Cambridge IGCSE IGCSE & A-Level Sheet preview

Energy and Momentum of Rotating Systems

AP Physics 1 · Topic 6

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6.1

Rotational Kinetic Energy
Syllabus
Learning ObjectiveEssential 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.

Source: College Board AP Course and Exam Description

A spinning object has rotational kinetic energy 转动动能, the rotational twin of $\tfrac12 mv^2$:

$$K_{\text{rot}}=\tfrac{1}{2}I\omega^2.$$
An object that both moves and spins (like a rolling ball) has both translational and rotational kinetic energy, and its total is $K=\tfrac12 mv^2+\tfrac12 I\omega^2$.

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English Chinese Pinyin
rotational kinetic energy 转动动能 zhuǎn dòng dòng néng
Exercise sheet
6.2

Torque and Work
Syllabus
Learning ObjectiveEssential 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.

Source: College Board AP Course and Exam Description

A torque acting through an angular displacement does work, changing rotational kinetic energy:

$$W=\tau\,\Delta\theta,\qquad P=\tau\omega.$$
This is the rotational form of $W=Fd$ and $P=Fv$, and it extends the work–energy theorem to rotation.

Worked example. A motor applies a steady torque of $8.0\ \text{N m}$ to a flywheel while it turns through $10\ \text{rad}$. The work done is $W=\tau\,\Delta\theta=8.0\times10=80\ \text{J}$, and if the flywheel started from rest this all becomes rotational kinetic energy.

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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.

Exercise sheet
6.3

Angular Momentum and Angular Impulse
Syllabus
Learning ObjectiveEssential 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.

Source: College Board AP Course and Exam Description

Angular momentum 角动量 is the rotational version of linear momentum:

$$L=I\omega.$$
A net torque acting over time delivers an angular impulse 角冲量 that changes it: $\tau\,\Delta t=\Delta L$ – 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 ObjectiveEssential 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.

Source: College Board AP Course and Exam Description

If the net external torque on a system is zero, its total angular momentum is conserved 守恒:

$$I_1\omega_1=I_2\omega_2.$$
So if $I$ decreases, $\omega$ increases to keep $L$ constant – this is why a spinning skater speeds up when pulling their arms in. It applies to collisions and explosions of rotating systems too.

Worked example. A skater spins at $2.0\ \text{rev/s}$ with rotational inertia $I_1=4.0\ \text{kg m}^2$. She pulls her arms in, dropping her rotational inertia to $I_2=1.6\ \text{kg m}^2$. With no external torque, angular momentum is conserved:

$$\omega_2=\frac{I_1}{I_2}\,\omega_1=\frac{4.0}{1.6}\times2.0=5.0\ \text{rev/s}.$$
Her kinetic energy actually rises – the extra energy comes from the work her muscles do pulling her arms in against the outward pull.

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

Vocabulary Train
English Chinese Pinyin
conserved 守恒 shǒu héng
Exercise sheet
6.5

Rolling
Syllabus
Learning ObjectiveEssential 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.

Source: College Board AP Course and Exam Description

Rolling without slipping 纯滚动 links the translational and rotational motions: the contact point is momentarily at rest, so

$$v=r\omega \qquad\text{and}\qquad a=r\alpha.$$
A rolling object's energy splits between translation and rotation, so on an incline it accelerates more slowly than a frictionless sliding object – some energy goes into spin.

Worked example. For a solid disk ($I=\tfrac12 mR^2$) that rolls without slipping, what fraction of its kinetic energy is rotational? Using $v=R\omega$, the rotational part is $\tfrac12 I\omega^2=\tfrac12(\tfrac12 mR^2)\omega^2=\tfrac14 mv^2$, while the translational part is $\tfrac12 mv^2$. So the total is $\tfrac34 mv^2$ and the rotational share is $\tfrac{1/4}{3/4}=\tfrac13$. A hoop, with its mass farther out, stores half its energy in spin and rolls down a ramp even more slowly.

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

Vocabulary Train
English Chinese Pinyin
Rolling without slipping 纯滚动 chún gǔn dòng
Exercise sheet
6.6

Motion of Orbiting Satellites
Syllabus
Learning ObjectiveEssential 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): $v_{\text{esc}} = \sqrt{\dfrac{2GM}{r}}$

Source: College Board AP Course and Exam Description

Orbital motion (Kepler's 2nd law)

A satellite 卫星 in orbit is in free fall: gravity provides the exact centripetal force needed to curve its path into an orbit. Setting gravity equal to the centripetal requirement,

$$\frac{GMm}{r^2}=\frac{mv^2}{r}\ \Rightarrow\ v=\sqrt{\frac{GM}{r}}.$$
So a larger orbit means a slower speed. For a circular orbit, angular momentum and mechanical energy are both constant; for an elliptical orbit, angular momentum is conserved (no torque about the planet) while speed varies – fastest when closest.

Gravity provides the centripetal force that keeps a satellite in orbit Gravity provides the centripetal force that keeps a satellite in orbit

Worked example. Find the speed of a satellite in a low orbit just above the Earth, radius $r=6.4\times10^{6}\ \text{m}$, with $GM=4.0\times10^{14}\ \text{m}^3/\text{s}^2$:

$$v=\sqrt{\frac{GM}{r}}=\sqrt{\frac{4.0\times10^{14}}{6.4\times10^{6}}}=\sqrt{6.25\times10^{7}}\approx 7.9\times10^{3}\ \text{m/s},$$
about $7.9\ \text{km/s}$ – and notice that the satellite's mass cancels, so all low orbits share this speed.

A photograph of the International Space Station orbiting Earth The International Space Station: a real satellite held in orbit by gravity alone

Escape velocity 逃逸速度 is the launch speed that just lets an object leave for good: its total mechanical energy is exactly zero, so it slows to zero speed only at infinite distance. Setting $\tfrac12 mv^2 - \dfrac{GMm}{r}=0$ (using the general gravitational PE $U_g=-GMm/r$) and solving for $v$,

$$v_{\text{esc}}=\sqrt{\frac{2GM}{r}}.$$
It is $\sqrt2$ times the circular-orbit speed at the same radius, and (like orbital speed) does not depend on the escaping object's mass.

Explore

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.

Vocabulary Train
English Chinese Pinyin
satellite 卫星 wèi xīng
Escape velocity 逃逸速度 táo yì sù dù
Exercise sheet
6.6

Exam tips
  • Conserve angular momentum $L=I\omega$ when no external torque acts: a smaller $I$ (arms pulled in) gives a larger $\omega$.
  • A rolling object splits its energy between $\tfrac12 mv^2$ and $\tfrac12 I\omega^2$, linked by $v=r\omega$ — so it accelerates down a ramp more slowly than a sliding one.
  • For a circular orbit set gravity equal to the centripetal requirement: $v=\sqrt{GM/r}$, so a larger orbit is slower and the satellite's mass cancels.
  • Pulling in raises the spin and the kinetic energy — the extra energy comes from the work done pulling inward; $L$ is unchanged.
  • Watch which rotational quantity is conserved: $L$ (no torque) versus energy (no friction) are different conditions.

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