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AP Physics C: Mechanics · ⁨AP 물리학 C: 역학⁩

Tips · ⁨팁⁩

AP Physics C: Mechanics is the calculus-based version: kinematics, forces, work, energy and power, momentum, torque and rotation, and oscillations. It covers less ground than Physics 1 but goes much deeper, and calculus is not decoration — it is how the results are derived and how the exam expects them to be used.

In practice: integrate to get a total from a varying quantity, and differentiate from position to velocity to acceleration, comfortably and without being prompted.

Rotational dynamics is where time goes, because moments of inertia must be set up as integrals rather than recalled.

The notes take each unit in CED order with the calculus set out in full. Released past papers and scoring guidelines sit in the library. Because so many marks ride on the setup rather than the arithmetic, the worked examples deliberately show the integral being built before it is evaluated.

  • 1

    Kinematics

    Watch lesson · ⁨수업 보기⁩
    1.1

    Scalars and Vectors · ⁨스칼라와 벡터⁩

    Syllabus
    English

    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.4 Vectors can be expressed in unit vector notation or as a magnitude and a direction.
      • 1.1.A.4.i Unit vector notation can be used to represent vectors as the sum of their constituent components in the $x$-, $y$-, and $z$-directions, denoted by $\hat{i}$, $\hat{j}$, and $\hat{k}$, respectively.
        • Equation: $\vec{r} = \left( A\hat{i} + B\hat{j} + C\hat{k} \right)$
      • 1.1.A.4.ii The position vector of a point is given by $\vec{r}$, and the unit vector in the direction of the position vector is denoted $\hat{r}$.
      • 1.1.A.4.iii A resultant vector is the vector sum of the addend vectors' components.
        • Equation: $\vec{C} = \vec{A} + \vec{B}$
        • Equation: $\vec{C} = \left( A_x + B_x \right)\hat{i} + \left( A_y + B_y \right)\hat{j}$
    • 1.1.A.5 In a given one-dimensional coordinate system, opposite directions are denoted by opposite signs.
    한국어

    학습 목표 1.1.A: 크기와 방향을 사용하여 스칼라 또는 벡터 양을 적절히 기술하시오.

    • 1.1.A.1 스칼라는 크기만으로 서술되는 양이며, 벡터는 크기와 방향 모두로 서술되는 양입니다.
    • 1.1.A.2 벡터는 적절한 방향과 크기에 비례하는 길이를 가진 화살표로 시각적으로 모델링할 수 있습니다.
    • 1.1.A.3 거리와 속도는 스칼라 양의 예시이며, 위치, 변위, 속도, 가속도는 벡터 양의 예시입니다.
    • 1.1.A.4 벡터는 단위 벡터 표기법이나 크기와 방향으로 표현할 수 있습니다.
      • 1.1.A.4.i 단위 벡터 표기법을 사용하여 벡터를 constituent components인 ⟨$x$⟩, ⟨$y$⟩, ⟨$z$⟩ 방향의 합으로 나타낼 수 있으며, 이를 각각 ⟨$\hat{i}$⟩, ⟨$\hat{j}$⟩, ⟨$\hat{k}$⟩로 표시한다.
        • 식: $\vec{r} = \left( A\hat{i} + B\hat{j} + C\hat{k} \right)$
      • 1.1.A.4.ii 점의 위치 벡터는 $\vec{r}$로 주어지며, 위치 벡터 방향의 단위 벡터는 $\hat{r}$로 표시됩니다.
      • 1.1.A.4.iii 합성 벡터는 덧셈 벡터들의 성분의 합입니다.
        • 식: $\vec{C} = \vec{A} + \vec{B}$
        • 식: $\vec{C} = \left( A_x + B_x \right)\hat{i} + \left( A_y + B_y \right)\hat{j}$
    • 1.1.A.5 주어진 1차원 좌표계에서 반대 방향은 반대 부호로 표시됩니다.

    Source: College Board AP Course and Exam Description · ⁨출처: College Board AP Course and Exam Description⁩

    English

    Kinematics 运动学 describes motion. Two kinds of quantity:

    • A scalar 标量 has only size (magnitude 大小): distance, speed, mass, time.
    • A vector 矢量 has magnitude and direction: displacement, velocity, acceleration, force.

    In this calculus-based course you resolve vectors into components 分量 and add them component by component; a vector's magnitude is $\sqrt{v_x^2+v_y^2+\cdots}$. Physics C also writes vectors with unit vectors 单位矢量: $\vec{v}=v_x\hat{i}+v_y\hat{j}$, where $\hat{i}$ and $\hat{j}$ are directions of length one along $x$ and $y$ – handy because calculus can then act on each component separately.

    한국어

    운동학은 운동을 기술합니다. 두 종류의 물리량:

    • 스칼라는 크기(절댓값)만 가집니다: 거리, 속도, 질량, 시간.
    • 벡터는 크기와 방향을 모두 가집니다: 변위, 속도, 가속도, 힘.

    이 미적분 기반 과정에서는 벡터를 성분으로 분해하여 성분별로 더합니다; 벡터의 크기는 $\sqrt{v_x^2+v_y^2+\cdots}$입니다. Physics C는 또한 단위 벡터를 사용하여 벡터를 표기합니다: $\vec{v}=v_x\hat{i}+v_y\hat{j}$, 여기서 $\hat{i}$와 $\hat{j}$는 $x$와 $y$ 방향으로 길이가 1인 방향입니다 – 이것이 유용한 이유는 미적분이 각 성분에 개별적으로 작용할 수 있기 때문입니다.

    벡터를 x 및 y 성분으로 분해함
    벡터를 x 및 y 성분으로 분해함
    자동차 속도계: 속도는 속도의 크기—스칼라 운동 속도입니다
    자동차 속도계: 속도는 속도의 크기—스칼라 운동 속도입니다
    Explore · ⁨탐색하기⁩

    Explore the dot product · ⁨내적(dot product) 탐구하기⁩

    Drag the components of $\vec{A}$ and $\vec{B}$. The dot product $\vec{A}\cdot\vec{B}=A_xB_x+A_yB_y$ reaches zero exactly when the two vectors are perpendicular — the geometry behind $\cos\theta$. · ⁨$\vec{A}$과 $\vec{B}$의 성분을 드래그하세요. 두 터가 **수직(perpendicular)**일 때 내적(dot product) $\vec{A}\cdot\vec{B}=A_xB_x+A_yB_y$이 정확히 **영(zero)**이 됩니다. 이는 $\cos\theta$의 기하학적 배경입니다.⁩

    Vocabulary · ⁨어휘⁩ Train · ⁨연습하기⁩
    English 한국어
    vector/ˈvektə/ 벡터
    components/kəmˈpəʊnənts/ 구성 요소
    unit vectors/ˈjuːnɪt ˈvektəz/ 단위 벡터
    1.2

    Displacement, Velocity, and Acceleration · ⁨변위, 속도, 그리고 가속도⁩

    Syllabus
    English

    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.
      • Equation: $\vec{v}_{\text{avg}} = \dfrac{\Delta \vec{x}}{\Delta t}$
    • 1.2.B.3 Average acceleration is the change in velocity divided by the interval of time in which that change in velocity occurs.
      • Equation: $\vec{a}_{\text{avg}} = \dfrac{\Delta \vec{v}}{\Delta t}$
    • 1.2.B.4 An object is accelerating if either the magnitude and/or direction of the object's velocity are changing.
    • 1.2.B.5 Calculating average velocity or average acceleration over a very small time interval yields a value that is very close to the instantaneous velocity or instantaneous acceleration.

    Learning Objective 1.2.C: Describe the instantaneous position, velocity, and acceleration of an object as a function of time.

    • 1.2.C.1 As the time interval used to calculate the average value of a quantity approaches zero, the average value of that quantity approaches the value of the quantity at that instant, called the instantaneous value.
      • 1.2.C.1.i Instantaneous velocity is the rate of change of the object's position, which is equal to the derivative of position with respect to time.
        • Equation: $\vec{v} = \dfrac{d\vec{r}}{dt}$
        • Equation: $v_x = \dfrac{dx}{dt}$
      • 1.2.C.1.ii Instantaneous acceleration is the rate of change of the object's velocity, which is equal to the derivative of velocity with respect to time.
        • Equation: $\vec{a} = \dfrac{d\vec{v}}{dt}$
        • Equation: $a_x = \dfrac{dv_x}{dt}$
    • 1.2.C.2 Time-dependent functions and instantaneous values of position, velocity, and acceleration can be determined using differentiation and integration.
    한국어

    학습 목표 1.2.A: 물체의 위치 변화를 기술하시오.

    • 1.2.A.1 물체 모델을 사용할 때는 크기, 형태, 내부 구성을 무시합니다. 물체는 질량과 전하와 같은 광범위한 특성을 가진 단일 점으로 취급될 수 있습니다.
    • 1.2.A.2 변위는 물체의 위치 변화입니다.
      • 식: $\Delta x = x - x_0$

    학습 목표 1.2.B: 물체의 평균 속도와 평균 가속도를 기술하시오.

    • 1.2.B.1 속도와 가속도의 평균값은 시간 구간 동안 물체의 초기 상태와 최종 상태를 고려하여 계산됩니다.
    • 1.2.B.2 평균 속도는 해당 변위가 발생하는 시간 구간으로 나눈 물체의 변위입니다.
      • 식: $\vec{v}_{\text{avg}} = \dfrac{\Delta \vec{x}}{\Delta t}$
    • 1.2.B.3 평균 가속도는 속도 변화량을 해당 속도 변화가 발생하는 시간 구간으로 나눈 값입니다.
      • 식: $\vec{a}_{\text{avg}} = \dfrac{\Delta \vec{v}}{\Delta t}$
    • 1.2.B.4 물체의 속도의 크기와/또는 방향이 변화하고 있다면 해당 물체는 가속하고 있습니다.
    • 1.2.B.5 매우 짧은 시간 구간 동안 평균 속도 또는 평균 가속도를 계산하면 instantaneous velocity 또는 instantaneous acceleration에 매우 근접한 값을 얻게 됩니다.

    학습 목표 1.2.C: 시간의 함수로서의 물체의 순간 위치, 속도 및 가속도를 설명하십시오.

    • 1.2.C.1 양의 평균값을 계산하는 데 사용되는 시간 간격이 0에 수렴할 때, 해당 양의 평균값은 그 순간의 양의 값인 순간값에 수렴한다.
      • 1.2.C.1.i 순간속도는 물체의 위치 변화율이며, 이는 시간에 대한 위치의 미분값과 같다.
        • 식: $\vec{v} = \dfrac{d\vec{r}}{dt}$
        • 식: $v_x = \dfrac{dx}{dt}$
      • 1.2.C.1.ii 순간가속도는 물체의 속도 변화율이며, 이는 시간에 대한 속도의 미분값과 같다.
        • 식: $\vec{a} = \dfrac{d\vec{v}}{dt}$
        • 식: $a_x = \dfrac{dv_x}{dt}$
    • 1.2.C.2 시간 의존 함수와 위치, 속도, 가속도의 순간값은 미분 및 적분을 통해 구할 수 있다.

    Source: College Board AP Course and Exam Description · ⁨출처: College Board AP Course and Exam Description⁩

    English
    Projectile motion is two independent motions

    Because motion can change continuously, define the rates as derivatives 导数:

    $$v=\frac{dx}{dt},\qquad a=\frac{dv}{dt}=\frac{d^2x}{dt^2}.$$
    Reversing this, integrate to recover velocity and position:
    $$v(t)=v_0+\int_0^t a\,dt,\qquad x(t)=x_0+\int_0^t v\,dt.$$
    Distinguish the average from the instantaneous: average velocity 平均速度 is $\Delta x/\Delta t$ over an interval, while instantaneous velocity 瞬时速度 is the derivative at one moment – the slope of the tangent line, and what a speedometer shows.

    For constant acceleration these give the familiar kinematic equations ($v=v_0+at$, $x=x_0+v_0t+\tfrac12at^2$, $v^2=v_0^2+2a\,\Delta x$). The most important constant-$a$ case is free fall 自由落体: near Earth's surface every object, heavy or light, accelerates downward at $g=9.8\ \text{m/s}^2$ once air resistance is negligible. When $a$ or $v$ varies with time, use the integrals directly.

    Worked example. A particle moves with $x(t)=2t^3-3t^2$ (metres). Differentiate for velocity and acceleration:

    $$v=\frac{dx}{dt}=6t^2-6t,\qquad a=\frac{dv}{dt}=12t-6.$$
    At $t=2\ \text{s}$, $v=24-12=12\ \text{m/s}$ and $a=24-6=18\ \text{m/s}^2$. This calculus link – not just the constant-$a$ formulas – is what distinguishes Physics C.

    Worked example. A particle starts from rest at the origin with $a(t)=6t$. Integrate: $v=\int 6t\,dt=3t^2$ and $x=\int 3t^2\,dt=t^3$. At $t=2\ \text{s}$, $v=12\ \text{m/s}$ and $x=8\ \text{m}$.

    한국어
    포물선 운동은 두 개의 독립된 운동입니다

    운동은 continuously(계속적으로) 변할 수 있으므로, 이를 미분으로 정의합니다:

    $$v=\frac{dx}{dt},\qquad a=\frac{dv}{dt}=\frac{d^2x}{dt^2}.$$
    이를 역으로 적용하면 적분을 통해 속도와 위치를 복원할 수 있습니다:
    $$v(t)=v_0+\int_0^t a\,dt,\qquad x(t)=x_0+\int_0^t v\,dt.$$
    평균과 즉시를 구분하십시오: 평균 속도는 구간에 대해 $\Delta x/\Delta t$이며, 즉시 속도는 한 순간의 미분입니다 – 접선의 기울기이며, 시속계가 나타내는 값입니다.

    일정한 가속도에 대해 이들은 친숙한 운동학 방정식을 제공합니다 ($v=v_0+at$, $x=x_0+v_0t+\tfrac12at^2$, $v^2=v_0^2+2a\,\Delta x$). 가장 중요한 상수-$a$ 경우는 자유 낙하: 지구 표면 근처에서 모든 물체는 공기 저항이 무시할 수 있을 때 무거우나 가벼우나 아래쪽 방향으로 $g=9.8\ \text{m/s}^2$로 가속합니다. $a$ 또는 $v$가 시간에 따라 변할 경우, 적분을 직접 사용하십시오.

    해설 예제. 입자가 $x(t)=2t^3-3t^2$(미터)로 움직입니다. 속도와 가속도로 미분하기:

    $$v=\frac{dx}{dt}=6t^2-6t,\qquad a=\frac{dv}{dt}=12t-6.$$
    $t=2\ \text{s}$에서, $v=24-12=12\ \text{m/s}$ 및 $a=24-6=18\ \text{m/s}^2$입니다. 이 미적분 연계성 – 단순히 상수-$a$ 공식뿐만 아니라 – 이것이 Physics C를 구분하는 특징입니다.

    해설 예제. 입자가 원점에서 정지 상태에서 시작하여 $a(t)=6t$로 움직입니다. 적분하기: $v=\int 6t\,dt=3t^2$ 및 $x=\int 3t^2\,dt=t^3$. $t=2\ \text{s}$에서, $v=12\ \text{m/s}$ 및 $x=8\ \text{m}$입니다.

    속도-시간 그래프에서 기울기는 가속도이며, 면적은 변위이다
    속도-시간 그래프에서 기울기는 가속도이고 면적은 변위입니다
    고속 열차: 운동학은 시간에 따른 변위, 속도, 가속도를 설명합니다
    고속 열차: 운동학은 시간에 따른 변위, 속도, 가속도를 설명합니다
    Vocabulary · ⁨어휘⁩ Train · ⁨연습하기⁩
    English 한국어
    Kinematics/ˌkɪnɪˈmætɪks/ 운동학
    scalar/ˈskeɪlə/ 스칼라
    magnitude/ˈmæɡnɪtjuːd/ 크기
    1.3

    Representing Motion · ⁨운동의 표현⁩

    Syllabus
    English

    Learning Objective 1.3.A: Describe the position, velocity, and acceleration of an object using representations of that object's motion.

    • 1.3.A.1 Motion can be represented by motion diagrams, figures, graphs, equations, and narrative descriptions.
    • 1.3.A.2 For constant acceleration, three kinematic equations can be used to describe instantaneous linear motion in one dimension:
      • Equation: $v_x = v_{x0} + a_x t$
      • Equation: $x = x_0 + v_{x0} t + \dfrac{1}{2} a_x t^2$
      • Equation: $v_x^2 = v_{x0}^2 + 2a_x \left( x - x_0 \right)$
      • Note: The equations above are written to indicate motion in the $x$-direction, but these equations can be used in any single dimension as appropriate.
    • 1.3.A.3 Near the surface of Earth, the vertical acceleration caused by the force of gravity is downward, constant, and has a measured value approximately equal to
      • Equation: $a_g = g \approx 10 \ \text{m/s}^2$.
    • 1.3.A.4 Graphs of position, velocity, and acceleration as functions of time can be used to find the relationships between those quantities.
      • 1.3.A.4.i An object's instantaneous velocity is the rate of change of the object's position, which is equal to the slope of a line tangent to a point on a graph of the object's position as a function of time.
        • Equation: $v_x = \dfrac{dx}{dt}$
      • 1.3.A.4.ii An object's instantaneous acceleration is the rate of change of the object's velocity, which is equal to the slope of a line tangent to a point on a graph of the object's velocity as a function of time.
        • Equation: $a_x = \dfrac{dv_x}{dt}$
      • 1.3.A.4.iii The displacement of an object during a time interval is equal to the area under the curve of a graph of the object's velocity as a function of time (i.e., the area bounded by the function and the horizontal axis for the appropriate interval).
        • Equation: $\Delta x = \displaystyle\int_{t_1}^{t_2} v_x(t) \ dt$
      • 1.3.A.4.iv The change in velocity of an object during a time interval is equal to the area under the curve of a graph of the acceleration of the object as a function of time.
        • Equation: $\Delta v_x = \displaystyle\int_{t_1}^{t_2} a_x(t) \ dt$

    Boundary statement: AP Physics C: Mechanics and AP Physics C: Electricity and Magnetism expects that for all situations in which a numerical quantity is required for g, the value $g \approx 10 \ \text{m/s}^2$ will be used. However, students will not be penalized for correctly using the more precise commonly accepted values of $g = 9.81 \ \text{m/s}^2$ or $g = 9.8 \ \text{m/s}^2$.

    한국어

    학습 목표 1.3.A: 운동 도식을 사용하여 물체의 위치, 속도, 가속도를 기술하시오.

    • 1.3.A.1 운동은 운동 도식, 그림, 그래프, 식, 서술적 설명으로 표현할 수 있습니다.
    • 1.3.A.2 등가속도인 경우 1차원의 순간 선형 운동을 기술하기 위해 세 가지 운동학 식을 사용할 수 있습니다:
      • 식: $v_x = v_{x0} + a_x t$
      • 식: $x = x_0 + v_{x0} t + \dfrac{1}{2} a_x t^2$
      • 식: $v_x^2 = v_{x0}^2 + 2a_x \left( x - x_0 \right)$
      • 참고: 위 식들은 $x$ 방향의 운동을 나타내도록 작성되었으나, 이러한 식들은 적절한 단일 차원에서의 운동에도 적용될 수 있다.
    • 1.3.A.3 지구 표면 근처에서 중력 힘에 의해 발생하는 수직 가속도는 아래쪽 방향이며 일정하고, 그 측정값은 대략적으로
      • 식: $a_g = g \approx 10 \ \text{m/s}^2$.
    • 1.3.A.4 시간의 함수로서 위치, 속도, 가속도의 그래프를 사용하여 이러한 양들 사이의 관계를 찾을 수 있습니다.
      • 1.3.A.4.i 물체의 순간 속도는 물체의 위치의 변화율로, 시간의 함수로서의 물체 위치 그래프 상 한 점에 접한 직선의 기울기와 같습니다.
        • 식: $v_x = \dfrac{dx}{dt}$
      • 1.3.A.4.ii 물체의 순간 가속도는 물체의 속도의 변화율로, 시간의 함수로서의 물체 속도 그래프 상 한 점에 접한 직선의 기울기와 같습니다.
        • 식: $a_x = \dfrac{dv_x}{dt}$
      • 1.3.A.4.iii 시간 구간 동안 물체의 변위는 시간의 함수로서의 물체 속도 그래프 하단 면적(즉, 해당 구간에 대한 함수와 수평축으로 둘러싸인 면적)과 같습니다.
        • 식: $\Delta x = \displaystyle\int_{t_1}^{t_2} v_x(t) \ dt$
      • 1.3.A.4.iv 시간 구간 동안 물체의 속도 변화량은 시간의 함수로서의 물체 가속도 그래프 하단 면적과 같습니다.
        • 식: $\Delta v_x = \displaystyle\int_{t_1}^{t_2} a_x(t) \ dt$

    경계 설명: AP Physics C: Mechanics 및 AP Physics C: Electricity and Magnetism에서는 g에 필요한 수치적 양이 있는 모든 상황에서 $g \approx 10 \ \text{m/s}^2$ 값을 사용할 것을 기대한다. 그러나 학생들은 더 정밀하게 일반적으로 수용되는 $g = 9.81 \ \text{m/s}^2$ 또는 $g = 9.8 \ \text{m/s}^2$ 값을 올바르게 사용하는 경우 감점되지 않는다.

