Learning Objective 1.1.A: Describe a scalar or vector quantity using magnitude and direction, as appropriate.
1.1.A.1 Scalars are quantities described by magnitude only; vectors are quantities described by both magnitude and direction.
1.1.A.2 Vectors can be visually modeled as arrows with appropriate direction and lengths proportional to their magnitude.
1.1.A.3 Distance and speed are examples of scalar quantities, while position, displacement, velocity, and acceleration are examples of vector quantities.
1.1.A.3.i Vectors are notated with an arrow above the symbol for that quantity.
Equation:$\vec{v} = \vec{v}_0 + \vec{a}t$
1.1.A.3.ii Vector notation is not required for vector components along an axis. In one dimension, the sign of the component completely describes the direction of that component.
Derived equation:$v_x = v_{x0} + a_x t$
Learning Objective 1.1.B: Describe a vector sum in one dimension.
1.1.B.1 When determining a vector sum in a given one-dimensional coordinate system, opposite directions are denoted by opposite signs.
Source: College Board AP Course and Exam Description · 出典: College Board AP コースおよび試験説明書
English
Kinematics 运动学 describes how objects move, without asking why. First, two kinds of quantity:
A scalar 标量 has only size (magnitude 大小): distance 距离, speed, time, mass.
A vector 矢量 has magnitude and direction: displacement, velocity, acceleration, force.
The difference matters. Distance is the total path length travelled – a scalar that only grows. Displacement is the straight-line change in position, with a direction. Walk $3\ \text{m}$ east then $1\ \text{m}$ back west: the distance is $4\ \text{m}$, but the displacement is only $2\ \text{m}$ east.
In one dimension, direction is just a sign (+ or −) along a chosen axis. Choosing the positive direction first is essential – every vector's sign depends on it. A velocity of $-5\ \text{m/s}$ does not mean "slow"; it means $5\ \text{m/s}$ in the negative direction.
日本語
A car speedometer: speed is the magnitude of velocity — a scalar rate of motion
Kinematics 运动学 describes how objects move, without asking why. First, two kinds of quantity:
A scalar 标量 has only size (magnitude 大小): distance 距离, speed, time, mass.
A vector 矢量 has magnitude and direction: displacement, velocity, acceleration, force.
The difference matters. Distance is the total path length travelled – a scalar that only grows. Displacement is the straight-line change in position, with a direction. Walk $3\ \text{m}$ east then $1\ \text{m}$ back west: the distance is $4\ \text{m}$, but the displacement is only $2\ \text{m}$ east.
In one dimension, direction is just a sign (+ or −) along a chosen axis. Choosing the positive direction first is essential – every vector's sign depends on it. A velocity of $-5\ \text{m/s}$ does not mean "slow"; it means $5\ \text{m/s}$ in the negative direction.
Learning Objective 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:$\Delta x = x - x_0$
Learning Objective 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.
Learning Objective 1.2.B: Describe the velocity and acceleration of an object.
1.2.B.4 An object is accelerating if 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.
Source: College Board AP Course and Exam Description · 出典: College Board AP コースおよび試験説明書
English
Projectile motion is two independent motions
Three linked vectors describe motion along a line:
Displacement 位移$\Delta x$ is the change in position – a vector from start to end (not the total path length, which is distance).
Velocity 速度 is the rate of change of position, $v=\dfrac{\Delta x}{\Delta t}$. Its sign gives direction; its magnitude is speed 速率.
Acceleration 加速度 is the rate of change of velocity, $a=\dfrac{\Delta v}{\Delta t}$.
Be careful to separate average velocity 平均速度 (total displacement over total time) from instantaneous velocity 瞬时速度 (the velocity at one instant, the slope of the position–time graph at that point). They are equal only when the velocity is constant.
An object speeds up when $v$ and $a$ have the same sign, and slows down (deceleration 减速) when they have opposite signs. Note that a negative acceleration does not always mean slowing down – a ball falling faster and faster has negative velocity and negative acceleration.
For constant acceleration, the four kinematic equations (often called SUVAT) apply:
Pick the equation that contains the three quantities you know plus the one you want, so only one unknown is left. They apply only while $a$ is constant.
Worked example. A car starts from rest and accelerates uniformly at $2.0\ \text{m/s}^2$ for $6.0\ \text{s}$. Find its final velocity and the distance it travels.
List what you know: $v_0=0$, $a=2.0\ \text{m/s}^2$, $t=6.0\ \text{s}$.
Worked example (free fall). A ball is thrown straight up at $15\ \text{m/s}$. Taking $g=9.8\ \text{m/s}^2$ and up as positive, how high does it rise, and how long is it in the air before returning to the thrower's hand?
At the highest point the velocity is momentarily zero, and $a=-g=-9.8\ \text{m/s}^2$ throughout (this is free fall 自由落体, ignoring air resistance 空气阻力):
$$v^2=v_0^2+2a\,\Delta x \;\Rightarrow\; 0=15^2+2(-9.8)\Delta x \;\Rightarrow\; \Delta x=\frac{225}{19.6}=11.5\ \text{m}.$$
Time to the top: $0=15-9.8\,t \Rightarrow t=1.53\ \text{s}$. By symmetry the fall takes the same time, so the total is $3.1\ \text{s}$.
日本語
A high-speed train: kinematics describes displacement, velocity and acceleration over timeProjectile motion is two independent motions
Three linked vectors describe motion along a line:
Displacement 位移 $\Delta x$ is the change in position – a vector from start to end (not the total path length, which is distance).
Velocity 速度 is the rate of change of position, $v=\dfrac{\Delta x}{\Delta t}$. Its sign gives direction; its magnitude is speed 速率.
Acceleration 加速度 is the rate of change of velocity, $a=\dfrac{\Delta v}{\Delta t}$.
Be careful to separate average velocity 平均速度 (total displacement over total time) from instantaneous velocity 瞬时速度 (the velocity at one instant, the slope of the position–time graph at that point). They are equal only when the velocity is constant.
An object speeds up when $v$ and $a$ have the same sign, and slows down (deceleration 减速) when they have opposite signs. Note that a negative acceleration does not always mean slowing down – a ball falling faster and faster has negative velocity and negative acceleration.
For constant acceleration, the four kinematic equations (often called SUVAT) apply:
Pick the equation that contains the three quantities you know plus the one you want, so only one unknown is left. They apply only while $a$ is constant.
Worked example. A car starts from rest and accelerates uniformly at $2.0\ \text{m/s}^2$ for $6.0\ \text{s}$. Find its final velocity and the distance it travels.
List what you know: $v_0=0$, $a=2.0\ \text{m/s}^2$, $t=6.0\ \text{s}$.
Worked example (free fall). A ball is thrown straight up at $15\ \text{m/s}$. Taking $g=9.8\ \text{m/s}^2$ and up as positive, how high does it rise, and how long is it in the air before returning to the thrower's hand?
At the highest point the velocity is momentarily zero, and $a=-g=-9.8\ \text{m/s}^2$ throughout (this is free fall 自由落体, ignoring air resistance 空气阻力):
$$v^2=v_0^2+2a\,\Delta x \;\Rightarrow\; 0=15^2+2(-9.8)\Delta x \;\Rightarrow\; \Delta x=\frac{225}{19.6}=11.5\ \text{m}.$$
Time to the top: $0=15-9.8\,t \Rightarrow t=1.53\ \text{s}$. By symmetry the fall takes the same time, so the total is $3.1\ \text{s}$.
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 $a_g = g \approx 10 \ m/s^2$.
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.
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.
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).
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.
Boundary statement: AP Physics 1 does not expect students to quantitatively analyze nonuniform acceleration. However, students will be expected to be able to qualitatively analyze, sketch appropriate graphs of, and discuss situations in which acceleration is nonuniform.
Boundary statement: For all situations in which a numerical quantity is required for $g$, the value $g \approx 10 \ m/s^2$ 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 · 出典: College Board AP コースおよび試験説明書
The same motion appears as a description, a graph, a table, or an equation, and you should move between them:
Reading a distance-time graph: flat means at rest, a straight slope means constant speed
On a position–time graph, the slope is velocity (steeper = faster; a curve = changing velocity).
On a velocity–time graph, the slope is acceleration, and the area under the line is displacement.
Reading slopes and areas off graphs is a core exam skill. To get displacement from a velocity–time graph, split the area into triangles and rectangles and add them up; area below the time axis counts as negative displacement (motion the other way).
On a velocity-time graph the slope is the acceleration and the shaded area is the displacement
Worked example. A cyclist speeds up uniformly from rest to $8.0\ \text{m/s}$ in $4.0\ \text{s}$, then holds $8.0\ \text{m/s}$ for $6.0\ \text{s}$. Find the total distance from the velocity–time graph.
Change the start velocity $u$ and the acceleration $a$. The gradient (slope) of the line is the acceleration; the area between the line and the time axis is the displacement. · 初期速度$u$ と加速度$a$ を変更せよ。直線の傾き(勾配)は加速度であり、直線と時間軸の間の面積は変位である。
1.4
Reference Frames and Relative Motion
Syllabus · シラバス
English
Learning Objective 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.
Learning Objective 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.
Boundary statement: Adding or subtracting vectors to find relative velocities is restricted to motion along one dimension for AP Physics 1.
境界声明: AP Physics 1 において、相対速度を求めるためのベクトルの加減算は、一維運動に限定される。
Source: College Board AP Course and Exam Description · 出典: College Board AP コースおよび試験説明書
English
All motion is measured against a reference frame 参考系. Velocities measured in different frames differ, and you combine them by vector addition. The velocity of A relative to C is
$$\vec{v}_{A/C}=\vec{v}_{A/B}+\vec{v}_{B/C}.$$
A person walking on a moving train has one velocity relative to the train and another relative to the ground – this is relative motion 相对运动. A useful shortcut: the velocity of A relative to B is $\vec{v}_{A/B}=\vec{v}_A-\vec{v}_B$ (subtract B's velocity).
A frame that is not accelerating is an inertial reference frame 惯性参考系: one in which a free object (no net force) obeys Newton's first law, staying at rest or moving at constant velocity. A frame that accelerates – a braking car – is non-inertial, where objects seem to speed up with no force acting on them.
Worked example. A boat points straight across a river and moves at $3.0\ \text{m/s}$ relative to the water. The current flows at $4.0\ \text{m/s}$ along the river. Find the boat's speed and direction relative to the bank.
The two velocities are perpendicular, so add them as a right triangle:
$$v=\sqrt{3.0^2+4.0^2}=5.0\ \text{m/s},\qquad \theta=\tan^{-1}\!\frac{4.0}{3.0}=53^{\circ}\ \text{downstream from straight across.}$$
日本語
All motion is measured against a reference frame 参考系. Velocities measured in different frames differ, and you combine them by vector addition. The velocity of A relative to C is
$$\vec{v}_{A/C}=\vec{v}_{A/B}+\vec{v}_{B/C}.$$
A person walking on a moving train has one velocity relative to the train and another relative to the ground – this is relative motion 相对运动. A useful shortcut: the velocity of A relative to B is $\vec{v}_{A/B}=\vec{v}_A-\vec{v}_B$ (subtract B's velocity).
A frame that is not accelerating is an inertial reference frame 惯性参考系: one in which a free object (no net force) obeys Newton's first law, staying at rest or moving at constant velocity. A frame that accelerates – a braking car – is non-inertial, where objects seem to speed up with no force acting on them.
Worked example. A boat points straight across a river and moves at $3.0\ \text{m/s}$ relative to the water. The current flows at $4.0\ \text{m/s}$ along the river. Find the boat's speed and direction relative to the bank.
The two velocities are perpendicular, so add them as a right triangle:
$$v=\sqrt{3.0^2+4.0^2}=5.0\ \text{m/s},\qquad \theta=\tan^{-1}\!\frac{4.0}{3.0}=53^{\circ}\ \text{downstream from straight across.}$$
Learning Objective 1.5.A: Describe the perpendicular components of a vector.
