circles apply to an ellipse.

Explore

Compare orbits at different radii

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

Vocabulary Train

English	Chinese	Pinyin
centripetal force	向心力	xiàng xīn lì
satellite	卫星	wèi xīng
gravitational potential energy	引力势能	yǐn lì shì néng
total mechanical energy	总机械能	zǒng jī xiè néng
escape velocity	逃逸速度	táo yì sù dù
elliptical orbit	椭圆轨道	tuǒ yuán guǐ dào
Kepler's second law	开普勒第二定律	kāi pǔ lēi dì èr dìng lǜ

Exercise sheet

6.6

Exam tips

Compute the moment of inertia $I=\int r^2\,dm$ and shift axes with the parallel-axis theorem $I=I_{cm}+Md^2$.
Rotational kinetic energy is $\tfrac12 I\omega^2$; a rolling body has both translational and rotational KE.
Conserve angular momentum $L=I\omega$ when net external torque is zero (a spinning skater pulling in).
Use energy conservation for rolling-without-slipping problems ($v=r\omega$ ties the two motions).
Know standard $I$ values (hoop, disk, rod, sphere) and where the axis is.

Rotational Kinetic Energy

Torque and Work

Balance torques on a beam

Angular Momentum and Angular Impulse

Conservation of Angular Momentum

Rolling

Motion of Orbiting Satellites

Compare orbits at different radii

Exam tips

Log in or create account

Feedback & help