    Source: College Board AP Course and Exam Description · ⁨출처: College Board AP Course and Exam Description⁩

    English

    Move fluently between description, graph, table, and equation:

    • On a position–time graph, the slope ($dx/dt$) is velocity.
    • On a velocity–time graph, the slope is acceleration and the area ($\int v\,dt$) is displacement.
    • On an acceleration–time graph, the area ($\int a\,dt$) is the change in velocity.

    Slopes are derivatives; areas are integrals – the two are inverse operations here.

    한국어

    기술, 그래프, 표, 방정식 간에 유창하게 전환하십시오:

    변위-시간 그래프: 특정 시점에서의 접선의 기울기가 속도이다
    위치-시간 그래프: 어떤 순간의 기울기는 속도입니다
    • 위치–시간 그래프에서 기울기($dx/dt$)는 속도입니다.
    • 속도–시간 그래프에서 기울기는 가속도이며 면적($\int v\,dt$)은 변위입니다.
    • 가속도–시간 그래프에서 면적($\int a\,dt$)은 속도 변화량입니다.

    기울기는 미분이고 면적은 적분입니다 – 여기서는 두 연산이 서로 역연산입니다.

    위치, 속도, 가속도는 미분에 의해 연결됨
    위치, 속도, 가속도는 미분법으로 연결됨
    Explore · ⁨탐색하기⁩

    Explore the velocity–time graph · ⁨속도-시간 그래프 탐구하기⁩

    Change the start velocity and acceleration. The slope of a $v$–$t$ line is the acceleration $a=\tfrac{dv}{dt}$, and the area underneath is the displacement $\Delta x=\int v\,dt$ — slope and area are the graphical faces of the derivative and the integral. · ⁨초기 속도와 가속도를 변경하세요. $v$-$t$ 그래프의 **기울기(slope)**는 가속도 $a=\tfrac{dv}{dt}$이며, 아래쪽 **면적(area)**는 변위(displacement) $\Delta x=\int v\,dt$입니다. 기울기와 면적은 미분과 적분의 시각적 형태입니다.⁩

    1.4

    Reference Frames and Relative Motion · ⁨기준 좌표계와 상대 운동⁩

    Syllabus
    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.

    한국어

    학습 목표 1.4.A: 특정 관측자의 기준 좌표계를 기술하시오.

    • 1.4.A.1 기준 좌표계의 선택은 해당 좌표계에서 관측자가 측정하는 물리량의 방향과 크기를 결정한다.

    학습 목표 1.4.B: 서로 다른 관성 기준 좌표계에서 관측자가 측정하는 물체의 운동을 설명한다.

    • 1.4.B.1 특정 기준 좌표계에서 측정한 값은 다른 기준 좌표계에서의 측정값으로 변환할 수 있다.
    • 1.4.B.2 물체의 관측된 속도는 물체 자체의 속도와 관측자의 기준 좌표계의 속도의 합성 결과이다.
      • 1.4.B.2.i 주어진 기준 좌표계 내에서 물체의 운동과 관측자의 운동을 합성하기 위해서는 벡터의 가산 또는 감산을 수행해야 한다.
      • 1.4.B.2.ii 모든 관성 기준 좌표계에서 측정할 때 임의의 물체의 가속도는 동일하다.

    경계 조항: 별다른 명시가 없는 한 모든 문제의 기준 좌표계는 관성 좌표계라고 가정할 수 있다.

    Source: College Board AP Course and Exam Description · ⁨출처: College Board AP Course and Exam Description⁩

    English

    All motion is relative to a reference frame 参考系. Combine velocities by vector addition: $\vec{v}_{A/C}=\vec{v}_{A/B}+\vec{v}_{B/C}$. This handles boats crossing rivers and passengers on moving vehicles – relative motion 相对运动 problems. Read the subscripts as a chain: "A relative to B" plus "B relative to C" gives "A relative to C".

    Worked example. A boat that moves at $4.0\ \text{m/s}$ in still water heads straight across a river flowing at $3.0\ \text{m/s}$. Relative to the ground the boat moves at $\sqrt{4.0^2+3.0^2}=5.0\ \text{m/s}$, angled downstream. If the river is $80\ \text{m}$ wide, the crossing still takes $t=\dfrac{80}{4.0}=20\ \text{s}$ – only the across-stream component crosses the river; the current just carries the boat $60\ \text{m}$ downstream.

    한국어

    모든 운동은 표준 좌표계에 대해 상대적이다. 속도를 벡터 합성하여 결합한다: $\vec{v}_{A/C}=\vec{v}_{A/B}+\vec{v}_{B/C}$. 이는 강을 건너는 배나 이동하는 차량의 승객과 같은 상대 운동 문제를 처리한다. 아래첨자를 사슬처럼 읽는다: "A relative to B" plus "B relative to C" gives "A relative to C".

    풀이 예제. 정수면에서 $4.0\ \text{m/s}$의 속도로 움직이는 보가 $3.0\ \text{m/s}$로 흐르는 강에 수직으로 진입한다. 지상에 대한 보의 운동 속도는 $\sqrt{4.0^2+3.0^2}=5.0\ \text{m/s}$이며 하류 방향으로 기우져 있다. 강 너비가 $80\ \text{m}$라면, 건널时间是仍需$t=\dfrac{80}{4.0}=20\ \text{s}$ – 오직 수직 방향 성분이 강을 건너는 데 기여하며, 유속은 단순히 보를 $60\ \text{m}$만큼 하류로 밀어내일 뿐이다.

    1.5

    Motion in Two or Three Dimensions · ⁨2차원 또는 3차원에서의 운동⁩

    Syllabus
    English

    Learning Objective 1.5.A: Describe the motion of an object moving in two or three dimensions.

    • 1.5.A.1 Motion in two or three dimensions can be analyzed using one-dimensional kinematic relationships if the motion is separated into components.
    • 1.5.A.2 Velocity and acceleration may be different in each dimension and may be nonuniform.
    • 1.5.A.3 Motion in one dimension may be changed without causing a change in a perpendicular dimension.
    • 1.5.A.4 Projectile motion is a special case of two-dimensional motion that has zero acceleration in one dimension and constant, nonzero acceleration in the second dimension.

    Boundary statement: AP Physics C: Mechanics only expects students to quantitatively analyze the motion of an object in two dimensions. AP Physics C: Electricity and Magnetism expects students to also qualitatively describe the motion of a particle in three dimensions.

    한국어

    학습 목표 1.5.A: 두 차원 또는 세 차원에서 움직이는 물체의 운동을 설명하라.

    • 1.5.A.1 운동이 성분으로 분리될 경우, 두 차원 또는 세 차원의 운동은 일차원 운동학 관계식을 사용하여 분석할 수 있다.
    • 1.5.A.2 속도와 가속도는 각 차원에서 다를 수 있으며 비균일일 수 있다.
    • 1.5.A.3 일차원의 운동은 수직인 차원에 변화를 주지 않고 변경될 수 있다.
    • 1.5.A.4 포물선 운동은 한 차원에서는 가속도가 없고 두 번째 차원에서는 일정하고 비영인 가속도를 가지는 2차원 운동의 특수한 경우이다.

    경계 설명: AP Physics C: Mechanics에서는 학생들에게 두 차원에서의 물체 운동을 정량적으로 분석할 것을 기대한다. AP Physics C: Electricity and Magnetism에서는 학생들에게 입자의 세 차원 운동을 정성적으로 기술할 것을 기대한다.

    Source: College Board AP Course and Exam Description · ⁨출처: College Board AP Course and Exam Description⁩

    English

    In two or three dimensions, position is a vector $\vec{r}(t)=\langle x(t),y(t),z(t)\rangle$, and velocity and acceleration are its successive derivatives – each component handled independently.

    Worked example. $\vec{r}(t)=\big(2t^2\big)\hat{i}+\big(4t-t^3\big)\hat{j}$ (metres). Differentiating each component: $\vec{v}=4t\,\hat{i}+(4-3t^2)\,\hat{j}$ and $\vec{a}=4\,\hat{i}-6t\,\hat{j}$. At $t=1\ \text{s}$: $\vec{v}=4\hat{i}+1\hat{j}$, so the speed is $\sqrt{17}\approx4.1\ \text{m/s}$ – no new physics, just one derivative per component.

    For projectile motion 抛体运动, horizontal and vertical motions are independent, linked only by time $t$: horizontally $a_x=0$ (constant velocity), vertically $a_y=-g$. The path is a parabola. This component method extends to any two-dimensional motion where the accelerations along each axis are known.

    Worked example. A ball is launched at $20\ \text{m/s}$, $30^{\circ}$ above the horizontal ($g=9.8\ \text{m/s}^2$). The components are $v_{0x}=20\cos30^{\circ}=17.3\ \text{m/s}$ and $v_{0y}=20\sin30^{\circ}=10\ \text{m/s}$. The vertical motion sets the time: total flight $=\dfrac{2v_{0y}}{g}=\dfrac{20}{9.8}=2.0\ \text{s}$, maximum height $=\dfrac{v_{0y}^2}{2g}=5.1\ \text{m}$, and range $=v_{0x}\times2.0=35\ \text{m}$.

    한국어

    2차원 또는 3차원에서 위치는 벡터 $\vec{r}(t)=\langle x(t),y(t),z(t)\rangle$이며, 속도와 가속도는 각각의 성분을 독립적으로 다루는 successive derivatives이다.

    풀이 예제. $\vec{r}(t)=\big(2t^2\big)\hat{i}+\big(4t-t^3\big)\hat{j}$ (미터). 각 성분을 미분하면: $\vec{v}=4t\,\hat{i}+(4-3t^2)\,\hat{j}$ 및 $\vec{a}=4\,\hat{i}-6t\,\hat{j}$. $t=1\ \text{s}$ 시점: $\vec{v}=4\hat{i}+1\hat{j}$, 따라서 속도는 $\sqrt{17}\approx4.1\ \text{m/s}$ – 새로운 물리 법칙은 없으며, 각 성분에 대해 하나의 미분만 수행했다.

    각도를 이루며 발사된 포물선: 수평 및 수직 운동은 서로 독립
    각도를 이루며 발사된 포물선: 수평 및 수직 운동은 서로 독립

    포물선 운동에서 수평 및 수직 운동은 서로 독립적이지만 시간 $t$에 의해 연결됩니다. 수평으로는 $a_x=0$(등속), 수직으로는 $a_y=-g$입니다. 경로는 포물선 형태입니다. 이 성분 분해 방법은 각 축의 가속도가 알려진 모든 2차원 운동으로 확장됩니다.

    해설 예제. 공을 $20\ \text{m/s}$, 수평선 above $30^{\circ}$ ($g=9.8\ \text{m/s}^2$)의 각도로 발사합니다. 성분은 $v_{0x}=20\cos30^{\circ}=17.3\ \text{m/s}$과 $v_{0y}=20\sin30^{\circ}=10\ \text{m/s}$입니다. 수직 운동이 시간을 결정하므로 총 비행 시간은 $=\dfrac{2v_{0y}}{g}=\dfrac{20}{9.8}=2.0\ \text{s}$, 최대 높이는 $=\dfrac{v_{0y}^2}{2g}=5.1\ \text{m}$, 사거리는 $=v_{0x}\times2.0=35\ \text{m}$입니다.

    각도로 발사된 포사 운동: 속도 성분, 최대 높이, 및 사거리
    각도로 발사된 포사 운동: 속도 성분, 최대 높이, 및 사거리
    Explore · ⁨탐색하기⁩

    Explore projectile motion · ⁨포물선 운동 탐구하기⁩

    Fire the projectile, then vary the angle and speed. The horizontal motion is steady while gravity acts only downward — together they trace a parabola. Find the angle that gives the greatest range (it peaks near $45^\circ$), and try the Moon. · ⁨포물질을 발사한 후 **각도(angle)**와 **속도(speed)**를 변경하세요. 수평 운동은 일정하고 중력은 수직 방향으로만 작용하므로 **포물선(parabola)**를 그립니다. **사거리(range)**가 가장 길어지는 각도(약 $45^\circ$附近)를 찾아보고, 달(Moon) 환경에서도 시도해 보세요.⁩

    Vocabulary · ⁨어휘⁩ Train · ⁨연습하기⁩
    English 한국어
    derivatives/dɪˈrɪvətɪvz/ 미분
    average velocity/ˈævrɪdʒ vəˈlɒsɪti/ 평균 속도
    instantaneous velocity/ˌɪnstənˈteɪnɪəs vəˈlɒsɪti/ 즉각 속도
    free fall/friː fɔːl/ 자유 낙하
    reference frame/ˈrefrəns freɪm/ 기준 좌표계
    relative motion/ˈrelətɪv ˈməʊʃn/ 相대 운동
    projectile motion/prəˈdʒektaɪl ˈməʊʃn/ 포물선 운동
    1.5

    Exam tips · ⁨시험 팁⁩

    English
    • Resolve every vector into components before adding — never add magnitudes at an angle directly.
    • On Physics C you are expected to use calculus: velocity is $\vec v=\tfrac{d\vec r}{dt}$ and acceleration $\vec a=\tfrac{d\vec v}{dt}$; reverse with integration.
    • Carry and check units and treat direction with signs (choose a positive axis and stick to it).
    • Use the dot product for work-type quantities and the cross product for torque and angular momentum.
    • Sketch the vectors — a diagram catches sign and direction errors the algebra hides.
    한국어
    • 벡터를 더하기 전에 반드시 성분을 분해하라 – 각도 있는状态下 직접 크기를 더하지 마라.
    • Physics C에서는 미적분 사용을 기대한다: 속도는 $\vec v=\tfrac{d\vec r}{dt}$이고 가속도는 $\vec a=\tfrac{d\vec v}{dt}$; 역연산은 적분으로 행한다.
    • 단위를 유지하고 확인하며, 방향은 부호로 다룬다 (양(+)의 축을 정하고 일관되게 사용하라).
    • 일-type 물리량은 점곱(dot product)을, 토크 및 각운동량은 곱(cross product)을 사용하라.
    • 벡터 도형을 그려라 – 도식이 대수식에서 숨겨진 부호와 방향 오류를 잡아준다.
  • 2

    Force and Translational Dynamics

    Watch lesson · ⁨수업 보기⁩
    2.1

    Systems and Center of Mass

    Syllabus
    English

    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 objects or systems with symmetrical mass distributions, the center of mass is located on lines of symmetry.
    • 2.1.B.2 The location of a system's center of mass along a given axis can be calculated using the equation
      • Equation: $\vec{x}_{\text{cm}} = \dfrac{\sum m_i \vec{x}_i}{\sum m_i}$
    • 2.1.B.3 For a nonuniform solid that can be considered as a collection of differential masses, $dm$, the solid's center of mass can be calculated using the equation
      • Equation: $\vec{r}_{\text{cm}} = \dfrac{\int \vec{r}\, dm}{\int dm}$
      • 2.1.B.3.i The linear mass density of a rod or other linear rigid body is the derivative of the rod's mass with respect to the position of the differential mass element on the rigid body.
        • Equation: $\lambda = \dfrac{d}{d\ell} m(\ell)$
      • 2.1.B.3.ii If a function of mass density is given for a solid, the total mass can be determined by integrating the mass density over the length (one dimension), area (two dimensions), or volume (three dimensions) of the solid. For example:
        • Equation: $M_{\text{total}} = \int \rho(r)\, dV$
    • 2.1.B.4 A system can be modeled as a singular object that is located at the system's center of mass.
    한국어

    학습 목표 2.1.A: 계의 특성과 상호작용을 설명한다.

    • 2.1.A.1 계의 특성은 계 내 구성 요소 간의 상호작용에 의해 결정된다.
    • 2.1.A.2 계를 거시적 현상으로 모델링할 때 구성 요소의 특성이나 상호작용이 중요하지 않다면, 계 전체를 단일 물체로 취급할 수 있다.
    • 2.1.A.3 계는 구성 요소와 환경 간 상호작용을 허용하며, 이는 에너지나 물질의 이동을 초래할 수 있다.
    • 2.1.A.4 선택된 계 내 개별 물체들은 서로 그리고 계 전체와도 다르게 행동할 수 있다.
    • 2.1.A.5 계의 내부 구조는 해당 계의 분석에 영향을 미친다.
    • 2.1.A.6 계 외부 변수가 변화하면 계의 하위 구조가 변경될 수 있다.

    학습 목표 2.1.B: 계의 질량 중심 위치를 계의 구성 요소에 대해 설명한다.

    • 2.1.B.1 질량 분포가 대칭인 물체나 시스템의 경우, 질량 중심은 대칭축 위에 위치한다.
    • 2.1.B.2 계의 질량 중심 위치는 주어진 축 상에서 다음 식을 사용하여 계산할 수 있다.
      • 식: $\vec{x}_{\text{cm}} = \dfrac{\sum m_i \vec{x}_i}{\sum m_i}$
    • 2.1.B.3 미소 질량의 집합으로 간주할 수 있는 불균일 고체의 경우, $dm$, 고체의 질량 중심은 다음 식을 사용하여 계산할 수 있다.
      • 식: $\vec{r}_{\text{cm}} = \dfrac{\int \vec{r}\, dm}{\int dm}$
      • 2.1.B.3.i 막대나 기타 선형 강체 선형 질량 밀도는 강체 위의 미소 질량 요소의 위치에 대한 막대의 질량의 미분값이다.
        • 식: $\lambda = \dfrac{d}{d\ell} m(\ell)$
      • 2.1.B.3.ii 고체에 대한 질량 밀도 함수가 주어졌다면, 고체의 전체 질량은 고체의 길이(일차원), 면적(이차원) 또는 부피(삼차원)에 대해 질량 밀도를 적분하여 결정할 수 있다. 예를 들어:
        • 식: $M_{\text{total}} = \int \rho(r)\, dV$
    • 2.1.B.4 시스템은 시스템의 질량 중심에 위치한 단일 물체로 모델링할 수 있다.

    Source: College Board AP Course and Exam Description · ⁨출처: College Board AP Course and Exam Description⁩

    A system 系统 is the object or objects you analyze, treated as a point at its center of mass 质心. For a set of masses, $x_{\text{cm}}=\dfrac{\sum m_i x_i}{\sum m_i}$; for a continuous body, $x_{\text{cm}}=\dfrac{1}{M}\int x\,dm$. Only external forces move the center of mass.

    Vocabulary · ⁨어휘⁩ Train · ⁨연습하기⁩
    English 한국어
    system/ˈsɪstəm/ system
    center of mass/ˈsentə ɒv mæs/ center of mass
    2.2

    Forces and Free-Body Diagrams

    Syllabus
    English

    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 or system with another object or system.
      • 2.2.A.1.ii An object or system cannot exert a net force on itself.
    • 2.2.A.2 Contact forces describe the interaction of an object or system touching another object or system and are macroscopic effects of interatomic electric forces.

    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 or system by the environment.
    • 2.2.B.3 Forces exerted on an object or system are represented as vectors originating from the representation of the center of mass, such as a dot. A system is treated as though all of its mass is located at the center of mass.
    • 2.2.B.4 A coordinate system with one axis parallel to the direction of acceleration of the object or system simplifies the translation from free-body diagram to algebraic representation. For example, in a free-body diagram of an object on an inclined plane, it is useful to set one axis parallel to the surface of the incline.

    Boundary statement: AP Physics C: Mechanics and AP Physics C: Electricity and Magnetism only expect students to depict the forces exerted on objects, not the force components on free-body diagrams. On the AP Physics exams, individual forces represented on a free-body diagram must be drawn as individual straight arrows, originating on the dot and pointing in the direction of the force. Individual forces that are in the same direction must be drawn side by side, not overlapping.

    한국어

    학습 목표 2.2.A: 힘을 두 개物体的物体 또는 시스템 간의 상호작용으로 설명합니다.

    • 2.2.A.1 힘은 물체나 계 간의 상호작용을 나타내는 벡터량이다.
      • 2.2.A.1.i 어떤 물체나 시스템에 가해지는 힘은 항상 해당 물체나 시스템이 다른 물체나 시스템과 상호작용한 결과입니다.
      • 2.2.A.1.ii 물체나 계는 자신에게서 순힘을 가할 수 없다.
    • 2.2.A.2 접촉력은 물체나 계가 다른 물체나 계에 닿아 상호작용하는 것을 기술하며, 이는 원자 간 전기력의 거시적 효과이다.

    학습 목표 2.2.B: 자유체도를 사용하여 물체나 계에 가해지는 힘을 설명한다.

    • 2.2.B.1 자유체도는 단일 물체나 계에 가해지는 힘을 시각화하고 물리적 상황을 나타내는 방정식을 도출하는 데 유용한 도구이다.
    • 2.2.B.2 자유도 도식(free-body diagram)은 환경이 물체나 시스템에 가하는 모든 힘을 보여줍니다.
    • 2.2.B.3 물체나 계에 가해지는 힘은 질량 중심(예: 점)에서 시작하는 벡터로 표현된다. 계는 모든 질량이 질량 중심에 집중되어 있는 것처럼 취급된다.
    • 2.2.B.4 물체나 계의 가속도 방향과 평행한 축을 갖는 좌표계를 설정하면 자유체도를 알기식 표현으로 변환하는 것이 용이하다. 예를 들어, 경사면 위의 물체에 대한 자유체도에서는 축 중 하나를 경사면 표면에 평행하게 설정하는 것이 유리하다.