1.5.A.1 Vectors can be mathematically modeled as the resultant of two perpendicular components.
1.5.A.2 Vectors can be resolved into components using a chosen coordinate system.
1.5.A.3 Vectors can be resolved into perpendicular components using trigonometric functions and relationships.
Equation:$\sin \theta = \dfrac{a}{c}$
Equation:$\cos \theta = \dfrac{b}{c}$
Equation:$\tan \theta = \dfrac{a}{b}$
Equation:$a^2 + b^2 = c^2$
Learning Objective 1.5.B: Describe the motion of an object moving in two dimensions.
1.5.B.1 Motion in two dimensions can be analyzed using one-dimensional kinematic relationships if the motion is separated into components.
1.5.B.2 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.
Source: College Board AP Course and Exam Description · 出典: College Board AP コースおよび試験説明書
English
In two dimensions, resolve each vector into components 分量 along perpendicular axes ($x$ and $y$), handle each axis separately, then recombine. A velocity $v$ at angle $\theta$ to the horizontal has components $v_x=v\cos\theta$ and $v_y=v\sin\theta$.
For projectile motion 抛体运动 (an object moving under gravity alone): the horizontal and vertical motions are independent. Horizontally, velocity is constant ($a_x=0$); vertically, acceleration is $-g$ (down). The two motions share only the time. So a projectile's path (its trajectory 轨迹) is a parabola, and you solve it as two one-dimensional problems joined by $t$.
Worked example. A ball is kicked at $20\ \text{m/s}$, $30^{\circ}$ above the horizontal. Taking $g=9.8\ \text{m/s}^2$, find the time of flight, the maximum height, and the horizontal range 射程 (assume it lands at launch height).
Vertical motion sets the time. At the top $v_y=0$, so $0=10-9.8\,t \Rightarrow t_{\text{up}}=1.02\ \text{s}$, and the total flight is $2t_{\text{up}}=2.0\ \text{s}$. The maximum height is
A common trap: at the top of the flight the vertical velocity is zero, but the ball is not at rest – its horizontal velocity $v_{0x}$ never changes. The speed at the top equals $v_{0x}=17.3\ \text{m/s}$.
日本語
In two dimensions, resolve each vector into components 分量 along perpendicular axes ($x$ and $y$), handle each axis separately, then recombine. A velocity $v$ at angle $\theta$ to the horizontal has components $v_x=v\cos\theta$ and $v_y=v\sin\theta$.
A velocity vector resolved into its horizontal and vertical components
For projectile motion 抛体运动 (an object moving under gravity alone): the horizontal and vertical motions are independent. Horizontally, velocity is constant ($a_x=0$); vertically, acceleration is $-g$ (down). The two motions share only the time. So a projectile's path (its trajectory 轨迹) is a parabola, and you solve it as two one-dimensional problems joined by $t$.
A projectile launched at an angle: the horizontal and vertical motions are independentA dropped ball and a horizontally launched ball fall together – the vertical motions are identical
Worked example. A ball is kicked at $20\ \text{m/s}$, $30^{\circ}$ above the horizontal. Taking $g=9.8\ \text{m/s}^2$, find the time of flight, the maximum height, and the horizontal range 射程 (assume it lands at launch height).
Vertical motion sets the time. At the top $v_y=0$, so $0=10-9.8\,t \Rightarrow t_{\text{up}}=1.02\ \text{s}$, and the total flight is $2t_{\text{up}}=2.0\ \text{s}$. The maximum height is
A common trap: at the top of the flight the vertical velocity is zero, but the ball is not at rest – its horizontal velocity $v_{0x}$ never changes. The speed at the top equals $v_{0x}=17.3\ \text{m/s}$.
Explore · 探索
Explore projectile motion · 投射運動を探索する
Fire the ball, then change the angle and speed. The horizontal motion stays steady while gravity pulls it down — together they trace a parabola. Find the launch angle that gives the longest range, and try the Moon. · ボールを打ち出し、その後角度と速さを変更せよ。水平方向の運動は一定のまま、重力が下へ引く——これらが組み合わさって放物線を描く。最も長い飛距離を与える発射角度を見つけ、月でも試してみよ。
Explore · 探索
Explore vectors and their components · ベクトルとその成分を探索する
Drag the vectors to change their $x$- and $y$-components. See how a single vector is built from independent horizontal and vertical parts, and how two vectors add tip-to-tail into a resultant. · ベクトルをドラッグして $x$ 成分と $y$ 成分を変更せよ。単一のベクトルが独立した水平成分と垂直成分からどのように構成されるか、そして2つのベクトルが先尾相连(tip-to-tail)によって合成ベクトル(resultant)にどう加算されるかを確認せよ。
Choose the right kinematic equation by listing the three quantities you know plus the one you want, so only one unknown remains; the SUVAT equations apply only while acceleration is constant.
Fix a positive direction first — every displacement, velocity, and acceleration then carries a sign; a negative velocity means "moving the other way", not "slow".
Treat a projectile as two independent 1-D problems sharing only the time $t$: constant velocity horizontally, $a=-g$ vertically. At the top $v_y=0$ but $v_x$ is unchanged.
On a velocity–time graph the gradient is the acceleration and the area is the displacement (area below the axis is negative).
Distinguish distance (scalar, total path) from displacement (vector, start-to-end), and speed from velocity.
Learning Objective 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.
Learning Objective 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 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
2.1.B.3 A system can be modeled as a singular object that is located at the system's center of mass.
Boundary statement: AP Physics 1 only expects students to calculate the center of mass for systems of five or fewer particles arranged in a two-dimensional configuration or for systems that are highly symmetrical.
境界文: AP Physics 1では、学生が2次元配置で5個以下の粒子からなるシステム、または極めて対称性が高いシステムの重心の計算のみを行うことが期待されている。
Source: College Board AP Course and Exam Description · 出典: College Board AP コースおよび試験説明書
English
A system 系统 is the object or group of objects you choose to analyze. A system can be treated as a single point at its center of mass 质心 – the average position of its mass. External forces change the motion of the center of mass; internal forces (between parts of the system) do not.
This is why a wrench spinning across a table still has its center of mass move in a straight line: the spinning is internal, and only the (near-zero) external force matters for the center of mass. For two masses $m_1,m_2$ on a line at positions $x_1,x_2$, the center of mass sits at $x_{\text{cm}}=\dfrac{m_1x_1+m_2x_2}{m_1+m_2}$ – always closer to the heavier mass.
日本語
A system 系统 is the object or group of objects you choose to analyze. A system can be treated as a single point at its center of mass 质心 – the average position of its mass. External forces change the motion of the center of mass; internal forces (between parts of the system) do not.
This is why a wrench spinning across a table still has its center of mass move in a straight line: the spinning is internal, and only the (near-zero) external force matters for the center of mass. For two masses $m_1,m_2$ on a line at positions $x_1,x_2$, the center of mass sits at $x_{\text{cm}}=\dfrac{m_1x_1+m_2x_2}{m_1+m_2}$ – always closer to the heavier mass.
Learning Objective 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 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.
Learning Objective 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 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 1 only expects 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.
境界文: AP Physics 1では、学生が自由体図に描くのは物体に作用する力のみであり、自由体図上の力の成分を描くことは求めません。AP Physics試験において、自由体図上に表される個別の力は、点から始まって力の方向へ向かう個々の直線矢印として描かなければなりません。同じ方向にある個別の力は、重ならないように横並びに描きます。
Source: College Board AP Course and Exam Description · 出典: College Board AP コースおよび試験説明書
English
A force 力 is a push or pull – a vector, measured in newtons (N). A free-body diagram 受力图 shows one object as a dot with arrows for every force acting on it (weight 重力, normal, tension 张力, friction, applied), each labelled and pointing the right way. Draw it before any dynamics problem; it is where most marks are won or lost.
Two rules keep free-body diagrams honest: draw only forces acting on the chosen object (not forces it exerts on other things), and draw only real, physical forces (a rope, a surface, gravity, a hand) – never an "$ma$" arrow, which is the result of the forces, not a force itself.
日本語
A skydiver: free-body diagrams balance gravity and air resistance at terminal speedA rocket launch: net force changes momentum — F_net = ma for translational dynamics
A force 力 is a push or pull – a vector, measured in newtons (N). A free-body diagram 受力图 shows one object as a dot with arrows for every force acting on it (weight 重力, normal, tension 张力, friction, applied), each labelled and pointing the right way. Draw it before any dynamics problem; it is where most marks are won or lost.
A free-body diagram shows every force acting on one object
Two rules keep free-body diagrams honest: draw only forces acting on the chosen object (not forces it exerts on other things), and draw only real, physical forces (a rope, a surface, gravity, a hand) – never an "$ma$" arrow, which is the result of the forces, not a force itself.
Explore · 探索
Balance the forces on a free-body diagram · 自由体図上の力のつり合いを確認する
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 $a=F/m$. · 自由体図は、ある物体に働くすべての力を矢印で示す。力が相殺されなければ物体は加速し、合力が $a=F/m$ を決定する。
Learning Objective 2.3.A: Describe the interaction of two objects using Newton's third law and a representation of paired forces exerted on each object.
2.3.A.1 Newton's third law describes the interaction of two objects 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 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.
Boundary statement: AP Physics 1 only expects students to describe tension qualitatively in a string, cable, chain, or similar system with mass. For example, students might note that the tension in a hanging chain is greater toward the top of the chain.
Boundary statement: The interaction between objects or systems at a distance is limited to gravitational forces in AP Physics 1. In AP Physics 2, gravitational, electric, and magnetic forces may be considered.
境界文: AP Physics 1では、質量を持つ紐、ケーブル、鎖、または類似したシステムにおける張力を定性的に説明することを学生に期待している。例えば、学生は吊り下げられた鎖の張力が鎖の上側に向かうほど大きくなることに言及するかもしれない。
境界文: AP Physics 1における遠隔の物体やシステム間の相互作用は重力のみを制限する。AP Physics 2では、重力、静電気電力、磁力を考慮することが可能である。
Source: College Board AP Course and Exam Description · 出典: College Board AP コースおよび試験説明書
English
Newton's third law 牛顿第三定律: if object A pushes on object B, then B pushes back on A with a force equal in size and opposite in direction. These two forces act on different objects, so they never cancel each other. Identify third-law pairs by the "A on B / B on A" wording.
A classic trap: the weight of a book and the normal force from the table are not a third-law pair – they act on the same object (the book). The partner of the book's weight is the pull the book exerts on the Earth; the partner of the table's push is the push the book makes on the table.
日本語
Newton's third law 牛顿第三定律: if object A pushes on object B, then B pushes back on A with a force equal in size and opposite in direction. These two forces act on different objects, so they never cancel each other. Identify third-law pairs by the "A on B / B on A" wording.
A Newton's third-law pair: equal and opposite forces on two different objects
A classic trap: the weight of a book and the normal force from the table are not a third-law pair – they act on the same object (the book). The partner of the book's weight is the pull the book exerts on the Earth; the partner of the table's push is the push the book makes on the table.
Learning Objective 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 a configuration of forces such that the net force exerted on a system is zero.
Derived equation:$\sum_i \vec{F}_i = 0$
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 · 出典: College Board AP コースおよび試験説明書
English
Newton's first law 牛顿第一定律 (the law of inertia 惯性): an object's velocity stays constant unless a net force 合力 acts on it. So zero net force means constant velocity (including rest) – the object is in translational equilibrium 平衡. Inertia is the tendency to resist changes in motion, measured by mass.
Worked example. A $1200\ \text{kg}$ car cruises at a steady $25\ \text{m/s}$ on a level road. What is the net force on it? Because the velocity is constant, the acceleration is zero, so by the first law the net force is zero – the forward drive force exactly balances drag and friction. "Steady speed" always means balanced forces.
日本語
Newton's first law 牛顿第一定律 (the law of inertia 惯性): an object's velocity stays constant unless a net force 合力 acts on it. So zero net force means constant velocity (including rest) – the object is in translational equilibrium 平衡. Inertia is the tendency to resist changes in motion, measured by mass.
