| Learning Objective | Essential Knowledge |
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1.1.A |
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AP Physics C: Mechanics
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1 Kinematics
1.1
Scalars and Vectors
Syllabus
Source: College Board AP Course and Exam Description
Kinematics 运动学 describes motion. Two kinds of quantity:
- A scalar 标量 has only size (magnitude 大小): distance, speed, mass, time.
- A vector 矢量 has magnitude and direction: displacement, velocity, acceleration, force.
In this calculus-based course you resolve vectors into components 分量 and add them component by component; a vector's magnitude is $\sqrt{v_x^2+v_y^2+\cdots}$. Physics C also writes vectors with unit vectors 单位矢量: $\vec{v}=v_x\hat{i}+v_y\hat{j}$, where $\hat{i}$ and $\hat{j}$ are directions of length one along $x$ and $y$ – handy because calculus can then act on each component separately.
Resolving a vector into its x and y componentsExploreExplore the dot product
Drag the components of $\vec{A}$ and $\vec{B}$. The dot product $\vec{A}\cdot\vec{B}=A_xB_x+A_yB_y$ reaches zero exactly when the two vectors are perpendicular — the geometry behind $\cos\theta$.
Vocabulary TrainEnglish Chinese Pinyin Kinematics 运动学 yùn dòng xué scalar 标量 biāo liàng magnitude 大小 dà xiǎo vector 矢量 shǐ liàng components 分量 fèn liàng unit vectors 单位矢量 dān wèi shǐ liàng 1.2
Displacement, Velocity, and Acceleration
Syllabus
Learning Objective Essential Knowledge 1.2.A
Describe a change in an object's position.- 1.2.A.1 When using the object model, the size, shape, and internal configuration are ignored. The object may be treated as a single point with extensive properties such as mass and charge.
- 1.2.A.2 Displacement is the change in an object's position.
- Equation: $\Delta x = x - x_0$
1.2.B
Describe the average velocity and acceleration of an object.- 1.2.B.1 Averages of velocity and acceleration are calculated considering the initial and final states of an object over an interval of time.
- 1.2.B.2 Average velocity is the displacement of an object divided by the interval of time in which that displacement occurs.
- Equation: $\vec{v}_{\text{avg}} = \dfrac{\Delta \vec{x}}{\Delta t}$
- 1.2.B.3 Average acceleration is the change in velocity divided by the interval of time in which that change in velocity occurs.
- Equation: $\vec{a}_{\text{avg}} = \dfrac{\Delta \vec{v}}{\Delta t}$
- 1.2.B.4 An object is accelerating if either the magnitude and/or direction of the object's velocity are changing.
- 1.2.B.5 Calculating average velocity or average acceleration over a very small time interval yields a value that is very close to the instantaneous velocity or instantaneous acceleration.
1.2.C
Describe the instantaneous position, velocity, and acceleration of an object as a function of time.- 1.2.C.1 As the time interval used to calculate the average value of a quantity approaches zero, the average value of that quantity approaches the value of the quantity at that instant, called the instantaneous value.
- 1.2.C.1.i Instantaneous velocity is the rate of change of the object's position, which is equal to the derivative of position with respect to time.
- Equation: $\vec{v} = \dfrac{d\vec{r}}{dt}$
- Equation: $v_x = \dfrac{dx}{dt}$
- 1.2.C.1.ii Instantaneous acceleration is the rate of change of the object's velocity, which is equal to the derivative of velocity with respect to time.
- Equation: $\vec{a} = \dfrac{d\vec{v}}{dt}$
- Equation: $a_x = \dfrac{dv_x}{dt}$
- 1.2.C.1.i Instantaneous velocity is the rate of change of the object's position, which is equal to the derivative of position with respect to time.
- 1.2.C.2 Time-dependent functions and instantaneous values of position, velocity, and acceleration can be determined using differentiation and integration.
Source: College Board AP Course and Exam Description
Because motion can change continuously, define the rates as derivatives 导数:
$$v=\frac{dx}{dt},\qquad a=\frac{dv}{dt}=\frac{d^2x}{dt^2}.$$Reversing this, integrate to recover velocity and position:$$v(t)=v_0+\int_0^t a\,dt,\qquad x(t)=x_0+\int_0^t v\,dt.$$Distinguish the average from the instantaneous: average velocity 平均速度 is $\Delta x/\Delta t$ over an interval, while instantaneous velocity 瞬时速度 is the derivative at one moment – the slope of the tangent line, and what a speedometer shows.For constant acceleration these give the familiar kinematic equations ($v=v_0+at$, $x=x_0+v_0t+\tfrac12at^2$, $v^2=v_0^2+2a\,\Delta x$). The most important constant-$a$ case is free fall 自由落体: near Earth's surface every object, heavy or light, accelerates downward at $g=9.8\ \text{m/s}^2$ once air resistance is negligible. When $a$ or $v$ varies with time, use the integrals directly.
Worked example. A particle moves with $x(t)=2t^3-3t^2$ (metres). Differentiate for velocity and acceleration:
$$v=\frac{dx}{dt}=6t^2-6t,\qquad a=\frac{dv}{dt}=12t-6.$$At $t=2\ \text{s}$, $v=24-12=12\ \text{m/s}$ and $a=24-6=18\ \text{m/s}^2$. This calculus link – not just the constant-$a$ formulas – is what distinguishes Physics C.Worked example. A particle starts from rest at the origin with $a(t)=6t$. Integrate: $v=\int 6t\,dt=3t^2$ and $x=\int 3t^2\,dt=t^3$. At $t=2\ \text{s}$, $v=12\ \text{m/s}$ and $x=8\ \text{m}$.
On a velocity-time graph the gradient is the acceleration and the area is the displacementVocabulary TrainEnglish Chinese Pinyin derivatives 导数 dǎo shù average velocity 平均速度 píng jūn sù dù instantaneous velocity 瞬时速度 shùn shí sù dù free fall 自由落体 zì yóu luò tǐ 1.3
Representing Motion
Syllabus
Learning Objective Essential Knowledge 1.3.A
Describe the position, velocity, and acceleration of an object using representations of that object's motion.- 1.3.A.1 Motion can be represented by motion diagrams, figures, graphs, equations, and narrative descriptions.
- 1.3.A.2 For constant acceleration, three kinematic equations can be used to describe instantaneous linear motion in one dimension:
- Equation: $v_x = v_{x0} + a_x t$
- Equation: $x = x_0 + v_{x0} t + \dfrac{1}{2} a_x t^2$
- Equation: $v_x^2 = v_{x0}^2 + 2a_x \left( x - x_0 \right)$
- Note: The equations above are written to indicate motion in the $x$-direction, but these equations can be used in any single dimension as appropriate.
- 1.3.A.3 Near the surface of Earth, the vertical acceleration caused by the force of gravity is downward, constant, and has a measured value approximately equal to
- Equation: $a_g = g \approx 10 \ \text{m/s}^2$.
- 1.3.A.4 Graphs of position, velocity, and acceleration as functions of time can be used to find the relationships between those quantities.
- 1.3.A.4.i An object's instantaneous velocity is the rate of change of the object's position, which is equal to the slope of a line tangent to a point on a graph of the object's position as a function of time.
- Equation: $v_x = \dfrac{dx}{dt}$
- 1.3.A.4.ii An object's instantaneous acceleration is the rate of change of the object's velocity, which is equal to the slope of a line tangent to a point on a graph of the object's velocity as a function of time.
- Equation: $a_x = \dfrac{dv_x}{dt}$
- 1.3.A.4.iii The displacement of an object during a time interval is equal to the area under the curve of a graph of the object's velocity as a function of time (i.e., the area bounded by the function and the horizontal axis for the appropriate interval).
- Equation: $\Delta x = \displaystyle\int_{t_1}^{t_2} v_x(t) \ dt$
- 1.3.A.4.iv The change in velocity of an object during a time interval is equal to the area under the curve of a graph of the acceleration of the object as a function of time.
- Equation: $\Delta v_x = \displaystyle\int_{t_1}^{t_2} a_x(t) \ dt$
- 1.3.A.4.i An object's instantaneous velocity is the rate of change of the object's position, which is equal to the slope of a line tangent to a point on a graph of the object's position as a function of time.
Boundary statement: AP Physics C: Mechanics and AP Physics C: Electricity and Magnetism expects that for all situations in which a numerical quantity is required for g, the value $g \approx 10 \ \text{m/s}^2$ will be used. However, students will not be penalized for correctly using the more precise commonly accepted values of $g = 9.81 \ \text{m/s}^2$ or $g = 9.8 \ \text{m/s}^2$.
Source: College Board AP Course and Exam Description
Move fluently between description, graph, table, and equation:
A displacement-time graph: the slope at any instant is the velocity- On a position–time graph, the slope ($dx/dt$) is velocity.
- On a velocity–time graph, the slope is acceleration and the area ($\int v\,dt$) is displacement.
- On an acceleration–time graph, the area ($\int a\,dt$) is the change in velocity.
Slopes are derivatives; areas are integrals – the two are inverse operations here.
Position, velocity, and acceleration are linked by differentiationExploreExplore the velocity–time graph
Change the start velocity and acceleration. The slope of a $v$–$t$ line is the acceleration $a=\tfrac{dv}{dt}$, and the area underneath is the displacement $\Delta x=\int v\,dt$ — slope and area are the graphical faces of the derivative and the integral.
1.4
Reference Frames and Relative Motion
Syllabus
Learning Objective Essential Knowledge 1.4.A
Describe the reference frame of a given observer.- 1.4.A.1 The choice of reference frame will determine the direction and magnitude of quantities measured by an observer in that reference frame.
1.4.B
Describe the motion of objects as measured by observers in different inertial reference frames.- 1.4.B.1 Measurements from a given reference frame may be converted to measurements from another reference frame.
- 1.4.B.2 The observed velocity of an object results from the combination of the object's velocity and the velocity of the observer's reference frame.
- 1.4.B.2.i Combining the motion of an object and the motion of an observer in a given reference frame involves the addition or subtraction of vectors.
- 1.4.B.2.ii The acceleration of any object is the same as measured from all inertial reference frames.
Boundary statement: Unless otherwise stated, the frame of reference of any problem may be assumed to be inertial.
Source: College Board AP Course and Exam Description
All motion is relative to a reference frame 参考系. Combine velocities by vector addition: $\vec{v}_{A/C}=\vec{v}_{A/B}+\vec{v}_{B/C}$. This handles boats crossing rivers and passengers on moving vehicles – relative motion 相对运动 problems. Read the subscripts as a chain: "A relative to B" plus "B relative to C" gives "A relative to C".
Worked example. A boat that moves at $4.0\ \text{m/s}$ in still water heads straight across a river flowing at $3.0\ \text{m/s}$. Relative to the ground the boat moves at $\sqrt{4.0^2+3.0^2}=5.0\ \text{m/s}$, angled downstream. If the river is $80\ \text{m}$ wide, the crossing still takes $t=\dfrac{80}{4.0}=20\ \text{s}$ – only the across-stream component crosses the river; the current just carries the boat $60\ \text{m}$ downstream.
Vocabulary TrainEnglish Chinese Pinyin reference frame 参考系 cān kǎo xì relative motion 相对运动 xiāng duì yùn dòng 1.5
Motion in Two or Three Dimensions
Syllabus
Learning Objective Essential Knowledge 1.5.A
Describe the motion of an object moving in two or three dimensions.- 1.5.A.1 Motion in two or three dimensions can be analyzed using one-dimensional kinematic relationships if the motion is separated into components.
- 1.5.A.2 Velocity and acceleration may be different in each dimension and may be nonuniform.
- 1.5.A.3 Motion in one dimension may be changed without causing a change in a perpendicular dimension.
- 1.5.A.4 Projectile motion is a special case of two-dimensional motion that has zero acceleration in one dimension and constant, nonzero acceleration in the second dimension.
Boundary statement: AP Physics C: Mechanics only expects students to quantitatively analyze the motion of an object in two dimensions. AP Physics C: Electricity and Magnetism expects students to also qualitatively describe the motion of a particle in three dimensions.
Source: College Board AP Course and Exam Description
In two or three dimensions, position is a vector $\vec{r}(t)=\langle x(t),y(t),z(t)\rangle$, and velocity and acceleration are its successive derivatives – each component handled independently.
Worked example. $\vec{r}(t)=\big(2t^2\big)\hat{i}+\big(4t-t^3\big)\hat{j}$ (metres). Differentiating each component: $\vec{v}=4t\,\hat{i}+(4-3t^2)\,\hat{j}$ and $\vec{a}=4\,\hat{i}-6t\,\hat{j}$. At $t=1\ \text{s}$: $\vec{v}=4\hat{i}+1\hat{j}$, so the speed is $\sqrt{17}\approx4.1\ \text{m/s}$ – no new physics, just one derivative per component.
A projectile launched at an angle: the horizontal and vertical motions are independentFor projectile motion 抛体运动, horizontal and vertical motions are independent, linked only by time $t$: horizontally $a_x=0$ (constant velocity), vertically $a_y=-g$. The path is a parabola. This component method extends to any two-dimensional motion where the accelerations along each axis are known.