    경계 문구: AP 물리학 C: 역학 및 AP 물리학 C: 전기와 자기에서 학생들은 물체에 가해지는 힘만을 표시하고, 자유도 도식 상에서의 힘의 성분을 나타낼 필요는 없습니다. AP 물리학 시험에서 자유도 도식에 표시된 개별 힘은 도트(dot)에서 시작하여 힘의 방향을 향하는 개별 직선 화살표로 그려야 합니다. 같은 방향의 개별 힘은 겹치지 않도록 나란히 그려야 합니다.

    Source: College Board AP Course and Exam Description · ⁨출처: College Board AP Course and Exam Description⁩

    A force 力 is a push or pull (a vector, in newtons). A free-body diagram 受力图 draws one object with an arrow for every force on it – weight, normal, tension, friction, applied, drag. Draw it first; it sets up every dynamics equation.

    A free-body diagram shows every force acting on one object
    A free-body diagram shows every force acting on one object
    Explore · ⁨탐색하기⁩

    Balance the forces on a free-body diagram · ⁨자유체 도상의 힘 평형 맞추기⁩

    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 $a=F/m$. · ⁨**자유체 도상(free-body diagram)**은 하나의 물체에 작용하는 모든 힘을 화살표로 보여줍니다. 물체가 가속되려면 힘이 상쇄되지 않아야 하며, **순힘(net force)**이 $a=F/m$를 결정합니다.⁩

    Vocabulary · ⁨어휘⁩ Train · ⁨연습하기⁩
    English 한국어
    force/fɔːs/ force
    free-body diagram/friː ˈbɒdi ˈdaɪəɡræm/ free-body diagram
    2.3

    Newton's Third Law

    Syllabus
    Learning ObjectiveEssential Knowledge

    2.3.A
    Describe the interaction of two objects or systems using Newton's third law and a representation of paired forces exerted on each object or system.

    • 2.3.A.1 Newton's third law describes the interaction of two objects or systems in terms of the paired forces that each exerts on the other.
      • Equation: $\vec{F}_{\text{A on B}} = -\vec{F}_{\text{B on A}}$
    • 2.3.A.2 Interactions between objects within a system (internal forces) do not influence the motion of a system's center of mass.
    • 2.3.A.3 Tension is the macroscopic net result of forces that infinitesimal segments of a string, cable, chain, or similar system exert on each other in response to an external force.
      • 2.3.A.3.i An ideal string has negligible mass and does not stretch when under tension.
      • 2.3.A.3.ii The tension in an ideal string is the same at all points within the string.
      • 2.3.A.3.iii In a string with nonnegligible mass, tension may not be the same at all points within the string.
      • 2.3.A.3.iv An ideal pulley is a pulley that has negligible mass and rotates about an axle through its center of mass with negligible friction.

    Source: College Board AP Course and Exam Description · ⁨출처: College Board AP Course and Exam Description⁩

    Newton's third law 牛顿第三定律: A pushes B, B pushes back equally and oppositely. The pair acts on different objects, so it never cancels within one free-body diagram.

    A Newton's third-law pair: equal and opposite forces on two different objects
    A Newton's third-law pair: equal and opposite forces on two different objects
    Vocabulary · ⁨어휘⁩ Train · ⁨연습하기⁩
    English 한국어
    Newton's third law/ˈnjuːtnz θɜːd lɔː/ Newton's third law
    2.4

    Newton's First Law

    Syllabus
    Learning ObjectiveEssential Knowledge

    2.4.A
    Describe the conditions under which a system's velocity remains constant.

    • 2.4.A.1 The net force on a system is the vector sum of all forces exerted on the system.
    • 2.4.A.2 Translational equilibrium is the configuration of forces such that the net force exerted on a system is zero.
      • Derived equation: $\sum \vec{F}_i = 0$
    • 2.4.A.3 Newton's first law states that if the net force exerted on a system is zero, the velocity of that system will remain constant.
    • 2.4.A.4 Forces may be balanced in one dimension but unbalanced in another. The system's velocity will change only in the direction of the unbalanced force.
    • 2.4.A.5 An inertial reference frame is one from which an observer would verify Newton's first law of motion.

    Source: College Board AP Course and Exam Description · ⁨출처: College Board AP Course and Exam Description⁩

    Newton's first law (inertia 惯性): with zero net force 合力, velocity stays constant – the object is in translational equilibrium 平动平衡.

    Vocabulary · ⁨어휘⁩ Train · ⁨연습하기⁩
    English 한국어
    inertia/ɪˈnɜːʃə/ inertia
    net force/net fɔːs/ net force
    translational equilibrium/trænˈsleɪʃənl ˌiːkwɪˈlɪbrɪəm/ translational equilibrium
    2.5

    Newton's Second Law

    Syllabus
    Learning ObjectiveEssential Knowledge

    2.5.A
    Describe the conditions under which a system's velocity changes.

    • 2.5.A.1 Unbalanced forces are a configuration of forces such that the net force exerted on a system is not equal to zero.
    • 2.5.A.2 Newton's second law of motion states that the acceleration of a system's center of mass has a magnitude proportional to the magnitude of the net force exerted on the system and is in the same direction as that net force.
      • Equation: $\vec{a}_{\text{sys}} = \dfrac{\sum \vec{F}}{m_{\text{sys}}} = \dfrac{\vec{F}_{\text{net}}}{m_{\text{sys}}}$
    • 2.5.A.3 The velocity of a system's center of mass will only change if a nonzero net external force is exerted on that system.

    Source: College Board AP Course and Exam Description · ⁨출처: College Board AP Course and Exam Description⁩

    The general form uses momentum:

    $$\sum\vec{F}=\frac{d\vec{p}}{dt}=m\vec{a}\ \ (\text{for constant mass}).$$
    Apply it one axis at a time. When forces depend on velocity or position, this becomes a differential equation 微分方程 to solve.

    Worked example. A $2.0\ \text{kg}$ block slides down a frictionless incline 斜面 at $30^{\circ}$. Only the along-ramp component of gravity drives it, so $a=g\sin30^{\circ}=9.8(0.5)=4.9\ \text{m/s}^2$, independent of the mass.

    Worked example (two bodies). A $6.0\ \text{kg}$ cart on a frictionless table is pulled by a string over a light pulley to a hanging $2.0\ \text{kg}$ mass. Treat the pair as one system: only the hanging weight drives it, so $a=\dfrac{2.0(9.8)}{8.0}=2.45\ \text{m/s}^2$. Then isolate the cart alone to find the string tension 张力: $T=6.0(2.45)\approx15\ \text{N}$. System first for $a$, single body for internal forces – that two-step is the standard pattern.

    A rocket launch: net force changes momentum — F_net = ma for translational dynamics
    A rocket launch: net force changes momentum — F_net = ma for translational dynamics
    Vocabulary · ⁨어휘⁩ Train · ⁨연습하기⁩
    English 한국어
    differential equation/ˌdɪfəˈrenʃl ɪˈkweɪʒn/ differential equation
    incline/ɪnˈklaɪn/ incline
    tension/ˈtenʃn/ tension
    2.6

    Gravitational Force

    Syllabus
    English

    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.
      • Equation: $\left| \vec{F}_g \right| = G \dfrac{m_1 m_2}{r^2}$
      • 2.6.A.1.i The gravitational force is attractive.
      • 2.6.A.1.ii The gravitational force is always exerted along the line connecting the center of mass of the two interacting systems.
      • 2.6.A.1.iii The gravitational force on a system can be considered to be exerted on the system's center of mass.
    • 2.6.A.2 A field models the effects of a noncontact force exerted on an object at various positions in space.
      • 2.6.A.2.i The magnitude of the gravitational field created by a system of mass $M$ at a point in space is equal to the ratio of the gravitational force exerted by the system on a test object of mass $m$ to the mass of the test object.
        • Derived equation: $\left| \vec{g} \right| = \dfrac{\left| \vec{F}_g \right|}{m} = G \dfrac{M}{r^2}$
      • 2.6.A.2.ii If the gravitational force is the only force exerted on an object, the observed acceleration of the object (in $\text{m/s}^2$) is numerically equal to the magnitude of the gravitational field strength (in $\text{N/kg}$) at that location.
    • 2.6.A.3 The gravitational force exerted by an astronomical body on a relatively small nearby object is called weight.
      • Derived equation: $\text{Weight} = F_g = mg$

    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
      • Equation: $g \approx 10\ \text{N/kg}$

    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.

    Learning Objective 2.6.E: Describe the gravitational force exerted on an object by a uniform spherical distribution of mass.

    • 2.6.E.1 The net gravitational force exerted on an object by a uniform spherical distribution of mass is the sum of the individual forces from small differential masses that comprise the distribution.
    • 2.6.E.2 Newton's shell theorem describes the net gravitational force exerted on an object by a uniform spherical shell of mass.
      • 2.6.E.2.i The net gravitational force exerted on an object inside a thin spherical shell is zero.
      • 2.6.E.2.ii The net gravitational force exerted on an object outside a thin spherical shell can be determined by treating the shell as a single massive object located at the center of the shell.
      • 2.6.E.2.iii An object inside a sphere of uniform density experiences a net gravitational force from only a partial mass of the sphere.
      • 2.6.E.2.iv The partial mass of a sphere that contributes to the net gravitational force exerted on an object within that sphere is the portion of the sphere's mass located a distance less than or equal to the object's distance from the center of the sphere and can be calculated using the density of the sphere.
        • Derived equation: $m_{\text{partial}} = \rho \dfrac{4}{3} \pi \left( r_{\text{partial}} \right)^3$
    • 2.6.E.3 The gravitational force exerted on an object within a uniform sphere can be shown to be proportional to the object's distance from the sphere's center.
      • Derived equation: $F_{g,\text{partial}} = -k r_{\text{partial}}$

    Boundary statement: AP Physics C: Mechanics does not expect students to mathematically prove or derive Newton's shell theorem.

    한국어

    학습 목표 2.6.A: 질량을 가진 두 물체나 시스템 간의 중력 상호작용을 설명할 수 있다.

    • 2.6.A.1 만유인력 법칙은 두 물체나 시스템 간의 중력 힘을 각 질량의 곱에 직접 비례하고 두 시스템의 질량 중심 사이의 거리의 제곱에 반비례한다고 설명한다.
      • 식: $\left| \vec{F}_g \right| = G \dfrac{m_1 m_2}{r^2}$
      • 2.6.A.1.i 중력력은 인력(attractive)이다.
      • 2.6.A.1.ii 중력은 항상 상호작용하는 두 시스템의 질량 중심을 연결하는 선을 따라 가해집니다.
      • 2.6.A.1.iii 시스템에 작용하는 중력력은 시스템의 질량 중심에 작용하는 것으로 간주할 수 있다.
    • 2.6.A.2 장은 공간상의 다양한 위치에서 물체에 가해지는 비접촉력의 효과를 모델링한다.
      • 2.6.A.2.i 질량 $M$인 시스템이 생성한 중력장의 세기는 특정 지점에서 시스템이 질량 $m$의 시험 물체에 가하는 중력힘과 시험 물체의 질량의 비율과 같다.
        • 파생 식: $\left| \vec{g} \right| = \dfrac{\left| \vec{F}_g \right|}{m} = G \dfrac{M}{r^2}$
      • 2.6.A.2.ii 중력이 물체에 작용하는 유일한 힘이라면, 관측된 물체의 가속도(단위: ⟨$\text{m/s}^2$⟩)는 해당 위치에서의 중력장 세기(단위: ⟨$\text{N/kg}$⟩)의 크기와 수치적으로 같다.
    • 2.6.A.3 천체가 상대적으로 작은 근접 물체에 가하는 중력력을 무게(weight)라고 한다.
      • 파생 식: $\text{Weight} = F_g = mg$

    학습 목표 2.6.B: 중력력을 일정하게 간주할 수 있는 상황을 설명할 수 있다.

    • 2.6.B.1 두 시스템의 질량 중심 간 중력력이 두 시스템의 상대적 위치 변화에 따라 무시할 만큼 변하지 않는다면, 초기 및 최종 위치 사이 모든 지점에서 중력력을 일정하다고 간주할 수 있다.
    • 2.6.B.2 지구 표면 근처에서 중력장의 세기는
      • 식: $g \approx 10\ \text{N/kg}$

    학습 목표 2.6.C: 시스템의 겉보기 무게(apparent weight) 크기가该系统에 가해지는 중력힘 크기와 다른 조건을 설명할 수 있다.

    • 2.6.C.1 시스템의 겉보기 무게 크기는该系统에 가해지는 법선력의 크기와 같다.
    • 2.6.C.2 시스템이 가속 중이라면该系统의 겉보기 무게 크기는该系统에 가해지는 중력힘 크기와 다르다.
    • 2.6.C.3该系统에 작용하는 힘이 없거나 중력만이该系统에 작용하는 경우该系统은 무중력 상태(weightlessness)로 보인다.
    • 2.6.C.4 등가원리는 비관성 기준계에 있는 관측자가 물체의 겉보기 무게와 중력장이 물체에 가하는 중력 사이의 차이를 구별할 수 없음을 의미한다.

    학습 목표 2.6.D: 관성 질량과 중력 질량을 설명하라.

    • 2.6.D.1 물체는 관성 질량, 즉 관성을 가지는데, 이는 다른 물체와 상호작용할 때 물체의 운동이 변하는 데 얼마나 저항하는지를 결정하는 성질이다.
    • 2.6.D.2 중력 질량은 질량을 가진 두 시스템 간의 인력他与 관련된다.
    • 2.6.D.3 관성 질량과 중력 질량은 실험적으로 등가임이 입증되었다.

    학습 목표 2.6.E: 균일한 구대칭 질량 분포가 물체에 가하는 중력을 설명합니다.

    • 2.6.E.1 균일한 구대칭 질량 분포가 물체에 가하는 합중력은 분포를 구성하는 작은 미소 질량들로부터 가해지는 개별 힘들의 합입니다.
    • 2.6.E.2 뉴턴의 껍질 정리(shell theorem)는 균일한 구껍질 질량이 물체에 가하는 합중력을 설명합니다.
      • 2.6.E.2.i 얇은 구껍질 내부에 있는 물체에 가하는 합중력은 0입니다.
      • 2.6.E.2.ii 얇은 구껍질 외부에 있는 물체에 가하는 합중력은 껍질을 껍질의 중심에 위치한 단일 질량체로 간주하여 결정할 수 있습니다.
      • 2.6.E.2.iii 균일 밀도의 구 내부에 있는 물체는 구의 일부 질량으로부터만 합중력을 받습니다.
      • 2.6.E.2.iv 구 내부의 물체에 가하는 합중력에 기여하는 구의 일부 질량은 구의 중심으로부터 물체의 거리보다 작거나 동일한 거리에 있는 구의 질량 부분이며, 구의 밀도를 사용하여 계산할 수 있습니다.
        • 파생 식: $m_{\text{partial}} = \rho \dfrac{4}{3} \pi \left( r_{\text{partial}} \right)^3$
    • 2.6.E.3 균일한 구 내부의 물체에 가하는 중력은 물체가 구의 중심으로부터의 거리에 비례함을 보일 수 있습니다.
      • 파생 식: $F_{g,\text{partial}} = -k r_{\text{partial}}$

    경계 문구: AP 물리학 C: 역학은纽顿의 껍질 정리를 수학적으로 증명하거나 유도할 것을 기대하지 않습니다.

    Source: College Board AP Course and Exam Description · ⁨출처: College Board AP Course and Exam Description⁩

    Orbital motion (Kepler's 2nd law)

    Near a surface, weight is $F_g=mg$. In general, Newton's law of gravitation 万有引力定律:

    $$F_g=\frac{Gm_1m_2}{r^2},$$
    attractive and inverse-square. The gravitational field is $g=\dfrac{GM}{r^2}$.

    Two masses attract each other with equal, opposite, inverse-square forces along the line joining them
    Two masses attract each other with equal, opposite, inverse-square forces along the line joining them
    Vocabulary · ⁨어휘⁩ Train · ⁨연습하기⁩
    English 한국어
    Newton's law of gravitation/ˈnjuːtnz lɔː ɒv ˌɡrævɪˈteɪʃn/ Newton's law of gravitation
    2.7

    Kinetic and Static Friction

    Syllabus
    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 the motion of each surface relative to the other surface.
      • 2.7.A.1.ii The force of friction between two surfaces does not depend on the size of the surface area of contact.
    • 2.7.A.2 The magnitude of the kinetic friction force exerted on an object is the product of the normal force the surface exerts on the object and the coefficient of kinetic friction.
      • Equation: $\left| \vec{F}_{f,k} \right| = \left| \mu_k \vec{F}_N \right|$
      • 2.7.A.2.i The coefficient of kinetic friction depends on the material properties of the surfaces that are in contact.
      • 2.7.A.2.ii Normal force is the perpendicular component of the force exerted on an object by the surface with which it is in contact; it is directed away from the surface.

    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.
      • Equation: $\left| \vec{F}_{f,s} \right| \leq \left| \mu_s \vec{F}_n \right|$
      • 2.7.B.2.i Slipping and sliding refer to situations in which two surfaces are moving relative to each other.
      • 2.7.B.2.ii There exists a maximum value for which static friction will prevent an object from slipping on a given surface.
        • Derived equation: $F_{f,s,\text{max}} = \mu_s F_N$
    • 2.7.B.3 The coefficient of static friction is typically greater than the coefficient of kinetic friction for a given pair of surfaces.
    한국어

    학습 목표 2.7.A: 두 표면 사이의 운동 마찰력을 설명합니다.

    • 2.7.A.1 접촉된 두 표면이 서로에 대해 상대 운동을 할 때 운동 마찰력이 발생한다.
      • 2.7.A.1.i 운동 마찰력은 각 표면이 다른 표면에 대해 움직이는 방향과 반대 방향으로 가해집니다.
      • 2.7.A.1.ii 두 표면 사이의 마찰력은 접촉 면적의 크기에 의존하지 않는다.
    • 2.7.A.2 물체에 작용하는 운동 마찰력의 세기는 표면에 의해 물체에 가해지는 수직 반력(정반력)과 운동 마찰 계수의 곱으로 주어진다.
      • 식: $\left| \vec{F}_{f,k} \right| = \left| \mu_k \vec{F}_N \right|$
      • 2.7.A.2.i 운동 마찰 계수는 접촉된 표면의 재질 특성에 따라 달라진다.
      • 2.7.A.2.ii 수직 반력은 물체에接触的한 표면에 의해 물체에 가해지는 힘의 수직 성분이며, 표면으로부터外向로 향한다.

    학습 목표 2.7.B: 두 표면 사이의 정지 마찰력을 설명하라.

    • 2.7.B.1 정지 마찰력은 서로에 대해 정지해 있는 두 물체의 접촉면 사이에서 발생할 수 있다.
    • 2.7.B.2 정지 마찰력은 물체가表面上을 미끄러지거나 미끄러지는 것을 방지하기 위해 필요한 값과 방향을 취한다.
      • 식: $\left| \vec{F}_{f,s} \right| \leq \left| \mu_s \vec{F}_n \right|$
      • 2.7.B.2.i 미끄러짐(slipping)과 미끄러짐(sliding)은 두 표면이 서로에 대해 상대 운동을 하는 상황을 의미한다.
      • 2.7.B.2.ii 주어진表面上에서 물체의 미끄러짐을 방지하는 정지 마찰력에는 최대값이 존재한다.
        • 파생 식: $F_{f,s,\text{max}} = \mu_s F_N$
    • 2.7.B.3 특정 쌍의 표면에 대해 정지 마찰 계수는 일반적으로 운동 마찰 계수보다 크다.

    Source: College Board AP Course and Exam Description · ⁨출처: College Board AP Course and Exam Description⁩

    Friction 摩擦力 opposes sliding along a surface: kinetic 动摩擦 $f_k=\mu_k N$ while sliding, and static 静摩擦 $f_s\le\mu_s N$ up to a maximum before sliding. $N$ is the normal force 法向力. Note the inequality: static friction is only as large as it needs to be, and $\mu_s N$ is its ceiling, not its value.

    Worked example. The same block on the same $30^{\circ}$ incline, now with $\mu_k=0.20$. Perpendicular to the ramp: $N=mg\cos30^{\circ}$. Along the ramp: $ma=mg\sin30^{\circ}-\mu_k mg\cos30^{\circ}$, so $a=g(\sin30^{\circ}-0.20\cos30^{\circ})=9.8(0.500-0.173)=3.2\ \text{m/s}^2$ – the mass cancels again.