Worked example. A $1200\ \text{kg}$ car cruises at a steady $25\ \text{m/s}$ on a level road. What is the net force on it? Because the velocity is constant, the acceleration is zero, so by the first law the net force is zero – the forward drive force exactly balances drag and friction. "Steady speed" always means balanced forces.
Learning Objective 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.
Apply it one axis at a time: add the force components along each axis and set the sum equal to $ma$ for that axis. Acceleration points the same way as the net force.
Worked example. A $4.0\ \text{kg}$ box is pulled along the floor by a horizontal force of $18\ \text{N}$. Friction on the box is $6.0\ \text{N}$. Find its acceleration. Along the direction of motion the net force is $18-6.0=12\ \text{N}$, so
Worked example (incline 斜面). A block of mass $m$ slides down a frictionless ramp tilted at angle $\theta$. Find its acceleration. Resolve gravity into components along and perpendicular to the ramp; only the along-ramp part, $mg\sin\theta$, drives the motion, so
$$a=\frac{mg\sin\theta}{m}=g\sin\theta.$$
The steeper the ramp, the larger $\sin\theta$ and the faster it accelerates; at $\theta=90^{\circ}$ it is free fall.
日本語
Newton's second law 牛顿第二定律 relates net force to acceleration:
Apply it one axis at a time: add the force components along each axis and set the sum equal to $ma$ for that axis. Acceleration points the same way as the net force.
Worked example. A $4.0\ \text{kg}$ box is pulled along the floor by a horizontal force of $18\ \text{N}$. Friction on the box is $6.0\ \text{N}$. Find its acceleration. Along the direction of motion the net force is $18-6.0=12\ \text{N}$, so
Worked example (incline 斜面). A block of mass $m$ slides down a frictionless ramp tilted at angle $\theta$. Find its acceleration. Resolve gravity into components along and perpendicular to the ramp; only the along-ramp part, $mg\sin\theta$, drives the motion, so
$$a=\frac{mg\sin\theta}{m}=g\sin\theta.$$
The steeper the ramp, the larger $\sin\theta$ and the faster it accelerates; at $\theta=90^{\circ}$ it is free fall.
Learning Objective 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.
2.6.A.1.ii The gravitational force is always exerted along the line connecting the centers 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.
2.6.A.2.ii If the gravitational force is the only force exerted on an object, the observed acceleration of the object (in m/s$^2$) is numerically equal to the magnitude of the gravitational field strength (in N/kg) at that location.
2.6.A.3 The gravitational force exerted by an astronomical body on a relatively small nearby object is called weight.
Equation:$\text{Weight} = F_g = mg$
Learning Objective 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 $g \approx 10 \text{ N/kg}$
Learning Objective 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.
Learning Objective 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.
Source: College Board AP Course and Exam Description · 出典: College Board AP コースおよび試験説明書
English
Orbital motion (Kepler's 2nd law)
Near a planet's surface, the gravitational force (weight) is $F_g=mg$, directed down, where $g$ is the gravitational field strength 重力场强度. More generally, Newton's law of gravitation 万有引力定律 gives the attraction between any two masses:
$$F_g=\frac{G m_1 m_2}{r^2},$$
directed along the line joining them, weaker as the distance $r$ grows (an inverse-square law). Doubling the separation quarters the force.
Worked example. A $2.0\ \text{kg}$ object weighs $19.6\ \text{N}$ on Earth ($g=9.8\ \text{m/s}^2$). On the Moon $g_{\text{Moon}}=1.6\ \text{m/s}^2$. Its mass is unchanged ($2.0\ \text{kg}$), but its weight becomes $F_g=mg=2.0\times1.6=3.2\ \text{N}$. Mass measures inertia; weight is a force that depends on where you are.
Mass actually plays two distinct roles. Inertial mass 惯性质量 sets how strongly an object resists acceleration ($F=ma$); gravitational mass 引力质量 sets how strongly it attracts other masses ($F=Gm_1m_2/r^2$). Experiments confirm the two are equal – which is exactly why every object, heavy or light, falls with the same $g$.
日本語
Orbital motion (Kepler's 2nd law)
Near a planet's surface, the gravitational force (weight) is $F_g=mg$, directed down, where $g$ is the gravitational field strength 重力场强度. More generally, Newton's law of gravitation 万有引力定律 gives the attraction between any two masses:
$$F_g=\frac{G m_1 m_2}{r^2},$$
directed along the line joining them, weaker as the distance $r$ grows (an inverse-square law). Doubling the separation quarters the force.
Two masses attract each other with equal, opposite, inverse-square forces along the line joining them
Worked example. A $2.0\ \text{kg}$ object weighs $19.6\ \text{N}$ on Earth ($g=9.8\ \text{m/s}^2$). On the Moon $g_{\text{Moon}}=1.6\ \text{m/s}^2$. Its mass is unchanged ($2.0\ \text{kg}$), but its weight becomes $F_g=mg=2.0\times1.6=3.2\ \text{N}$. Mass measures inertia; weight is a force that depends on where you are.
Mass actually plays two distinct roles. Inertial mass 惯性质量 sets how strongly an object resists acceleration ($F=ma$); gravitational mass 引力质量 sets how strongly it attracts other masses ($F=Gm_1m_2/r^2$). Experiments confirm the two are equal – which is exactly why every object, heavy or light, falls with the same $g$.
gravitational field strength/ˌɡrævɪˈteɪʃənl fiːld streŋθ/
重力場の強さ
Newton's law of gravitation/ˈnjuːtnz lɔː ɒv ˌɡrævɪˈteɪʃn/
万有引力の法則
Inertial mass/ɪˈnɜːʃl mæs/
慣性質量
gravitational mass/ˌɡrævɪˈteɪʃənl mæs/
重力質量
2.7
Kinetic and Static Friction
Syllabus · シラバス
English
Learning Objective 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 to 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.
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.
Learning Objective 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.
Static friction 静摩擦 (while not yet sliding): $f_s\le \mu_s N$ – it adjusts up to a maximum to prevent motion.
Here $N$ is the normal force 法向力 (surface push, perpendicular to the surface) and $\mu$ is the coefficient of friction 摩擦系数.
Worked example. A $5.0\ \text{kg}$ crate sits on a level floor with $\mu_s=0.40$. Will a horizontal push of $15\ \text{N}$ move it? On a level floor $N=mg=5.0\times9.8=49\ \text{N}$, so the largest static friction is $f_{s,\max}=\mu_s N=0.40\times49=19.6\ \text{N}$. The $15\ \text{N}$ push is smaller than $19.6\ \text{N}$, so friction rises to match it and the crate stays still.
日本語
Friction 摩擦力 acts along a surface, opposing relative sliding (or the tendency to slide):
Static friction 静摩擦 (while not yet sliding): $f_s\le \mu_s N$ – it adjusts up to a maximum to prevent motion.
Here $N$ is the normal force 法向力 (surface push, perpendicular to the surface) and $\mu$ is the coefficient of friction 摩擦系数.
Worked example. A $5.0\ \text{kg}$ crate sits on a level floor with $\mu_s=0.40$. Will a horizontal push of $15\ \text{N}$ move it? On a level floor $N=mg=5.0\times9.8=49\ \text{N}$, so the largest static friction is $f_{s,\max}=\mu_s N=0.40\times49=19.6\ \text{N}$. The $15\ \text{N}$ push is smaller than $19.6\ \text{N}$, so friction rises to match it and the crate stays still.
Explore · 探索
Slide a block down a slope with friction · 摩擦のある斜面の上をブロックを滑らせる
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. · 摩擦力は最大$\mu N$ まで運動に逆らう。斜面の傾斜を変えて、重力の斜面方向の成分が静止摩擦力を上回り、ブロックが動き出す点を探せ。
Source: College Board AP Course and Exam Description · 出典: College Board AP コースおよび試験説明書
English
Hooke's law & the elastic limit
An ideal spring exerts a restoring force 回复力 proportional to its stretch or compression – Hooke's law 胡克定律:
$$F_s=-kx,$$
where $k$ is the spring constant 弹簧常数 (stiffness) and $x$ is the displacement from the spring's natural length. The minus sign means the force points back toward equilibrium.
Worked example. A spring with $k=200\ \text{N/m}$ hangs vertically and a $0.50\ \text{kg}$ mass is hung on it. How far does it stretch at rest? At rest the spring force balances the weight, $kx=mg$, so
An ideal spring exerts a restoring force 回复力 proportional to its stretch or compression – Hooke's law 胡克定律:
$$F_s=-kx,$$
where $k$ is the spring constant 弹簧常数 (stiffness) and $x$ is the displacement from the spring's natural length. The minus sign means the force points back toward equilibrium.
Hooke's law: extension is proportional to load up to the limit of proportionality
Worked example. A spring with $k=200\ \text{N/m}$ hangs vertically and a $0.50\ \text{kg}$ mass is hung on it. How far does it stretch at rest? At rest the spring force balances the weight, $kx=mg$, so
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. · バネの力は伸びに比例し、$F=kx$ (フックの法則)。強く引っ張ると伸びも按比例に増加する——バネの限度に達するまで。
Learning Objective 2.9.A: Describe the motion of an object traveling in a circular path.
2.9.A.1 Centripetal acceleration is the component of an object's acceleration directed toward the center of the object's circular path.
2.9.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:$a_c = \dfrac{v^2}{r}$
2.9.A.1.ii Centripetal acceleration is directed toward the center of an object's circular path.
2.9.A.2 Centripetal acceleration can result from a single force, more than one force, or components of forces exerted on an object in circular motion.
2.9.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 speed, the gravitational force is the only force that causes the centripetal acceleration.
Equation:$v = \sqrt{gr}$
2.9.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.
2.9.A.2.iii A component of tension contributes to the net force producing centripetal acceleration experienced by a conical pendulum.
2.9.A.3 Tangential acceleration is the rate at which an object's speed changes and is directed tangent to the object's circular path.
2.9.A.4 The net acceleration of an object moving in a circle is the vector sum of the centripetal acceleration and tangential acceleration.
2.9.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.
2.9.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, $T$.
2.9.A.5.ii The rate at which an object is completing revolutions is defined as frequency, $f$.
Equation:$T = \dfrac{1}{f}$
2.9.A.5.iii For an object traveling at a constant speed in a circular path, the period is given by the derived equation
Equation:$T = \dfrac{2\pi r}{v}$
Learning Objective 2.9.B: Describe circular orbits using Kepler's third law.
2.9.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.
Equation:$T^2 = \dfrac{4\pi^2}{GM} R^3$
Boundary statement: AP Physics 1 only expects students to quantitatively analyze banked curves in which no friction is required to maintain uniform circular motion. Analysis of situations in which friction is required on a banked curve is limited to qualitative descriptions.
Boundary statement: AP Physics 1 does not expect students to know Kepler's first or second laws of planetary motion.
境界文: AP Physics 1では、学生が定量的に解析するのは摩擦なしで等速円運動を維持できる傾斜曲線のみである。傾斜曲線上で摩擦が必要な状況の解析は、定性的な説明に限定されている。
範囲の声明: AP Physics 1では、学生がケプラーの惑星運動の第一法則または第二法則を知っていることを期待しない。
Source: College Board AP Course and Exam Description · 出典: College Board AP コースおよび試験説明書
English
Uniform circular motion
An object moving in a circle at constant speed still accelerates, because its velocity direction keeps changing. This centripetal acceleration 向心加速度 points toward the center:
$$a_c=\frac{v^2}{r}.$$
It is produced by a net inward (centripetal) force 向心力$F_c=\dfrac{mv^2}{r}$ – supplied by whatever real force points inward (tension, gravity, friction, normal). The time for one full revolution is the period 周期$T$, and the number of revolutions per second is the frequency 频率$f$; they are reciprocals, $f = 1/T$, and the same definitions describe any rotation or oscillation.There is no separate outward force; "centrifugal" is only an apparent effect.