Worked example. A ball is launched at $20\ \text{m/s}$, $30^{\circ}$ above the horizontal ($g=9.8\ \text{m/s}^2$). The components are $v_{0x}=20\cos30^{\circ}=17.3\ \text{m/s}$ and $v_{0y}=20\sin30^{\circ}=10\ \text{m/s}$. The vertical motion sets the time: total flight $=\dfrac{2v_{0y}}{g}=\dfrac{20}{9.8}=2.0\ \text{s}$, maximum height $=\dfrac{v_{0y}^2}{2g}=5.1\ \text{m}$, and range $=v_{0x}\times2.0=35\ \text{m}$.
A projectile launched at an angle: velocity components, maximum height, and rangeExploreExplore projectile motion
Fire the projectile, then vary the angle and speed. The horizontal motion is steady while gravity acts only downward — together they trace a parabola. Find the angle that gives the greatest range (it peaks near $45^\circ$), and try the Moon.
Vocabulary TrainEnglish Chinese Pinyin projectile motion 抛体运动 pāo tǐ yùn dòng 1.5
Exam tips
- Resolve every vector into components before adding — never add magnitudes at an angle directly.
- On Physics C you are expected to use calculus: velocity is $\vec v=\tfrac{d\vec r}{dt}$ and acceleration $\vec a=\tfrac{d\vec v}{dt}$; reverse with integration.
- Carry and check units and treat direction with signs (choose a positive axis and stick to it).
- Use the dot product for work-type quantities and the cross product for torque and angular momentum.
- Sketch the vectors — a diagram catches sign and direction errors the algebra hides.
-
2 Force and Translational Dynamics
2.1
Systems and Center of Mass
Syllabus
Learning Objective Essential Knowledge 2.1.A
Describe the properties and interactions of a system.- 2.1.A.1 System properties are determined by the interactions between objects within the system.
- 2.1.A.2 If the properties or interactions of the constituent objects within a system are not important in modeling the behavior of the macroscopic system, the system can itself be treated as a single object.
- 2.1.A.3 Systems may allow interactions between constituent parts of the system and the environment, which may result in the transfer of energy or mass.
- 2.1.A.4 Individual objects within a chosen system may behave differently from each other as well as from the system as a whole.
- 2.1.A.5 The internal structure of a system affects the analysis of that system.
- 2.1.A.6 As variables external to a system are changed, the system's substructure may change.
2.1.B
Describe the location of a system's center of mass with respect to the system's constituent parts.- 2.1.B.1 For objects or systems with symmetrical mass distributions, the center of mass is located on lines of symmetry.
- 2.1.B.2 The location of a system's center of mass along a given axis can be calculated using the equation
- Equation: $\vec{x}_{\text{cm}} = \dfrac{\sum m_i \vec{x}_i}{\sum m_i}$
- 2.1.B.3 For a nonuniform solid that can be considered as a collection of differential masses, $dm$, the solid's center of mass can be calculated using the equation
- Equation: $\vec{r}_{\text{cm}} = \dfrac{\int \vec{r}\, dm}{\int dm}$
- 2.1.B.3.i The linear mass density of a rod or other linear rigid body is the derivative of the rod's mass with respect to the position of the differential mass element on the rigid body.
- Equation: $\lambda = \dfrac{d}{d\ell} m(\ell)$
- 2.1.B.3.ii If a function of mass density is given for a solid, the total mass can be determined by integrating the mass density over the length (one dimension), area (two dimensions), or volume (three dimensions) of the solid. For example:
- Equation: $M_{\text{total}} = \int \rho(r)\, dV$
- 2.1.B.4 A system can be modeled as a singular object that is located at the system's center of mass.
Source: College Board AP Course and Exam Description
A system 系统 is the object or objects you analyze, treated as a point at its center of mass 质心. For a set of masses, $x_{\text{cm}}=\dfrac{\sum m_i x_i}{\sum m_i}$; for a continuous body, $x_{\text{cm}}=\dfrac{1}{M}\int x\,dm$. Only external forces move the center of mass.
Vocabulary TrainEnglish Chinese Pinyin system 系统 xì tǒng center of mass 质心 zhì xīn 2.2
Forces and Free-Body Diagrams
Syllabus
Learning Objective Essential Knowledge 2.2.A
Describe a force as an interaction between two objects or systems- 2.2.A.1 Forces are vector quantities that describe the interactions between objects or systems.
- 2.2.A.1.i A force exerted on an object or system is always due to the interaction of that object or system with another object or system.
- 2.2.A.1.ii An object or system cannot exert a net force on itself.
- 2.2.A.2 Contact forces describe the interaction of an object or system touching another object or system and are macroscopic effects of interatomic electric forces.
2.2.B
Describe the forces exerted on an object or system using a free-body diagram.- 2.2.B.1 Free-body diagrams are useful tools for visualizing forces being exerted on a single object or system and for determining the equations that represent a physical situation.
- 2.2.B.2 The free-body diagram of an object or system shows each of the forces exerted on the object or system by the environment.
- 2.2.B.3 Forces exerted on an object or system are represented as vectors originating from the representation of the center of mass, such as a dot. A system is treated as though all of its mass is located at the center of mass.
- 2.2.B.4 A coordinate system with one axis parallel to the direction of acceleration of the object or system simplifies the translation from free-body diagram to algebraic representation. For example, in a free-body diagram of an object on an inclined plane, it is useful to set one axis parallel to the surface of the incline.
Boundary statement: AP Physics C: Mechanics and AP Physics C: Electricity and Magnetism only expect students to depict the forces exerted on objects, not the force components on free-body diagrams. On the AP Physics exams, individual forces represented on a free-body diagram must be drawn as individual straight arrows, originating on the dot and pointing in the direction of the force. Individual forces that are in the same direction must be drawn side by side, not overlapping.
Source: College Board AP Course and Exam Description
A force 力 is a push or pull (a vector, in newtons). A free-body diagram 受力图 draws one object with an arrow for every force on it – weight, normal, tension, friction, applied, drag. Draw it first; it sets up every dynamics equation.
A free-body diagram shows every force acting on one objectExploreBalance the forces on a free-body diagram
A free-body diagram shows every force on one object as an arrow. The object accelerates only if the forces don't cancel — the net force sets $a=F/m$.
Vocabulary TrainEnglish Chinese Pinyin force 力 lì free-body diagram 受力图 shòu lì tú 2.3
Newton's Third Law
Syllabus
Learning Objective Essential Knowledge 2.3.A
Describe the interaction of two objects or systems using Newton's third law and a representation of paired forces exerted on each object or system.- 2.3.A.1 Newton's third law describes the interaction of two objects or systems in terms of the paired forces that each exerts on the other.
- Equation: $\vec{F}_{\text{A on B}} = -\vec{F}_{\text{B on A}}$
- 2.3.A.2 Interactions between objects within a system (internal forces) do not influence the motion of a system's center of mass.
- 2.3.A.3 Tension is the macroscopic net result of forces that infinitesimal segments of a string, cable, chain, or similar system exert on each other in response to an external force.
- 2.3.A.3.i An ideal string has negligible mass and does not stretch when under tension.
- 2.3.A.3.ii The tension in an ideal string is the same at all points within the string.
- 2.3.A.3.iii In a string with nonnegligible mass, tension may not be the same at all points within the string.
- 2.3.A.3.iv An ideal pulley is a pulley that has negligible mass and rotates about an axle through its center of mass with negligible friction.
Source: College Board AP Course and Exam Description
Newton's third law 牛顿第三定律: A pushes B, B pushes back equally and oppositely. The pair acts on different objects, so it never cancels within one free-body diagram.
A Newton's third-law pair: equal and opposite forces on two different objectsVocabulary TrainEnglish Chinese Pinyin Newton's third law 牛顿第三定律 niú dùn dì sān dìng lǜ 2.4
Newton's First Law
Syllabus
Learning Objective Essential Knowledge 2.4.A
Describe the conditions under which a system's velocity remains constant.- 2.4.A.1 The net force on a system is the vector sum of all forces exerted on the system.
- 2.4.A.2 Translational equilibrium is the configuration of forces such that the net force exerted on a system is zero.
- Derived equation: $\sum \vec{F}_i = 0$
- 2.4.A.3 Newton's first law states that if the net force exerted on a system is zero, the velocity of that system will remain constant.
- 2.4.A.4 Forces may be balanced in one dimension but unbalanced in another. The system's velocity will change only in the direction of the unbalanced force.
- 2.4.A.5 An inertial reference frame is one from which an observer would verify Newton's first law of motion.
Source: College Board AP Course and Exam Description
Newton's first law (inertia 惯性): with zero net force 合力, velocity stays constant – the object is in translational equilibrium 平动平衡.
Vocabulary TrainEnglish Chinese Pinyin inertia 惯性 guàn xìng net force 合力 hé lì translational equilibrium 平动平衡 píng dòng píng héng 2.5
Newton's Second Law
Syllabus
Learning Objective Essential Knowledge 2.5.A
Describe the conditions under which a system's velocity changes.- 2.5.A.1 Unbalanced forces are a configuration of forces such that the net force exerted on a system is not equal to zero.
- 2.5.A.2 Newton's second law of motion states that the acceleration of a system's center of mass has a magnitude proportional to the magnitude of the net force exerted on the system and is in the same direction as that net force.
- Equation: $\vec{a}_{\text{sys}} = \dfrac{\sum \vec{F}}{m_{\text{sys}}} = \dfrac{\vec{F}_{\text{net}}}{m_{\text{sys}}}$
- 2.5.A.3 The velocity of a system's center of mass will only change if a nonzero net external force is exerted on that system.
Source: College Board AP Course and Exam Description
The general form uses momentum:
$$\sum\vec{F}=\frac{d\vec{p}}{dt}=m\vec{a}\ \ (\text{for constant mass}).$$Apply it one axis at a time. When forces depend on velocity or position, this becomes a differential equation 微分方程 to solve.Worked example. A $2.0\ \text{kg}$ block slides down a frictionless incline 斜面 at $30^{\circ}$. Only the along-ramp component of gravity drives it, so $a=g\sin30^{\circ}=9.8(0.5)=4.9\ \text{m/s}^2$, independent of the mass.
Worked example (two bodies). A $6.0\ \text{kg}$ cart on a frictionless table is pulled by a string over a light pulley to a hanging $2.0\ \text{kg}$ mass. Treat the pair as one system: only the hanging weight drives it, so $a=\dfrac{2.0(9.8)}{8.0}=2.45\ \text{m/s}^2$. Then isolate the cart alone to find the string tension 张力: $T=6.0(2.45)\approx15\ \text{N}$. System first for $a$, single body for internal forces – that two-step is the standard pattern.
Vocabulary TrainEnglish Chinese Pinyin differential equation 微分方程 wēi fēn fāng chéng incline 斜面 xié miàn tension 张力 zhāng lì 2.6
Gravitational Force
Syllabus
Learning Objective Essential Knowledge 2.6.A
Describe the gravitational interaction between two objects or systems with mass.- 2.6.A.1 Newton's law of universal gravitation describes the gravitational force between two objects or systems as directly proportional to each of their masses and inversely proportional to the square of the distance between the systems' centers of mass.
- Equation: $\left| \vec{F}_g \right| = G \dfrac{m_1 m_2}{r^2}$
- 2.6.A.1.i The gravitational force is attractive.
- 2.6.A.1.ii The gravitational force is always exerted along the line connecting the center of mass of the two interacting systems.
- 2.6.A.1.iii The gravitational force on a system can be considered to be exerted on the system's center of mass.
- 2.6.A.2 A field models the effects of a noncontact force exerted on an object at various positions in space.
- 2.6.A.2.i The magnitude of the gravitational field created by a system of mass $M$ at a point in space is equal to the ratio of the gravitational force exerted by the system on a test object of mass $m$ to the mass of the test object.
- Derived equation: $\left| \vec{g} \right| = \dfrac{\left| \vec{F}_g \right|}{m} = G \dfrac{M}{r^2}$
- 2.6.A.2.ii If the gravitational force is the only force exerted on an object, the observed acceleration of the object (in $\text{m/s}^2$) is numerically equal to the magnitude of the gravitational field strength (in $\text{N/kg}$) at that location.
- 2.6.A.2.i The magnitude of the gravitational field created by a system of mass $M$ at a point in space is equal to the ratio of the gravitational force exerted by the system on a test object of mass $m$ to the mass of the test object.
- 2.6.A.3 The gravitational force exerted by an astronomical body on a relatively small nearby object is called weight.
- Derived equation: $\text{Weight} = F_g = mg$
2.6.B
Describe situations in which the gravitational force can be considered constant.- 2.6.B.1 If the gravitational force between two systems' centers of mass has a negligible change as the relative position of the two systems changes, the gravitational force can be considered constant at all points between the initial and final positions of the systems.
- 2.6.B.2 Near the surface of Earth, the strength of the gravitational field is
- Equation: $g \approx 10\ \text{N/kg}$
2.6.C
Describe the conditions under which the magnitude of a system's apparent weight is different from the magnitude of the gravitational force exerted on that system.- 2.6.C.1 The magnitude of the apparent weight of a system is the magnitude of the normal force exerted on the system.