    Explore · ⁨탐색하기⁩

    Slide a block down a slope with friction · ⁨마찰이 있는 경사면을 따라 블록 미끄러지기⁩

    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. · ⁨마찰은 최대 $\mu N$까지 운동을 방해합니다. 경사면을 기울여 중력의 경사 방향 분력이 정지 마찰력을 넘어서도록 하면 블록이 미끄러지기 시작합니다.⁩

    Vocabulary · ⁨어휘⁩ Train · ⁨연습하기⁩
    English 한국어
    Friction/ˈfrɪkʃn/ Friction
    kinetic/kɪˈnetɪk/ kinetic
    static/ˈstætɪk/ static
    normal force/ˈnɔːml fɔːs/ normal force
    2.8

    Spring Forces

    Syllabus
    English

    Learning Objective 2.8.A: Describe the force exerted on an object by an ideal spring.

    • 2.8.A.1 An ideal spring has negligible mass and exerts a force that is proportional to the change in its length as measured from its relaxed length. A nonideal spring either has nonnegligible mass or exerts a force that is not proportional to the change in its length as measured from its relaxed length.
    • 2.8.A.2 The magnitude of the force exerted by an ideal spring on an object is given by Hooke's law:
      • Equation: $\vec{F}_s = -k \Delta \vec{x}$
    • 2.8.A.3 The force exerted on an object by a spring is always directed toward the equilibrium position of the object–spring system.

    Learning Objective 2.8.B: Describe the equivalent spring constant of a combination of springs exerting forces on an object.

    • 2.8.B.1 A collection of springs that exert forces on an object may behave as though they were a single spring with an equivalent spring constant $k_{\text{eq}}$.
      • 2.8.B.1.i The inverse of the equivalent spring constant of a set of springs in series is equal to the sum of the inverses of the individual spring constants.
        • Derived equation: $\dfrac{1}{k_{\text{eq, series}}} = \sum_i \dfrac{1}{k_i} = \dfrac{1}{k_1} + \dfrac{1}{k_2} + \dots$
      • 2.8.B.1.ii The equivalent spring constant of a set of springs arranged in series is smaller than the smallest constituent spring constant.
      • 2.8.B.1.iii The equivalent spring constant of a set of springs arranged in parallel is the sum of the individual spring constants.
        • Derived equation: $k_{\text{eq, parallel}} = \sum_i k_i = k_1 + k_2 + \dots$

    Boundary statement: AP Physics C: Mechanics only expects students to find the effective spring constant of systems of springs that are arranged either in series or in parallel and does not expect students to find the effective spring constant of a system in which springs are arranged in both series and parallel.

    한국어

    학습 목표 2.8.A: 이상적인 스프링이 물체에 가하는 힘을 설명합니다.

    • 2.8.A.1 이상적인 스프링은 질량이 무시할 만큼 작으며, 휴지 상태(rest length)로부터 측정된 길이 변화에 비례하는 힘을 가합니다. 비이상적인 스프링은 무시할 수 없는 질량을 가지거나, 휴지 상태로부터 측정된 길이 변화에 비례하지 않는 힘을 가합니다.
    • 2.8.A.2 이상적인 스프링이 물체에 가하는 힘의 세기는 훅의 법칙에 의해 주어진다:
      • 식: $\vec{F}_s = -k \Delta \vec{x}$
    • 2.8.A.3 스프링이 물체에 가하는 힘은 항상 물체-스프링 시스템의 평형 위치를 향한다.

    학습 목표 2.8.B: 물체에 힘을 가하는 스프링 조합의 등가 스프링 상수(equivalent spring constant)를 설명합니다.

    • 2.8.B.1 물체에 힘을 가하는 스프링 모음은 등가 스프링 상수 $k_{\text{eq}}$를 가진 단일 스프링처럼 동작할 수 있습니다.
      • 2.8.B.1.i 직렬로 연결된 스프링 세트의 등가 프링 상수의 역수는 개별 스프링 상수들의 역수들의 합과 같습니다.
        • 파생 식: $\dfrac{1}{k_{\text{eq, series}}} = \sum_i \dfrac{1}{k_i} = \dfrac{1}{k_1} + \dfrac{1}{k_2} + \dots$
      • 2.8.B.1.ii 직렬로 배열된 스프링 세트의 등가 스프링 상수는 구성 스프링 상수 중 가장 작은 값보다 작습니다.
      • 2.8.B.1.iii 병렬로 배열된 스프링 세트의 등가 스프링 상수는 개별 스프링 상수들의 합입니다.
        • 파생 식: $k_{\text{eq, parallel}} = \sum_i k_i = k_1 + k_2 + \dots$

    경계 문구: AP 물리학 C: 역학에서는 스프링이 직렬 또는 병렬로 배열된 시스템의 유효 스프링 상수만 구하도록 하며, 직렬과 병렬이 혼합된 시스템의 유효 스프링 상수를 구하는 것은 기대하지 않습니다.

    Source: College Board AP Course and Exam Description · ⁨출처: College Board AP Course and Exam Description⁩

    Hooke's law & the elastic limit

    An ideal spring obeys Hooke's law 胡克定律 $F_s=-kx$, a restoring force set by the spring constant 弹簧劲度系数 $k$. Its stored energy is $U=\tfrac12 kx^2$.

    Hooke's law: extension is proportional to load up to the limit of proportionality
    Hooke's law: extension is proportional to load up to the limit of proportionality

    Combined springs act as one equivalent spring of constant $k_{\text{eq}}$. In parallel (side by side, sharing the load) the constants add, $k_{\text{eq}}=k_1+k_2$ – stiffer than either. In series (end to end) they combine as $\dfrac{1}{k_{\text{eq}}}=\dfrac{1}{k_1}+\dfrac{1}{k_2}$ – softer than the softest one. Two springs pulling a mass from opposite sides also act in parallel, giving a restoring constant $k_1+k_2$. Whichever you have, the SHM period uses $k_{\text{eq}}$: $T=2\pi\sqrt{m/k_{\text{eq}}}$.

    Explore · ⁨탐색하기⁩

    Stretch a spring (Hooke's law) · ⁨스프링 늘리기 (후크의 법칙)⁩

    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. · ⁨스프링의 힘은 신도에 비례하며, $F=kx$(후크의 법칙)입니다. 더 세게 당길수록 신도가 비례하여 증가—but 스프링의 한계까지.⁩

    Vocabulary · ⁨어휘⁩ Train · ⁨연습하기⁩
    English 한국어
    Hooke's law/hʊks lɔː/ Hooke's law
    spring constant/sprɪŋ ˈkɒnstənt/ spring constant
    2.9

    Resistive Forces

    Syllabus
    Learning ObjectiveEssential Knowledge

    2.9.A
    Describe the motion of an object subject to a resistive force.

    • 2.9.A.1 A resistive force is defined as a velocity-dependent force in the opposite direction of an object's velocity, for example:
      • Equation: $\vec{F}_r = -k\vec{v}$
    • 2.9.A.2 Applying Newton's second law to an object upon which a resistive force is exerted results in a differential equation for velocity.
      • 2.9.A.2.i Using the method of separation of variables, the velocity can be determined by integrating over the proper limits of integration.
      • 2.9.A.2.ii The acceleration or position of a moving object that is subject to a velocity-dependent force may be determined using initial conditions of the object and methods of calculus, once a function for velocity is determined.
      • 2.9.A.2.iii The position, velocity, and acceleration as functions of time of an object under the influence of a resistive force of the form $\vec{F}_r = -k\vec{v}$ are exponential and have asymptotes that are determined by the initial conditions of the object and the forces exerted on the object.
    • 2.9.A.3 Terminal velocity is defined as the maximum speed achieved by an object moving under the influence of a constant force and a resistive force that are exerted on the object in opposite directions. The terminal condition is reached when the net force exerted on the object is zero.

    Source: College Board AP Course and Exam Description · ⁨출처: College Board AP Course and Exam Description⁩

    A resistive force 阻力 (drag) opposes motion through a fluid and grows with speed, often modeled as $F=-bv$ or $F=-cv^2$. Newton's second law then gives a differential equation, e.g. $m\dfrac{dv}{dt}=mg-bv$ for a falling object. As speed rises, drag builds until it balances the driving force; the object then stops accelerating and moves at a constant terminal velocity 终极速度, found by setting the net force (and $dv/dt$) to zero.

    Worked example. For a falling object with $m\dfrac{dv}{dt}=mg-bv$, the terminal velocity is where $\dfrac{dv}{dt}=0$: $mg=bv_T$, so $v_T=\dfrac{mg}{b}$. With $m=0.10\ \text{kg}$ and $b=0.50\ \text{kg/s}$, $v_T=\dfrac{0.10(9.8)}{0.50}=2.0\ \text{m/s}$.

    To get the full motion $v(t)$, separate the variables 分离变量 and integrate from rest:

    $$\int_0^{v}\frac{dv'}{mg-bv'}=\int_0^{t}\frac{dt'}{m}\;\Rightarrow\;-\frac{1}{b}\ln\!\frac{mg-bv}{mg}=\frac{t}{m},$$
    which rearranges to $v(t)=v_T\left(1-e^{-bt/m}\right)$ – the speed rising exponentially toward $v_T$. Showing this integration is the calculus skill a Physics C free-response rewards, not just quoting the final formula.

    An object falling through a fluid speeds up to a terminal velocity
    An object falling through a fluid speeds up to a terminal velocity
    Skydivers in freefall: drag grows with speed until it balances weight at terminal velocity
    Skydivers in freefall: drag grows with speed until it balances weight at terminal velocity
    Vocabulary · ⁨어휘⁩ Train · ⁨연습하기⁩
    English 한국어
    resistive force/rɪˈzɪstɪv fɔːs/ resistive force
    terminal velocity/ˈtɜːmɪnl vəˈlɒsɪti/ terminal velocity
    separate the variables/ˈseprət ðə ˈveərɪəblz/ separate the variables
    2.10

    Circular Motion

    Syllabus
    Learning ObjectiveEssential Knowledge

    2.10.A
    Describe the motion of an object traveling in a circular path.

    • 2.10.A.1 Centripetal acceleration is the component of an object's acceleration directed toward the center of the object's circular path.
      • 2.10.A.1.i The magnitude of centripetal acceleration for an object moving in a circular path is the ratio of the object's tangential speed squared to the radius of the circular path.
        • Equation: $a_c = \dfrac{v^2}{r}$
      • 2.10.A.1.ii Centripetal acceleration is directed toward the center of an object's circular path.
    • 2.10.A.2 Centripetal acceleration can result from a single force, more than one force, or components of forces that are exerted on an object in circular motion.
      • 2.10.A.2.i At the top of a vertical, circular loop, an object requires a minimum speed to maintain circular motion. At this point, and with this minimum velocity, the gravitational force is the only force that causes the centripetal acceleration.
        • Derived equation: $v = \sqrt{gr}$
      • 2.10.A.2.ii Components of the static friction force and the normal force can contribute to the net force producing centripetal acceleration of an object traveling in a circle on a banked surface.
      • 2.10.A.2.iii A component of tension contributes to the net force producing centripetal acceleration experienced by a conical pendulum.
    • 2.10.A.3 Tangential acceleration is the rate at which an object's speed changes and is directed tangent to the object's circular path.
    • 2.10.A.4 The net acceleration of an object moving in a circle is the vector sum of the centripetal acceleration and tangential acceleration.
    • 2.10.A.5 The revolution of an object traveling in a circular path at a constant speed (uniform circular motion) can be described using period and frequency.
      • 2.10.A.5.i The time to complete one full circular path, one full rotation, or a full cycle of oscillatory motion is defined as period, $T$.
      • 2.10.A.5.ii The rate at which an object is completing revolutions is defined as frequency, $f$.
        • Equation: $T = \dfrac{1}{f}$
      • 2.10.A.5.iii For an object traveling at a constant speed in a circular path, the period is given by the derived equation
        • Derived equation: $T = \dfrac{2\pi r}{v}$

    2.10.B
    Describe circular orbits using Kepler's third law.

    • 2.10.B.1 For a satellite in circular orbit around a central body, the satellite's centripetal acceleration is caused only by gravitational attraction. The period and radius of the circular orbit are related to the mass of the central body.
      • Derived equation: $T^2 = \dfrac{4\pi^2}{GM} R^3$

    Boundary statement: AP Physics C: Mechanics does not expect students to know Kepler's first or second laws of planetary motion.

    Source: College Board AP Course and Exam Description · ⁨출처: College Board AP Course and Exam Description⁩

    Uniform circular motion

    Uniform circular motion has a centripetal acceleration 向心加速度 $a_c=\dfrac{v^2}{r}$ pointing to the center, requiring a net inward force $F_c=\dfrac{mv^2}{r}$ supplied by a real force (tension, gravity, friction, normal). There is no outward force. For vertical circles and banked curves, decompose the real forces to find which provides the centripetal requirement.

    Worked example. A $0.50\ \text{kg}$ ball is whirled on a $1.0\ \text{m}$ string in a horizontal circle at $3.0\ \text{m/s}$. The string tension supplies the whole centripetal force: $T=\dfrac{mv^2}{r}=\dfrac{0.50(3.0)^2}{1.0}=4.5\ \text{N}$.

    Worked example (vertical circle). At the top of a vertical loop of radius $r$, gravity and the normal force both point toward the center: $N+mg=\dfrac{mv^2}{r}$. The slowest possible speed at the top is where the track pushes with nothing at all ($N=0$): $v_{\min}=\sqrt{gr}$. For $r=2.5\ \text{m}$: $v_{\min}=\sqrt{9.8(2.5)}=4.9\ \text{m/s}$. Any slower and the cart leaves the track before the top.

    The velocity points along the tangent; the centripetal force points to the centre
    The velocity points along the tangent; the centripetal force points to the centre
    Vocabulary · ⁨어휘⁩ Train · ⁨연습하기⁩
    English 한국어
    centripetal acceleration/senˈtrɪpɪtl əkˌseləˈreɪʃn/ centripetal acceleration
    2.10

    Exam tips

    • Locate the center of mass with $x_{cm}=\tfrac{1}{M}\int x\,dm$ (or $\tfrac{\sum m_i x_i}{\sum m_i}$ for point masses) and use $\lambda,\sigma,\rho$ for the mass element.
    • The net external force moves the center of mass as if all mass sat there: $\vec F_{net}=M\vec a_{cm}$.
    • Define your system clearly — internal forces cancel, so only external forces change its momentum.
    • Exploit symmetry to shortcut a center-of-mass integral.
    • Distinguish center of mass from center of gravity (identical in a uniform field).
  • 3

    Work, Energy, and Power

    Watch lesson · ⁨수업 보기⁩
    3.1

    Translational Kinetic Energy

    Syllabus
    English

    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.
    한국어

    학습 목표 3.1.A: 물체의 질량과 속도를 통해 물체의 병진 운동 에너지를 설명한다.

    • 3.1.A.1 물체의 병진 운동 에너지는 다음 식으로 주어진다.
      • 식: $K = \dfrac{1}{2}mv^2$
    • 3.1.A.2 병진 운동 에너지는 스칼라량이다.
    • 3.1.A.3 관측자의 기준계에 따라 물체의 병진 운동 에너지 측정값이 다를 수 있다.

    Source: College Board AP Course and Exam Description · ⁨출처: College Board AP Course and Exam Description⁩

    Kinetic energy 动能 is the energy of motion, a scalar 标量 measured in joules 焦耳 (J):

    $$K=\tfrac{1}{2}mv^2.$$

    It grows with the square of speed – doubling the speed quadruples $K$. One subtlety worth knowing: kinetic energy depends on the observer's reference frame 参考系. A passenger walking down a train has a small $K$ measured inside the train and a huge one measured from the ground – both observers are right, each in their own frame.

    Vocabulary · ⁨어휘⁩ Train · ⁨연습하기⁩
    English 한국어
    Kinetic energy/kɪˈnetɪk ˈenədʒi/ 운동 에너지
    scalar/ˈskeɪlə/ 스칼라
    joules/dʒuːlz/ 줄
    reference frame/ˈrefrəns freɪm/ 기준 좌표계
    3.2

    Work

    Syllabus
    English

    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 The most common nonconservative forces are friction and air resistance.
    • 3.2.A.2 Work is a scalar quantity that may be positive, negative, or zero.
    • 3.2.A.3 The work done on an object by a variable force is calculated using
      • Equation: $W = \displaystyle\int_a^b \vec{F}(r) \cdot d\vec{r}$, where the integral is taken over the path from point $a$ to point $b$.
      • 3.2.A.3.i The dot product between two vectors, $\vec{A}$ and $\vec{B}$, results in a scalar quantity of magnitude $\vec{A} \cdot \vec{B} = AB\cos\theta$.
      • 3.2.A.3.ii Only the component of the force exerted on a system that is parallel to the displacement of the point of application of the force will change the system's total energy.
      • 3.2.A.3.iii If the component of the force exerted on a system that is parallel to the displacement is constant, the work done on the system by the force is given by the derived equation $W = F_{\parallel}d = Fd\cos\theta$.
      • 3.2.A.3.iv The component of the force exerted on a system perpendicular to the direction of the displacement of the system's center of mass can change the direction of the system's motion without changing the system's kinetic energy.
    • 3.2.A.4 The work–energy theorem states that the change in an object's kinetic energy is equal to the sum of the work (net work) being done by all forces exerted on the object.
      • Equation: $\Delta K = \displaystyle\sum W_i = \sum F_{\parallel,i}\, d_i$
      • 3.2.A.4.i An external force may change the configuration of a system. The component of the external force parallel to the displacement times the displacement of the point of application of the force gives the change in kinetic energy of the system.
      • 3.2.A.4.ii If the system's center of mass and the point of application of the force move the same distance when a force is exerted on a system, then the system may be modeled as an object, and only the system's kinetic energy can change.
      • 3.2.A.4.iii The energy dissipated by friction is typically equated to the force of friction times the length of the path over which the force is exerted.
        • Equation: $\Delta E_{\text{mech}} = F_f d\cos\theta$
      • 3.2.A.5 Work is equal to the area under the curve of a graph of $F_{\parallel}$ as a function of displacement.

    Boundary statement: AP Physics C: Mechanics only expects students to analyze the transfer of mechanical energy, although students should be aware that mechanical energy may be dissipated in the form of thermal energy or sound.

    한국어

    학습 목표 3.2.A: 주어진 힘이나 힘의 집합에 의해 물체나 계에 가해진 일을 설명한다.

    • 3.2.A.1 일은 한 힘이 그系统进行 거리 동안 가할 때 시스템으로 들어오거나 나가는 에너지의 양이다.
      • 3.2.A.1.i 보존력에 의한 시스템에 대한 일은 경로와 무관하며 오직该系统의 초기 및 최종 구성에만 의존한다.
      • 3.2.A.1.ii 보존력에 의한 시스템에 대한 일, 혹은 시스템의ポテンシャル 에너지의 변화량은该系统가 초기 구성으로 돌아갈 경우 0이 된다.
      • 3.2.A.1.iii 포텐셜 에너지는 오직 보존력과 관련된다.
      • 3.2.A.1.iv 비보존력에 의한 일은 경로에 의존한다.
      • 3.2.A.1.v 가장 일반적인 비보존력은 마찰력과 공기저항입니다.
    • 3.2.A.2 일은 양수, 음수, 또는 0이 될 수 있는 스칼라량이다.
    • 3.2.A.3 가변력에 의한 물체에게 행해지는 일은 다음을 사용하여 계산합니다:
      • 식: $W = \displaystyle\int_a^b \vec{F}(r) \cdot d\vec{r}$, 여기서 적분은 점 $a$부터 점 $b$까지의 경로에 대해 수행됩니다.
      • 3.2.A.3.i 두 터 $\vec{A}$과 $\vec{B}$ 사이의 내적은 크기가 $\vec{A} \cdot \vec{B} = AB\cos\theta$인 스칼라 양을 산출합니다.
      • 3.2.A.3.ii 시스템에 작용하는 힘 중 힘의 작용점의 변위와 평행한 성분만이 시스템의 총 에너지를 변화시니다.
      • 3.2.A.3.iii 시스템에 작용하는 힘 중 변위와 평행한 성분이 일정할 경우, 힘에 의한 시스템에게 행해지는 일은 유도된 식 $W = F_{\parallel}d = Fd\cos\theta$로 주어집니다.
      • 3.2.A.3.iv 시스템 중심축의 변위 방향에 수직인 힘의 성분은 시스템의 운동 방향을 변경할 수 있으나 시스템의 운동 에너지는 변경하지 않습니다.
    • 3.2.A.4 일-에너지 정리란 물체의 운동 에너지 변화량은 물체에 작용하는 모든 힘에 의해 행해지는 일(합일)의 합과 같다는 것입니다.
      • 식: $\Delta K = \displaystyle\sum W_i = \sum F_{\parallel,i}\, d_i$
      • 3.2.A.4.i 외부 힘은 시스템의 구성을 변경시킬 수 있다. 변위에 평행한 외부 힘의 성분과 힘의 적용점의 변위를 곱하면 시스템의 운동 에너지 변화량을 구할 수 있다.
      • 3.2.A.4.ii 시스템의 질량 중심과 힘의 적용점이 같은 거리를 이동한다면,该系统는 물체로 모델링될 수 있으며 시스템의 운동 에너지만 변경될 수 있다.
      • 3.2.A.4.iii 마찰력에 의해 소산되는 에너지는 일반적으로 마찰력의 크기에 해당 힘이 작용하는 경로의 길이를 곱한 값과 같습니다.
        • 식: $\Delta E_{\text{mech}} = F_f d\cos\theta$
      • 3.2.A.5 일은 변위를 변수로 한 $F_{\parallel}$ 그래프 아래 면적과 같습니다.