Worked example. A $0.30\ \text{kg}$ ball on a string is whirled in a horizontal circle of radius $0.80\ \text{m}$ at $4.0\ \text{m/s}$. Find the tension in the string. The tension supplies the whole centripetal force:
If the string can take at most $6.0\ \text{N}$, this is the fastest the ball can go at that radius – spin any faster and the string breaks.
日本語
Uniform circular motion
An object moving in a circle at constant speed still accelerates, because its velocity direction keeps changing. This centripetal acceleration 向心加速度 points toward the center:
$$a_c=\frac{v^2}{r}.$$
It is produced by a net inward (centripetal) force 向心力 $F_c=\dfrac{mv^2}{r}$ – supplied by whatever real force points inward (tension, gravity, friction, normal). The time for one full revolution is the period 周期 $T$, and the number of revolutions per second is the frequency 频率 $f$; they are reciprocals, $f = 1/T$, and the same definitions describe any rotation or oscillation.There is no separate outward force; "centrifugal" is only an apparent effect.
The velocity points along the tangent; the centripetal force and acceleration point to the centre
Worked example. A $0.30\ \text{kg}$ ball on a string is whirled in a horizontal circle of radius $0.80\ \text{m}$ at $4.0\ \text{m/s}$. Find the tension in the string. The tension supplies the whole centripetal force:
Learning Objective 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 · 出典: College Board AP コースおよび試験説明書
English
Energy 能量 is the capacity to do work, measured in joules 焦耳 (J). A moving object has kinetic energy 动能:
$$K=\tfrac{1}{2}mv^2.$$
It depends on the square of the speed, so doubling the speed quadruples the kinetic energy. Kinetic energy is a scalar and is never negative.
Worked example. A $1500\ \text{kg}$ car travels at $20\ \text{m/s}$. Its kinetic energy is $K=\tfrac12\times1500\times20^2=3.0\times10^{5}\ \text{J}=300\ \text{kJ}$. If it speeds up to $40\ \text{m/s}$ (double), the kinetic energy becomes $4\times$ larger, $1200\ \text{kJ}$ – which is why stopping distance grows so fast with speed.
日本語
Energy 能量 is the capacity to do work, measured in joules 焦耳 (J). A moving object has kinetic energy 动能:
$$K=\tfrac{1}{2}mv^2.$$
It depends on the square of the speed, so doubling the speed quadruples the kinetic energy. Kinetic energy is a scalar and is never negative.
Worked example. A $1500\ \text{kg}$ car travels at $20\ \text{m/s}$. Its kinetic energy is $K=\tfrac12\times1500\times20^2=3.0\times10^{5}\ \text{J}=300\ \text{kJ}$. If it speeds up to $40\ \text{m/s}$ (double), the kinetic energy becomes $4\times$ larger, $1200\ \text{kJ}$ – which is why stopping distance grows so fast with speed.
Learning Objective 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 Examples of 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 amount of work done on a system by a constant force is related to the components of that force and the displacement of the point at which that force is exerted.
3.2.A.3.i 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.
Equation:$W = F_{\parallel}d = Fd\cos\theta$
3.2.A.3.ii 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:$\Delta K = \sum_{i} W_i = \sum_{i} F_{\parallel,i}\,d$
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
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 1 only expects students to analyze the transfer of mechanical energy (as defined in Unit 3, Topic 4: Conservation of Energy), although students should be aware that mechanical energy may be dissipated in the form of thermal energy or sound. In AP Physics 2, students will also study how thermal energy can be transferred between systems through heating or cooling.
範囲の声明: AP Physics 1では、学生が第3単元第4トピック「エネルギー保存」で定義された機械的エネルギーの移転のみを解析することを期待している。ただし、学生には機械的エネルギーが熱エネルギーや音として散逸しうることも知っておくべきである。AP Physics 2では、学生はさらに、熱エネルギーが加熱または冷却によって系間で移転する仕組みも学ぶ。
Source: College Board AP Course and Exam Description · 出典: College Board AP コースおよび試験説明書
English
Work 功 is energy transferred by a force acting over a displacement:
$$W=F\,d\cos\theta,$$
where $\theta$ is the angle between the force and the displacement. Work is positive when the force has a component along the motion (adds energy), negative when it opposes the motion (removes energy), and zero when the force is perpendicular. On a force–position graph, work is the area under the curve. The work–energy theorem 动能定理 states that the net work equals the change in kinetic energy: $W_{\text{net}}=\Delta K$.
Worked example. A $2.0\ \text{kg}$ block moving at $3.0\ \text{m/s}$ on a frictionless floor is pushed by a $5.0\ \text{N}$ force over $4.0\ \text{m}$ in the direction of motion. Find its final speed. The net work is $W=Fd=5.0\times4.0=20\ \text{J}$, and by the work–energy theorem $W=\tfrac12 m(v^2-v_0^2)$:
Work 功 is energy transferred by a force acting over a displacement:
$$W=F\,d\cos\theta,$$
where $\theta$ is the angle between the force and the displacement. Work is positive when the force has a component along the motion (adds energy), negative when it opposes the motion (removes energy), and zero when the force is perpendicular. On a force–position graph, work is the area under the curve. The work–energy theorem 动能定理 states that the net work equals the change in kinetic energy: $W_{\text{net}}=\Delta K$.
Only the force component along the displacement does work
Worked example. A $2.0\ \text{kg}$ block moving at $3.0\ \text{m/s}$ on a frictionless floor is pushed by a $5.0\ \text{N}$ force over $4.0\ \text{m}$ in the direction of motion. Find its final speed. The net work is $W=Fd=5.0\times4.0=20\ \text{J}$, and by the work–energy theorem $W=\tfrac12 m(v^2-v_0^2)$:
Learning Objective 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 potential energy of common physical systems can be described using the physical properties of that system.
3.3.A.4.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.4.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.4.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.5 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 · 出典: College Board AP コースおよび試験説明書
English
Potential energy 势能 is stored energy that depends on position or configuration:
Gravitational potential energy 重力势能 near the surface: $U_g=mgh$ (height $h$ above a reference level).
Elastic potential energy 弹性势能 in a spring: $U_s=\tfrac{1}{2}kx^2$.
Potential energy is defined only for conservative forces 保守力 (gravity, springs), for which the stored energy depends on position, not path. Only changes in potential energy matter, so you may put the zero level wherever is convenient.
The $U_g=mgh$ form only holds near a surface where $g$ is roughly constant. The general form, for two spherical masses a distance $r$ apart, is
$$U_g=-\frac{G m_1 m_2}{r}.$$
It is negative and defined to be zero at infinite separation, so gravitational PE rises toward zero as the masses move apart. $mgh$ is simply its near-surface approximation. (This is the form you need for satellites and escape speed.)
日本語
Potential energy 势能 is stored energy that depends on position or configuration:
Gravitational potential energy 重力势能 near the surface: $U_g=mgh$ (height $h$ above a reference level).
Elastic potential energy 弹性势能 in a spring: $U_s=\tfrac{1}{2}kx^2$.
Potential energy is defined only for conservative forces 保守力 (gravity, springs), for which the stored energy depends on position, not path. Only changes in potential energy matter, so you may put the zero level wherever is convenient.
The $U_g=mgh$ form only holds near a surface where $g$ is roughly constant. The general form, for two spherical masses a distance $r$ apart, is
$$U_g=-\frac{G m_1 m_2}{r}.$$
It is negative and defined to be zero at infinite separation, so gravitational PE rises toward zero as the masses move apart. $mgh$ is simply its near-surface approximation. (This is the form you need for satellites and escape speed.)
Explore · 探索
Store elastic potential energy in a spring · バネに弾性位置エネルギーを蓄える
Stretching a spring stores elastic potential energy$\tfrac12 kx^2$ — the area under the force-extension line. Release it and that energy becomes kinetic. · スプリングを伸ばすと弾性ポテンシャルエネルギー$\tfrac12 kx^2$ — 力-伸長曲線下面的積 — が蓄えられます。解放するとそのエネルギーは運動エネルギーに変化します。
Learning Objective 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.
Learning Objective 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.
Learning Objective 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 1 expects students to know that mechanical energy can be dissipated as thermal energy or sound by nonconservative forces.
境界文: AP Physics 1では、学生は非保存的な力によって力学的エネルギーが熱エネルギーや音として散逸することを理解していることが期待されている。
Source: College Board AP Course and Exam Description · 出典: College Board AP コースおよび試験説明書
English
Energy conservation: KE ⇄ PE
The total mechanical energy 机械能 is $E=K+U$. When only conservative forces do work, mechanical energy is conserved 守恒:
$$K_1+U_1=K_2+U_2.$$
When friction or other non-conservative forces act, they transfer mechanical energy to thermal energy 热能; then the general statement is that total energy (including thermal) is conserved. Energy bar charts are a good way to track where the energy goes.
Worked example. A ball is released from rest at the top of a frictionless ramp $5.0\ \text{m}$ high. Find its speed at the bottom. All the gravitational potential energy becomes kinetic energy:
The mass cancels, so every object reaches the same speed – exactly the free-fall result, now got from energy. If instead $30\ \text{J}$ were lost to friction, you would subtract it: $mgh-30=\tfrac12 mv^2$.
日本語
Energy conservation: KE ⇄ PEA roller coaster trades energy back and forth: it is highest (most potential energy) at the top and fastest (most kinetic energy) at the bottom
The total mechanical energy 机械能 is $E=K+U$. When only conservative forces do work, mechanical energy is conserved 守恒:
$$K_1+U_1=K_2+U_2.$$
When friction or other non-conservative forces act, they transfer mechanical energy to thermal energy 热能; then the general statement is that total energy (including thermal) is conserved. Energy bar charts are a good way to track where the energy goes.
A swinging pendulum trades gravitational potential energy for kinetic energy and back
Worked example. A ball is released from rest at the top of a frictionless ramp $5.0\ \text{m}$ high. Find its speed at the bottom. All the gravitational potential energy becomes kinetic energy:
The mass cancels, so every object reaches the same speed – exactly the free-fall result, now got from energy. If instead $30\ \text{J}$ were lost to friction, you would subtract it: $mgh-30=\tfrac12 mv^2$.
Explore · 探索
Watch energy convert as an object falls · 物体が落下する際のエネルギー変換を観察する
With no friction, mechanical energy is conserved: as an object falls, gravitational potential energy turns into kinetic energy while the total stays fixed. · 摩擦がない場合、力学的エネルギーは保存される: 物体が落下する際、重力の位置エネルギーは運動エネルギーに変換され、合計は一定に保たれる。
Learning Objective 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.
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 the component of a constant force parallel to the object's velocity can be described with the derived equation.
So the same job done faster requires more power. On an energy–time graph, power is the slope.
Worked example. A motor lifts a $50\ \text{kg}$ load at a steady $2.0\ \text{m/s}$. Because it moves at constant speed, the lifting force equals the weight, so
$$P=Fv=mgv=50\times9.8\times2.0=980\ \text{W}.$$
Real machines waste some energy, so we quote efficiency 效率 – useful output power divided by total input power. If this motor draws $1400\ \text{W}$ of electrical power to deliver $980\ \text{W}$ of useful lifting, its efficiency is $980/1400=0.70$, or $70\%$; the other $30\%$ becomes heat and sound.
日本語
Power 功率 is the rate of doing work or transferring energy, measured in watts 瓦特 (W):
So the same job done faster requires more power. On an energy–time graph, power is the slope.
Power is the slope of the work-time graph: the same work in less time means more power
Worked example. A motor lifts a $50\ \text{kg}$ load at a steady $2.0\ \text{m/s}$. Because it moves at constant speed, the lifting force equals the weight, so
$$P=Fv=mgv=50\times9.8\times2.0=980\ \text{W}.$$
Real machines waste some energy, so we quote efficiency 效率 – useful output power divided by total input power. If this motor draws $1400\ \text{W}$ of electrical power to deliver $980\ \text{W}$ of useful lifting, its efficiency is $980/1400=0.70$, or $70\%$; the other $30\%$ becomes heat and sound.