- 2.6.C.2 If the system is accelerating, the apparent weight of the system is not equal to the magnitude of the gravitational force exerted on the system.
- 2.6.C.3 A system appears weightless when there are no forces exerted on the system or when the force of gravity is the only force exerted on the system.
- 2.6.C.4 The equivalence principle states that an observer in a noninertial reference frame is unable to distinguish between an object's apparent weight and the gravitational force exerted on the object by a gravitational field.
2.6.D
Describe inertial and gravitational mass.- 2.6.D.1 Objects have inertial mass, or inertia, a property that determines how much an object's motion resists changes when interacting with another object.
- 2.6.D.2 Gravitational mass is related to the force of attraction between two systems with mass.
- 2.6.D.3 Inertial mass and gravitational mass have been experimentally verified to be equivalent.
2.6.E
Describe the gravitational force exerted on an object by a uniform spherical distribution of mass.- 2.6.E.1 The net gravitational force exerted on an object by a uniform spherical distribution of mass is the sum of the individual forces from small differential masses that comprise the distribution.
- 2.6.E.2 Newton's shell theorem describes the net gravitational force exerted on an object by a uniform spherical shell of mass.
- 2.6.E.2.i The net gravitational force exerted on an object inside a thin spherical shell is zero.
- 2.6.E.2.ii The net gravitational force exerted on an object outside a thin spherical shell can be determined by treating the shell as a single massive object located at the center of the shell.
- 2.6.E.2.iii An object inside a sphere of uniform density experiences a net gravitational force from only a partial mass of the sphere.
- 2.6.E.2.iv The partial mass of a sphere that contributes to the net gravitational force exerted on an object within that sphere is the portion of the sphere's mass located a distance less than or equal to the object's distance from the center of the sphere and can be calculated using the density of the sphere.
- Derived equation: $m_{\text{partial}} = \rho \dfrac{4}{3} \pi \left( r_{\text{partial}} \right)^3$
- 2.6.E.3 The gravitational force exerted on an object within a uniform sphere can be shown to be proportional to the object's distance from the sphere's center.
- Derived equation: $F_{g,\text{partial}} = -k r_{\text{partial}}$
Boundary statement: AP Physics C: Mechanics does not expect students to mathematically prove or derive Newton's shell theorem.
Source: College Board AP Course and Exam Description
Near a surface, weight is $F_g=mg$. In general, Newton's law of gravitation 万有引力定律:
$$F_g=\frac{Gm_1m_2}{r^2},$$attractive and inverse-square. The gravitational field is $g=\dfrac{GM}{r^2}$.
Two masses attract each other with equal, opposite, inverse-square forces along the line joining themVocabulary TrainEnglish Chinese Pinyin Newton's law of gravitation 万有引力定律 wàn yǒu yǐn lì dìng lǜ 2.7
Kinetic and Static Friction
Syllabus
Learning Objective Essential Knowledge 2.7.A
Describe kinetic friction between two surfaces.- 2.7.A.1 Kinetic friction occurs when two surfaces in contact move relative to each other.
- 2.7.A.1.i The kinetic friction force is exerted in a direction opposite the motion of each surface relative to the other surface.
- 2.7.A.1.ii The force of friction between two surfaces does not depend on the size of the surface area of contact.
- 2.7.A.2 The magnitude of the kinetic friction force exerted on an object is the product of the normal force the surface exerts on the object and the coefficient of kinetic friction.
- Equation: $\left| \vec{F}_{f,k} \right| = \left| \mu_k \vec{F}_N \right|$
- 2.7.A.2.i The coefficient of kinetic friction depends on the material properties of the surfaces that are in contact.
- 2.7.A.2.ii Normal force is the perpendicular component of the force exerted on an object by the surface with which it is in contact; it is directed away from the surface.
2.7.B
Describe static friction between two surfaces.- 2.7.B.1 Static friction may occur between the contacting surfaces of two objects that are not moving relative to each other.
- 2.7.B.2 Static friction adopts the value and direction required to prevent an object from slipping or sliding on a surface.
- Equation: $\left| \vec{F}_{f,s} \right| \leq \left| \mu_s \vec{F}_n \right|$
- 2.7.B.2.i Slipping and sliding refer to situations in which two surfaces are moving relative to each other.
- 2.7.B.2.ii There exists a maximum value for which static friction will prevent an object from slipping on a given surface.
- Derived equation: $F_{f,s,\text{max}} = \mu_s F_N$
- 2.7.B.3 The coefficient of static friction is typically greater than the coefficient of kinetic friction for a given pair of surfaces.
Source: College Board AP Course and Exam Description
Friction 摩擦力 opposes sliding along a surface: kinetic 动摩擦 $f_k=\mu_k N$ while sliding, and static 静摩擦 $f_s\le\mu_s N$ up to a maximum before sliding. $N$ is the normal force 法向力. Note the inequality: static friction is only as large as it needs to be, and $\mu_s N$ is its ceiling, not its value.
Worked example. The same block on the same $30^{\circ}$ incline, now with $\mu_k=0.20$. Perpendicular to the ramp: $N=mg\cos30^{\circ}$. Along the ramp: $ma=mg\sin30^{\circ}-\mu_k mg\cos30^{\circ}$, so $a=g(\sin30^{\circ}-0.20\cos30^{\circ})=9.8(0.500-0.173)=3.2\ \text{m/s}^2$ – the mass cancels again.
ExploreSlide a block down a slope with friction
Friction opposes motion up to a maximum $\mu N$. Tilt the slope until gravity's pull along it beats static friction and the block starts to slide.
Vocabulary TrainEnglish Chinese Pinyin Friction 摩擦力 mó cā lì kinetic 动摩擦 dòng mó cā static 静摩擦 jìng mó cā normal force 法向力 fǎ xiàng lì 2.8
Spring Forces
Syllabus
Learning Objective Essential Knowledge 2.8.A
Describe the force exerted on an object by an ideal spring.- 2.8.A.1 An ideal spring has negligible mass and exerts a force that is proportional to the change in its length as measured from its relaxed length. A nonideal spring either has nonnegligible mass or exerts a force that is not proportional to the change in its length as measured from its relaxed length.
- 2.8.A.2 The magnitude of the force exerted by an ideal spring on an object is given by Hooke's law:
- Equation: $\vec{F}_s = -k \Delta \vec{x}$
- 2.8.A.3 The force exerted on an object by a spring is always directed toward the equilibrium position of the object–spring system.
2.8.B
Describe the equivalent spring constant of a combination of springs exerting forces on an object.- 2.8.B.1 A collection of springs that exert forces on an object may behave as though they were a single spring with an equivalent spring constant $k_{\text{eq}}$.
- 2.8.B.1.i The inverse of the equivalent spring constant of a set of springs in series is equal to the sum of the inverses of the individual spring constants.
- Derived equation: $\dfrac{1}{k_{\text{eq, series}}} = \sum_i \dfrac{1}{k_i} = \dfrac{1}{k_1} + \dfrac{1}{k_2} + \dots$
- 2.8.B.1.ii The equivalent spring constant of a set of springs arranged in series is smaller than the smallest constituent spring constant.
- 2.8.B.1.iii The equivalent spring constant of a set of springs arranged in parallel is the sum of the individual spring constants.
- Derived equation: $k_{\text{eq, parallel}} = \sum_i k_i = k_1 + k_2 + \dots$
- 2.8.B.1.i The inverse of the equivalent spring constant of a set of springs in series is equal to the sum of the inverses of the individual spring constants.
Boundary statement: AP Physics C: Mechanics only expects students to find the effective spring constant of systems of springs that are arranged either in series or in parallel and does not expect students to find the effective spring constant of a system in which springs are arranged in both series and parallel.
Source: College Board AP Course and Exam Description
An ideal spring obeys Hooke's law 胡克定律 $F_s=-kx$, a restoring force set by the spring constant 弹簧劲度系数 $k$. Its stored energy is $U=\tfrac12 kx^2$.
Hooke's law: extension is proportional to load up to the limit of proportionalityExploreStretch a spring (Hooke's law)
A spring's force is proportional to its extension, $F=kx$ (Hooke's law). Pull harder and the extension grows in step — until the spring's limit.
Vocabulary TrainEnglish Chinese Pinyin Hooke's law 胡克定律 hú kè dìng lǜ spring constant 弹簧劲度系数 tán huáng jìn dù xì shù 2.9
Resistive Forces
Syllabus
Learning Objective Essential Knowledge 2.9.A
Describe the motion of an object subject to a resistive force.- 2.9.A.1 A resistive force is defined as a velocity-dependent force in the opposite direction of an object's velocity, for example:
- Equation: $\vec{F}_r = -k\vec{v}$
- 2.9.A.2 Applying Newton's second law to an object upon which a resistive force is exerted results in a differential equation for velocity.
- 2.9.A.2.i Using the method of separation of variables, the velocity can be determined by integrating over the proper limits of integration.
- 2.9.A.2.ii The acceleration or position of a moving object that is subject to a velocity-dependent force may be determined using initial conditions of the object and methods of calculus, once a function for velocity is determined.
- 2.9.A.2.iii The position, velocity, and acceleration as functions of time of an object under the influence of a resistive force of the form $\vec{F}_r = -k\vec{v}$ are exponential and have asymptotes that are determined by the initial conditions of the object and the forces exerted on the object.
- 2.9.A.3 Terminal velocity is defined as the maximum speed achieved by an object moving under the influence of a constant force and a resistive force that are exerted on the object in opposite directions. The terminal condition is reached when the net force exerted on the object is zero.
Source: College Board AP Course and Exam Description
A resistive force 阻力 (drag) opposes motion through a fluid and grows with speed, often modeled as $F=-bv$ or $F=-cv^2$. Newton's second law then gives a differential equation, e.g. $m\dfrac{dv}{dt}=mg-bv$ for a falling object. As speed rises, drag builds until it balances the driving force; the object then stops accelerating and moves at a constant terminal velocity 终极速度, found by setting the net force (and $dv/dt$) to zero.
Worked example. For a falling object with $m\dfrac{dv}{dt}=mg-bv$, the terminal velocity is where $\dfrac{dv}{dt}=0$: $mg=bv_T$, so $v_T=\dfrac{mg}{b}$. With $m=0.10\ \text{kg}$ and $b=0.50\ \text{kg/s}$, $v_T=\dfrac{0.10(9.8)}{0.50}=2.0\ \text{m/s}$. (Solving the full equation gives $v(t)=v_T(1-e^{-bt/m})$, approaching $v_T$ exponentially.)
An object falling through a fluid speeds up to a terminal velocityVocabulary TrainEnglish Chinese Pinyin resistive force 阻力 zǔ lì terminal velocity 终极速度 zhōng jí sù dù 2.10
Circular Motion
Syllabus
Learning Objective Essential Knowledge 2.10.A
Describe the motion of an object traveling in a circular path.- 2.10.A.1 Centripetal acceleration is the component of an object's acceleration directed toward the center of the object's circular path.
- 2.10.A.1.i The magnitude of centripetal acceleration for an object moving in a circular path is the ratio of the object's tangential speed squared to the radius of the circular path.
- Equation: $a_c = \dfrac{v^2}{r}$
- 2.10.A.1.ii Centripetal acceleration is directed toward the center of an object's circular path.
- 2.10.A.1.i The magnitude of centripetal acceleration for an object moving in a circular path is the ratio of the object's tangential speed squared to the radius of the circular path.
- 2.10.A.2 Centripetal acceleration can result from a single force, more than one force, or components of forces that are exerted on an object in circular motion.
- 2.10.A.2.i At the top of a vertical, circular loop, an object requires a minimum speed to maintain circular motion. At this point, and with this minimum velocity, the gravitational force is the only force that causes the centripetal acceleration.
- Derived equation: $v = \sqrt{gr}$
- 2.10.A.2.ii Components of the static friction force and the normal force can contribute to the net force producing centripetal acceleration of an object traveling in a circle on a banked surface.
- 2.10.A.2.iii A component of tension contributes to the net force producing centripetal acceleration experienced by a conical pendulum.
- 2.10.A.2.i At the top of a vertical, circular loop, an object requires a minimum speed to maintain circular motion. At this point, and with this minimum velocity, the gravitational force is the only force that causes the centripetal acceleration.
- 2.10.A.3 Tangential acceleration is the rate at which an object's speed changes and is directed tangent to the object's circular path.
- 2.10.A.4 The net acceleration of an object moving in a circle is the vector sum of the centripetal acceleration and tangential acceleration.
- 2.10.A.5 The revolution of an object traveling in a circular path at a constant speed (uniform circular motion) can be described using period and frequency.
- 2.10.A.5.i The time to complete one full circular path, one full rotation, or a full cycle of oscillatory motion is defined as period, $T$.
- 2.10.A.5.ii The rate at which an object is completing revolutions is defined as frequency, $f$.