    경계 문구: AP Physics C: Mechanics에서는 학생들이 기계 에너지의 전이만 분석하도록 요구하지만, 기계 에너지가 열에너지 또는 소리 형태로 소산될 수 있음을 인지하고 있어야 합니다.

    Source: College Board AP Course and Exam Description · ⁨출처: College Board AP Course and Exam Description⁩

    Work 功 is energy transferred into or out of a system by a force acting over a distance. For a variable force it is an integral 积分 along the path:

    $$W=\int_a^b \vec{F}\cdot d\vec{r}=\int F\cos\theta\,dr.$$

    For a constant force this reduces to $W=Fd\cos\theta$; on a force–position graph, work is the area under the curve. Work is a scalar with a sign, and the sign is physics, not bookkeeping:

    • Positive – force has a component along the motion (it speeds the object up).
    • Negative – force opposes the motion (friction 摩擦力 and air drag do negative work).
    • Zero – force perpendicular to the motion (the normal force 法向力 on a sliding block, gravity on a horizontal move, tension in a circular swing).
    Only the force component along the displacement does work
    Only the force component along the displacement does work

    The work–energy theorem 动能定理 collects every force's contribution: the net work equals the change in kinetic energy,

    $$W_{\text{net}}=\sum W_i=\Delta K.$$

    Worked example. A variable force $F(x)=3x^2\ \text{N}$ (along the motion) acts from $x=0$ to $x=2\ \text{m}$: $W=\displaystyle\int_0^2 3x^2\,dx=\big[x^3\big]_0^2=8\ \text{J}$. Acting alone on a body starting from rest, it would raise the kinetic energy to exactly $8\ \text{J}$.

    Vocabulary · ⁨어휘⁩ Train · ⁨연습하기⁩
    English 한국어
    Work/wɜːk/ 일
    integral/ˈɪntɪɡrəl/ 적분Unless integral.
    friction/ˈfrɪkʃn/ 마찰력
    normal force/ˈnɔːml fɔːs/ 반작용력
    work–energy theorem/wɜːk ˈenədʒi ˈθɪərəm/ 일-에너지 정리
    3.3

    Potential Energy

    Syllabus
    English

    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 relationship between conservative forces exerted on a system and the system's potential energy is
      • Equation: $\Delta U = -\displaystyle\int_a^b \vec{F}_{cf}(r) \cdot d\vec{r}$
    • 3.3.A.5 The conservative forces exerted on a system in a single dimension can be determined using the slope of the system's potential energy with respect to position in that dimension; these forces point in the direction of decreasing potential energy.
      • Equation: $F_x = -\dfrac{dU(x)}{dx}$
    • 3.3.A.6 Graphs of a system's potential energy as a function of its position can be useful in determining physical properties of that system.
      • 3.3.A.6.i Stable equilibrium is a location at which a small displacement in an object's position results in a force exerted on the object opposite to the direction of the small displacement, accelerating the object back toward the equilibrium position.
      • 3.3.A.6.ii Unstable equilibrium is a location at which a small displacement in an object's position results in a force exerted on the object in the same direction as the small displacement, accelerating the object away from the equilibrium position.
      • 3.3.A.6.iii In a given dimension, stable equilibrium positions exist at locations where the potential energy as a function of position in that dimension has a local minimum.
      • 3.3.A.6.iv In a given dimension, unstable equilibrium positions occur at locations where the potential energy as a function of position in that dimension has a local maximum.
    • 3.3.A.7 The potential energy of common physical systems can be described using the physical properties of that system.
      • 3.3.A.7.i The elastic potential energy of an ideal spring is given by the following equation, where $\Delta x$ is the distance the spring has been stretched or compressed from its equilibrium length.
        • Equation: $U_s = \dfrac{1}{2}k(\Delta x)^2$
      • 3.3.A.7.ii The general form for the gravitational potential energy of a system consisting of two approximately spherical distributions of mass (e.g., moons, planets, or stars) is given by the equation
        • Equation: $U_g = -G\dfrac{m_1 m_2}{r}$
      • 3.3.A.7.iii Because the gravitational field near the surface of a planet is nearly constant, the change in gravitational potential energy in a system consisting of an object with mass $m$ and a planet with gravitational field of magnitude $g$ when the object is near the surface of the planet may be approximated by the equation
        • Equation: $\Delta U_g = mg\Delta y$
    • 3.3.A.8 The total potential energy of a system containing more than two objects is the sum of the potential energy of each pair of objects within the system.
    한국어

    학습 목표 3.3.A: 시스템의 포텐셜 에너지를 설명한다.

    • 3.3.A.1 두 개 이상의 물체로 구성된 시스템은该系统 내의 물체들이 서로 보존력만으로 상호작용할 경우 포텐셜 에너지를 가진다.
    • 3.3.A.2 포텐셜 에너지는该系统 내의 물체의 위치와 관련된 스칼라량이다.
    • 3.3.A.3 특정 시스템에 대한 포텐셜 에너지의 0점 정의는 상황을 고려하여 분석을 단순화하거나 돕기 위해 관측자가 결정하는 사항이다.
    • 3.3.A.4 시스템에 작용하는 보존력과 시스템의ポテン셜 에너지 사이의 관계는 다음과 같습니다.
      • 식: $\Delta U = -\displaystyle\int_a^b \vec{F}_{cf}(r) \cdot d\vec{r}$
    • 3.3.A.5 단일 차원에서의 시스템에 작용하는 보존력은 해당 차원의 포지션에 대한 시스템의ポテン셜 에너지의 기울기를 사용하여 결정할 수 있으며, 이 힘들은ポテン셜 에너지가 감소하는 방향으로 향합니다.
      • 식: $F_x = -\dfrac{dU(x)}{dx}$
    • 3.3.A.6 시스템의 포지션에 따른ポテン셜 에너지 그래프는 해당 시스템의 물리적 특성을 파악하는 데 유용할 수 있습니다.
      • 3.3.A.6.i 안정 평형은 물체의 위치에 작은 변위가 있을 때 그 변위와 반대 방향으로 물체에 힘이 작용하여 물체를 평형 위치로 다시 가속시키는 지점입니다.
      • 3.3.A.6.ii 불안정 평형은 물체의 위치에 작은 변위가 있을 때 그 변위와 같은 방향으로 물체에 힘이 작용하여 물체를 평형 위치에서 멀어지게 가속시키는 지점입니다.
      • 3.3.A.6.iii 특정 차원에서 안정 평형 위치는 해당 차원에 대한 포지션에 따른ポテン셜 에너지가 국소 최소값을 갖는 지점에 존재합니다.
      • 3.3.A.6.iv 특정 차원에서 불안정 평형 위치는 해당 차원에 대한 포지션에 따른ポテン셜 에너지가 국소 최대값을 갖는 지점에 발생합니다.
    • 3.3.A.7 일반적인 물리 시스템의ポテン셜 에너지는 해당 시스템의 물리적 특성을 사용하여 설명할 수 있습니다.
      • 3.3.A.7.i 이상 스프링의 탄성포텐셜 에너지는 다음 식으로 주며, 여기서 $\Delta x$는 스프링이 평형 길이로부터 늘어나거나 압축된 거리입니다.
        • 식: $U_s = \dfrac{1}{2}k(\Delta x)^2$
      • 3.3.A.7.ii 대략적인 구형 질량 분포(예: 위성, 행성 또는 별)로 구성된 시스템의 중력포텐셜 에너지의 일반 형태는 다음 식으로 주어집니다.
        • 식: $U_g = -G\dfrac{m_1 m_2}{r}$
      • 3.3.A.7.iii 행성 표면 근처의 중력장은 거의 일정하므로, 질량이 $m$인 물체와 중력장 세기가 $g$인 행성이 있는 시스템에서 물체가 행성 표면 근처에 있을 때 중력포텐셜 에너지의 변화량은 다음 식으로 근사할 수 있습니다.
        • 식: $\Delta U_g = mg\Delta y$
    • 3.3.A.8 두 개 이상의 물체를 포함하는 시스템의 총ポテン셜 에너지는 시스템 내 각 물체 쌍의ポテン셜 에너지의 합입니다.

    Source: College Board AP Course and Exam Description · ⁨출처: College Board AP Course and Exam Description⁩

    Potential energy 势能 is energy a system stores by the positions of its parts – it exists only for conservative forces 保守力, whose work is path independent. It is defined through work:

    $$\Delta U=-\int_a^b\vec{F}\cdot d\vec{r},$$

    and you are free to choose where $U=0$ – only changes in $U$ matter, so pick the zero that makes the problem simplest. The standard results:

    • Gravity near a surface: $\Delta U_g=mg\,\Delta y$.
    • Gravity in general: $U_g=-\dfrac{Gm_1m_2}{r}$ (zero at infinite separation).
    • Spring: $U_s=\tfrac12k(\Delta x)^2$, with $\Delta x$ measured from natural length.

    For systems of several objects, add the potential energy of each pair. Turning the definition around, a conservative force is minus the derivative 导数 of its potential energy:

    $$F_x=-\frac{dU}{dx}.$$

    The force points "downhill" on the $U(x)$ curve. Equilibrium sits where the slope is zero: a minimum is a stable equilibrium 稳定平衡 (displaced, the force pushes back), a maximum is an unstable equilibrium 不稳定平衡 (displaced, the force pushes away).

    On a potential-energy curve the force points downhill and E = U marks the turning points
    On a potential-energy curve the force points downhill and E = U marks the turning points

    Worked example. Given $U(x)=2x^3-6x$ (joules), the force is $F=-\dfrac{dU}{dx}=6-6x^2$. Equilibria sit at $F=0$: $x=\pm1$. Since $\dfrac{d^2U}{dx^2}=12x$ is positive at $x=+1$ (a minimum – stable) and negative at $x=-1$ (a maximum – unstable), the two points behave oppositely.

    Explore · ⁨탐색하기⁩

    Store elastic potential energy in a spring · ⁨스프링에 탄성 위치에너지 저장하기⁩

    Stretching a spring stores elastic potential energy $\tfrac12 kx^2$ — the area under the force-extension line. Release it and that energy becomes kinetic. · ⁨스프링을 늘리면 탄력 Potential 에너지 $\tfrac12 kx^2$가 저장되는데 — 이는 힘-신도 선 아랫면적입니다. 놓아주면 이 에너지가 운동 에너지로 전환됩니다.⁩

    Vocabulary · ⁨어휘⁩ Train · ⁨연습하기⁩
    English 한국어
    Potential energy/pəˈtenʃl ˈenədʒi/ 위치 에너지
    conservative forces/kənˈsɜːvətɪv ˈfɔːsɪz/ 보존력
    derivative/dɪˈrɪvətɪv/ 미분Unless derivative.
    stable equilibrium/ˈsteɪbl ˌiːkwɪˈlɪbrɪəm/ 안정 평형
    unstable equilibrium/ʌnˈsteɪbl ˌiːkwɪˈlɪbrɪəm/ 불안정 평형
    3.4

    Conservation of Energy

    Syllabus
    English

    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 C: Mechanics expects students to know that mechanical energy can be dissipated as thermal energy or sound by nonconservative forces.

    한국어

    학습 목표 3.4.A: 시스템에 존재하는 에너지를 설명한다.

    • 3.4.A.1 단일 물체만으로 구성된 시스템은 운동 에너지만 가질 수 있다.
    • 3.4.A.2 보존력을 통해 상호작용하거나 형상을 가역적으로 변경할 수 있는 물체를 포함하는 시스템은 운동 에너지와 포텐셜 에너지를 모두 가질 수 있다.

    학습 목표 3.4.B: 역학적 에너지 보존 원리를 사용하여 시스템의 거동을 설명한다.

    • 3.4.B.1 역학적 에너지는 시스템의 운동에너지와 위치에너지의 합이다.
    • 3.4.B.2 시스템 내 한 형태의 에너지가 변화할 경우,该系统 내 다른 형태의 에너지에 동일한 크기의 변화가 일어나거나 시스템과 주변 환경 간에 에너지가 이동해야 한다.
    • 3.4.B.3 특정 시스템의 총 에너지를 일정하게 만들 수 있도록 시스템을 선택할 수 있다.
    • 3.4.B.4 시스템의 총 에너지가 변화할 경우, 그 변화량은 시스템으로 유입되거나 시스템에서 유출된 에너지와 동일하다.

    학습 목표 3.4.C: 시스템의 선택이 해당 시스템의 에너지 변화 여부에 미치는 영향을 설명하라.

    • 3.4.C.1 모든 상호작용에서 에너지는 보존된다.
    • 3.4.C.2 선택된 시스템에 가해진仕事が 0이며 시스템 내에 비보존적 상호작용이 없으면, 시스템의 총 역학적 에너지는 일정하다.
    • 3.4.C.3 선택한 물체에 가해진 일이 영이 아닌 경우, 에너지가 물체와 환경 사이를 이동한다.

    경계 문구: AP Physics C: Mechanics에서는 학생들이 비보존력에 의해 기계 에너지가 열에너지 또는 소리로 소산될 수 있음을 알고 있어야 한다고 요구합니다.

    Source: College Board AP Course and Exam Description · ⁨출처: College Board AP Course and Exam Description⁩

    Energy conservation: KE ⇄ PE

    Energy is conserved in all interactions – the question is only where it goes. Mechanical energy 机械能 is the sum $E=K+U$. If the external work on a system is zero and nothing inside it acts through nonconservative forces 非保守力, then

    $$K_1+U_1=K_2+U_2.$$

    When friction or drag act, they convert mechanical energy into thermal energy 热能 ($\Delta E_{\text{mech}}=-F_fd$), and the balance must include that term. If external work is done, the system's total energy changes by exactly the energy transferred: $W_{\text{ext}}=\Delta E_{\text{sys}}$. Choosing the system is choosing the bookkeeping – a single object can only have kinetic energy; include the Earth or the spring and the system can store potential energy too.

    A pendulum trades gravitational potential energy for kinetic energy and back
    A pendulum trades gravitational potential energy for kinetic energy and back

    A potential-energy graph is a complete motion map: the horizontal line at height $E$ is the total energy, the gap $E-U(x)$ is the kinetic energy at each $x$, and the crossings $E=U$ are the turning points 转折点 where the object momentarily stops and reverses.

    Worked example. A $2.0\ \text{kg}$ block slides down a ramp from rest at height $1.5\ \text{m}$, arriving at the bottom at $4.0\ \text{m/s}$. Energy accounting: $mgh=29.4\ \text{J}$ available; $\tfrac12mv^2=16\ \text{J}$ arrives as kinetic energy; so friction converted $29.4-16=13\ \text{J}$ into thermal energy along the way.

    A roller coaster trades energy back and forth: highest (most PE) at the top, fastest (most KE) at the bottom
    A roller coaster trades energy back and forth: highest (most PE) at the top, fastest (most KE) at the bottom
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    Watch energy convert as an object falls · ⁨物体 falling 시 에너지 변환 Watching⁩

    With no friction, mechanical energy is conserved: as an object falls, gravitational potential energy turns into kinetic energy while the total stays fixed. · ⁨마찰이 없을 때, 기계적 에너지 보존: Object falling 시 중력 **위치에너지(potential energy)**가 **운동에너지(kinetic energy)**로 전환되며 총량은 일정합니다.⁩

    Vocabulary · ⁨어휘⁩ Train · ⁨연습하기⁩
    English 한국어
    Mechanical energy/mɪˈkænɪkl ˈenədʒi/ 역학적 에너지
    nonconservative forces/ˌnɒŋkənˈsɜːvətɪv ˈfɔːsɪz/ 비보존력
    thermal energy/ˈθɜːml ˈenədʒi/ 열에너지
    turning points/ˈtɜːnɪŋ pɔɪnts/ 회전 지점
    Watch lesson · ⁨수업 보기⁩
    3.5

    Power

    Syllabus
    English

    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.
      • Equation: $P_{\text{avg}} = \dfrac{\Delta E}{\Delta t}$
    • 3.5.A.3 Because work is the change in energy of an object or system due to a force, average power is the total work done, divided by the time during which that work was done.
      • Equation: $P_{\text{avg}} = \dfrac{W}{\Delta t}$
    • 3.5.A.4 The instantaneous power delivered to an object by a force is given by the equation
      • Equation: $P_{\text{inst}} = \dfrac{dW}{dt}$
    • 3.5.A.5 The instantaneous power delivered to an object by the component of a constant force parallel to the object's velocity can be described with the derived equation
      • Equation: $P_{\text{inst}} = F_{\parallel}v = Fv\cos\theta$
    한국어

    학습 목표 3.5.A: 출력을 사용하여 시스템 안으로, 밖으로, 혹은 내부로 에너지가 이동하는 과정을 설명하라.

    • 3.5.A.1 출력은 시간당 에너지의 변화율로, 시스템으로 유입되거나 유출되는 경우이거나 시스템 내에서 한 형태에서 다른 형태로 전환되는 경우를 의미한다.
    • 3.5.A.2 평균 출력은 이동하거나 전환된 에너지의 양을, 해당 이동이나 전환이发生的所需时间로 나눈 값이다.
      • 식: $P_{\text{avg}} = \dfrac{\Delta E}{\Delta t}$
    • 3.5.A.3 일은 힘에 의해 물체나 시스템의 에너지가 변한 것이므로, 평균 출력은 행해진 총 일량을 그 일이 행해진 시간으로 나눈 값이다.
      • 식: $P_{\text{avg}} = \dfrac{W}{\Delta t}$
    • 3.5.A.4 힘에 의한 물체에 전달되는 순간 출력은 다음 식으로 주어집니다.
      • 식: $P_{\text{inst}} = \dfrac{dW}{dt}$
    • 3.5.A.5 물체의 속도와 평행한 상수힘의 성분에 의한 물체에 전달되는 순간 출력은 유도된 식으로 설명할 수 있습니다.
      • 식: $P_{\text{inst}} = F_{\parallel}v = Fv\cos\theta$

    Source: College Board AP Course and Exam Description · ⁨출처: College Board AP Course and Exam Description⁩

    Power 功率 is the rate of energy transfer, in watts 瓦特 (W):

    $$P_{\text{avg}}=\frac{\Delta E}{\Delta t}=\frac{W}{\Delta t},\qquad P_{\text{inst}}=\frac{dW}{dt}=\vec{F}\cdot\vec{v}.$$

    It measures how fast work is done, not how much. The dot product matters: only the force component along the velocity delivers power.

    Power is the slope of the work-time graph: the same work in less time means more power
    Power is the slope of the work-time graph: the same work in less time means more power

    Worked example. A block released from rest at the top of a frictionless $3.0\ \text{m}$-high ramp reaches the bottom at $v=\sqrt{2gh}=7.7\ \text{m/s}$ (from $mgh=\tfrac12mv^2$). A motor that then drives it at a steady $7.7\ \text{m/s}$ against a $20\ \text{N}$ resistance delivers $P=Fv=20(7.7)\approx150\ \text{W}$.

    Worked example. A $1200\ \text{kg}$ car climbs a hill that rises $1.0\ \text{m}$ for every $20\ \text{m}$ of road, at a steady $15\ \text{m/s}$. The engine must supply gravity's power drain: $P=mg\,v\sin\theta=1200(9.8)(15)\big(\tfrac{1}{20}\big)\approx8.8\ \text{kW}$ – before adding air resistance.

    Exam skill. Energy FRQs reward the accounting sentence: name your system, state which forces do work on it, and write the balance ($W_{\text{ext}}=\Delta K+\Delta U+\Delta E_{\text{thermal}}$) before plugging in numbers. "Friction is present, so mechanical energy is not conserved" is a scored statement.

    Vocabulary · ⁨어휘⁩ Train · ⁨연습하기⁩
    English 한국어
    Power/ˈpaʊə/ 출력
    watts/wɒts/ 와트(watts)
    3.5

    Exam tips

    • Use the work–energy theorem $W_{net}=\Delta KE$ and compute work as $W=\int \vec F\cdot d\vec r$ for a variable force.
    • Read work off a force–position graph as the area under the curve.
    • Power is $P=\tfrac{dW}{dt}=\vec F\cdot\vec v$; watch instantaneous vs average.
    • Split forces into conservative (define a potential energy) and non-conservative (dissipate energy).
    • Choose energy methods over kinematics when the force varies or the path is complex.
  • 4

    Linear Momentum

    Watch lesson · ⁨수업 보기⁩
    4.1

    Linear Momentum

    Syllabus
    English

    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 $\vec{p} = m\vec{v}$.
    • 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.
    한국어

    학습 목표 4.1.A: 물체나 시스템의 선형 운동량을 설명하라.