A waterfall: power is how quickly energy is transferred — the same drop in less time is more power
Use $W=Fd\cos\theta$: work is zero when the force is perpendicular to the motion, and negative when it opposes it.
Reach for the work–energy theorem ($W_{\text{net}}=\Delta K$) or energy conservation ($K_1+U_1=K_2+U_2$) instead of forces whenever the path is complicated.
When friction acts, mechanical energy is not conserved — subtract the energy lost to heat.
Remember $K\propto v^2$: doubling the speed quadruples the kinetic energy (and the stopping distance).
Use $P=Fv$ for power at a steady speed; at constant velocity the net force is zero but the power is not.
Learning Objective 4.1.A: Describe the linear momentum of an object or system.
4.1.A.1 Linear momentum is defined by the equation $\vec{p} = m\vec{v}$.
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.
Boundary statement: Unless otherwise stated, the general term "momentum" will refer specifically to linear momentum.
Source: College Board AP Course and Exam Description · 出典: College Board AP コースおよび試験説明書
English
Linear momentum 动量 is mass times velocity – a vector pointing the same way as the velocity:
$$\vec{p}=m\vec{v}.$$
It measures "how hard it is to stop" a moving object. A heavy slow truck and a light fast ball can have the same momentum. Its unit is $\text{kg m/s}$, the same as $\text{N s}$.
4.2.A.2 Impulse is defined as the product of the average force exerted on a system and the time interval during which that force is exerted on the system.
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.
Learning Objective 4.2.B: Describe the relationship between the impulse exerted on an object or a 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 exerted on a system and the system's change in momentum.
Equation:$\vec{J} = \vec{F}_{\text{avg}} \Delta t = \Delta \vec{p}$
4.2.B.3 Newton's second law of motion is a direct result of the impulse–momentum theorem applied to systems with constant mass.
Boundary statement: AP Physics 1 does not require students to quantitatively analyze systems in which the mass of the system changes with respect to time.
境界文: AP Physics 1では、質量が時間とともに変化する系に関する定量的な解析は学生に求められない。
Source: College Board AP Course and Exam Description · 出典: College Board AP コースおよび試験説明書
English
A net force acting over time changes momentum. The impulse 冲量 delivered is
$$\vec{J}=\vec{F}\,\Delta t=\Delta\vec{p}.$$
This is the impulse–momentum theorem: impulse equals the change in momentum. On a force–time graph, impulse is the area under the curve. It explains why airbags and follow-through help – spreading the same momentum change over a longer time reduces the force.
Worked example. A $0.15\ \text{kg}$ ball hits a wall at $20\ \text{m/s}$ and bounces straight back at $15\ \text{m/s}$. The contact lasts $0.020\ \text{s}$. Find the average force on the ball. Take the rebound direction as positive, so $u=-20\ \text{m/s}$ and $v=+15\ \text{m/s}$:
Learning Objective 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 $\vec{v}_{\text{cm}} = \dfrac{\sum \vec{p}_i}{\sum m_i} = \dfrac{\sum m_i \vec{v}_i}{\sum m_i}$.
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.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 1 includes a quantitative and qualitative treatment of conservation of momentum in one dimension and a semiquantitative treatment of conservation of momentum in two dimensions. Exam questions involving solution of simultaneous equations are not included in AP Physics 1, but the AP Physics 1 Exam may include questions that assess whether students can set up the equations properly and reason about how changing a given mass, speed, or angle would affect other quantities. AP Physics 2 includes a full treatment of conservation of momentum in two dimensions for problems that include one unknown final velocity.
Learning Objective 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.
Internal forces (like the push between two colliding carts) come in third-law pairs and cancel, so they cannot change the total momentum. This is the key tool for collisions 碰撞 and explosions – apply it separately in the $x$ and $y$ directions.
Worked example (recoil 反冲). A $60\ \text{kg}$ skater, initially at rest on frictionless ice, throws a $2.0\ \text{kg}$ ball at $8.0\ \text{m/s}$. Find her recoil speed. The total momentum starts at zero and stays zero:
Since total momentum is conserved when no net external force acts, $\vec{v}_{\text{cm}}$ then stays constant – a $v_{\text{cm}}$-versus-time graph is flat unless an outside force acts. An internal collision or explosion never changes it, however violently the parts fly apart.
Collide two carts and conserve momentum · 2台の cart を衝突させて運動量を保存する
In any collision the total momentum$\sum mv$ before equals the total after. Set the masses and speeds and check the momentum bookkeeping. · いかなる衝突でも、衝突前の総運動量$\sum mv$ は衝突後のそれと等しい。質量と速度を設定して、運動量の計算を確認せよ。
4.4
Elastic and Inelastic Collisions · 弾性衝突と非弾性衝突
Syllabus · シラバス
English
Learning Objective 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 · 出典: College Board AP コースおよび試験説明書
English
Conservation of momentum in a collision
Momentum is conserved in every collision (with no external force). Kinetic energy is not:
In an elastic collision 弹性碰撞, kinetic energy is also conserved (objects bounce apart cleanly).
In an inelastic collision 非弹性碰撞, some kinetic energy becomes heat or deformation. In a perfectly inelastic collision the objects stick together and move with one common velocity afterward.
Strategy: always write momentum conservation; add energy conservation only if the collision is stated to be elastic. One elastic fact worth memorising: in a 1D elastic collision between equal masses, the two objects simply swap velocities (a moving ball striking an identical stationary one stops dead, and the target flies off at the incoming speed).
Worked example. A $1000\ \text{kg}$ car moving at $20\ \text{m/s}$ runs into a stationary $1500\ \text{kg}$ car and they lock together. Find their common speed, and the kinetic energy lost. Momentum conservation gives
Kinetic energy before is $\tfrac12(1000)(20^2)=2.0\times10^{5}\ \text{J}$; after is $\tfrac12(2500)(8.0^2)=8.0\times10^{4}\ \text{J}$. So $1.2\times10^{5}\ \text{J}$ (about $60\%$) is lost to crumpling and heat – momentum is still conserved, but kinetic energy is not.
Compare elastic and inelastic collisions · 弾性衝突と非弾性衝突を比較する
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. · 運動量は常に保存されるが、運動エネルギー は 弾性衝突 のみで保存される。非弾性衝突 では cart がくっつき、一部のエネルギーが熱となる。
4.4
Exam tips · 試験対策
English
Momentum is a vector — assign $+$/$-$ signs before adding; a ball that rebounds reverses its velocity, giving a large $\Delta p$.
Momentum is conserved in every collision (no external force); kinetic energy is conserved only if the collision is stated to be elastic.
In a perfectly inelastic collision the objects stick and move with one common velocity.
Impulse $=F\,\Delta t=\Delta p$ = the area under a force–time graph; spreading a collision over a longer time reduces the force (airbags, bending knees).
For recoil/explosions, set the total momentum equal before and after (often zero before).
Learning Objective 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 Average angular velocity is the average rate at which angular position changes with respect to time.
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:
Equation:$\omega = \omega_0 + \alpha t$
Equation:$\theta = \theta_0 + \omega_0 t + \dfrac{1}{2}\alpha t^2$
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.
Boundary statement: Descriptions of the directions of rotation for a point or object are limited to clockwise and counterclockwise with respect to a given axis of rotation.
For constant $\alpha$, the rotational kinematic equations have the same form as the linear ones, with $\theta,\omega,\alpha$ replacing $x,v,a$: $\omega=\omega_0+\alpha t$, $\theta=\omega_0 t+\tfrac12\alpha t^2$, and $\omega^2=\omega_0^2+2\alpha\theta$.
Worked example. A wheel starts from rest and speeds up uniformly to $30\ \text{rad/s}$ in $6.0\ \text{s}$. Find its angular acceleration and the total angle turned:
A crash-test collision: large forces and torques produce rapid changes in linear and angular motionA spring under load: Hooke's law links force and extension — useful context for energy in rotating systems
Rotation is described by angular quantities that mirror the linear ones:
One radian is the angle whose arc length equals the radius
angular displacement 角位移 $\theta$ (in radians 弧度),
For constant $\alpha$, the rotational kinematic equations have the same form as the linear ones, with $\theta,\omega,\alpha$ replacing $x,v,a$: $\omega=\omega_0+\alpha t$, $\theta=\omega_0 t+\tfrac12\alpha t^2$, and $\omega^2=\omega_0^2+2\alpha\theta$.
Worked example. A wheel starts from rest and speeds up uniformly to $30\ \text{rad/s}$ in $6.0\ \text{s}$. Find its angular acceleration and the total angle turned:
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:
Equation:$s = r\theta$
Equation:$v = r\omega$
Equation:$a_T = r\alpha$
5.2.A.3 For a rigid system, all points within that system have the same angular velocity and angular acceleration.
Boundary statement: Descriptions of the directions of rotation for a point or object are limited to clockwise and counterclockwise with respect to a given axis of rotation.
Source: College Board AP Course and Exam Description · 出典: College Board AP コースおよび試験説明書
A point at radius $r$ from the axis has linear quantities tied to the angular ones:
Points farther from the axis move faster. This link lets you switch between "how fast the wheel spins" and "how fast a point on its rim moves."
As the radius turns through an angle, a point moves along an arc at speed v
Worked example. A bicycle wheel of radius $0.35\ \text{m}$ spins at $12\ \text{rad/s}$. A point on the rim (and so the bike) moves at $v=r\omega=0.35\times12=4.2\ \text{m/s}$. A point halfway to the axis moves at half that speed.
5.3
Torque
Syllabus · シラバス
English
Learning Objective 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.
Learning Objective 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 magnitude of the torque exerted on a rigid system by a force is described by the following equation, where $\theta$ is the angle between the force vector and the position vector from the axis of rotation to the point of application of the force.
Equation:$\tau = rF_{\perp} = rF\sin\theta$
Boundary statement: While AP Physics 1 expects students to mathematically manipulate the magnitude of torque using vector conventions, the direction of torque is beyond the scope of the course.
境界文: AP Physics 1では、学生がベクトル規則を用いてトルクの大きさを数学的に操作することが期待されるが、トルクの方向については課程の範囲外である。
Source: College Board AP Course and Exam Description · 出典: College Board AP コースおよび試験説明書
English
The principle of moments (torque)
Torque 力矩 is the rotational effect of a force – how effectively it turns an object about an axis:
$$\tau=r F\sin\theta = F\cdot r_\perp,$$
where $r_\perp$ is the moment arm 力臂 (the perpendicular distance from the axis to the force's line of action). A force applied farther out, or more perpendicular, produces more torque. Torque has a sign (clockwise vs counterclockwise).
Worked example. You push with $20\ \text{N}$ at the end of a $0.30\ \text{m}$ wrench. Perpendicular to the wrench the torque is $\tau=rF=0.30\times20=6.0\ \text{N m}$. If you push at $60^{\circ}$ to the wrench instead, only the perpendicular part counts: $\tau=rF\sin 60^{\circ}=0.30\times20\times0.87=5.2\ \text{N m}$ – which is why you push at a right angle for the most turning effect.
日本語
The principle of moments (torque)
Torque 力矩 is the rotational effect of a force – how effectively it turns an object about an axis:
$$\tau=r F\sin\theta = F\cdot r_\perp,$$
where $r_\perp$ is the moment arm 力臂 (the perpendicular distance from the axis to the force's line of action). A force applied farther out, or more perpendicular, produces more torque. Torque has a sign (clockwise vs counterclockwise).
The moment of a force depends on the perpendicular distance from the pivot
Worked example. You push with $20\ \text{N}$ at the end of a $0.30\ \text{m}$ wrench. Perpendicular to the wrench the torque is $\tau=rF=0.30\times20=6.0\ \text{N m}$. If you push at $60^{\circ}$ to the wrench instead, only the perpendicular part counts: $\tau=rF\sin 60^{\circ}=0.30\times20\times0.87=5.2\ \text{N m}$ – which is why you push at a right angle for the most turning effect.