- Equation: $T = \dfrac{1}{f}$
- 2.10.A.5.iii For an object traveling at a constant speed in a circular path, the period is given by the derived equation
- Derived equation: $T = \dfrac{2\pi r}{v}$
2.10.B
Describe circular orbits using Kepler's third law.- 2.10.B.1 For a satellite in circular orbit around a central body, the satellite's centripetal acceleration is caused only by gravitational attraction. The period and radius of the circular orbit are related to the mass of the central body.
- Derived equation: $T^2 = \dfrac{4\pi^2}{GM} R^3$
Boundary statement: AP Physics C: Mechanics does not expect students to know Kepler's first or second laws of planetary motion.
Source: College Board AP Course and Exam Description
Uniform circular motion has a centripetal acceleration 向心加速度 $a_c=\dfrac{v^2}{r}$ pointing to the center, requiring a net inward force $F_c=\dfrac{mv^2}{r}$ supplied by a real force (tension, gravity, friction, normal). There is no outward force. For vertical circles and banked curves, decompose the real forces to find which provides the centripetal requirement.
Worked example. A $0.50\ \text{kg}$ ball is whirled on a $1.0\ \text{m}$ string in a horizontal circle at $3.0\ \text{m/s}$. The string tension supplies the whole centripetal force: $T=\dfrac{mv^2}{r}=\dfrac{0.50(3.0)^2}{1.0}=4.5\ \text{N}$.
Worked example (vertical circle). At the top of a vertical loop of radius $r$, gravity and the normal force both point toward the center: $N+mg=\dfrac{mv^2}{r}$. The slowest possible speed at the top is where the track pushes with nothing at all ($N=0$): $v_{\min}=\sqrt{gr}$. For $r=2.5\ \text{m}$: $v_{\min}=\sqrt{9.8(2.5)}=4.9\ \text{m/s}$. Any slower and the cart leaves the track before the top.
The velocity points along the tangent; the centripetal force points to the centreVocabulary TrainEnglish Chinese Pinyin centripetal acceleration 向心加速度 xiàng xīn jiā sù dù 2.10
Exam tips
- Locate the center of mass with $x_{cm}=\tfrac{1}{M}\int x\,dm$ (or $\tfrac{\sum m_i x_i}{\sum m_i}$ for point masses) and use $\lambda,\sigma,\rho$ for the mass element.
- The net external force moves the center of mass as if all mass sat there: $\vec F_{net}=M\vec a_{cm}$.
- Define your system clearly — internal forces cancel, so only external forces change its momentum.
- Exploit symmetry to shortcut a center-of-mass integral.
- Distinguish center of mass from center of gravity (identical in a uniform field).
-
3 Work, Energy, and Power
3.1
Translational Kinetic Energy
Syllabus
Learning Objective Essential Knowledge 3.1.A
Describe the translational kinetic energy of an object in terms of the object's mass and velocity.- 3.1.A.1 An object's translational kinetic energy is given by the equation
- Equation: $K = \dfrac{1}{2}mv^2$
- 3.1.A.2 Translational kinetic energy is a scalar quantity.
- 3.1.A.3 Different observers may measure different values of the translational kinetic energy of an object, depending on the observer's frame of reference.
Source: College Board AP Course and Exam Description
Kinetic energy 动能 is the energy of motion, a scalar 标量 measured in joules 焦耳 (J):
$$K=\tfrac{1}{2}mv^2.$$It grows with the square of speed – doubling the speed quadruples $K$. One subtlety worth knowing: kinetic energy depends on the observer's reference frame 参考系. A passenger walking down a train has a small $K$ measured inside the train and a huge one measured from the ground – both observers are right, each in their own frame.
Vocabulary TrainEnglish Chinese Pinyin Kinetic energy 动能 dòng néng scalar 标量 biāo liàng joules 焦耳 jiāo ěr reference frame 参考系 cān kǎo xì 3.2
Work
Syllabus
Learning Objective Essential Knowledge 3.2.A
Describe the work done on an object or system by a given force or collection of forces.- 3.2.A.1 Work is the amount of energy transferred into or out of a system by a force exerted on that system over a distance.
- 3.2.A.1.i The work done by a conservative force exerted on a system is path-independent and only depends on the initial and final configurations of that system.
- 3.2.A.1.ii The work done by a conservative force on a system—or the change in the potential energy of the system—will be zero if the system returns to its initial configuration.
- 3.2.A.1.iii Potential energies are associated only with conservative forces.
- 3.2.A.1.iv The work done by a nonconservative force is path-dependent.
- 3.2.A.1.v The most common nonconservative forces are friction and air resistance.
- 3.2.A.2 Work is a scalar quantity that may be positive, negative, or zero.
- 3.2.A.3 The work done on an object by a variable force is calculated using
- Equation: $W = \displaystyle\int_a^b \vec{F}(r) \cdot d\vec{r}$, where the integral is taken over the path from point $a$ to point $b$.
- 3.2.A.3.i The dot product between two vectors, $\vec{A}$ and $\vec{B}$, results in a scalar quantity of magnitude $\vec{A} \cdot \vec{B} = AB\cos\theta$.
- 3.2.A.3.ii Only the component of the force exerted on a system that is parallel to the displacement of the point of application of the force will change the system's total energy.
- 3.2.A.3.iii If the component of the force exerted on a system that is parallel to the displacement is constant, the work done on the system by the force is given by the derived equation $W = F_{\parallel}d = Fd\cos\theta$.
- 3.2.A.3.iv The component of the force exerted on a system perpendicular to the direction of the displacement of the system's center of mass can change the direction of the system's motion without changing the system's kinetic energy.
- 3.2.A.4 The work–energy theorem states that the change in an object's kinetic energy is equal to the sum of the work (net work) being done by all forces exerted on the object.
- Equation: $\Delta K = \displaystyle\sum W_i = \sum F_{\parallel,i}\, d_i$
- 3.2.A.4.i An external force may change the configuration of a system. The component of the external force parallel to the displacement times the displacement of the point of application of the force gives the change in kinetic energy of the system.
- 3.2.A.4.ii If the system's center of mass and the point of application of the force move the same distance when a force is exerted on a system, then the system may be modeled as an object, and only the system's kinetic energy can change.
- 3.2.A.4.iii The energy dissipated by friction is typically equated to the force of friction times the length of the path over which the force is exerted.
- Equation: $\Delta E_{\text{mech}} = F_f d\cos\theta$
- 3.2.A.5 Work is equal to the area under the curve of a graph of $F_{\parallel}$ as a function of displacement.
Boundary statement: AP Physics C: Mechanics only expects students to analyze the transfer of mechanical energy, although students should be aware that mechanical energy may be dissipated in the form of thermal energy or sound.
Source: College Board AP Course and Exam Description
Work 功 is energy transferred into or out of a system by a force acting over a distance. For a variable force it is an integral 积分 along the path:
$$W=\int_a^b \vec{F}\cdot d\vec{r}=\int F\cos\theta\,dr.$$For a constant force this reduces to $W=Fd\cos\theta$; on a force–position graph, work is the area under the curve. Work is a scalar with a sign, and the sign is physics, not bookkeeping:
- Positive – force has a component along the motion (it speeds the object up).
- Negative – force opposes the motion (friction 摩擦力 and air drag do negative work).
- Zero – force perpendicular to the motion (the normal force 法向力 on a sliding block, gravity on a horizontal move, tension in a circular swing).
Only the force component along the displacement does workThe work–energy theorem 动能定理 collects every force's contribution: the net work equals the change in kinetic energy,
$$W_{\text{net}}=\sum W_i=\Delta K.$$Worked example. A variable force $F(x)=3x^2\ \text{N}$ (along the motion) acts from $x=0$ to $x=2\ \text{m}$: $W=\displaystyle\int_0^2 3x^2\,dx=\big[x^3\big]_0^2=8\ \text{J}$. Acting alone on a body starting from rest, it would raise the kinetic energy to exactly $8\ \text{J}$.
Vocabulary TrainEnglish Chinese Pinyin Work 功 gōng integral 积分 jī fēn friction 摩擦力 mó cā lì normal force 法向力 fǎ xiàng lì work–energy theorem 动能定理 dòng néng dìng lǐ 3.3
Potential Energy
Syllabus
Learning Objective Essential Knowledge 3.3.A
Describe the potential energy of a system.- 3.3.A.1 A system composed of two or more objects has potential energy if the objects within that system only interact with each other through conservative forces.
- 3.3.A.2 Potential energy is a scalar quantity associated with the position of objects within a system.
- 3.3.A.3 The definition of zero potential energy for a given system is a decision made by the observer considering the situation to simplify or otherwise assist in analysis.
- 3.3.A.4 The relationship between conservative forces exerted on a system and the system's potential energy is
- Equation: $\Delta U = -\displaystyle\int_a^b \vec{F}_{cf}(r) \cdot d\vec{r}$
- 3.3.A.5 The conservative forces exerted on a system in a single dimension can be determined using the slope of the system's potential energy with respect to position in that dimension; these forces point in the direction of decreasing potential energy.
- Equation: $F_x = -\dfrac{dU(x)}{dx}$
- 3.3.A.6 Graphs of a system's potential energy as a function of its position can be useful in determining physical properties of that system.
- 3.3.A.6.i Stable equilibrium is a location at which a small displacement in an object's position results in a force exerted on the object opposite to the direction of the small displacement, accelerating the object back toward the equilibrium position.
- 3.3.A.6.ii Unstable equilibrium is a location at which a small displacement in an object's position results in a force exerted on the object in the same direction as the small displacement, accelerating the object away from the equilibrium position.
- 3.3.A.6.iii In a given dimension, stable equilibrium positions exist at locations where the potential energy as a function of position in that dimension has a local minimum.
- 3.3.A.6.iv In a given dimension, unstable equilibrium positions occur at locations where the potential energy as a function of position in that dimension has a local maximum.
- 3.3.A.7 The potential energy of common physical systems can be described using the physical properties of that system.
- 3.3.A.7.i The elastic potential energy of an ideal spring is given by the following equation, where $\Delta x$ is the distance the spring has been stretched or compressed from its equilibrium length.
- Equation: $U_s = \dfrac{1}{2}k(\Delta x)^2$
- 3.3.A.7.ii The general form for the gravitational potential energy of a system consisting of two approximately spherical distributions of mass (e.g., moons, planets, or stars) is given by the equation
- Equation: $U_g = -G\dfrac{m_1 m_2}{r}$
- 3.3.A.7.iii Because the gravitational field near the surface of a planet is nearly constant, the change in gravitational potential energy in a system consisting of an object with mass $m$ and a planet with gravitational field of magnitude $g$ when the object is near the surface of the planet may be approximated by the equation
- Equation: $\Delta U_g = mg\Delta y$
- 3.3.A.7.i The elastic potential energy of an ideal spring is given by the following equation, where $\Delta x$ is the distance the spring has been stretched or compressed from its equilibrium length.
- 3.3.A.8 The total potential energy of a system containing more than two objects is the sum of the potential energy of each pair of objects within the system.
Source: College Board AP Course and Exam Description
Potential energy 势能 is energy a system stores by the positions of its parts – it exists only for conservative forces 保守力, whose work is path independent. It is defined through work:
$$\Delta U=-\int_a^b\vec{F}\cdot d\vec{r},$$and you are free to choose where $U=0$ – only changes in $U$ matter, so pick the zero that makes the problem simplest. The standard results:
- Gravity near a surface: $\Delta U_g=mg\,\Delta y$.
- Gravity in general: $U_g=-\dfrac{Gm_1m_2}{r}$ (zero at infinite separation).
- Spring: $U_s=\tfrac12k(\Delta x)^2$, with $\Delta x$ measured from natural length.
For systems of several objects, add the potential energy of each pair. Turning the definition around, a conservative force is minus the derivative 导数 of its potential energy:
$$F_x=-\frac{dU}{dx}.$$The force points "downhill" on the $U(x)$ curve. Equilibrium sits where the slope is zero: a minimum is a stable equilibrium 稳定平衡 (displaced, the force pushes back), a maximum is an unstable equilibrium 不稳定平衡 (displaced, the force pushes away).
On a potential-energy curve the force points downhill and E = U marks the turning pointsWorked example. Given $U(x)=2x^3-6x$ (joules), the force is $F=-\dfrac{dU}{dx}=6-6x^2$. Equilibria sit at $F=0$: $x=\pm1$. Since $\dfrac{d^2U}{dx^2}=12x$ is positive at $x=+1$ (a minimum – stable) and negative at $x=-1$ (a maximum – unstable), the two points behave oppositely.
ExploreStore elastic potential energy in a spring
Stretching a spring stores elastic potential energy $\tfrac12 kx^2$ — the area under the force-extension line. Release it and that energy becomes kinetic.
Vocabulary TrainEnglish Chinese Pinyin Potential energy 势能 shì néng conservative forces 保守力 bǎo shǒu lì derivative 导数 dǎo shù stable equilibrium 稳定平衡 wěn dìng píng héng unstable equilibrium 不稳定平衡 bù wěn dìng píng héng 3.4
Conservation of Energy
Syllabus
Learning Objective Essential Knowledge 3.4.A
Describe the energies present in a system.- 3.4.A.1 A system composed of only a single object can only have kinetic energy.
- 3.4.A.2 A system that contains objects that interact via conservative forces or that can change its shape reversibly may have both kinetic and potential energies.