    • 4.1.A.1 선형 운동량은 식 $\vec{p} = m\vec{v}$에 의해 정의된다.
    • 4.1.A.2 운동량은 벡터량이며 속도와 같은 방향을 가진다.
    • 4.1.A.3 운동량은 충돌 및 폭발 현상을 분석하는 데 사용될 수 있다.
      • 4.1.A.3.i 충돌은 시스템 내 관련 물체들 사이에 작용하는 힘이 상호작용 중 물체에 가해지는 외부 합력에 비해 훨씬 큰 상호작용을 모델링하는 것이다.
      • 4.1.A.3.ii 충돌의 초기 상태와 최종 상태만 분석되므로, 물체 모델을 사용하여 충돌을 분석할 수 있다.
      • 4.1.A.3.iii 폭발은 시스템 내부의 힘들이 시스템 내 물체들을 서로 멀어지게 하는 상호작용을 모델링하는 것이다.

    Source: College Board AP Course and Exam Description · ⁨출처: College Board AP Course and Exam Description⁩

    Linear momentum 动量 is mass times velocity:

    $$\vec{p}=m\vec{v}.$$

    It is a vector 矢量 pointing along the velocity, and it is the quantity for analysing collisions 碰撞 and explosions 爆炸. A collection of objects can be treated as one system moving with the velocity of its center of mass 质心:

    $$\vec{v}_{\text{cm}}=\frac{\sum m_i\vec{v}_i}{\sum m_i},$$

    so the system's total momentum is its total mass times $\vec{v}_{\text{cm}}$ – one object's worth of bookkeeping for any number of parts.

    Vocabulary · ⁨어휘⁩ Train · ⁨연습하기⁩
    English 한국어
    Linear momentum/ˈlɪnɪə məʊˈmentəm/ Linear momentum
    vector/ˈvektə/ 벡터
    collisions/kəˈlɪʒnz/ 충돌
    explosions/ekˈspləʊʒnz/ 폭발
    center of mass/ˈsentə ɒv mæs/ 질량 중심
    4.2

    Change in Momentum and Impulse

    Syllabus
    Learning ObjectiveEssential Knowledge

    4.2.A
    Describe the impulse delivered to an object or system.

    • 4.2.A.1 The rate of change of a system's momentum is equal to the net external force exerted on that system.
      • Equation: $\vec{F}_{\text{net}} = \dfrac{d\vec{p}}{dt}$
    • 4.2.A.2 Impulse is defined as the integral of a force exerted on an object or system over a time interval.
      • Equation: $\vec{J} = \displaystyle\int_{t_1}^{t_2} \vec{F}_{\text{net}}(t)\,dt$
    • 4.2.A.3 Impulse is a vector quantity and has the same direction as the net force exerted on the system.
    • 4.2.A.4 The impulse delivered to a system by a net external force is equal to the area under the curve of a graph of the net external force exerted on the system as a function of time.
    • 4.2.A.5 The net external force exerted on a system is equal to the slope of a graph of the momentum of the system as a function of time.

    4.2.B
    Describe the relationship between the impulse exerted on an object or system and the change in momentum of the object or system.

    • 4.2.B.1 Change in momentum is the difference between a system's final momentum and its initial momentum.
      • Equation: $\Delta\vec{p} = \vec{p} - \vec{p}_0$
    • 4.2.B.2 The impulse–momentum theorem relates the impulse delivered to an object and the object's change in momentum.
      • 4.2.B.2.i The impulse exerted on an object is equal to the object's change in momentum.
        • Equation: $\vec{J} = \displaystyle\int_{t_1}^{t_2} \vec{F}_{\text{net}}(t)\,dt = \Delta\vec{p}$
      • 4.2.B.2.ii Newton's second law of motion is a direct result of the impulse–momentum theorem applied to systems with constant mass.
        • Equation: $\vec{F}_{\text{net}} = \dfrac{d\vec{p}}{dt} = m\dfrac{d\vec{v}}{dt} = m\vec{a}$
      • 4.2.B.2.iii The impulse–momentum theorem also describes the behavior of a system in which the velocity is constant but the mass changes with respect to time.
        • Equation: $\vec{F}_{\text{net}} = \dfrac{d\vec{p}}{dt} = \dfrac{dm}{dt}\vec{v}$

    Source: College Board AP Course and Exam Description · ⁨출처: College Board AP Course and Exam Description⁩

    Newton's second law is really a statement about momentum:

    $$\vec{F}_{\text{net}}=\frac{d\vec{p}}{dt},$$

    which reduces to $m\vec{a}$ when the mass is constant – and handles a changing mass when it is not (the special case $\vec{F}=\dfrac{dm}{dt}\vec{v}$ holds only when the velocity is constant, e.g. a chain piling onto a scale or sand landing on a belt moving at fixed speed; a rocket does not fit, since it accelerates). The impulse 冲量 delivered by a force is its time integral, and the impulse–momentum theorem 冲量-动量定理 says it equals the change in momentum:

    $$\vec{J}=\int_{t_1}^{t_2}\vec{F}_{\text{net}}\,dt=\Delta\vec{p}.$$

    Impulse is a vector along the net force. Read it off graphs both ways:

    • On a force–time graph, impulse is the area under the curve.
    • On a momentum–time graph, the net force is the slope 斜率 at each instant.
    Impulse is the area under the force-time curve, equal to the average force times the contact time
    Impulse is the area under the force-time curve, equal to the average force times the contact time

    Spreading the same $\Delta p$ over a longer time lowers the force – that is why airbags, crumple zones, and soft landings work, and why you bend your knees when you land.

    Worked example. A time-varying force $F(t)=10t\ \text{N}$ acts on a $2.0\ \text{kg}$ object for $2.0\ \text{s}$. The impulse is $J=\displaystyle\int_0^2 10t\,dt=\big[5t^2\big]_0^2=20\ \text{N}\cdot\text{s}$, so the speed changes by $\Delta v=J/m=10\ \text{m/s}$.

    Vocabulary · ⁨어휘⁩ Train · ⁨연습하기⁩
    English 한국어
    impulse/ˈɪmpʌls/ 신호
    impulse–momentum theorem/ˈɪmpʌls məʊˈmentəm ˈθɪərəm/ 임펄스-운동량 정리
    slope/sləʊp/ 기울기
    4.3

    Conservation of Linear Momentum

    Syllabus
    Learning ObjectiveEssential Knowledge

    4.3.A
    Describe the behavior of a system using conservation of linear momentum.

    • 4.3.A.1 A collection of objects with individual momenta can be described as one system with one center-of-mass velocity.
      • 4.3.A.1.i For a collection of objects, the velocity of a system's center of mass can be calculated using the equation
        • Equation: $\vec{v}_{\text{cm}} = \dfrac{\sum \vec{p}_i}{\sum m_i} = \dfrac{\sum (m_i \vec{v}_i)}{\sum m_i}$
      • 4.3.A.1.ii The velocity of a system's center of mass is constant in the absence of a net external force.
    • 4.3.A.2 The total momentum of a system is the sum of the momenta of the system's constituent parts.
    • 4.3.A.3 In the absence of net external forces, any change to the momentum of an object within a system must be balanced by an equivalent and opposite change of momentum elsewhere within the system. Any change to the momentum of a system is due to a transfer of momentum between the system and its surroundings.
      • 4.3.A.3.i The impulse exerted by one object on a second object is equal and opposite to the impulse exerted by the second object on the first. This is a direct result of Newton's third law.
      • 4.3.A.3.ii A system may be selected so that the total momentum of that system is constant.
      • 4.3.A.3.iii If the total momentum of a system changes, that change will be equivalent to the impulse exerted on the system.
        • Equation: $\vec{J} = \Delta\vec{p}$
    • 4.3.A.4 Correct application of conservation of momentum can be used to determine the velocity of a system immediately before and immediately after collisions or explosions.

    Boundary statement: AP Physics C: Mechanics only expects students to quantitatively analyze collisions and interactions in one or two dimensions. Three-dimensional collisions may be analyzed qualitatively.

    4.3.B
    Describe how the selection of a system determines whether the momentum of that system changes.

    • 4.3.B.1 Momentum is conserved in all interactions.
    • 4.3.B.2 If the net external force on the selected system is zero, the total momentum of the system is constant.
    • 4.3.B.3 If the net external force on the selected system is nonzero, momentum is transferred between the system and the environment.

    Source: College Board AP Course and Exam Description · ⁨출처: College Board AP Course and Exam Description⁩

    Internal forces come in Newton's-third-law pairs, so they cancel inside any system: they can shuffle momentum between parts but never change the total. Momentum only enters or leaves a system through a net external force ($\vec{J}=\Delta\vec{p}$). So, with zero net external force, total momentum is conserved 守恒:

    $$\sum\vec{p}_{\text{before}}=\sum\vec{p}_{\text{after}}.$$

    Momentum is conserved in every collision, however violent – choose the system large enough that the collision forces are internal. Apply conservation separately along each axis; AP asks for quantitative work in one or two dimensions.

    A head-on collision: total momentum before equals total momentum after
    A head-on collision: total momentum before equals total momentum after

    Worked example (explosion). A $6.0\ \text{kg}$ shell at rest splits into a $2.0\ \text{kg}$ piece moving at $9.0\ \text{m/s}$ east and a $4.0\ \text{kg}$ piece. Total momentum stays zero, so the heavy piece moves west at $v=\dfrac{2.0(9.0)}{4.0}=4.5\ \text{m/s}$. The kinetic energy came from stored (chemical or spring) energy – momentum conservation does not require kinetic-energy conservation.

    Explore · ⁨탐색하기⁩

    Collide two carts and conserve momentum · ⁨두 카트 충돌 및 운동량 보존⁩

    In any collision the total momentum $\sum mv$ before equals the total after. Set the masses and speeds and check the momentum bookkeeping. · ⁨모든 충돌에서 충돌 전 총운동량 $\sum mv$는 충돌 후와 같습니다. 질량과 속도를 설정하여 운동량 계산이 맞는지 확인해 보십시오.⁩

    Vocabulary · ⁨어휘⁩ Train · ⁨연습하기⁩
    English 한국어
    conserved/kənˈsɜːvd/ 보존된다
    4.4

    Elastic and Inelastic Collisions

    Syllabus
    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.
    한국어

    학습 목표 4.4.A: 물체 간의 상호작용이 탄성인지 비탄성인지 서술한다.

    • 4.4.A.1 탄성 충돌이란 시스템의 초기 운동 에너지가 최종 운동 에너지와 동일한 충돌이다.
    • 4.4.A.2 탄성 충돌에서 시스템 내 각 물체의 최종 운동 에너지는 초기 운동 에너지와 다를 수 있다.
    • 4.4.A.3 비탄성 충돌이란 시스템의 총 운동 에너지가 감소하는 충돌이다.
    • 4.4.A.4 비탄성 충돌에서 일부 초기 운동 에너지는 운동 에너지로 회복되지 않고 비보존력에 의해 다른 형태의 에너지로 변환된다.
    • 4.4.A.5 완벽 비탄성 충돌에서는 물체가 서로 붙어 충돌 후 동일한 속도로 움직인다.

    Source: College Board AP Course and Exam Description · ⁨출처: College Board AP Course and Exam Description⁩

    Conservation of momentum in a collision

    All collisions conserve momentum; they differ in what happens to the kinetic energy 动能:

    type momentum kinetic energy
    elastic collision 弹性碰撞 conserved total conserved (individual shares may change)
    inelastic collision 非弹性碰撞 conserved decreases – some becomes heat, sound, deformation 形变
    perfectly inelastic collision 完全非弹性碰撞 conserved largest possible loss – the objects stick and share one velocity

    Strategy: always write momentum conservation first; add the kinetic-energy equation only when the problem says "elastic". Two useful elastic facts: equal masses in a 1D elastic collision simply exchange velocities, and in the center-of-mass frame each object just reverses its velocity.

    Worked example. A $1000\ \text{kg}$ car at $20\ \text{m/s}$ strikes a stationary $1500\ \text{kg}$ car and they lock together – perfectly inelastic. Momentum: $v=\dfrac{1000(20)}{2500}=8.0\ \text{m/s}$. Kinetic energy falls from $2.0\times10^5\ \text{J}$ to $\tfrac12(2500)(8.0)^2=8.0\times10^4\ \text{J}$: about $60\%$ is lost, even though momentum is exactly conserved.

    Worked example (2D). A puck moving east at $4.0\ \text{m/s}$ strikes an identical puck at rest; after the glancing hit, one moves at $2.0\ \text{m/s}$ at $60^\circ$ north of east. Conserve each axis: east–west, $m(4.0)=m(2.0)\cos60^\circ+mv_x$, so $v_x=3.0\ \text{m/s}$; north–south, $0=m(2.0)\sin60^\circ-mv_y$, so $v_y=1.7\ \text{m/s}$. The second puck moves at $\sqrt{3.0^2+1.7^2}=3.5\ \text{m/s}$, about $30^\circ$ south of east.

    A glancing collision, resolved along two perpendicular axes
    A glancing collision, resolved along two perpendicular axes

    Exam skill. On FRQs, justify with the condition, not the slogan: "the net external force on the two-puck system is zero during the collision, so its total momentum is constant." If asked whether the collision is elastic, compute the kinetic energy before and after and compare – never assume.

    A Newton's cradle: five steel balls hanging in a row
    A Newton's cradle shows momentum and kinetic energy passing through a near-elastic collision
    Explore · ⁨탐색하기⁩

    Compare elastic and inelastic collisions · ⁨탄성 충돌과 비탄성 충돌 비교⁩

    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. · ⁨운동량은 항상 보존되지만, 운동 에너지는 탄성 충돌에서만 보존됩니다. 비탄성 충돌에서는 카트가 붙게 되며 일부 에너지가 열에너지로 변환됩니다.⁩

    Vocabulary · ⁨어휘⁩ Train · ⁨연습하기⁩
    English 한국어
    kinetic energy/kɪˈnetɪk ˈenədʒi/ 운동 에너지Unless kinetic energy.
    elastic collision/ɪˈlæstɪk kəˈlɪʒn/ 탄성 충돌
    inelastic collision/ɪnɪˈlæstɪk kəˈlɪʒn/ 비탄성 충돌
    deformation/ˌdiːfɔːˈmeɪʃn/ 변형
    perfectly inelastic collision/ˈpɜːfektlɪ ɪnɪˈlæstɪk kəˈlɪʒn/ 완전 비탄성 충돌
    4.4

    Exam tips

    • Impulse equals the momentum change: $\vec J=\int \vec F\,dt=\Delta\vec p$, and it is the area under a force–time graph.
    • Momentum is conserved whenever the net external force is zero — always the go-to for collisions.
    • Distinguish elastic (kinetic energy conserved) from inelastic (kinetic energy decreases) collisions; in a perfectly inelastic collision the objects stick and move with one velocity.
    • Apply conservation to each component (x and y) separately in 2-D.
    • Connect to center of mass: total momentum $=M\vec v_{cm}$.
  • 5

    Torque and Rotational Dynamics

    Watch lesson · ⁨수업 보기⁩
    5.1

    Rotational Kinematics

    Syllabus
    English

    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 Angular velocity is the rate at which angular position changes with respect to time.
      • Equation: $\omega = \dfrac{d\theta}{dt}$
    • 5.1.A.3 Angular acceleration is the rate at which angular velocity changes with respect to time.
      • Equation: $\alpha = \dfrac{d\omega}{dt}$
    • 5.1.A.4 Angular displacement, angular velocity, and angular acceleration around one axis are analogous to linear displacement, velocity, and acceleration in one dimension and demonstrate the same mathematical relationships.
      • 5.1.A.4.i For constant angular acceleration, the mathematical relationships between angular displacement, angular velocity, and angular acceleration can be described with the following equations:
        • $\omega = \omega_0 + \alpha t$
        • $\theta = \theta_0 + \omega_0 t + \dfrac{1}{2}\alpha t^2$
        • $\omega^2 = \omega_0^2 + 2\alpha(\theta - \theta_0)$
      • 5.1.A.4.ii Graphs of angular displacement, angular velocity, and angular acceleration as functions of time can be used to find the relationships between those quantities.

    Boundary statement: AP Physics C: Mechanics expects students to be able to mathematically manipulate the magnitudes of angular displacement, angular velocity, and angular acceleration using vector conventions. However, the directions of said vectors will not be assessed on the exam.

    Descriptions of the directions of rotational kinematics quantities for a point or rigid body are limited to clockwise and counterclockwise with respect to a given axis of rotation.

    한국어

    학습 목표 5.1.A: 각변위, 각속도 및 각가속도를 사용하여 시스템의 회전을 시간에 대해 서술한다.

    • 5.1.A.1 각변위는 경질 시스템 상의 한 점이 특정 축을 중심으로 회전하는 각도의 라디안 단위로 측정된 값이다.
      • 식: $\Delta\theta = \theta - \theta_0$
      • 5.1.A.1.i 경질 시스템은 형태를 유지하지만 회전 중 시스템의 다른 점들이 서로 다른 방향으로 움직이는 시스템이다. 경질 시스템은 단일 물체로 모델링할 수 없다.
      • 5.1.A.1.ii 회전축에 대한 각변위의 한 방향(시계 방향 또는 반시계 방향)은 일반적으로 수학적으로 양수(+), 다른 방향은 음수(-)로 표시된다.
      • 5.1.A.1.iii 시스템의 축에 대한 회전이 시스템의 질량 중심의 운동으로 잘 묘사될 수 있다면, 시스템을 단일 물체로 취급할 수 있다. 예를 들어, 지구-태양 질량 중심에 대한 지구의 공전을 고려할 때 지구의 자전은 무시할 수 있다.
    • 5.1.A.2 각속도는 각도 위치가 시간에 따라 변하는 비율이다.
      • 식: $\omega = \dfrac{d\theta}{dt}$
    • 5.1.A.3 각가속도는 각속도가 시간에 따라 변하는 비율이다.
      • 식: $\alpha = \dfrac{d\omega}{dt}$
    • 5.1.A.4 한 축에 대한 각변위, 각속도, 각가속도는 한 차원의 선형 변위, 속도, 가속도와 유사하며 동일한 수학적 관계를 나타낸다.
      • 5.1.A.4.i 일정 각가속도일 때, 각변위, 각속도, 각가속도 사이의 수학적 관계는 다음 식으로 설명할 수 있다:
        • $\omega = \omega_0 + \alpha t$
        • $\theta = \theta_0 + \omega_0 t + \dfrac{1}{2}\alpha t^2$
        • $\omega^2 = \omega_0^2 + 2\alpha(\theta - \theta_0)$
      • 5.1.A.4.ii 시간 함수로서의 각변위, 각속도, 각가속도의 그래프를 사용하여 해당 물리량 간의 관계를 찾을 수 있다.

    범위 규정: AP 물리 C: 역학에서는 학생들에게 터 규약을 사용하여 각변위, 각속도, 각가속도의 크기를 수학적으로 조작할 수 있어야 한다. 그러나 이러한 벡터들의 방향은 시험에서 평가되지 않는다.

    점이나 강체에 대한 회전 운동학 양의 방향 설명은 주어진 회전 축에 대해 시계 방향과 반시계 방향으로 제한된다.

    Source: College Board AP Course and Exam Description · ⁨출처: College Board AP Course and Exam Description⁩

    Rotation mirrors linear motion with angular quantities, defined as derivatives:

    $$\omega=\frac{d\theta}{dt},\qquad \alpha=\frac{d\omega}{dt}=\frac{d^2\theta}{dt^2}.$$

    Here $\theta$ is angular displacement 角位移 in radians 弧度, $\omega$ the angular velocity 角速度, and $\alpha$ the angular acceleration 角加速度. (AP works with their magnitudes and signs; the 3D vector directions are not assessed.) For constant $\alpha$, the kinematic equations are the linear ones re-lettered:

    $$\omega=\omega_0+\alpha t,\qquad \Delta\theta=\omega_0t+\tfrac12\alpha t^2,\qquad \omega^2=\omega_0^2+2\alpha\,\Delta\theta.$$
    One radian is the angle whose arc length equals the radius
    One radian is the angle whose arc length equals the radius

    Worked example. A fan blade slows from $30\ \text{rad/s}$ to rest with $\alpha=-6.0\ \text{rad/s}^2$: it takes $t=5.0\ \text{s}$ and turns through $\Delta\theta=\dfrac{0-30^2}{2(-6.0)}=75\ \text{rad}$ – about $12$ revolutions.

    Vocabulary · ⁨어휘⁩ Train · ⁨연습하기⁩
    English 한국어
    angular displacement/ˈæŋɡjʊlə dɪˈspleɪsmənt/ 각변위
    radians/ˈreɪdɪənz/ 라디안(radians)
    angular velocity/ˈæŋɡjʊlə vəˈlɒsɪti/ 각속도
    angular acceleration/ˈæŋɡjʊlə əkˌseləˈreɪʃn/ 각가속도
    5.2

    Connecting Linear and Rotational Motion

    Syllabus
    Learning ObjectiveEssential Knowledge

    5.2.A
    Describe the linear motion of a point on a rotating rigid system that corresponds to the rotational motion of that point, and vice versa.

    • 5.2.A.1 For a point at a distance $r$ from a fixed axis of rotation, the linear distance $s$ traveled by the point as the system rotates through an angle $\Delta\theta$ is given by the equation $\Delta s = r\Delta\theta$.
    • 5.2.A.2 Derived relationships of linear velocity and of the tangential component of acceleration to their respective angular quantities are given by the following equations:
      • $s = r\theta$
      • $v = r\omega$
      • $a_T = r\alpha$
    • 5.2.A.3 For a rigid system, all points within that system have the same angular velocity and angular acceleration.