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Balance torques on a beam · ビーム上のモーメントを釣り合わせる
Torque is force times perpendicular distance, $\tau=Fd$. The beam is in rotational equilibrium when the torques on each side are equal. · モーメント は力に垂直距離を掛けたもの、$\tau=Fd$。両側のモーメントが等しいとき、ビームは 回転平衡状態 にある。
Learning Objective 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
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:
Learning Objective 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 1 only expects students to calculate the rotational inertia for systems of five or fewer objects arranged in a two-dimensional configuration.
Boundary statement: Students do not need to know the rotational inertia of extended rigid systems, as these will be provided within the exam. 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 disk of the same mass and radius.
Source: College Board AP Course and Exam Description · 出典: College Board AP コースおよび試験説明書
English
Rotational inertia 转动惯量 (moment of inertia) $I$ measures how hard it is to change an object's rotation – the rotational version of mass. It depends on both the mass and how far that mass sits from the axis: mass spread farther out gives a larger $I$. For a point mass, $I=mr^2$; for extended bodies, standard formulas are provided (a hoop is $mR^2$, a solid disk $\tfrac12 mR^2$). This is why a figure skater spins faster when she pulls her arms in – she reduces $I$.
日本語
Rotational inertia 转动惯量 (moment of inertia) $I$ measures how hard it is to change an object's rotation – the rotational version of mass. It depends on both the mass and how far that mass sits from the axis: mass spread farther out gives a larger $I$. For a point mass, $I=mr^2$; for extended bodies, standard formulas are provided (a hoop is $mR^2$, a solid disk $\tfrac12 mR^2$). This is why a figure skater spins faster when she pulls her arms in – she reduces $I$.
Learning Objective 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 1 does not expect students to simultaneously analyze rotation in multiple planes.
Source: College Board AP Course and Exam Description · 出典: College Board AP コースおよび試験説明書
English
An object is in rotational equilibrium 转动平衡 when the net torque is zero, so its angular velocity stays constant. For a balanced (static) object, both the net force and the net torque are zero. Choosing the axis at an unknown force's location removes it from the torque equation – a useful trick for beam and ladder problems.
Worked example. A $30\ \text{kg}$ child sits $2.0\ \text{m}$ from the pivot of a seesaw. Where must a $40\ \text{kg}$ child sit on the other side to balance it? Set the clockwise torque equal to the anticlockwise torque (the $g$'s cancel):
The heavier child sits closer to the pivot – less distance, same torque.
日本語
An object is in rotational equilibrium 转动平衡 when the net torque is zero, so its angular velocity stays constant. For a balanced (static) object, both the net force and the net torque are zero. Choosing the axis at an unknown force's location removes it from the torque equation – a useful trick for beam and ladder problems.
At balance the clockwise and anticlockwise moments about the pivot are equal
Worked example. A $30\ \text{kg}$ child sits $2.0\ \text{m}$ from the pivot of a seesaw. Where must a $40\ \text{kg}$ child sit on the other side to balance it? Set the clockwise torque equal to the anticlockwise torque (the $g$'s cancel):
The heavier child sits closer to the pivot – less distance, same torque.
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Find the balance point · 釣り合う点を見つける
For rotational equilibrium the total clockwise torque equals the total anticlockwise torque. Move the forces and distances until the beam balances. · 回転平衡状態 では、時計回りの総モーメントが反時計回りの総モーメントに等しい。力が釣り合うまで、力や距離を移動させろ。
Learning Objective 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.
Source: College Board AP Course and Exam Description · 出典: College Board AP コースおよび試験説明書
Net torque produces angular acceleration, in direct analogy with $F=ma$:
$$\sum\tau = I\alpha.$$
So a larger net torque, or a smaller rotational inertia, gives a larger angular acceleration. Solve rotation problems just like translation problems, with $\tau\leftrightarrow F$, $I\leftrightarrow m$, and $\alpha\leftrightarrow a$.
Worked example. A net torque of $12\ \text{N m}$ acts on a wheel with rotational inertia $I=3.0\ \text{kg m}^2$. Its angular acceleration is $\alpha=\tau/I=12/3.0=4.0\ \text{rad/s}^2$ – the exact rotational twin of $a=F/m$.
5.6
Exam tips
Torque $\tau=Fr_\perp$ uses the perpendicular distance from the pivot; a force through the pivot gives zero torque.
For balance, set clockwise torque = anticlockwise torque; choosing the pivot at an unknown force removes it from the equation.
Use the rotational analogues: $\tau\leftrightarrow F$, $I\leftrightarrow m$, $\alpha\leftrightarrow a$, so $\sum\tau=I\alpha$ mirrors $\sum F=ma$.
Rotational inertia depends on how far the mass sits from the axis, not just its amount — a hoop resists spinning more than a disc of equal mass.
Convert angles to radians and link linear to angular with $v=r\omega$, $a_t=r\alpha$.
6
Energy and Momentum of Rotating Systems · 回転系エネルギーと運動量
Learning Objective 6.1.A: Describe the rotational kinetic energy of a rigid system in terms of the rotational inertia and angular velocity of that rigid system.
6.1.A.1 The rotational kinetic energy of an object or rigid system is related to the rotational inertia and angular velocity of the rigid system and is given by the equation $K = \frac{1}{2} I \omega^2$.
6.1.A.1.i The rotational inertia of an object about a fixed axis can be used to show that the rotational kinetic energy of that object is equivalent to its translational kinetic energy, which is its total kinetic energy.
6.1.A.1.ii The total kinetic energy of a rigid system is the sum of its rotational kinetic energy due to its rotation about its center of mass and the translational kinetic energy due to the linear motion of its center of mass.
6.1.A.2 A rigid system can have rotational kinetic energy while its center of mass is at rest due to the individual points within the rigid system having linear speed and, therefore, kinetic energy.
6.1.A.3 Rotational kinetic energy is a scalar quantity.
日本語
学習目標 6.1.A: 回転慣性と角速度を用いて、剛体系の回転運動エネルギーを説明する。
6.1.A.1 物体または剛体系の回転運動エネルギーは、剛体系の回転慣性と角速度に関連しており、以下の式で与えられる $K = \frac{1}{2} I \omega^2$。
Source: College Board AP Course and Exam Description · 出典: College Board AP コースおよび試験説明書
English
A spinning object has rotational kinetic energy 转动动能, the rotational twin of $\tfrac12 mv^2$:
$$K_{\text{rot}}=\tfrac{1}{2}I\omega^2.$$
An object that both moves and spins (like a rolling ball) has both translational and rotational kinetic energy, and its total is $K=\tfrac12 mv^2+\tfrac12 I\omega^2$.
Learning Objective 6.2.A: Describe the work done on a rigid system by a given torque or collection of torques.
6.2.A.1 A torque can transfer energy into or out of an object or rigid system if the torque is exerted over an angular displacement.
6.2.A.2 The amount of work done on a rigid system by a torque is related to the magnitude of that torque and the angular displacement through which the rigid system rotates during the interval in which that torque is exerted.
Equation:$W = \tau \Delta\theta$
6.2.A.3 Work done on a rigid system by a given torque can be found from the area under the curve of a graph of torque as a function of angular position.
Source: College Board AP Course and Exam Description · 出典: College Board AP コースおよび試験説明書
English
A torque acting through an angular displacement does work, changing rotational kinetic energy:
$$W=\tau\,\Delta\theta,\qquad P=\tau\omega.$$
This is the rotational form of $W=Fd$ and $P=Fv$, and it extends the work–energy theorem to rotation.
Worked example. A motor applies a steady torque of $8.0\ \text{N m}$ to a flywheel while it turns through $10\ \text{rad}$. The work done is $W=\tau\,\Delta\theta=8.0\times10=80\ \text{J}$, and if the flywheel started from rest this all becomes rotational kinetic energy.
Balance torques on a seesaw · こけびの上でのモーメントを釣り合わせる
Torque is force times perpendicular distance, $\tau = Fd$. The beam balances when the torques on each side are equal — move the forces and distances to find the balance point. · モーメント は力に垂直距離を掛けたもの、$\tau = Fd$。両側のモーメントが等しくなればビームは釣り合い、力を動かして釣り合い点を見つけろ。
6.3
Angular Momentum and Angular Impulse · 角運動量と角インパルス
Syllabus · シラバス
English
Learning Objective 6.3.A: Describe the angular momentum of an object or rigid system.
6.3.A.1 The magnitude of the angular momentum of a rigid system about a specific axis can be described with the equation $L = I\omega$.
6.3.A.2 The magnitude of the angular momentum of an object about a given point is $L = rmv \sin\theta$.
6.3.A.2.i The selection of the axis about which an object is considered to rotate influences the determination of the angular momentum of that object.
6.3.A.2.ii The measured angular momentum of an object traveling in a straight line depends on the distance between the reference point and the object, the mass of the object, the speed of the object, and the angle between the radial distance and the velocity of the object.
Learning Objective 6.3.B: Describe the angular impulse delivered to an object or rigid system by a torque.
6.3.B.1 Angular impulse is defined as the product of the torque exerted on an object or rigid system and the time interval during which the torque is exerted.
Equation:$\text{angular impulse} = \tau \Delta t$
6.3.B.2 Angular impulse has the same direction as the torque exerted on the object or system.
6.3.B.3 The angular impulse delivered to an object or rigid system by a torque can be found from the area under the curve of a graph of the torque as a function of time.
Learning Objective 6.3.C: Relate the change in angular momentum of an object or rigid system to the angular impulse given to that object or rigid system.
6.3.C.1 The magnitude of the change in angular momentum can be described by comparing the magnitudes of the final and initial angular momenta of the object or rigid system: $\Delta L = L - L_0$
6.3.C.2 A rotational form of the impulse–momentum theorem relates the angular impulse delivered to an object or rigid system and the change in angular momentum of that object or rigid system.
6.3.C.2.i The angular impulse exerted on an object or rigid system is equal to the change in angular momentum of that object or rigid system.
Equation:$\Delta L = \tau \Delta t$
6.3.C.2.ii The rotational form of the impulse–momentum theorem is a direct result of the rotational form of Newton's second law of motion for cases in which rotational inertia is constant: $\tau_{\text{net}} = \dfrac{\Delta L}{\Delta t} = I \dfrac{\Delta\omega}{\Delta t} = I\alpha$
6.3.C.3 The net torque exerted on an object is equal to the slope of the graph of the angular momentum of an object as a function of time.
6.3.C.4 The angular impulse delivered to an object is equal to the area under the curve of a graph of the net external torque exerted on an object as a function of time.
Boundary statement: While AP Physics 1 expects that students can mathematically manipulate the magnitude of angular momentum using one-dimensional vector conventions, the direction of angular momentum and angular impulse is beyond the scope of the course.
Learning Objective 6.4.A: Describe the behavior of a system using conservation of angular momentum.
6.4.A.1 The total angular momentum of a system about a rotational axis is the sum of the angular momenta of the system's constituent parts about that axis.
6.4.A.2 Any change to a system's angular momentum must be due to an interaction between the system and its surroundings.
6.4.A.2.i The angular impulse exerted by one object or system on a second object or system is equal and opposite to the angular impulse exerted by the second object or system on the first. This is a direct result of Newton's third law.
6.4.A.2.ii A system may be selected so that the total angular momentum of that system is constant.
6.4.A.2.iii The angular speed of a nonrigid system may change without the angular momentum of the system changing if the system changes shape by moving mass closer to or further from the rotational axis.
6.4.A.2.iv If the total angular momentum of a system changes, that change will be equivalent to the angular impulse exerted on the system.
Learning Objective 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 · 出典: College Board AP コースおよび試験説明書
English
If the net external torque on a system is zero, its total angular momentum is conserved 守恒:
$$I_1\omega_1=I_2\omega_2.$$
So if $I$ decreases, $\omega$ increases to keep $L$ constant – this is why a spinning skater speeds up when pulling their arms in. It applies to collisions and explosions of rotating systems too.