3.4.B
Describe the behavior of a system using conservation of mechanical energy principles.- 3.4.B.1 Mechanical energy is the sum of a system's kinetic and potential energies.
- 3.4.B.2 Any change to a type of energy within a system must be balanced by an equivalent change of other types of energies within the system or by a transfer of energy between the system and its surroundings.
- 3.4.B.3 A system may be selected so that the total energy of that system is constant.
- 3.4.B.4 If the total energy of a system changes, that change will be equivalent to the energy transferred into or out of the system.
3.4.C
Describe how the selection of a system determines whether the energy of that system changes.- 3.4.C.1 Energy is conserved in all interactions.
- 3.4.C.2 If the work done on a selected system is zero and there are no nonconservative interactions within the system, the total mechanical energy of the system is constant.
- 3.4.C.3 If the work done on a selected system is nonzero, energy is transferred between the system and the environment.
Boundary statement: AP Physics C: Mechanics expects students to know that mechanical energy can be dissipated as thermal energy or sound by nonconservative forces.
Source: College Board AP Course and Exam Description
Energy is conserved in all interactions – the question is only where it goes. Mechanical energy 机械能 is the sum $E=K+U$. If the external work on a system is zero and nothing inside it acts through nonconservative forces 非保守力, then
$$K_1+U_1=K_2+U_2.$$When friction or drag act, they convert mechanical energy into thermal energy 热能 ($\Delta E_{\text{mech}}=-F_fd$), and the balance must include that term. If external work is done, the system's total energy changes by exactly the energy transferred: $W_{\text{ext}}=\Delta E_{\text{sys}}$. Choosing the system is choosing the bookkeeping – a single object can only have kinetic energy; include the Earth or the spring and the system can store potential energy too.
A pendulum trades gravitational potential energy for kinetic energy and backA potential-energy graph is a complete motion map: the horizontal line at height $E$ is the total energy, the gap $E-U(x)$ is the kinetic energy at each $x$, and the crossings $E=U$ are the turning points 转折点 where the object momentarily stops and reverses.
Worked example. A $2.0\ \text{kg}$ block slides down a ramp from rest at height $1.5\ \text{m}$, arriving at the bottom at $4.0\ \text{m/s}$. Energy accounting: $mgh=29.4\ \text{J}$ available; $\tfrac12mv^2=16\ \text{J}$ arrives as kinetic energy; so friction converted $29.4-16=13\ \text{J}$ into thermal energy along the way.
ExploreWatch energy convert as an object falls
With no friction, mechanical energy is conserved: as an object falls, gravitational potential energy turns into kinetic energy while the total stays fixed.
Vocabulary TrainEnglish Chinese Pinyin Mechanical energy 机械能 jī xiè néng nonconservative forces 非保守力 fēi bǎo shǒu lì thermal energy 热能 rè néng turning points 转折点 zhuǎn zhé diǎn 3.5
Power
Syllabus
Learning Objective Essential Knowledge 3.5.A
Describe the transfer of energy into, out of, or within a system in terms of power.- 3.5.A.1 Power is the rate at which energy changes with respect to time, either by transfer into or out of a system or by conversion from one type to another within a system.
- 3.5.A.2 Average power is the amount of energy being transferred or converted, divided by the time it took for that transfer or conversion to occur.
- Equation: $P_{\text{avg}} = \dfrac{\Delta E}{\Delta t}$
- 3.5.A.3 Because work is the change in energy of an object or system due to a force, average power is the total work done, divided by the time during which that work was done.
- Equation: $P_{\text{avg}} = \dfrac{W}{\Delta t}$
- 3.5.A.4 The instantaneous power delivered to an object by a force is given by the equation
- Equation: $P_{\text{inst}} = \dfrac{dW}{dt}$
- 3.5.A.5 The instantaneous power delivered to an object by the component of a constant force parallel to the object's velocity can be described with the derived equation
- Equation: $P_{\text{inst}} = F_{\parallel}v = Fv\cos\theta$
Source: College Board AP Course and Exam Description
Power 功率 is the rate of energy transfer, in watts 瓦特 (W):
$$P_{\text{avg}}=\frac{\Delta E}{\Delta t}=\frac{W}{\Delta t},\qquad P_{\text{inst}}=\frac{dW}{dt}=\vec{F}\cdot\vec{v}.$$It measures how fast work is done, not how much. The dot product matters: only the force component along the velocity delivers power.
Power is the slope of the work-time graph: the same work in less time means more powerWorked example. A block released from rest at the top of a frictionless $3.0\ \text{m}$-high ramp reaches the bottom at $v=\sqrt{2gh}=7.7\ \text{m/s}$ (from $mgh=\tfrac12mv^2$). A motor that then drives it at a steady $7.7\ \text{m/s}$ against a $20\ \text{N}$ resistance delivers $P=Fv=20(7.7)\approx150\ \text{W}$.
Worked example. A $1200\ \text{kg}$ car climbs a hill that rises $1.0\ \text{m}$ for every $20\ \text{m}$ of road, at a steady $15\ \text{m/s}$. The engine must supply gravity's power drain: $P=mg\,v\sin\theta=1200(9.8)(15)\big(\tfrac{1}{20}\big)\approx8.8\ \text{kW}$ – before adding air resistance.
Exam skill. Energy FRQs reward the accounting sentence: name your system, state which forces do work on it, and write the balance ($W_{\text{ext}}=\Delta K+\Delta U+\Delta E_{\text{thermal}}$) before plugging in numbers. "Friction is present, so mechanical energy is not conserved" is a scored statement.
Vocabulary TrainEnglish Chinese Pinyin Power 功率 gōng lǜ watts 瓦特 wǎ tè 3.5
Exam tips
- Use the work–energy theorem $W_{net}=\Delta KE$ and compute work as $W=\int \vec F\cdot d\vec r$ for a variable force.
- Read work off a force–position graph as the area under the curve.
- Power is $P=\tfrac{dW}{dt}=\vec F\cdot\vec v$; watch instantaneous vs average.
- Split forces into conservative (define a potential energy) and non-conservative (dissipate energy).
- Choose energy methods over kinematics when the force varies or the path is complex.
- 3.1.A.1 An object's translational kinetic energy is given by the equation
-
4 Linear Momentum
4.1
Linear Momentum
Syllabus
Learning Objective Essential Knowledge 4.1.A
Describe the linear momentum of an object or system.- 4.1.A.1 Linear momentum is defined by the equation $\vec{p} = m\vec{v}$.
- 4.1.A.2 Momentum is a vector quantity and has the same direction as the velocity.
- 4.1.A.3 Momentum can be used to analyze collisions and explosions.
- 4.1.A.3.i A collision is a model for an interaction where the forces exerted between the involved objects in the system are much larger than the net external force exerted on those objects during the interaction.
- 4.1.A.3.ii As only the initial and final states of a collision are analyzed, the object model may be used to analyze collisions.
- 4.1.A.3.iii An explosion is a model for an interaction in which forces internal to the system move objects within that system apart.
Source: College Board AP Course and Exam Description
Linear momentum 动量 is mass times velocity:
$$\vec{p}=m\vec{v}.$$It is a vector 矢量 pointing along the velocity, and it is the quantity for analysing collisions 碰撞 and explosions 爆炸. A collection of objects can be treated as one system moving with the velocity of its center of mass 质心:
$$\vec{v}_{\text{cm}}=\frac{\sum m_i\vec{v}_i}{\sum m_i},$$so the system's total momentum is its total mass times $\vec{v}_{\text{cm}}$ – one object's worth of bookkeeping for any number of parts.
Vocabulary TrainEnglish Chinese Pinyin Linear momentum 动量 dòng liàng vector 矢量 shǐ liàng collisions 碰撞 pèng zhuàng explosions 爆炸 bào zhà center of mass 质心 zhì xīn 4.2
Change in Momentum and Impulse
Syllabus
Learning Objective Essential Knowledge 4.2.A
Describe the impulse delivered to an object or system.- 4.2.A.1 The rate of change of a system's momentum is equal to the net external force exerted on that system.
- Equation: $\vec{F}_{\text{net}} = \dfrac{d\vec{p}}{dt}$
- 4.2.A.2 Impulse is defined as the integral of a force exerted on an object or system over a time interval.
- Equation: $\vec{J} = \displaystyle\int_{t_1}^{t_2} \vec{F}_{\text{net}}(t)\,dt$
- 4.2.A.3 Impulse is a vector quantity and has the same direction as the net force exerted on the system.
- 4.2.A.4 The impulse delivered to a system by a net external force is equal to the area under the curve of a graph of the net external force exerted on the system as a function of time.
- 4.2.A.5 The net external force exerted on a system is equal to the slope of a graph of the momentum of the system as a function of time.
4.2.B
Describe the relationship between the impulse exerted on an object or system and the change in momentum of the object or system.- 4.2.B.1 Change in momentum is the difference between a system's final momentum and its initial momentum.
- Equation: $\Delta\vec{p} = \vec{p} - \vec{p}_0$
- 4.2.B.2 The impulse–momentum theorem relates the impulse delivered to an object and the object's change in momentum.
- 4.2.B.2.i The impulse exerted on an object is equal to the object's change in momentum.
- Equation: $\vec{J} = \displaystyle\int_{t_1}^{t_2} \vec{F}_{\text{net}}(t)\,dt = \Delta\vec{p}$
- 4.2.B.2.ii Newton's second law of motion is a direct result of the impulse–momentum theorem applied to systems with constant mass.
- Equation: $\vec{F}_{\text{net}} = \dfrac{d\vec{p}}{dt} = m\dfrac{d\vec{v}}{dt} = m\vec{a}$
- 4.2.B.2.iii The impulse–momentum theorem also describes the behavior of a system in which the velocity is constant but the mass changes with respect to time.
- Equation: $\vec{F}_{\text{net}} = \dfrac{d\vec{p}}{dt} = \dfrac{dm}{dt}\vec{v}$
- 4.2.B.2.i The impulse exerted on an object is equal to the object's change in momentum.
Source: College Board AP Course and Exam Description
Newton's second law is really a statement about momentum:
$$\vec{F}_{\text{net}}=\frac{d\vec{p}}{dt},$$which reduces to $m\vec{a}$ when the mass is constant – and handles a changing mass ($\vec{F}=\dfrac{dm}{dt}\vec{v}$, a rocket or a chain piling onto a scale) when it is not. The impulse 冲量 delivered by a force is its time integral, and the impulse–momentum theorem 冲量-动量定理 says it equals the change in momentum:
$$\vec{J}=\int_{t_1}^{t_2}\vec{F}_{\text{net}}\,dt=\Delta\vec{p}.$$Impulse is a vector along the net force. Read it off graphs both ways:
- On a force–time graph, impulse is the area under the curve.
- On a momentum–time graph, the net force is the slope 斜率 at each instant.
Impulse is the area under the force-time curve, equal to the average force times the contact timeSpreading the same $\Delta p$ over a longer time lowers the force – that is why airbags, crumple zones, and soft landings work, and why you bend your knees when you land.
Worked example. A time-varying force $F(t)=10t\ \text{N}$ acts on a $2.0\ \text{kg}$ object for $2.0\ \text{s}$. The impulse is $J=\displaystyle\int_0^2 10t\,dt=\big[5t^2\big]_0^2=20\ \text{N}\cdot\text{s}$, so the speed changes by $\Delta v=J/m=10\ \text{m/s}$.
Vocabulary TrainEnglish Chinese Pinyin impulse 冲量 chōng liàng impulse–momentum theorem 冲量-动量定理 chōng liàng - dòng liàng dìng lǐ slope 斜率 xié lǜ 4.3
Conservation of Linear Momentum
Syllabus
Learning Objective Essential Knowledge 4.3.A
Describe the behavior of a system using conservation of linear momentum.- 4.3.A.1 A collection of objects with individual momenta can be described as one system with one center-of-mass velocity.
- 4.3.A.1.i For a collection of objects, the velocity of a system's center of mass can be calculated using the equation
- Equation: $\vec{v}_{\text{cm}} = \dfrac{\sum \vec{p}_i}{\sum m_i} = \dfrac{\sum (m_i \vec{v}_i)}{\sum m_i}$
- 4.3.A.1.ii The velocity of a system's center of mass is constant in the absence of a net external force.
- 4.3.A.1.i For a collection of objects, the velocity of a system's center of mass can be calculated using the equation
- 4.3.A.2 The total momentum of a system is the sum of the momenta of the system's constituent parts.
- 4.3.A.3 In the absence of net external forces, any change to the momentum of an object within a system must be balanced by an equivalent and opposite change of momentum elsewhere within the system. Any change to the momentum of a system is due to a transfer of momentum between the system and its surroundings.
- 4.3.A.3.i The impulse exerted by one object on a second object is equal and opposite to the impulse exerted by the second object on the first. This is a direct result of Newton's third law.
- 4.3.A.3.ii A system may be selected so that the total momentum of that system is constant.
- 4.3.A.3.iii If the total momentum of a system changes, that change will be equivalent to the impulse exerted on the system.