    Boundary statement: AP Physics C: Mechanics expects students to be able to mathematically manipulate the magnitudes of angular displacement, angular velocity, and angular acceleration using vector conventions. However, the directions of the vectors will not be assessed on the exam.

    Descriptions of the directions of rotational kinematics quantities for a point or rigid body are limited to clockwise and counterclockwise with respect to a given axis of rotation.

    Source: College Board AP Course and Exam Description · ⁨출처: College Board AP Course and Exam Description⁩

    A point at radius $r$ from the axis moves along its arc with

    $$s=r\theta,\qquad v=r\omega,\qquad a_t=r\alpha,$$

    so points farther out move faster. A rotating point generally has two acceleration components at once: the tangential acceleration 切向加速度 $a_t=r\alpha$ (speeding up along the arc) and the centripetal acceleration 向心加速度 $a_c=\dfrac{v^2}{r}=r\omega^2$ (turning, towards the axis). They are perpendicular, so $a=\sqrt{a_t^2+a_c^2}$.

    As the radius turns through an angle, a point moves along an arc at speed v
    As the radius turns through an angle, a point moves along an arc at speed v
    Vocabulary · ⁨어휘⁩ Train · ⁨연습하기⁩
    English 한국어
    tangential acceleration/tænˈdʒenʃl əkˌseləˈreɪʃn/ 접선가속도를 가진다
    centripetal acceleration/senˈtrɪpɪtl əkˌseləˈreɪʃn/ 구심 가속도
    5.3

    Torque

    Syllabus
    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 torque exerted on a rigid system about a chosen pivot point by a given force is described by $\vec{\tau} = \vec{r} \times \vec{F}$.
      • 5.3.B.2.i The cross-product between two vectors, $\vec{A}$ and $\vec{B}$, results in a vector quantity of magnitude $\vec{A} \times \vec{B} = AB\sin\theta$.
      • 5.3.B.2.ii The direction of the vector resulting from the cross-product of vectors $\vec{A}$ and $\vec{B}$ is perpendicular to both vectors $\vec{A}$ and $\vec{B}$ and therefore is normal to the plane defined by vectors $\vec{A}$ and $\vec{B}$.
      • 5.3.B.2.iii The direction of the vector resulting from the cross-product of vectors $\vec{A}$ and $\vec{B}$ can be qualitatively determined by applying the appropriate right-hand rule.
    한국어

    학습 목표 5.3.A: 경질 시스템에 작용하는 토크를 식별한다.

    • 5.3.A.1 토크는 회전축에서 힘의 작용점까지의 위치 벡터에 수직인 힘 성분에서만 발생한다.
    • 5.3.A.2 레버암은 회전축에서 가해진 힘의 작용선까지의 수직 거리이다.

    학습 목표 5.3.B: 강체 시스템에 작용하는 토크를 설명하시오.

    • 5.3.B.1 토크는 힘 도표를 사용하여 설명할 수 있습니다.
      • 5.3.B.1.i 힘 도표는 자유체도(Free-body diagram)와 유사하며, 강체 시스템에 작용하는 토크를 분석하는 데 사용됩니다.
      • 5.3.B.1.ii 자유체도와 마찬가지로 힘 도표는 강체 시스템에 작용하는 힘의 상대적 크기와 방향을 나타냅니다. 또한 힘 도표는 회전 축에 대한 각 힘이 작용하는 위치도 함께 묘사합니다.
    • 5.3.B.2 선택한 지지점에 대해 주어진 힘이 강체 시스템에 가하는 토크는 $\vec{\tau} = \vec{r} \times \vec{F}$로 기술된다.
      • 5.3.B.2.i 두 벡터 $\vec{A}$과 $\vec{B}$의 외적은 크기 $\vec{A} \times \vec{B} = AB\sin\theta$인 벡터 양이 된다.
      • 5.3.B.2.ii 벡터 $\vec{A}$과 $\vec{B}$의 외적으로 생성된 벡터의 방향은 벡터 $\vec{A}$와 $\vec{B}$ 모두에 수직이므로, 벡터 $\vec{A}$와 $\vec{B}$가 정의하는 평면에 수직(정규)이다.
      • 5.3.B.2.iii 벡터 $\vec{A}$과 $\vec{B}$의 외적으로 생성된 벡터의 방향은 적절한 오른손 법칙을 적용하여 정성적으로 결정할 수 있다.

    Source: College Board AP Course and Exam Description · ⁨출처: College Board AP Course and Exam Description⁩

    The principle of moments (torque)

    Torque 力矩 is the turning effect of a force – the rotational cause, as force is the translational cause. It is a vector product 矢量积:

    $$\vec{\tau}=\vec{r}\times\vec{F},\qquad \tau=rF\sin\theta=F\,r_\perp,$$

    where $r_\perp$, the moment arm 力臂, is the perpendicular distance from the axis to the force's line of action. More force, applied farther from the axis, more perpendicular: more torque. A force pointing straight at (or away from) the axis has zero torque. In a plane, give counterclockwise and clockwise torques opposite signs and add.

    As a vector product, $\vec\tau=\vec r\times\vec F$ points perpendicular to the plane of $\vec r$ and $\vec F$, with its direction fixed by the right-hand rule 右手定则: curl your right fingers from $\vec r$ toward $\vec F$ and your thumb points along $\vec\tau$ – out of the page for a counterclockwise turn, into it for a clockwise one. The same rule gives the direction of angular momentum $\vec L=\vec r\times\vec p$.

    The torque of a force depends on the perpendicular distance from the axis
    The torque of a force depends on the perpendicular distance from the axis
    A gear train: torques and rotational inertia couple through the gear ratio
    A gear train: torques and rotational inertia couple through the gear ratio
    Explore · ⁨탐색하기⁩

    Balance torques on a beam · ⁨빔에 작용하는 토크 균형⁩

    Torque is force times perpendicular distance, $\tau=Fd$. The beam is in rotational equilibrium when the torques on each side are equal. · ⁨토크는 힘과 수직 거리의 곱으로 정의되며, $\tau=Fd$입니다. 양쪽의 토크가 같을 때 빔은 회전 평형 상태가 됩니다.⁩

    Vocabulary · ⁨어휘⁩ Train · ⁨연습하기⁩
    English 한국어
    Torque/tɔːk/ 토크
    vector product/ˈvektə ˈprɒdʌkt/ 내적(벡터 곱)
    moment arm/ˈməʊmənt ɑːm/ 모멘트 암
    right-hand rule/raɪt hænd ruːl/ 오른손 법칙
    Watch lesson · ⁨수업 보기⁩
    5.4

    Rotational Inertia

    Syllabus
    English

    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 $I = mr^2$.
    • 5.4.A.3 The total rotational inertia of a collection of objects about an axis is the sum of the rotational inertias of each object about that axis.
      • Equation: $I_{\text{tot}} = \sum I_i = \sum m_i r_i^2$
    • 5.4.A.4 For a solid that can be considered as a collection of differential masses, $dm$, the solid's rotational inertia can be calculated using the equation $I = \int r^2\, dm$, where $r$ is the perpendicular distance from $dm$ to the axis of rotation.

    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 C: Mechanics only expects students to use calculus in the derivations of the rotational inertia of thin rods of uniform or nonuniform density about an arbitrary axis perpendicular to the rod, as well as derivations of the rotational inertia of a thin cylindrical shell, disk, or rigid bodies that can be considered to be made up of coaxial rings or shells about an axis that passes through their centers (e.g., annular rings).

    Students should have a qualitative understanding of the factors that affect rotational inertia; for example, how rotational inertia is greater when mass is farther from the axis of rotation, which is why a hoop has more rotational inertia than a solid puck of the same mass and radius.

    한국어

    학습 목표 5.4.A: 주어진 회전 축에 대해 강체 시스템의 회전 관성을 설명하시오.

    • 5.4.A.1 회전 관성은 강체 시스템이 회전 상태 변화에 저항하는 정도를 측정하며, 시스템의 질량과 질량이 회전 축에 대해 어떻게 분포되어 있는지에 관련되어 있습니다.
    • 5.4.A.2 회전축으로부터 수직 거리 $r$에 있는 물체의 회전 관성은 다음 식 $I = mr^2$으로 기술된다.
    • 5.4.A.3 회전축에 대한 여러 물체 집합의 총 회전 관성은 각 물체가 해당 축에 대해 가진 회전 관성의 합이다.
      • 식: $I_{\text{tot}} = \sum I_i = \sum m_i r_i^2$
    • 5.4.A.4 미소 질량의 집합으로 간주할 수 있는 고체에서, $dm$은 고체의 회전 관성은 식 $I = \int r^2\, dm$을 사용하여 계산할 수 있으며, 여기서 $r$는 $dm$에서 회전축까지의 수직 거리이다.

    학습 목표 5.4.B: 중심的方法来 passes through the system's center of mass does not pass through the system's center of mass. (Original: 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 강체 시스템의 특정 평면에서의 회전 관성은 회전 축이 시스템의 중심을 통과할 때 최소값을 가집니다.
    • 5.4.B.2 평행축 정리는 중심을 통과하는 축과 평행한 임의의 축에 대한 강체 시스템의 회전 관성을 연결하기 위해 다음 식을 사용합니다:
      • 식: $I' = I_{\text{cm}} + Md^2$

    범위 규정: AP 물리 C: 역학에서는 학생들에게 균일하거나 비균일 밀도를 가진 얇은 막대, 원통 껍질, 디스크, 혹은 중심을 지나는 축에 대해 동심 원반이나 껍질로 구성될 수 있는 강체(예: 환형 원반)의 회전 관성을 유도할 때 미적분을 사용할 것을 기대한다.

    학생들은 회전 관성에 영향을 미치는 요인에 대해 정성적인 이해를 가져야 하며, 예를 들어 질량이 회전축에서 더 멀리 있을수록 회전 관성이 커지므로, 같은 질량과 반지름을 가진 고체 원판보다 후프(hoop)가 더 큰 회전 관성을 가지는 이유 등을 알아야 한다.

    Source: College Board AP Course and Exam Description · ⁨출처: College Board AP Course and Exam Description⁩

    Rotational inertia 转动惯量 $I$ measures resistance to angular acceleration – the rotational analog of mass, but it depends on where the mass sits. For point masses, $I=\sum m_ir_i^2$; for a continuous body,

    $$I=\int r^2\,dm.$$

    Mass far from the axis dominates, because of the $r^2$.

    Worked example (rod). A uniform rod (mass $M$, length $L$) about its centre: with linear density 线密度 $\lambda=M/L$, $dm=\lambda\,dx$, so $I=\displaystyle\int_{-L/2}^{L/2}x^2\,\frac{M}{L}\,dx=\frac{1}{12}ML^2$. AP also expects nonuniform rods: if $\lambda(x)=cx$ on $[0,L]$, first find $M=\int_0^L cx\,dx=\tfrac12cL^2$, then $I=\int_0^L x^2(cx)\,dx=\tfrac14cL^4=\tfrac12ML^2$ about the light end.

    Worked example (disk from rings). A solid disk is a nest of rings: a ring at radius $r$ of width $dr$ has $dm=\dfrac{M}{\pi R^2}\,2\pi r\,dr$, so

    $$I=\int_0^R r^2\,dm=\frac{2M}{R^2}\int_0^R r^3\,dr=\tfrac12MR^2.$$

    The same coaxial-shell method handles cylindrical shells and annular rings.

    The parallel-axis theorem 平行轴定理 shifts any known $I$ to a parallel axis a distance $d$ away:

    $$I'=I_{\text{cm}}+Md^2.$$

    For the rod about one end: $I=\tfrac1{12}ML^2+M\big(\tfrac{L}{2}\big)^2=\tfrac13ML^2$.

    Vocabulary · ⁨어휘⁩ Train · ⁨연습하기⁩
    English 한국어
    Rotational inertia/rəʊˈteɪʃənl ɪˈnɜːʃə/ 회전 관성
    linear density/ˈlɪnɪə ˈdensɪti/ 선밀도
    parallel-axis theorem/ˈpærəlel ˈæksɪs ˈθɪərəm/ 평행축 정리
    5.5

    Rotational Equilibrium

    Syllabus
    Learning ObjectiveEssential Knowledge

    5.5.A
    Describe the conditions under which a system's angular velocity remains constant.

    • 5.5.A.1 A system may exhibit rotational equilibrium (constant angular velocity) without being in translational equilibrium, and vice versa.
      • 5.5.A.1.i Free-body and force diagrams describe the nature of the forces and torques exerted on an object or rigid system.
      • 5.5.A.1.ii Rotational equilibrium is a configuration of torques such that the net torque exerted on the system is zero.
        • Equation: $\sum \tau_i = 0$
      • 5.5.A.1.iii The rotational analog of Newton's first law is that a system will have a constant angular velocity only if the net torque exerted on the system is zero.
    • 5.5.A.2 A rotational corollary to Newton's second law states that if the torques exerted on a rigid system are not balanced, the system's angular velocity must be changing.

    Boundary statement: AP Physics C: Mechanics does not expect students to simultaneously analyze rotation in multiple planes.

    Source: College Board AP Course and Exam Description · ⁨출처: College Board AP Course and Exam Description⁩

    Newton's first law has a rotational form: with zero net torque, angular velocity is constant – an object in rotational equilibrium 转动平衡 either does not rotate or rotates steadily. Full static equilibrium 静平衡 needs both $\sum F=0$ and $\sum\tau=0$ (about any axis). The standard move for beams and ladders: put the axis at the point where an unknown force acts, so that force drops out of the torque equation.

    At balance the clockwise and anticlockwise torques about the axis are equal
    At balance the clockwise and anticlockwise torques about the axis are equal

    Worked example. A uniform $20\ \text{kg}$ beam, $4.0\ \text{m}$ long, is pivoted at its left end and held horizontal by a vertical rope at its right end. A $40\ \text{kg}$ child sits $1.0\ \text{m}$ from the pivot. Torques about the pivot: $T(4.0)=20g(2.0)+40g(1.0)$, so $T=\dfrac{392+392}{4.0}=196\ \text{N}$. Then vertical force balance gives the pivot's upward push: $F_p=(60)(9.8)-196=392\ \text{N}$. Choosing the pivot as the axis removed $F_p$ from the torque equation entirely.

    A beam balance: rotational equilibrium when clockwise and anticlockwise torques match
    A beam balance: rotational equilibrium when clockwise and anticlockwise torques match
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    Find the balance point · ⁨균형점 balance point 찾기⁩

    For rotational equilibrium the total clockwise torque equals the total anticlockwise torque. Move the forces and distances until the beam balances. · ⁨회전 평형이 되기 위해서는 총 시계 방향 토크와 총 반시계 방향 토크가 서로 같아야 합니다. 빔이 균형을 이루도록 힘과 거리를 이동시키세요.⁩

    Vocabulary · ⁨어휘⁩ Train · ⁨연습하기⁩
    English 한국어
    rotational equilibrium/rəʊˈteɪʃənl ˌiːkwɪˈlɪbrɪəm/ 회전 평형
    static equilibrium/ˈstætɪk ˌiːkwɪˈlɪbrɪəm/ 정적 평형
    Watch lesson · ⁨수업 보기⁩
    5.6

    Newton's Second Law in Rotational Form

    Syllabus
    Learning ObjectiveEssential Knowledge

    5.6.A
    Describe the conditions under which a system's angular velocity changes.

    • 5.6.A.1 Angular velocity changes when the net torque exerted on the object or system is not equal to zero.
    • 5.6.A.2 The rate at which the angular velocity of a rigid system changes is directly proportional to the net torque exerted on the rigid system and is in the same direction. The angular acceleration of the rigid system is inversely proportional to the rotational inertia of the rigid system.
      • Equation: $\alpha_{\text{sys}} = \dfrac{\Sigma\tau}{I_{\text{sys}}} = \dfrac{\tau_{\text{net}}}{I_{\text{sys}}}$
    • 5.6.A.3 To fully describe a rotating rigid system, linear and rotational analyses may need to be performed independently.

    Source: College Board AP Course and Exam Description · ⁨출처: College Board AP Course and Exam Description⁩

    A net torque produces angular acceleration in proportion to the rotational inertia:

    $$\alpha=\frac{\sum\tau}{I}\qquad\Big(\text{more generally }\sum\vec{\tau}=\frac{d\vec{L}}{dt}\Big).$$

    Solve rotational dynamics exactly like translational dynamics, with $\tau\leftrightarrow F$, $I\leftrightarrow m$, $\alpha\leftrightarrow a$ – free-body diagram, equations, solve.

    A massive pulley: the two tensions differ because the pulley needs net torque
    A massive pulley: the two tensions differ because the pulley needs net torque

    Worked example (massive pulley). Masses $m_1=4.0\ \text{kg}$ and $m_2=2.0\ \text{kg}$ hang from a rope over a pulley 滑轮 of rotational inertia $I=0.50\ \text{kg}\cdot\text{m}^2$ and radius $R=0.20\ \text{m}$. Three Newton's-law equations, one per body:

    $$m_1g-T_1=m_1a,\qquad T_2-m_2g=m_2a,\qquad (T_1-T_2)R=I\alpha=\frac{Ia}{R}.$$

    Adding them (with $I/R^2=12.5\ \text{kg}$): $a=\dfrac{(m_1-m_2)g}{m_1+m_2+I/R^2}=\dfrac{19.6}{18.5}=1.1\ \text{m/s}^2$. Then $T_1=m_1(g-a)=35\ \text{N}$ and $T_2=m_2(g+a)=22\ \text{N}$. The tensions must differ – otherwise nothing would spin the pulley. (If a problem says the pulley is light, then $I\approx0$ and the tensions become equal again.)

    Exam skill. Rotational FRQs score the setup: separate free-body diagrams for each mass and the pulley, the string constraint $a=R\alpha$ stated, and the sign convention consistent. The single most common error is assuming one tension throughout a rope that passes over a massive pulley.

    Vocabulary · ⁨어휘⁩ Train · ⁨연습하기⁩
    English 한국어
    pulley/ˈpʊli/ 풀리
    5.6

    Exam tips

    • Use the rotational analogues: $\theta,\omega=\tfrac{d\theta}{dt},\alpha=\tfrac{d\omega}{dt}$, with the constant-$\alpha$ equations mirroring linear kinematics.
    • Convert between linear and angular with $v=r\omega$ and $a_t=r\alpha$ (plus centripetal $a_c=\tfrac{v^2}{r}=\omega^2 r$).
    • Keep radians throughout and fix a positive sense of rotation.
    • Relate torque to angular acceleration through $\tau=I\alpha$.
    • Draw the rotation axis — the moment arm is the perpendicular distance to it.
  • 6

    Energy and Momentum of Rotating Systems

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    6.1

    Rotational Kinetic Energy

    Syllabus
    English

    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 $K_{\text{rot}} = \dfrac{1}{2} I \omega^2$.
      • Equation: $K_{\text{rot}} = \dfrac{1}{2} I \omega^2$
      • 6.1.A.1.i The rotational inertia of an object about a fixed axis can be used to show that the rotational kinetic energy of that object is equivalent to its translational kinetic energy, which is its total kinetic energy.
      • 6.1.A.1.ii The total kinetic energy of a rigid system is the sum of its rotational kinetic energy due to its rotation about its center of mass and the translational kinetic energy due to the linear motion of its center of mass.
    • 6.1.A.2 A rigid system can have rotational kinetic energy while its center of mass is at rest due to the individual points within the rigid system having linear speed and, therefore, kinetic energy.
    • 6.1.A.3 Rotational kinetic energy is a scalar quantity.
    한국어

    학습 목표 6.1.A: 강체 시스템의 회전 운동 에너지를该系统의회전 관성과 각속도를 통해 설명하시오.

    • 6.1.A.1 물체나 강체 시스템의 회전 운동 에너지는该系统的回전 관성과 각속도에 관련되며, 다음 식으로 주어집니다: $K_{\text{rot}} = \dfrac{1}{2} I \omega^2$.
      • 식: $K_{\text{rot}} = \dfrac{1}{2} I \omega^2$
      • 6.1.A.1.i 고정 축에 대한 물체의 회전 관성을 사용하면,该系统의回전 운동 에너지가该系统的平动 운동 에너지(총 운동 에너지)와 동등함을 보여줄 수 있습니다.
      • 6.1.A.1.ii 강체 시스템의 총 운동 에너지는该系统的center of mass에 대한 회전 운동 에너지와该系统的center of mass의 선형 운동에 의한 평동 운동 에너지의 합입니다.
    • 6.1.A.2 강체 시스템은该系统的center of mass가 정지해 있더라도该系统의internal points가 선형 속도를 가지므로 운동 에너지를 가지고 있어回전 운동 에너지를 가질 수 있습니다.
    • 6.1.A.3回전 운동 에너지는 스칼라 양입니다.