Worked example. A skater spins at $2.0\ \text{rev/s}$ with rotational inertia $I_1=4.0\ \text{kg m}^2$. She pulls her arms in, dropping her rotational inertia to $I_2=1.6\ \text{kg m}^2$. With no external torque, angular momentum is conserved:
Learning Objective 6.5.B: Describe the motion of a system that is rolling without slipping.
6.5.B.1 While rolling without slipping, the translational motion of a system's center of mass is related to the rotational motion of the system itself with the equations:
Equation:$\Delta x_{\text{cm}} = r\Delta\theta$
Equation:$v_{\text{cm}} = r\omega$
Equation:$a_{\text{cm}} = r\alpha$
6.5.B.2 For ideal cases, rolling without slipping implies that the frictional force does not dissipate any energy from the rolling system.
Learning Objective 6.5.C: Describe the motion of a system that is rolling while slipping.
6.5.C.1 When slipping, the motion of a system's center of mass and the system's rotational motion cannot be directly related.
6.5.C.2 When a rotating system is slipping relative to another surface, the point of application of the force of kinetic friction exerted on the system moves with respect to the surface, so the force of kinetic friction will dissipate energy from the system.
Boundary statement: Rolling friction is beyond the scope of AP Physics 1.
Boundary statement: The precise mathematical relationships between linear and angular quantities while a rigid body is rolling while slipping are beyond the scope of AP Physics 1 and 2, and students will not be expected to model those relationships quantitatively. However, students are expected to qualitatively explain the changes to linear and angular quantities while a rigid body is rolling while slipping.
Source: College Board AP Course and Exam Description · 出典: College Board AP コースおよび試験説明書
English
Rolling without slipping 纯滚动 links the translational and rotational motions: the contact point is momentarily at rest, so
$$v=r\omega \qquad\text{and}\qquad a=r\alpha.$$
A rolling object's energy splits between translation and rotation, so on an incline it accelerates more slowly than a frictionless sliding object – some energy goes into spin.
Worked example. For a solid disk ($I=\tfrac12 mR^2$) that rolls without slipping, what fraction of its kinetic energy is rotational? Using $v=R\omega$, the rotational part is $\tfrac12 I\omega^2=\tfrac12(\tfrac12 mR^2)\omega^2=\tfrac14 mv^2$, while the translational part is $\tfrac12 mv^2$. So the total is $\tfrac34 mv^2$ and the rotational share is $\tfrac{1/4}{3/4}=\tfrac13$. A hoop, with its mass farther out, stores half its energy in spin and rolls down a ramp even more slowly.
Learning Objective 6.6.A: Describe the motions of a system consisting of two objects interacting only via gravitational forces.
6.6.A.1 In a system consisting only of a massive central object and an orbiting satellite with mass that is negligible in comparison to the central object's mass, the motion of the central object itself is negligible.
6.6.A.2 The motion of satellites in orbits is constrained by conservation laws.
6.6.A.2.i In circular orbits, the system's total mechanical energy, the system's gravitational potential energy, and the satellite's angular momentum and kinetic energy are constant.
6.6.A.2.ii In elliptical orbits, the system's total mechanical energy and the satellite's angular momentum are constant, but the system's gravitational potential energy and the satellite's kinetic energy can each change.
6.6.A.2.iii The gravitational potential energy of a system consisting of a satellite and a massive central object is defined to be zero when the satellite is an infinite distance from the central object.
Equation:$U_g = -G\dfrac{m_1 m_2}{r}$
6.6.A.3 The escape velocity of a satellite is the satellite's velocity such that the mechanical energy of the satellite–central-object system is equal to zero.
6.6.A.3.i When the only force exerted on a satellite is gravity from a central object, a satellite that reaches escape velocity will move away from the central body until its speed reaches zero at an infinite distance from the central body.
6.6.A.3.ii The escape velocity of a satellite from a central body of mass $M$ can be derived using conservation of energy laws.
Source: College Board AP Course and Exam Description · 出典: College Board AP コースおよび試験説明書
English
Orbital motion (Kepler's 2nd law)
A satellite 卫星 in orbit is in free fall: gravity provides the exact centripetal force needed to curve its path into an orbit. Setting gravity equal to the centripetal requirement,
So a larger orbit means a slower speed. For a circular orbit, angular momentum and mechanical energy are both constant; for an elliptical orbit, angular momentum is conserved (no torque about the planet) while speed varies – fastest when closest.
Worked example. Find the speed of a satellite in a low orbit just above the Earth, radius $r=6.4\times10^{6}\ \text{m}$, with $GM=4.0\times10^{14}\ \text{m}^3/\text{s}^2$:
about $7.9\ \text{km/s}$ – and notice that the satellite's mass cancels, so all low orbits share this speed.
Escape velocity 逃逸速度 is the launch speed that just lets an object leave for good: its total mechanical energy is exactly zero, so it slows to zero speed only at infinite distance. Setting $\tfrac12 mv^2 - \dfrac{GMm}{r}=0$ (using the general gravitational PE $U_g=-GMm/r$) and solving for $v$,
$$v_{\text{esc}}=\sqrt{\frac{2GM}{r}}.$$
It is $\sqrt2$ times the circular-orbit speed at the same radius, and (like orbital speed) does not depend on the escaping object's mass.
Compare orbits at different radii · 異なる半径での軌道を比較する
A satellite is in free fall, its gravity supplying the centripetal force. A larger orbit means a slower speed and a longer period — Kepler's third law. · 衛星は 自由落下 にあり、重力が向心力を供給している。大きな軌道ほど速度は遅く、周期も長い — ケプラーの第三法則。
Conserve angular momentum$L=I\omega$ when no external torque acts: a smaller $I$ (arms pulled in) gives a larger $\omega$.
A rolling object splits its energy between $\tfrac12 mv^2$ and $\tfrac12 I\omega^2$, linked by $v=r\omega$ — so it accelerates down a ramp more slowly than a sliding one.
For a circular orbit set gravity equal to the centripetal requirement: $v=\sqrt{GM/r}$, so a larger orbit is slower and the satellite's mass cancels.
Pulling in raises the spin and the kinetic energy — the extra energy comes from the work done pulling inward; $L$ is unchanged.
Watch which rotational quantity is conserved: $L$ (no torque) versus energy (no friction) are different conditions.
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.
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.
7.1.A.2.iii The motion of a pendulum with a small angular displacement can be modeled as simple harmonic motion because the restoring torque is proportional to the angular displacement.
Source: College Board AP Course and Exam Description · 出典: College Board AP コースおよび試験説明書
English
Simple harmonic motion
Simple harmonic motion 简谐运动 (SHM) is a back-and-forth oscillation 振动 caused by a restoring force 回复力 that is proportional to the displacement from equilibrium and always points back toward it: $F=-kx$. A mass on a spring and (for small angles) a pendulum are the standard examples. Because the force grows with displacement, the motion is smooth and repeating, tracing a sine curve in time.
The test for SHM is exactly this: acceleration proportional to displacement and opposite in direction, $a=-\dfrac{k}{m}x$. A pendulum only obeys it for small swings, where $\sin\theta\approx\theta$; large swings are not quite SHM.
日本語
Simple harmonic motion
Simple harmonic motion 简谐运动 (SHM) is a back-and-forth oscillation 振动 caused by a restoring force 回复力 that is proportional to the displacement from equilibrium and always points back toward it: $F=-kx$. A mass on a spring and (for small angles) a pendulum are the standard examples. Because the force grows with displacement, the motion is smooth and repeating, tracing a sine curve in time.
In SHM the acceleration always points back towards equilibrium, opposite the displacement
The test for SHM is exactly this: acceleration proportional to displacement and opposite in direction, $a=-\dfrac{k}{m}x$. A pendulum only obeys it for small swings, where $\sin\theta\approx\theta$; large swings are not quite SHM.
A spring under load: Hooke's law F = −kx is the restoring force that produces SHM
Notice the amplitude never entered – a wide or narrow swing keeps the same time, which is what makes pendulums good clocks.
Tuning forks: a longer, heavier fork vibrates more slowly, giving a lower frequency
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Time a pendulum's swing · 振り子の周期を計測する
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. · 振り子の 周期 は長さや重力加速度に依存し、質量や(小さな)振幅には依存しない:$T=2\pi\sqrt{L/g}$。長くすると各振り子の時間も長くなる。
Learning Objective 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 Changing the amplitude of a system exhibiting SHM will not change the period of that system.
7.3.A.3 Properties of SHM can be determined and analyzed using graphical representations.
Source: College Board AP Course and Exam Description · 出典: College Board AP コースおよび試験説明書
English
The displacement varies sinusoidally: $x(t)=A\cos(\omega t)$ (or sine), where $A$ is the amplitude 振幅 (maximum displacement) and $\omega=2\pi f$ is the angular frequency 角频率. Reading the motion:
At the extremes ($x=\pm A$): displacement and restoring force are maximum, so acceleration is maximum, but velocity is zero.
At equilibrium 平衡 ($x=0$): force and acceleration are zero, but speed is maximum.
Velocity and acceleration are also sinusoidal, shifted in phase 相位 from the displacement – velocity leads displacement by a quarter cycle, and acceleration is exactly opposite to displacement.
Read the figure as a story: where $x$ is largest (the turning point), $v$ has just fallen to zero and $a$ is at its most negative, hauling the mass back; a quarter-cycle later the mass races through the middle at top speed with zero acceleration.
日本語
The displacement varies sinusoidally: $x(t)=A\cos(\omega t)$ (or sine), where $A$ is the amplitude 振幅 (maximum displacement) and $\omega=2\pi f$ is the angular frequency 角频率. Reading the motion:
Displacement varies sinusoidally with time in simple harmonic motion
At the extremes ($x=\pm A$): displacement and restoring force are maximum, so acceleration is maximum, but velocity is zero.
At equilibrium 平衡 ($x=0$): force and acceleration are zero, but speed is maximum.
Velocity and acceleration are also sinusoidal, shifted in phase 相位 from the displacement – velocity leads displacement by a quarter cycle, and acceleration is exactly opposite to displacement.
Displacement, velocity, and acceleration in SHM, each a quarter-cycle apart
Read the figure as a story: where $x$ is largest (the turning point), $v$ has just fallen to zero and $a$ is at its most negative, hauling the mass back; a quarter-cycle later the mass races through the middle at top speed with zero acceleration.
Learning Objective 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.
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:$E_{\text{total}} = \dfrac{1}{2}kA^2$
At the extremes it is all potential; at equilibrium it is all kinetic (maximum speed). Because $E\propto A^2$, doubling the amplitude quadruples the energy.
Kinetic and potential energy swap over a cycle while the total energy stays constant
Worked example. A $0.50\ \text{kg}$ mass on a spring of stiffness $k=200\ \text{N/m}$ oscillates with amplitude $A=0.10\ \text{m}$. Find its maximum speed. All the energy is kinetic at the equilibrium point, so $\tfrac12 kA^2=\tfrac12 mv_{\max}^2$:
The energy method is faster than tracking the sine functions when you only need the greatest speed.
Explore · 探索
Trade kinetic and potential energy in SHM · SHM で運動エネルギーとポテンシャルエネルギーを交換する
In simple harmonic motion, energy sloshes between kinetic (fastest at the centre) and potential (greatest at the extremes) while the total stays constant. · 単振動 において、運動エネルギー(中心で最大)とポテンシャルエネルギー(端で最大)の間でエネルギーが往復し、総量は一定である。
7.4
Exam tips
Test for SHM: the acceleration must be proportional to the displacement and directed back toward the middle ($a=-\tfrac{k}{m}x$).
The period does not depend on amplitude — use $T=2\pi\sqrt{m/k}$ (spring) or $T=2\pi\sqrt{L/g}$ (pendulum, small angles).
Speed is maximum at the middle (all kinetic) and zero at the extremes (all potential); acceleration is largest at the extremes.
Use energy ($\tfrac12 kA^2 = \tfrac12 kx^2 + \tfrac12 mv^2$) to find the maximum speed quickly: $v_{\max}=A\sqrt{k/m}$.