- Equation: $\vec{J} = \Delta\vec{p}$
- 4.3.A.4 Correct application of conservation of momentum can be used to determine the velocity of a system immediately before and immediately after collisions or explosions.
Boundary statement: AP Physics C: Mechanics only expects students to quantitatively analyze collisions and interactions in one or two dimensions. Three-dimensional collisions may be analyzed qualitatively.
4.3.B
Describe how the selection of a system determines whether the momentum of that system changes.- 4.3.B.1 Momentum is conserved in all interactions.
- 4.3.B.2 If the net external force on the selected system is zero, the total momentum of the system is constant.
- 4.3.B.3 If the net external force on the selected system is nonzero, momentum is transferred between the system and the environment.
Source: College Board AP Course and Exam Description
Internal forces come in Newton's-third-law pairs, so they cancel inside any system: they can shuffle momentum between parts but never change the total. Momentum only enters or leaves a system through a net external force ($\vec{J}=\Delta\vec{p}$). So, with zero net external force, total momentum is conserved 守恒:
$$\sum\vec{p}_{\text{before}}=\sum\vec{p}_{\text{after}}.$$Momentum is conserved in every collision, however violent – choose the system large enough that the collision forces are internal. Apply conservation separately along each axis; AP asks for quantitative work in one or two dimensions.
A head-on collision: total momentum before equals total momentum afterWorked example (explosion). A $6.0\ \text{kg}$ shell at rest splits into a $2.0\ \text{kg}$ piece moving at $9.0\ \text{m/s}$ east and a $4.0\ \text{kg}$ piece. Total momentum stays zero, so the heavy piece moves west at $v=\dfrac{2.0(9.0)}{4.0}=4.5\ \text{m/s}$. The kinetic energy came from stored (chemical or spring) energy – momentum conservation does not require kinetic-energy conservation.
ExploreCollide two carts and conserve momentum
In any collision the total momentum $\sum mv$ before equals the total after. Set the masses and speeds and check the momentum bookkeeping.
Vocabulary TrainEnglish Chinese Pinyin conserved 守恒 shǒu héng 4.4
Elastic and Inelastic Collisions
Syllabus
Learning Objective Essential Knowledge 4.4.A
Describe whether an interaction between objects is elastic or inelastic.- 4.4.A.1 An elastic collision between objects is one in which the initial kinetic energy of the system is equal to the final kinetic energy of the system.
- 4.4.A.2 In an elastic collision, the final kinetic energies of each of the objects within the system may be different from their initial kinetic energies.
- 4.4.A.3 An inelastic collision between objects is one in which the total kinetic energy of the system decreases.
- 4.4.A.4 In an inelastic collision, some of the initial kinetic energy is not restored to kinetic energy but is transformed by nonconservative forces into other forms of energy.
- 4.4.A.5 In a perfectly inelastic collision, the objects stick together and move with the same velocity after the collision.
Source: College Board AP Course and Exam Description
All collisions conserve momentum; they differ in what happens to the kinetic energy 动能:
type momentum kinetic energy elastic collision 弹性碰撞 conserved total conserved (individual shares may change) inelastic collision 非弹性碰撞 conserved decreases – some becomes heat, sound, deformation 形变 perfectly inelastic collision 完全非弹性碰撞 conserved largest possible loss – the objects stick and share one velocity Strategy: always write momentum conservation first; add the kinetic-energy equation only when the problem says "elastic". Two useful elastic facts: equal masses in a 1D elastic collision simply exchange velocities, and in the center-of-mass frame each object just reverses its velocity.
Worked example. A $1000\ \text{kg}$ car at $20\ \text{m/s}$ strikes a stationary $1500\ \text{kg}$ car and they lock together – perfectly inelastic. Momentum: $v=\dfrac{1000(20)}{2500}=8.0\ \text{m/s}$. Kinetic energy falls from $2.0\times10^5\ \text{J}$ to $\tfrac12(2500)(8.0)^2=8.0\times10^4\ \text{J}$: about $60\%$ is lost, even though momentum is exactly conserved.
Worked example (2D). A puck moving east at $4.0\ \text{m/s}$ strikes an identical puck at rest; after the glancing hit, one moves at $2.0\ \text{m/s}$ at $60^\circ$ north of east. Conserve each axis: east–west, $m(4.0)=m(2.0)\cos60^\circ+mv_x$, so $v_x=3.0\ \text{m/s}$; north–south, $0=m(2.0)\sin60^\circ-mv_y$, so $v_y=1.7\ \text{m/s}$. The second puck moves at $\sqrt{3.0^2+1.7^2}=3.5\ \text{m/s}$, about $30^\circ$ south of east.
A glancing collision, resolved along two perpendicular axesExam skill. On FRQs, justify with the condition, not the slogan: "the net external force on the two-puck system is zero during the collision, so its total momentum is constant." If asked whether the collision is elastic, compute the kinetic energy before and after and compare – never assume.
ExploreCompare elastic and inelastic collisions
Momentum is always conserved, but kinetic energy is only conserved in an elastic collision. In an inelastic one the carts stick and some energy becomes heat.
Vocabulary TrainEnglish Chinese Pinyin kinetic energy 动能 dòng néng elastic collision 弹性碰撞 tán xìng pèng zhuàng inelastic collision 非弹性碰撞 fēi tán xìng pèng zhuàng deformation 形变 xíng biàn perfectly inelastic collision 完全非弹性碰撞 wán quán fēi tán xìng pèng zhuàng 4.4
Exam tips
- Impulse equals the momentum change: $\vec J=\int \vec F\,dt=\Delta\vec p$, and it is the area under a force–time graph.
- Momentum is conserved whenever the net external force is zero — always the go-to for collisions.
- Distinguish elastic (kinetic energy conserved) from inelastic (objects stick) collisions.
- Apply conservation to each component (x and y) separately in 2-D.
- Connect to center of mass: total momentum $=M\vec v_{cm}$.
-
5 Torque and Rotational Dynamics
5.1
Rotational Kinematics
Syllabus
Learning Objective Essential Knowledge 5.1.A
Describe the rotation of a system with respect to time using angular displacement, angular velocity, and angular acceleration.- 5.1.A.1 Angular displacement is the measurement of the angle, in radians, through which a point on a rigid system rotates about a specified axis.
- Equation: $\Delta\theta = \theta - \theta_0$
- 5.1.A.1.i A rigid system is one that holds its shape but in which different points on the system move in different directions during rotation. A rigid system cannot be modeled as an object.
- 5.1.A.1.ii One direction of angular displacement about an axis of rotation—clockwise or counterclockwise—is typically indicated as mathematically positive, with the other direction becoming mathematically negative.
- 5.1.A.1.iii If the rotation of a system about an axis may be well described using the motion of the system's center of mass, the system may be treated as a single object. For example, the rotation of Earth about its axis may be considered negligible when considering the revolution of Earth about the center of mass of the Earth–Sun system.
- 5.1.A.2 Angular velocity is the rate at which angular position changes with respect to time.
- Equation: $\omega = \dfrac{d\theta}{dt}$
- 5.1.A.3 Angular acceleration is the rate at which angular velocity changes with respect to time.
- Equation: $\alpha = \dfrac{d\omega}{dt}$
- 5.1.A.4 Angular displacement, angular velocity, and angular acceleration around one axis are analogous to linear displacement, velocity, and acceleration in one dimension and demonstrate the same mathematical relationships.
- 5.1.A.4.i For constant angular acceleration, the mathematical relationships between angular displacement, angular velocity, and angular acceleration can be described with the following equations:
- $\omega = \omega_0 + \alpha t$
- $\theta = \theta_0 + \omega_0 t + \dfrac{1}{2}\alpha t^2$
- $\omega^2 = \omega_0^2 + 2\alpha(\theta - \theta_0)$
- 5.1.A.4.ii Graphs of angular displacement, angular velocity, and angular acceleration as functions of time can be used to find the relationships between those quantities.
- 5.1.A.4.i For constant angular acceleration, the mathematical relationships between angular displacement, angular velocity, and angular acceleration can be described with the following equations:
Boundary statement: AP Physics C: Mechanics expects students to be able to mathematically manipulate the magnitudes of angular displacement, angular velocity, and angular acceleration using vector conventions. However, the directions of said vectors will not be assessed on the exam.
Descriptions of the directions of rotational kinematics quantities for a point or rigid body are limited to clockwise and counterclockwise with respect to a given axis of rotation.
Source: College Board AP Course and Exam Description
Rotation mirrors linear motion with angular quantities, defined as derivatives:
$$\omega=\frac{d\theta}{dt},\qquad \alpha=\frac{d\omega}{dt}=\frac{d^2\theta}{dt^2}.$$Here $\theta$ is angular displacement 角位移 in radians 弧度, $\omega$ the angular velocity 角速度, and $\alpha$ the angular acceleration 角加速度. (AP works with their magnitudes and signs; the 3D vector directions are not assessed.) For constant $\alpha$, the kinematic equations are the linear ones re-lettered:
$$\omega=\omega_0+\alpha t,\qquad \Delta\theta=\omega_0t+\tfrac12\alpha t^2,\qquad \omega^2=\omega_0^2+2\alpha\,\Delta\theta.$$
One radian is the angle whose arc length equals the radiusWorked example. A fan blade slows from $30\ \text{rad/s}$ to rest with $\alpha=-6.0\ \text{rad/s}^2$: it takes $t=5.0\ \text{s}$ and turns through $\Delta\theta=\dfrac{0-30^2}{2(-6.0)}=75\ \text{rad}$ – about $12$ revolutions.
Vocabulary TrainEnglish Chinese Pinyin angular displacement 角位移 jiǎo wèi yí radians 弧度 hú dù angular velocity 角速度 jiǎo sù dù angular acceleration 角加速度 jiǎo jiā sù dù 5.2
Connecting Linear and Rotational Motion
Syllabus
Learning Objective Essential Knowledge 5.2.A
Describe the linear motion of a point on a rotating rigid system that corresponds to the rotational motion of that point, and vice versa.- 5.2.A.1 For a point at a distance $r$ from a fixed axis of rotation, the linear distance $s$ traveled by the point as the system rotates through an angle $\Delta\theta$ is given by the equation $\Delta s = r\Delta\theta$.
- 5.2.A.2 Derived relationships of linear velocity and of the tangential component of acceleration to their respective angular quantities are given by the following equations:
- $s = r\theta$
- $v = r\omega$
- $a_T = r\alpha$
- 5.2.A.3 For a rigid system, all points within that system have the same angular velocity and angular acceleration.
Boundary statement: AP Physics C: Mechanics expects students to be able to mathematically manipulate the magnitudes of angular displacement, angular velocity, and angular acceleration using vector conventions. However, the directions of the vectors will not be assessed on the exam.
Descriptions of the directions of rotational kinematics quantities for a point or rigid body are limited to clockwise and counterclockwise with respect to a given axis of rotation.
Source: College Board AP Course and Exam Description
A point at radius $r$ from the axis moves along its arc with
$$s=r\theta,\qquad v=r\omega,\qquad a_t=r\alpha,$$so points farther out move faster. A rotating point generally has two acceleration components at once: the tangential acceleration 切向加速度 $a_t=r\alpha$ (speeding up along the arc) and the centripetal acceleration 向心加速度 $a_c=\dfrac{v^2}{r}=r\omega^2$ (turning, towards the axis). They are perpendicular, so $a=\sqrt{a_t^2+a_c^2}$.
As the radius turns through an angle, a point moves along an arc at speed vVocabulary TrainEnglish Chinese Pinyin tangential acceleration 切向加速度 qiè xiàng jiā sù dù centripetal acceleration 向心加速度 xiàng xīn jiā sù dù 5.3
Torque
Syllabus
Learning Objective Essential Knowledge 5.3.A
Identify the torques exerted on a rigid system.- 5.3.A.1 Torque results only from the force component perpendicular to the position vector from the axis of rotation to the point of application of the force.
- 5.3.A.2 The lever arm is the perpendicular distance from the axis of rotation to the line of action of the exerted force.
5.3.B
Describe the torques exerted on a rigid system.- 5.3.B.1 Torques can be described using force diagrams.
- 5.3.B.1.i Force diagrams are similar to free-body diagrams and are used to analyze the torques exerted on a rigid system.
- 5.3.B.1.ii Similar to free-body diagrams, force diagrams represent the relative magnitude and direction of the forces exerted on a rigid system. Force diagrams also depict the location at which those forces are exerted relative to the axis of rotation.
- 5.3.B.2 The torque exerted on a rigid system about a chosen pivot point by a given force is described by $\vec{\tau} = \vec{r} \times \vec{F}$.
- 5.3.B.2.i The cross-product between two vectors, $\vec{A}$ and $\vec{B}$, results in a vector quantity of magnitude $\vec{A} \times \vec{B} = AB\sin\theta$.
- 5.3.B.2.ii The direction of the vector resulting from the cross-product of vectors $\vec{A}$ and $\vec{B}$ is perpendicular to both vectors $\vec{A}$ and $\vec{B}$ and therefore is normal to the plane defined by vectors $\vec{A}$ and $\vec{B}$.