Total energy $\propto A^2$, so doubling the amplitude quadruples the energy.
Learning Objective 8.1.A: Describe the properties of a fluid.
8.1.A.1 Distinguishing properties of solids, liquids, and gases stem from the varying interactions between atoms and molecules.
8.1.A.2 A fluid is a substance that has no fixed shape.
8.1.A.3 Fluids can be characterized by their density. Density is defined as a ratio of mass to volume.
Equation:$\rho = \dfrac{m}{V}$
8.1.A.4 An ideal fluid is incompressible and has no viscosity.
日本語
学習目標 8.1.A: 流体の性質を記述する。
8.1.A.1 固体、液体、気体の区別される性質は、原子や分子間の相互作用の違いに由来する。
8.1.A.2 流体とは、固定された形状を持たない物質である。
8.1.A.3 流体は密度によって特徴づけられる。密度とは質量と体積の比として定義される。
式:$\rho = \dfrac{m}{V}$
8.1.A.4 理想流体は圧縮不可能であり、粘性を持たない。
Source: College Board AP Course and Exam Description · 出典: College Board AP コースおよび試験説明書
English
The differences between solids, liquids, and gases come from how strongly their particles interact. A fluid 流体 (liquid or gas) has no fixed shape – it flows because its particles move past one another. Density 密度 is mass per unit volume:
$$\rho=\frac{m}{V}.$$
It depends on the substance and its state. An object sinks or floats depending on how its density compares with the surrounding fluid's – less dense floats, more dense sinks. An ideal fluid 理想流体 is incompressible 不可压缩 (constant density, whatever the pressure) and has no viscosity 黏度 (internal friction) – the model AP uses throughout.
日本語
The differences between solids, liquids, and gases come from how strongly their particles interact. A fluid 流体 (liquid or gas) has no fixed shape – it flows because its particles move past one another. Density 密度 is mass per unit volume:
$$\rho=\frac{m}{V}.$$
It depends on the substance and its state. An object sinks or floats depending on how its density compares with the surrounding fluid's – less dense floats, more dense sinks. An ideal fluid 理想流体 is incompressible 不可压缩 (constant density, whatever the pressure) and has no viscosity 黏度 (internal friction) – the model AP uses throughout.
Explore · 探索
Float or sink by density · 密度によって浮上か沈下かを決める
An object floats if it is less dense than the fluid. Change the density and watch it ride higher or sink, displacing its own weight of fluid. · 物体が流体より 密度が小さい 場合、浮上する。密度を変えると、より高く乗ったり沈んだりして、自身の重量分の流体を排気する様子がわかる。
Learning Objective 8.2.A: Describe the pressure exerted on a surface by a given force.
8.2.A.1 Pressure is defined as the magnitude of the perpendicular force component exerted per unit area over a given surface area, as described by the equation
Equation:$P = \dfrac{F_\perp}{A}$
8.2.A.2 Pressure is a scalar quantity.
8.2.A.3 The volume and density of a given amount of an incompressible fluid is constant regardless of the pressure exerted on that fluid.
Learning Objective 8.2.B: Describe the pressure exerted by a fluid.
8.2.B.1 The pressure exerted by a fluid is the result of the entirety of the interactions between the fluid's constituent particles and the surface with which those particles interact.
8.2.B.2 The absolute pressure of a fluid at a given point is equal to the sum of a reference pressure $P_0$, such as the atmospheric pressure $P_{\text{atm}}$, and the gauge pressure $P_{\text{gauge}}$.
Equation:$P = P_0 + \rho g h$
8.2.B.3 The gauge pressure of a vertical column of fluid is described by the equation
Source: College Board AP Course and Exam Description · 出典: College Board AP コースおよび試験説明書
English
Pressure 压强 is the perpendicular force per unit area, a scalar 标量 measured in pascals (Pa):
$$P=\frac{F_\perp}{A}.$$
In a fluid at rest, pressure increases with depth 深度 because of the weight of fluid above:
$$P=P_0+\rho g h,$$
where $P_0$ is the pressure at the surface and $h$ is the depth. Pressure acts equally in all directions at a point and pushes perpendicular to any surface.
Distinguish the two pressures AP asks about: the gauge pressure 表压 is the extra pressure the fluid column adds, $P_{\text{gauge}}=\rho g h$, while the absolute pressure 绝对压强 is the total, $P=P_0+P_{\text{gauge}}$ (with $P_0$ usually atmospheric). A tyre gauge reading "$200\ \text{kPa}$" is gauge pressure; the air inside is really at about $300\ \text{kPa}$ absolute.
Worked example. Find the total pressure on a diver $10\ \text{m}$ below the surface of water ($\rho=1000\ \text{kg/m}^3$, surface pressure $P_0=1.0\times10^{5}\ \text{Pa}$):
$$P=P_0+\rho g h=1.0\times10^{5}+1000\times9.8\times10=1.98\times10^{5}\ \text{Pa}.$$
Every $10\ \text{m}$ of water adds roughly one extra atmosphere of pressure. Notice the pressure does not depend on the shape or width of the container, only on the depth.
日本語
Pressure 压强 is the perpendicular force per unit area, a scalar 标量 measured in pascals (Pa):
$$P=\frac{F_\perp}{A}.$$
In a fluid at rest, pressure increases with depth 深度 because of the weight of fluid above:
$$P=P_0+\rho g h,$$
where $P_0$ is the pressure at the surface and $h$ is the depth. Pressure acts equally in all directions at a point and pushes perpendicular to any surface.
Hydraulic jacks: pressure applied to a small piston is transmitted undiminished through the fluid to a large one, so a small force lifts a car
Distinguish the two pressures AP asks about: the gauge pressure 表压 is the extra pressure the fluid column adds, $P_{\text{gauge}}=\rho g h$, while the absolute pressure 绝对压强 is the total, $P=P_0+P_{\text{gauge}}$ (with $P_0$ usually atmospheric). A tyre gauge reading "$200\ \text{kPa}$" is gauge pressure; the air inside is really at about $300\ \text{kPa}$ absolute.
The weight of a liquid column sets the extra pressure at a depth
Worked example. Find the total pressure on a diver $10\ \text{m}$ below the surface of water ($\rho=1000\ \text{kg/m}^3$, surface pressure $P_0=1.0\times10^{5}\ \text{Pa}$):
$$P=P_0+\rho g h=1.0\times10^{5}+1000\times9.8\times10=1.98\times10^{5}\ \text{Pa}.$$
Every $10\ \text{m}$ of water adds roughly one extra atmosphere of pressure. Notice the pressure does not depend on the shape or width of the container, only on the depth.
An aneroid barometer measures air pressure with a sealed metal box that flexes as the pressure outside changes
Learning Objective 8.3.A: Describe the conditions under which a fluid's velocity changes.
8.3.A.1 Newton's laws can be used to describe the motion of particles within a fluid.
8.3.A.2 The macroscopic behavior of a fluid is a result of the internal interactions between the fluid's constituent particles and external forces exerted on the fluid.
Learning Objective 8.3.B: Describe the buoyant force exerted on an object interacting with a fluid.
8.3.B.1 The buoyant force is a net upward force exerted on an object by a fluid.
8.3.B.2 The buoyant force exerted on an object by a fluid is a result of the collective forces exerted on the object by the particles making up the fluid.
8.3.B.3 The magnitude of the buoyant force exerted on an object by a fluid is equivalent to the weight of the fluid displaced by the object.
Combine this with Newton's laws: the object floats when buoyancy balances weight, sinks when weight wins, and rises when buoyancy wins. A floating object displaces exactly its own weight of fluid.
Worked example. A block of density $600\ \text{kg/m}^3$ floats in water ($1000\ \text{kg/m}^3$). What fraction is under the surface? For floating, the buoyant force equals the weight, so $\rho_{\text{fluid}}\,g\,V_{\text{sub}}=\rho_{\text{object}}\,g\,V$:
Combine this with Newton's laws: the object floats when buoyancy balances weight, sinks when weight wins, and rises when buoyancy wins. A floating object displaces exactly its own weight of fluid.
Worked example. A block of density $600\ \text{kg/m}^3$ floats in water ($1000\ \text{kg/m}^3$). What fraction is under the surface? For floating, the buoyant force equals the weight, so $\rho_{\text{fluid}}\,g\,V_{\text{sub}}=\rho_{\text{object}}\,g\,V$:
Learning Objective 8.4.A: Describe the flow of an incompressible fluid through a cross-sectional area by using mass conservation.
8.4.A.1 A difference in pressure between two locations causes a fluid to flow.
8.4.A.1.i The rate at which matter enters a fluid-filled tube open at both ends must equal the rate at which matter exits the tube.
8.4.A.1.ii The rate at which matter flows into a location is proportional to the cross-sectional area of the flow and the speed at which the fluid flows.
Derived equation:$\dfrac{V}{t} = A v$
8.4.A.2 The continuity equation for fluid flow describes conservation of mass flow rate in incompressible fluids.
Source: College Board AP Course and Exam Description · 出典: College Board AP コースおよび試験説明書
English
For an ideal fluid flowing steadily, two conservation ideas apply:
Continuity 连续性 (conservation of mass): the volume flow rate is constant, so $A_1 v_1 = A_2 v_2$. A narrower pipe forces faster flow.
Bernoulli's equation 伯努利方程 (conservation of energy per volume): along a streamline,
$$P+\tfrac{1}{2}\rho v^2+\rho g y = \text{constant}.$$
Together they explain why fluid speeds up and its pressure drops where a pipe narrows or where flow is fastest.
Worked example. Water flows at $2.0\ \text{m/s}$ through a pipe of cross-section $0.010\ \text{m}^2$, then enters a narrower section of $0.0040\ \text{m}^2$. By continuity the speed there is
By Bernoulli's equation this faster stream is at lower pressure – the effect that lifts an aeroplane wing and pulls two passing ships together.
Exam skill. Choose the right law by what changes. If the pipe changes width, start with continuity ($A_1v_1=A_2v_2$) to get the speeds; if you then need a pressure, feed those speeds into Bernoulli. Watch the height term $\rho g y$ only when the pipe also changes level.
日本語
For an ideal fluid flowing steadily, two conservation ideas apply:
Upthrust arises because the pressure on the bottom of an object exceeds that on the top
Continuity 连续性 (conservation of mass): the volume flow rate is constant, so $A_1 v_1 = A_2 v_2$. A narrower pipe forces faster flow.
Bernoulli's equation 伯努利方程 (conservation of energy per volume): along a streamline,
$$P+\tfrac{1}{2}\rho v^2+\rho g y = \text{constant}.$$
Together they explain why fluid speeds up and its pressure drops where a pipe narrows or where flow is fastest.
Where the pipe narrows the fluid speeds up (continuity) and its pressure drops (Bernoulli)
Worked example. Water flows at $2.0\ \text{m/s}$ through a pipe of cross-section $0.010\ \text{m}^2$, then enters a narrower section of $0.0040\ \text{m}^2$. By continuity the speed there is
By Bernoulli's equation this faster stream is at lower pressure – the effect that lifts an aeroplane wing and pulls two passing ships together.
Exam skill. Choose the right law by what changes. If the pipe changes width, start with continuity ($A_1v_1=A_2v_2$) to get the speeds; if you then need a pressure, feed those speeds into Bernoulli. Watch the height term $\rho g y$ only when the pipe also changes level.
Pressure with depth is $P=P_0+\rho g h$ — it depends on depth only, not the container's shape or width.
Buoyant force = weight of fluid displaced ($\rho_{\text{fluid}}\,gV_{\text{disp}}$); a floating object displaces its own weight, so the fraction submerged is $\rho_{\text{object}}/\rho_{\text{fluid}}$.
Compare densities to predict floating vs sinking; a floating object is in equilibrium (buoyancy = weight), not weightless.
Use continuity$A_1v_1=A_2v_2$: a narrower pipe means faster flow.
Bernoulli: where a fluid flows faster its pressure is lower (wing lift, spray).
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