- 5.3.B.2.iii The direction of the vector resulting from the cross-product of vectors $\vec{A}$ and $\vec{B}$ can be qualitatively determined by applying the appropriate right-hand rule.
Source: College Board AP Course and Exam Description
Torque 力矩 is the turning effect of a force – the rotational cause, as force is the translational cause. It is a vector product 矢量积:
$$\vec{\tau}=\vec{r}\times\vec{F},\qquad \tau=rF\sin\theta=F\,r_\perp,$$where $r_\perp$, the moment arm 力臂, is the perpendicular distance from the axis to the force's line of action. More force, applied farther from the axis, more perpendicular: more torque. A force pointing straight at (or away from) the axis has zero torque. In a plane, give counterclockwise and clockwise torques opposite signs and add.
The torque of a force depends on the perpendicular distance from the axisExploreBalance torques on a beam
Torque is force times perpendicular distance, $\tau=Fd$. The beam is in rotational equilibrium when the torques on each side are equal.
Vocabulary TrainEnglish Chinese Pinyin Torque 力矩 lì jǔ vector product 矢量积 shǐ liàng jī moment arm 力臂 lì bì 5.4
Rotational Inertia
Syllabus
Learning Objective Essential Knowledge 5.4.A
Describe the rotational inertia of a rigid system relative to a given axis of rotation.- 5.4.A.1 Rotational inertia measures a rigid system's resistance to changes in rotation and is related to the mass of the system and the distribution of that mass relative to the axis of rotation.
- 5.4.A.2 The rotational inertia of an object rotating a perpendicular distance $r$ from an axis is described by the equation $I = mr^2$.
- 5.4.A.3 The total rotational inertia of a collection of objects about an axis is the sum of the rotational inertias of each object about that axis.
- Equation: $I_{\text{tot}} = \sum I_i = \sum m_i r_i^2$
- 5.4.A.4 For a solid that can be considered as a collection of differential masses, $dm$, the solid's rotational inertia can be calculated using the equation $I = \int r^2\, dm$, where $r$ is the perpendicular distance from $dm$ to the axis of rotation.
5.4.B
Describe the rotational inertia of a rigid system rotating about an axis that does not pass through the system's center of mass.- 5.4.B.1 A rigid system's rotational inertia in a given plane is at a minimum when the rotational axis passes through the system's center of mass.
- 5.4.B.2 The parallel axis theorem uses the following equation to relate the rotational inertia of a rigid system about any axis that is parallel to an axis through its center of mass:
- Equation: $I' = I_{\text{cm}} + Md^2$
Boundary statement: AP Physics C: Mechanics only expects students to use calculus in the derivations of the rotational inertia of thin rods of uniform or nonuniform density about an arbitrary axis perpendicular to the rod, as well as derivations of the rotational inertia of a thin cylindrical shell, disk, or rigid bodies that can be considered to be made up of coaxial rings or shells about an axis that passes through their centers (e.g., annular rings).
Students should have a qualitative understanding of the factors that affect rotational inertia; for example, how rotational inertia is greater when mass is farther from the axis of rotation, which is why a hoop has more rotational inertia than a solid puck of the same mass and radius.
Source: College Board AP Course and Exam Description
Rotational inertia 转动惯量 $I$ measures resistance to angular acceleration – the rotational analog of mass, but it depends on where the mass sits. For point masses, $I=\sum m_ir_i^2$; for a continuous body,
$$I=\int r^2\,dm.$$Mass far from the axis dominates, because of the $r^2$.
Worked example (rod). A uniform rod (mass $M$, length $L$) about its centre: with linear density 线密度 $\lambda=M/L$, $dm=\lambda\,dx$, so $I=\displaystyle\int_{-L/2}^{L/2}x^2\,\frac{M}{L}\,dx=\frac{1}{12}ML^2$. AP also expects nonuniform rods: if $\lambda(x)=cx$ on $[0,L]$, first find $M=\int_0^L cx\,dx=\tfrac12cL^2$, then $I=\int_0^L x^2(cx)\,dx=\tfrac14cL^4=\tfrac12ML^2$ about the light end.
Worked example (disk from rings). A solid disk is a nest of rings: a ring at radius $r$ of width $dr$ has $dm=\dfrac{M}{\pi R^2}\,2\pi r\,dr$, so
$$I=\int_0^R r^2\,dm=\frac{2M}{R^2}\int_0^R r^3\,dr=\tfrac12MR^2.$$The same coaxial-shell method handles cylindrical shells and annular rings.
The parallel-axis theorem 平行轴定理 shifts any known $I$ to a parallel axis a distance $d$ away:
$$I'=I_{\text{cm}}+Md^2.$$For the rod about one end: $I=\tfrac1{12}ML^2+M\big(\tfrac{L}{2}\big)^2=\tfrac13ML^2$.
Vocabulary TrainEnglish Chinese Pinyin Rotational inertia 转动惯量 zhuǎn dòng guàn liàng linear density 线密度 xiàn mì dù parallel-axis theorem 平行轴定理 píng xíng zhóu dìng lǐ 5.5
Rotational Equilibrium
Syllabus
Learning Objective Essential Knowledge 5.5.A
Describe the conditions under which a system's angular velocity remains constant.- 5.5.A.1 A system may exhibit rotational equilibrium (constant angular velocity) without being in translational equilibrium, and vice versa.
- 5.5.A.1.i Free-body and force diagrams describe the nature of the forces and torques exerted on an object or rigid system.
- 5.5.A.1.ii Rotational equilibrium is a configuration of torques such that the net torque exerted on the system is zero.
- Equation: $\sum \tau_i = 0$
- 5.5.A.1.iii The rotational analog of Newton's first law is that a system will have a constant angular velocity only if the net torque exerted on the system is zero.
- 5.5.A.2 A rotational corollary to Newton's second law states that if the torques exerted on a rigid system are not balanced, the system's angular velocity must be changing.
Boundary statement: AP Physics C: Mechanics does not expect students to simultaneously analyze rotation in multiple planes.
Source: College Board AP Course and Exam Description
Newton's first law has a rotational form: with zero net torque, angular velocity is constant – an object in rotational equilibrium 转动平衡 either does not rotate or rotates steadily. Full static equilibrium 静平衡 needs both $\sum F=0$ and $\sum\tau=0$ (about any axis). The standard move for beams and ladders: put the axis at the point where an unknown force acts, so that force drops out of the torque equation.
At balance the clockwise and anticlockwise torques about the axis are equalWorked example. A uniform $20\ \text{kg}$ beam, $4.0\ \text{m}$ long, is pivoted at its left end and held horizontal by a vertical rope at its right end. A $40\ \text{kg}$ child sits $1.0\ \text{m}$ from the pivot. Torques about the pivot: $T(4.0)=20g(2.0)+40g(1.0)$, so $T=\dfrac{392+392}{4.0}=196\ \text{N}$. Then vertical force balance gives the pivot's upward push: $F_p=(60)(9.8)-196=392\ \text{N}$. Choosing the pivot as the axis removed $F_p$ from the torque equation entirely.
ExploreFind the balance point
For rotational equilibrium the total clockwise torque equals the total anticlockwise torque. Move the forces and distances until the beam balances.
Vocabulary TrainEnglish Chinese Pinyin rotational equilibrium 转动平衡 zhuǎn dòng píng héng static equilibrium 静平衡 jìng píng héng 5.6
Newton's Second Law in Rotational Form
Syllabus
Learning Objective Essential Knowledge 5.6.A
Describe the conditions under which a system's angular velocity changes.- 5.6.A.1 Angular velocity changes when the net torque exerted on the object or system is not equal to zero.
- 5.6.A.2 The rate at which the angular velocity of a rigid system changes is directly proportional to the net torque exerted on the rigid system and is in the same direction. The angular acceleration of the rigid system is inversely proportional to the rotational inertia of the rigid system.
- Equation: $\alpha_{\text{sys}} = \dfrac{\Sigma\tau}{I_{\text{sys}}} = \dfrac{\tau_{\text{net}}}{I_{\text{sys}}}$
- 5.6.A.3 To fully describe a rotating rigid system, linear and rotational analyses may need to be performed independently.
Source: College Board AP Course and Exam Description
A net torque produces angular acceleration in proportion to the rotational inertia:
$$\alpha=\frac{\sum\tau}{I}\qquad\Big(\text{more generally }\sum\vec{\tau}=\frac{d\vec{L}}{dt}\Big).$$Solve rotational dynamics exactly like translational dynamics, with $\tau\leftrightarrow F$, $I\leftrightarrow m$, $\alpha\leftrightarrow a$ – free-body diagram, equations, solve.
A massive pulley: the two tensions differ because the pulley needs net torqueWorked example (massive pulley). Masses $m_1=4.0\ \text{kg}$ and $m_2=2.0\ \text{kg}$ hang from a rope over a pulley 滑轮 of rotational inertia $I=0.50\ \text{kg}\cdot\text{m}^2$ and radius $R=0.20\ \text{m}$. Three Newton's-law equations, one per body:
$$m_1g-T_1=m_1a,\qquad T_2-m_2g=m_2a,\qquad (T_1-T_2)R=I\alpha=\frac{Ia}{R}.$$Adding them (with $I/R^2=12.5\ \text{kg}$): $a=\dfrac{(m_1-m_2)g}{m_1+m_2+I/R^2}=\dfrac{19.6}{18.5}=1.1\ \text{m/s}^2$. Then $T_1=m_1(g-a)=35\ \text{N}$ and $T_2=m_2(g+a)=22\ \text{N}$. The tensions must differ – otherwise nothing would spin the pulley. (If a problem says the pulley is light, then $I\approx0$ and the tensions become equal again.)
Exam skill. Rotational FRQs score the setup: separate free-body diagrams for each mass and the pulley, the string constraint $a=R\alpha$ stated, and the sign convention consistent. The single most common error is assuming one tension throughout a rope that passes over a massive pulley.
Vocabulary TrainEnglish Chinese Pinyin pulley 滑轮 huá lún 5.6
Exam tips
- Use the rotational analogues: $\theta,\omega=\tfrac{d\theta}{dt},\alpha=\tfrac{d\omega}{dt}$, with the constant-$\alpha$ equations mirroring linear kinematics.
- Convert between linear and angular with $v=r\omega$ and $a_t=r\alpha$ (plus centripetal $a_c=\tfrac{v^2}{r}=\omega^2 r$).
- Keep radians throughout and fix a positive sense of rotation.
- Relate torque to angular acceleration through $\tau=I\alpha$.
- Draw the rotation axis — the moment arm is the perpendicular distance to it.
- 5.1.A.1 Angular displacement is the measurement of the angle, in radians, through which a point on a rigid system rotates about a specified axis.
-
6 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.
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 TrainEnglish Chinese Pinyin rotational kinetic energy 转动动能 zhuǎn dòng dòng néng rotational inertia 转动惯量 zhuǎn dòng guàn liàng 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.
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.
ExploreBalance torques on a beam
Torque is force times perpendicular distance, $\tau=Fd$. Equal torques on each side keep the beam in rotational equilibrium.
Vocabulary TrainEnglish Chinese Pinyin torque 力矩 lì jǔ angular displacement 角位移 jiǎo wèi yí 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.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.
- 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 TrainEnglish Chinese Pinyin Angular momentum 角动量 jiǎo dòng liàng angular impulse 角冲量 jiǎo chōng liàng 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.
Source: College Board AP Course and Exam Description
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 rodWorked 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.)
Vocabulary TrainEnglish Chinese Pinyin conserved 守恒 shǒu héng 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.
Source: College Board AP Course and Exam Description
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 = rωStatic 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 firstWorked 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 TrainEnglish Chinese Pinyin Rolling without slipping 无滑滚动 wú huá gǔn dòng disk 圆盘 yuán pán hoop 圆环 yuán huán 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: $v_{\text{esc}} = \sqrt{\dfrac{2GM}{r}}$
Source: College Board AP Course and Exam Description
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 orbitIn 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.
ExploreCompare 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 TrainEnglish 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 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.
- 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 $K_{\text{rot}} = \dfrac{1}{2} I \omega^2$.
-
7 Oscillations
7.1
Defining Simple Harmonic Motion
Syllabus
Learning Objective Essential Knowledge 7.1.A
Describe simple harmonic motion.- 7.1.A.1 Simple harmonic motion is a special case of periodic motion.
- 7.1.A.2 SHM results when the magnitude of the restoring force exerted on an object is proportional to that object's displacement from its equilibrium position.
- Derived equation: $ma_x = -k\Delta x$
- 7.1.A.2.i A restoring force is a force that is exerted in a direction opposite to the object's displacement from an equilibrium position.
- 7.1.A.2.ii An equilibrium position is a location at which the net force exerted on an object or system is zero.
Source: College Board AP Course and Exam Description
Simple harmonic motion 简谐运动 (SHM) is a special case of periodic motion 周期运动. It appears whenever two things are true: there is an equilibrium 平衡位置 position where the net force is zero, and displacing the object produces a restoring force 回复力 – a force pointing back toward equilibrium – whose magnitude is proportional to the displacement:
$$F=-k\,\Delta x.$$Newton's second law then gives the defining differential equation 微分方程:
$$\frac{d^2x}{dt^2}=-\frac{k}{m}\,x=-\omega^2 x,\qquad \omega=\sqrt{\frac{k}{m}}.$$Its solution is sinusoidal, $x(t)=A\cos(\omega t+\phi)$. You do not need to prove this solution – you need to recognise the equation: any system whose motion obeys "$\ddot{x}=-\omega^2x$" is a simple harmonic oscillator, whatever it is made of. (A mass hanging on a vertical spring works the same way: gravity only shifts the equilibrium point; the oscillation about it is unchanged.)
In SHM the acceleration always points back towards equilibrium, opposite the displacementVocabulary TrainEnglish Chinese Pinyin Simple harmonic motion 简谐运动 jiǎn xié yùn dòng periodic motion 周期运动 zhōu qī yùn dòng equilibrium 平衡位置 píng héng wèi zhì restoring force 回复力 huí fù lì differential equation 微分方程 wēi fēn fāng chéng 7.2
Frequency and Period of SHM
Syllabus
Learning Objective Essential Knowledge 7.2.A
Describe the frequency and period of an object exhibiting SHM.- 7.2.A.1 The period of SHM is related to the angular frequency, $\omega$, of the object's motion by the following equation:
- Equation: $T = \dfrac{2\pi}{\omega} = \dfrac{1}{f}$
- 7.2.A.1.i The period of an object–ideal-spring oscillator is given by the equation
- Equation: $T_s = 2\pi\sqrt{\dfrac{m}{k}}.$
- 7.2.A.1.ii The period of a simple pendulum displaced by a small angle is given by the equation
- Equation: $T_p = 2\pi\sqrt{\dfrac{l}{g}}.$
Source: College Board AP Course and Exam Description
The angular frequency 角频率 $\omega$ sets the period 周期 and frequency 频率:
$$T=\frac{2\pi}{\omega}=\frac{1}{f}.$$For the two standard systems:
$$T_{\text{spring}}=2\pi\sqrt{\frac{m}{k}},\qquad T_{\text{pendulum}}=2\pi\sqrt{\frac{\ell}{g}}.$$Two classic conceptual traps: the period of SHM never depends on the amplitude 振幅, and each system ignores one obvious variable – the pendulum's period does not involve its mass, and the spring's period does not involve $g$ (a spring–block oscillator keeps perfect time in orbit; a pendulum clock does not).
ExploreTime a pendulum's swing
A pendulum's period depends on its length and gravity, not its mass or (small) amplitude: $T=2\pi\sqrt{L/g}$. Lengthen it and each swing takes longer.
Vocabulary TrainEnglish Chinese Pinyin angular frequency 角频率 jiǎo pín lǜ period 周期 zhōu qī frequency 频率 pín lǜ amplitude 振幅 zhèn fú 7.3
Representing and Analyzing SHM
Syllabus
Learning Objective Essential Knowledge 7.3.A
Describe the displacement, velocity, and acceleration of an object exhibiting SHM.- 7.3.A.1 For an object exhibiting SHM, the displacement of that object measured from its equilibrium position can be represented by the equations $x = A\cos(2\pi ft)$ or $x = A\sin(2\pi ft)$.
- 7.3.A.1.i Minima, maxima, and zeros of displacement, velocity, and acceleration are features of harmonic motion.
- 7.3.A.1.ii Recognizing the positions or times at which the displacement, velocity, and acceleration for SHM have extrema or zeros can help in qualitatively describing the behavior of the motion.
- 7.3.A.2 The position as a function of time for an object exhibiting SHM is a solution of the second-order differential equation derived from the application of Newton's second law.
- Derived equation: $\dfrac{d^2 x}{dt^2} = -\omega^2 x$
- 7.3.A.3 Characteristics of SHM, such as velocity and acceleration, can be determined by or derived from the equation $x = A\cos(\omega t + \phi).$
- 7.3.A.3.i The acceleration of an object exhibiting SHM is related to the object's angular frequency and position.
- Derived equation: $a = -\omega^2 x$
- 7.3.A.3.ii It can be shown that the maximum velocity and acceleration of an object exhibiting SHM are related to the angular frequency of the object's motion.
- Derived equations: $v_{\max} = A\omega$
- $a_{\max} = A\omega^2$
- 7.3.A.3.i The acceleration of an object exhibiting SHM is related to the object's angular frequency and position.
- 7.3.A.4 In the presence of a sinusoidal external force, a system may exhibit resonance.
- 7.3.A.4.i Resonance occurs when an external force is exerted at the natural frequency of an oscillating system.
- 7.3.A.4.ii Resonance increases the amplitude of oscillating motion.
- 7.3.A.4.iii The natural frequency of a system is the frequency at which the system will oscillate when it is displaced from its equilibrium position.
- 7.3.A.5 Changing the amplitude of a system exhibiting SHM will not change its period.
- 7.3.A.6 Properties of SHM can be determined and analyzed using graphical representations.
Boundary statement: AP Physics C: Mechanics only expects students to know the solution to the second-order differential equation that describes SHM, as well as be able to identify SHM. AP Physics C: Mechanics does not expect students to mathematically prove that the solution is correct.
Source: College Board AP Course and Exam Description
Differentiate $x(t)=A\cos(\omega t+\phi)$ twice:
$$v=-A\omega\sin(\omega t+\phi),\qquad a=-A\omega^2\cos(\omega t+\phi)=-\omega^2x.$$So at the ends of the swing ($x=\pm A$) the speed is zero and the acceleration is largest; passing through equilibrium ($x=0$) the acceleration is zero and the speed is largest, $v_{\max}=A\omega$. The amplitude $A$ and the phase constant 相位常数 $\phi$ come from the initial conditions: where the object starts and how fast it is moving. Eliminating $t$ (or using energy, below) gives speed as a function of position:
$$v=\pm\,\omega\sqrt{A^2-x^2}.$$
Displacement varies sinusoidally with time in simple harmonic motionWorked example. A $0.50\ \text{kg}$ mass on a $k=200\ \text{N/m}$ spring: $\omega=\sqrt{k/m}=20\ \text{rad/s}$, $T=2\pi/\omega=0.31\ \text{s}$. With $A=0.10\ \text{m}$: $v_{\max}=A\omega=2.0\ \text{m/s}$ at equilibrium, and at $x=0.050\ \text{m}$ the speed is $v=\omega\sqrt{A^2-x^2}=20\sqrt{0.10^2-0.050^2}=1.7\ \text{m/s}$.
Vocabulary TrainEnglish Chinese Pinyin phase constant 相位常数 xiàng wèi cháng shù 7.4
Energy of Simple Harmonic Oscillators
Syllabus
Learning Objective Essential Knowledge 7.4.A
Describe the mechanical energy of a system exhibiting SHM.- 7.4.A.1 The total energy of a system exhibiting SHM is the sum of the system's kinetic and potential energies.
- Relevant equation: $E_{\text{total}} = U + K$
- 7.4.A.2 Conservation of energy indicates that the total energy of a system exhibiting SHM is constant.
- 7.4.A.3 The kinetic energy of a system exhibiting SHM is at a maximum when the system's potential energy is at a minimum.
- 7.4.A.4 The potential energy of a system exhibiting SHM is at a maximum when the system's kinetic energy is at a minimum.
- 7.4.A.4.i The minimum kinetic energy of a system exhibiting SHM is zero.
- 7.4.A.4.ii Changing the amplitude of a system exhibiting SHM will change the maximum potential energy of the system and, therefore, the total energy of the system.
- Relevant equation for a spring–object system: $E_{\text{total}} = \dfrac{1}{2}kA^2$
Source: College Board AP Course and Exam Description
The total energy of the oscillator is the sum $E=U+K$, and with no friction it is constant – energy just trades back and forth between the spring's potential energy and the mass's kinetic energy:
$$E=\tfrac12kA^2=\tfrac12kx^2+\tfrac12mv^2.$$All potential at the ends (where $K=0$), all kinetic at equilibrium (where $U$ is minimum). Because $E\propto A^2$, doubling the amplitude quadruples the energy.
Kinetic and potential energy swap over a cycle while the total energy stays constantWorked example. Where is the energy split evenly? Set $\tfrac12kx^2=\tfrac12E=\tfrac14kA^2$, so $x=A/\sqrt2\approx0.71A$ – much closer to the end than to the middle. For the system above ($A=0.10\ \text{m}$): $x=0.071\ \text{m}$.
ExploreTrade kinetic and potential energy in SHM
In simple harmonic motion, energy sloshes between kinetic (fastest at the centre) and potential (greatest at the extremes) while the total stays constant.
7.5
Simple and Physical Pendulums
Syllabus
Learning Objective Essential Knowledge 7.5.A
Describe the properties of a physical pendulum.- 7.5.A.1 A physical pendulum is a rigid body that undergoes oscillation about a fixed axis.
- 7.5.A.2 For small amplitudes of motion, the period of a physical pendulum is derived from the application of Newton's second law in rotational form.
- Relevant equation: $T_{\text{phys}} = 2\pi\sqrt{\dfrac{I}{mgd}}$
- 7.5.A.2.i When displaced from equilibrium, the gravitational force exerted on a physical pendulum's center of mass provides a restoring torque.
- Derived equation: $\tau = -mgd\sin\theta$
- 7.5.A.2.ii For small amplitudes of motion, the small-angle approximation can be applied to the restoring torque.
- Derived equation: $\sin\theta \approx \theta$
- $\tau = -mgd\theta = I\alpha$
- 7.5.A.2.iii The small-angle approximation and Newton's second law in rotational form yield a second-order differential equation that describes SHM:
- Equation: $\dfrac{d^2\theta}{dt^2} = -\omega^2\theta$
- 7.5.A.3 A simple pendulum is a special case of physical pendulums in which the hanging object can be modeled as a point mass at a distance, $l$, from the pivot point.
- Relevant equation: $T_p = 2\pi\sqrt{\dfrac{\ell}{g}}$
- 7.5.A.4 A torsion pendulum is a case of SHM where the restoring torque is proportional to the angular displacement of a rotating system. For example, a horizontal disk that is suspended from a wire attached to its center of mass may undergo rotational oscillations about the wire in the horizontal plane.
- Derived equation: $I\alpha = -k\Delta\theta$
Source: College Board AP Course and Exam Description
A physical pendulum 物理摆 is any rigid body swinging about a fixed pivot. Displace it by an angle $\theta$ and gravity, acting at the center of mass a distance $d$ from the pivot, supplies a restoring torque
$$\tau=-mgd\sin\theta.$$For small angles, apply the small-angle approximation 小角度近似 $\sin\theta\approx\theta$ and Newton's second law in rotational form ($\tau=I\alpha$):
$$\frac{d^2\theta}{dt^2}=-\frac{mgd}{I}\,\theta=-\omega^2\theta\quad\Rightarrow\quad T_{\text{phys}}=2\pi\sqrt{\frac{I}{mgd}}.$$This is the same "$\ddot\theta=-\omega^2\theta$" pattern as before – recognising it is the derivation. A simple pendulum 单摆 is the special case of a point mass on a light string: $I=m\ell^2$ and $d=\ell$ give $T=2\pi\sqrt{\ell/g}$.
Gravity acting at the center of mass provides a physical pendulum's restoring torqueWorked example. A uniform rod (mass $M$, length $L$) swings from one end: $I=\tfrac13ML^2$, $d=\tfrac{L}{2}$, so
$$T=2\pi\sqrt{\frac{\tfrac13ML^2}{Mg\,\tfrac{L}{2}}}=2\pi\sqrt{\frac{2L}{3g}}$$– shorter than a simple pendulum of length $L$, because the rod's mass sits nearer the pivot.
A torsion pendulum 扭摆 – a disk hanging from a wire that twists – is SHM one more time: the wire's restoring torque is proportional to the twist angle, $I\alpha=-\kappa\,\Delta\theta$, giving $T=2\pi\sqrt{I/\kappa}$.
Exam skill. Every pendulum FRQ wants the same three steps: write the restoring torque about the pivot, apply the small-angle approximation, and match the result to $\ddot\theta=-\omega^2\theta$ to read off $\omega$. State the small-angle step explicitly – it is a scored point, and it is why large-amplitude swings are not simple harmonic.
Vocabulary TrainEnglish Chinese Pinyin physical pendulum 物理摆 wù lǐ bǎi small-angle approximation 小角度近似 xiǎo jiǎo dù jìn sì simple pendulum 单摆 dān bǎi torsion pendulum 扭摆 niǔ bǎi 7.5
Exam tips
- Identify SHM from a linear restoring force $F=-kx$, which gives $\tfrac{d^2x}{dt^2}=-\omega^2 x$ with $\omega=\sqrt{k/m}$.
- Write the solution $x=A\cos(\omega t+\phi)$ and get $v,a$ by differentiating; period $T=\tfrac{2\pi}{\omega}$ is amplitude-independent.
- Energy trades between $\tfrac12 kx^2$ and $\tfrac12 mv^2$, with total $\tfrac12 kA^2$.
- For a pendulum, use the small-angle approximation $\sin\theta\approx\theta$ to reach SHM.
- Match phase to the start: released from rest at maximum displacement uses cosine.