    Source: College Board AP Course and Exam Description · ⁨출처: College Board AP Course and Exam Description⁩

    A spinning body has rotational kinetic energy 转动动能

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

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

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

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

    Vocabulary · ⁨어휘⁩ Train · ⁨연습하기⁩
    English 한국어
    rotational kinetic energy/rəʊˈteɪʃənl kɪˈnetɪk ˈenədʒi/ rotational kinetic energy
    rotational inertia/rəʊˈteɪʃənl ɪˈnɜːʃə/ rotational inertia
    6.2

    Torque and Work

    Syllabus
    English

    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: $W = \displaystyle\int_{\theta_1}^{\theta_2} \tau \, d\theta$
    • 6.2.A.3 Work done on a rigid system by a given torque can be found from the area under the curve of a graph of the torque as a function of angular position.
    한국어

    학습 목표 6.2.A: 주어진 토크 또는 일련의 토크가 강체 시스템에 가하는 일을 설명하시오.

    • 6.2.A.1 토크가 각 변위 동안 작용할 경우, 그 토크는 물체나 강체 시스템에 에너지를 전달하거나 빼낼 수 있다.
    • 6.2.A.2 토크에 의해 강체 시스템에 가해지는일의 양은 해당 토크의 크기와 토크가 작용하는 구간 동안 강체 시스템이 회전하는 각 변위에 관련되어 있다.
      • 식: $W = \displaystyle\int_{\theta_1}^{\theta_2} \tau \, d\theta$
    • 6.2.A.3 주어진 토크가 강체 시스템에 하는 일은 토크를 각도 위치의 함수로 나타낸 그래프 아래 면적으로 구할 수 있다.

    Source: College Board AP Course and Exam Description · ⁨출처: College Board AP Course and Exam Description⁩

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

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

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

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

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    Balance torques on a beam

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

    Vocabulary · ⁨어휘⁩ Train · ⁨연습하기⁩
    English 한국어
    torque/tɔːk/ torque
    angular displacement/ˈæŋɡjʊlə dɪˈspleɪsmənt/ angular displacement
    6.3

    Angular Momentum and Angular Impulse

    Syllabus
    Learning ObjectiveEssential Knowledge

    6.3.A
    Describe the angular momentum of an object or rigid system.

    • 6.3.A.1 The magnitude of the angular momentum of a rigid system about a specific axis can be described with the equation $L = I\omega$.
      • Equation: $L = I\omega$
    • 6.3.A.2 The angular momentum of an object about a given point is $\vec{L} = \vec{r} \times \vec{p}$.
      • Equation: $\vec{L} = \vec{r} \times \vec{p}$
      • 6.3.A.2.i The selection of the axis about which an object is considered to rotate influences the determination of the angular momentum of that object.
      • 6.3.A.2.ii The measured angular momentum of an object traveling in a straight line depends on the distance between the reference point and the object, the mass of the object, the speed of the object, and the angle between the radial distance and the velocity of the object.

    6.3.B
    Describe the angular impulse delivered to an object or rigid system by a torque.

    • 6.3.B.1 Angular impulse is defined as the product of the torque exerted on an object or rigid system and the time interval during which the torque is exerted.
      • Equation: $\text{angular impulse} = \displaystyle\int \tau \, dt$
    • 6.3.B.2 Angular impulse has the same direction as the torque imparting it.
    • 6.3.B.3 The angular impulse delivered to an object or rigid system by a torque can be found from the area under the curve of a graph of the torque as a function of time.

    6.3.C
    Relate the change in angular momentum of an object or rigid system to the angular impulse given to that object or rigid system.

    • 6.3.C.1 The magnitude of the change in angular momentum can be described by comparing the magnitudes of the final and initial momenta of the object or rigid system.
      • Equation: $\Delta L = L - L_0$
    • 6.3.C.2 A rotational form of the impulse–momentum theorem relates the angular impulse delivered to an object or rigid system and the change in angular momentum of that object or rigid system.
      • 6.3.C.2.i The angular impulse exerted on an object or rigid system is equal to the change in angular momentum of that object or rigid system.
        • Equation: $\Delta L = \displaystyle\int_{t_1}^{t_2} \tau \, dt$
      • 6.3.C.2.ii The rotational form of the impulse–momentum theorem is a direct result of Newton's second law of motion for cases in which rotational inertia is constant.
        • Equation: $\tau_{\text{net}} = \dfrac{dL}{dt} = I\dfrac{d\omega}{dt} = I\alpha$
    • 6.3.C.3 The net torque exerted on an object or rigid system is equal to the slope of the graph of the angular momentum of an object as a function of time.
    • 6.3.C.4 The angular impulse delivered to an object or rigid system is equal to the area under the curve of a graph of the net external torque exerted on an object as a function of time.

    Source: College Board AP Course and Exam Description · ⁨출처: College Board AP Course and Exam Description⁩

    Angular momentum 角动量 for a particle is

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

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

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

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

    Vocabulary · ⁨어휘⁩ Train · ⁨연습하기⁩
    English 한국어
    Angular momentum/ˈæŋɡjʊlə məʊˈmentəm/ Angular momentum
    angular impulse/ˈæŋɡjʊlə ˈɪmpʌls/ angular impulse
    6.4

    Conservation of Angular Momentum

    Syllabus
    Learning ObjectiveEssential Knowledge

    6.4.A
    Describe the behavior of a system using conservation of angular momentum.

    • 6.4.A.1 The total angular momentum of a system about a rotational axis is the sum of the angular momenta of the system's constituent parts about that rotational axis.
    • 6.4.A.2 Any change to a system's angular momentum must be due to an interaction between the system and its surroundings.
      • 6.4.A.2.i The angular impulse exerted by one object or system on a second object or system is equal and opposite to the angular impulse exerted by the second object or system on the first. This is a direct result of Newton's third law.
      • 6.4.A.2.ii A system may be selected so that the total angular momentum of that system is constant.
      • 6.4.A.2.iii The angular speed of a nonrigid system may change without the angular momentum of the system changing if the system changes shape by moving mass closer to or farther from the rotational axis.
      • 6.4.A.2.iv If the total angular momentum of a system changes, that change will be equivalent to the angular impulse exerted on the system.

    6.4.B
    Describe how the selection of a system determines whether the angular momentum of that system changes.

    • 6.4.B.1 Angular momentum is conserved in all interactions.
    • 6.4.B.2 If the net external torque exerted on a selected object or rigid system is zero, the total angular momentum of that system is constant.
    • 6.4.B.3 If the net external torque exerted on a selected object or rigid system is nonzero, angular momentum is transferred between the system and the environment.

    Source: College Board AP Course and Exam Description · ⁨출처: College Board AP Course and Exam Description⁩

    Angular momentum: pull in, spin faster

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

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

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

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

    Pulling mass inward lowers I, so ω rises to conserve L = Iω
    Pulling mass inward lowers I, so ω rises to conserve L = Iω
    Angular momentum about the pivot is conserved as the bullet embeds in the rod
    Angular momentum about the pivot is conserved as the bullet embeds in the rod

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

    Figure skaters performing on the ice
    Figure skating and angular momentum: a skater who pulls their arms in during a spin lowers their rotational inertia, so they spin faster
    Vocabulary · ⁨어휘⁩ Train · ⁨연습하기⁩
    English 한국어
    conserved/kənˈsɜːvd/ conserved
    6.5

    Rolling

    Syllabus
    English

    Learning Objective 6.5.A: Describe the kinetic energy of a system that has translational and rotational motion.

    • 6.5.A.1 The total kinetic energy of a system is the sum of the system's translational and rotational kinetic energies.
      • Equation: $K_{\text{tot}} = K_{\text{trans}} + K_{\text{rot}}$

    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 following equations:
      • Equation: $\Delta x_{\text{cm}} = r\Delta\theta$
      • Equation: $v_{\text{cm}} = r\omega$
      • Equation: $a_{\text{cm}} = r\alpha$
    • 6.5.B.2 For ideal cases, rolling without slipping implies that the frictional force does not dissipate any energy from the rolling system.

    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 C: Mechanics.

    한국어

    학습 목표 6.5.A: 병진 운동과 회전 운동을 모두 가지는 시스템의 운동 에너지를 설명하시오。

    • 6.5.A.1 시스템의 총 운동 에너지는该系统的平动动能和转动动能之和。
      • 식: $K_{\text{tot}} = K_{\text{trans}} + K_{\text{rot}}$

    학습 목표 6.5.B: 미끄러짐 없이 굴리는 시스템의 운동을 설명하시오。

    • 6.5.B.1 미끄러짐 없이 굴러가는 동안 시스템의 질량 중심의 병진 운동은 다음과 같은 방정식을 통해 시스템 자체의 회전 운동과 관련된다:
      • 식: $\Delta x_{\text{cm}} = r\Delta\theta$
      • 식: $v_{\text{cm}} = r\omega$
      • 식: $a_{\text{cm}} = r\alpha$
    • 6.5.B.2 이상적인 경우, 미끄러짐 없이 굴러가는 것은 마찰력이 굴리는 시스템으로부터 에너지를 소산시키지 않음을 의미한다.

    학습 목표 6.5.C: 미끄러지며 굴러가는 시스템의 운동을 설명하시오.

    • 6.5.C.1 미끄러질 때, 시스템의 질량 중심의 운동과 시스템의 회전 운동은 직접적으로 관련될 수 없다.
    • 6.5.C.2 회전하는 시스템이 다른 표면相对于로 미끄러질 때, 운동 마찰력의 작용점에Applied된 힘은 표면에 대해 이동하므로 운동 마찰력은 시스템으로부터 에너지를 소산시킨다.

    범위 명시: 굴림 마찰력은 AP Physics C: Mechanics의 학습 범위 밖이다.

    Source: College Board AP Course and Exam Description · ⁨출처: College Board AP Course and Exam Description⁩

    Rolling without slipping

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

    $$\Delta x_{\text{cm}}=r\,\Delta\theta,\qquad v_{\text{cm}}=r\omega,\qquad a_{\text{cm}}=r\alpha.$$
    In rolling without slipping the contact point is at rest, so v = rω
    In rolling without slipping the contact point is at rest, so v = rω

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

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

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

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

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

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

    Vocabulary · ⁨어휘⁩ Train · ⁨연습하기⁩
    English 한국어
    Rolling without slipping/ˈrəʊlɪŋ wɪˈðaʊt ˈslɪpɪŋ/ Rolling without slipping
    disk/dɪsk/ disk
    hoop/huːp/ hoop
    6.6

    Motion of Orbiting Satellites

    Syllabus
    English

    Learning Objective 6.6.A: Describe the motions of a system consisting of two objects or systems interacting only via gravitational forces.

    • 6.6.A.1 In a system consisting only of a massive central object and an orbiting satellite with mass that is negligible in comparison to the central object's mass, the motion of the central object itself is negligible.
    • 6.6.A.2 The motion of satellites in orbits is constrained by conservation laws.
      • 6.6.A.2.i In circular orbits, the system's total mechanical energy, the system's gravitational potential energy, and the satellite's angular momentum and kinetic energy are constant.
      • 6.6.A.2.ii In elliptical orbits, the system's total mechanical energy and the satellite's angular momentum are constant, but the system's gravitational potential energy and the satellite's kinetic energy can each change.
      • 6.6.A.2.iii The gravitational potential energy of a system consisting of a satellite and a massive central object is defined to be zero when the satellite is an infinite distance from the central object.
        • Equation: $U_g = -G\dfrac{m_1 m_2}{r}$
    • 6.6.A.3 The total energy of a system consisting of a satellite orbiting a central object in a circular path can be written in terms of the gravitational potential energy of that system or the kinetic energy of the satellite.
      • Equation: $K = -\dfrac{1}{2}U$
      • Equation: $E_{total} = \dfrac{1}{2}U = -\dfrac{GMm}{2r}$
    • 6.6.A.4 The escape velocity of a satellite is the satellite's velocity such that the mechanical energy of the satellite–central-object system is equal to zero.
      • 6.6.A.4.i When the only force exerted on a satellite is gravity from a central object, a satellite that reaches escape velocity will move away from the central body until its speed reaches zero at an infinite distance from the central body.
      • 6.6.A.4.ii The escape velocity of a satellite from a central body of mass $M$ can be derived using conservation of energy laws.
        • Equation: $v_{\text{esc}} = \sqrt{\dfrac{2GM}{r}}$
    한국어

    학습 목표 6.6.A: 중력 상호작용만 있는 두 개 또는 여러 개의 물체/시스템으로 구성된 시스템의 운동을 설명한다.

    • 6.6.A.1 질량이 매우 큰 중심 천체와 그 질량에 비해 질량이 무시할 수 있는 궤도 위 인공위성으로만 구성된 시스템에서, 중심 천체 자체의 운동은 무시할 수 있다.
    • 6.6.A.2 궤도 위 인공위성의 운동은 보존 법칙에 의해 제한받는다.
      • 6.6.A.2.i 원형 궤도에서 시스템의 총 역학적 에너지, 시스템의 중력ポテン셜에너지, 인공위성의 각운동량 및 운동에너지는 일정하다.
      • 6.6.A.2.ii 타원 궤도에서 시스템의 총 역학적 에너지와 인공위성의 각운동량은 일정하지만, 시스템의 중력ポテン셜에너지와 인공위성의 운동에너지는 각각 변할 수 있다.
      • 6.6.A.2.iii 인공위성과 질량이 매우 큰 중심 천체로 구성된 시스템의 중력ポテン셜에너지는 인공위성이 중심 천체로부터 무한한 거리에 있을 때 0으로 정의한다.
        • 식: $U_g = -G\dfrac{m_1 m_2}{r}$
    • 6.6.A.3 중심 천체를 원궤도로 도는 위성을 포함하는 시스템의 총 에너지는 해당 시스템의 중력 퍼텐셜 에너지나 위성의 운동 에너지로 표현할 수 있다.
      • 식: $K = -\dfrac{1}{2}U$
      • 식: $E_{total} = \dfrac{1}{2}U = -\dfrac{GMm}{2r}$
    • 6.6.A.4 위성의 탈출 속도는 위성-중심천체 시스템의 역학적 에너지가 0이 되는 위성의 속도이다.
      • 6.6.A.4.i 중심 천체로부터의 중력만이 위성에 작용할 때, 탈출 속도에 도달한 위성은 중심 천체로부터 멀어지다가 무한한 거리에서 속도가 0이 된다.
      • 6.6.A.4.ii 질량 $M$인 중심 천체로부터 위성이 탈출하는 속도는 에너지 보존 법칙을 사용하여 유도할 수 있다.
        • 식: $v_{\text{esc}} = \sqrt{\dfrac{2GM}{r}}$

    Source: College Board AP Course and Exam Description · ⁨출처: College Board AP Course and Exam Description⁩

    Orbital motion (Kepler's 2nd law)

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

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

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

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

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

    $$v_{\text{esc}}=\sqrt{\frac{2GM}{r}}.$$
    Gravity provides the centripetal force that keeps a satellite in orbit
    Gravity provides the centripetal force that keeps a satellite in orbit

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

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

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

    Explore · ⁨탐색하기⁩

    Compare orbits at different radii

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

    Vocabulary · ⁨어휘⁩ Train · ⁨연습하기⁩
    English 한국어
    centripetal force/senˈtrɪpɪtl fɔːs/ centripetal force
    satellite/ˈsætəlaɪt/ satellite
    gravitational potential energy/ˌɡrævɪˈteɪʃənl pəˈtenʃl ˈenədʒi/ gravitational potential energy
    total mechanical energy/ˈtəʊtl mɪˈkænɪkl ˈenədʒi/ total mechanical energy
    escape velocity/eˈskeɪp vəˈlɒsɪti/ escape velocity
    elliptical orbit/ɪˈlɪptɪkl ˈɔːbɪt/ elliptical orbit
    Kepler's second law/ˈkepləz ˈsekənd lɔː/ Kepler's second law
    6.6

    Exam tips

    • Compute the moment of inertia $I=\int r^2\,dm$ and shift axes with the parallel-axis theorem $I=I_{cm}+Md^2$.
    • Rotational kinetic energy is $\tfrac12 I\omega^2$; a rolling body has both translational and rotational KE.
    • Conserve angular momentum $L=I\omega$ when net external torque is zero (a spinning skater pulling in).
    • Use energy conservation for rolling-without-slipping problems ($v=r\omega$ ties the two motions).
    • Know standard $I$ values (hoop, disk, rod, sphere) and where the axis is.
  • 7

    Oscillations

    Watch lesson · ⁨수업 보기⁩
    7.1

    Defining Simple Harmonic Motion

    Syllabus
    English

    Learning Objective 7.1.A: Describe simple harmonic motion.

    • 7.1.A.1 Simple harmonic motion is a special case of periodic motion.
    • 7.1.A.2 SHM results when the magnitude of the restoring force exerted on an object is proportional to that object's displacement from its equilibrium position.
      • Derived equation: $ma_x = -k\Delta x$
      • 7.1.A.2.i A restoring force is a force that is exerted in a direction opposite to the object's displacement from an equilibrium position.
      • 7.1.A.2.ii An equilibrium position is a location at which the net force exerted on an object or system is zero.
    한국어

    학습 목표 7.1.A: 단순조화운동을 설명하시오.

    • 7.1.A.1 단순조화운동은 주기운동의 특별한 경우이다.
    • 7.1.A.2 평형 위치에서의 변위에 비례하는 복원력의 크기가 작용할 때 단순조화운동이 발생한다.
      • 파생 식: $ma_x = -k\Delta x$
      • 7.1.A.2.i 복원력은 평형 위치에서의 변위와 반대 방향으로 작용하는 힘이다.
      • 7.1.A.2.ii 평형 위치란 물체나 시스템에 작용하는 합력이 0인 위치를 말한다.

    Source: College Board AP Course and Exam Description · ⁨출처: College Board AP Course and Exam Description⁩

    Simple harmonic motion

    Simple harmonic motion 简谐运动 (SHM) is a special case of periodic motion 周期运动. It appears whenever two things are true: there is an equilibrium 平衡位置 position where the net force is zero, and displacing the object produces a restoring force 回复力 – a force pointing back toward equilibrium – whose magnitude is proportional to the displacement:

    $$F=-k\,\Delta x.$$

    Newton's second law then gives the defining differential equation 微分方程:

    $$\frac{d^2x}{dt^2}=-\frac{k}{m}\,x=-\omega^2 x,\qquad \omega=\sqrt{\frac{k}{m}}.$$

    Its solution is sinusoidal, $x(t)=A\cos(\omega t+\phi)$. You do not need to prove this solution – you need to recognise the equation: any system whose motion obeys "$\ddot{x}=-\omega^2x$" is a simple harmonic oscillator, whatever it is made of. (A mass hanging on a vertical spring works the same way: gravity only shifts the equilibrium point; the oscillation about it is unchanged.)

    In SHM the acceleration always points back towards equilibrium, opposite the displacement
    In SHM the acceleration always points back towards equilibrium, opposite the displacement
    Vocabulary · ⁨어휘⁩ Train · ⁨연습하기⁩
    English 한국어
    Simple harmonic motion/ˈsɪmpl hɑːˈmɒnɪk ˈməʊʃn/ 간단 진동
    periodic motion/ˌpɪərɪˈɒdɪk ˈməʊʃn/ 주기 운동
    equilibrium/ˌiːkwɪˈlɪbrɪəm/ 평형
    restoring force/rɪˈstɔːrɪŋ fɔːs/ 복원력
    differential equation/ˌdɪfəˈrenʃl ɪˈkweɪʒn/ 미분 방정식
    7.2

    Frequency and Period of SHM

    Syllabus
    English

    Learning Objective 7.2.A: Describe the frequency and period of an object exhibiting SHM.

    • 7.2.A.1 The period of SHM is related to the angular frequency, $\omega$, of the object's motion by the following equation:
      • Equation: $T = \dfrac{2\pi}{\omega} = \dfrac{1}{f}$
      • 7.2.A.1.i The period of an object–ideal-spring oscillator is given by the equation
        • Equation: $T_s = 2\pi\sqrt{\dfrac{m}{k}}.$
      • 7.2.A.1.ii The period of a simple pendulum displaced by a small angle is given by the equation
        • Equation: $T_p = 2\pi\sqrt{\dfrac{l}{g}}.$
    한국어

    학습 목표 7.2.A: SHM을 나타내는 물체의 주파수와 주기를 설명하시오.

    • 7.2.A.1 조화 진동(SHM)의 주기는 물체의 운동 각주파수 $\omega$와 다음 방정식으로 관련된다:
      • 식: $T = \dfrac{2\pi}{\omega} = \dfrac{1}{f}$
      • 7.2.A.1.i 물체-이상 스프링 오실레이터의 주기는 다음 방정식으로 주어진다
        • 식: $T_s = 2\pi\sqrt{\dfrac{m}{k}}.$
      • 7.2.A.1.ii 작은 각도만큼 변위된 단진자의 주기는 다음 방정식으로 주어진다
        • 식: $T_p = 2\pi\sqrt{\dfrac{l}{g}}.$

    Source: College Board AP Course and Exam Description · ⁨출처: College Board AP Course and Exam Description⁩

    The angular frequency 角频率 $\omega$ sets the period 周期 and frequency 频率:

    $$T=\frac{2\pi}{\omega}=\frac{1}{f}.$$

    For the two standard systems:

    $$T_{\text{spring}}=2\pi\sqrt{\frac{m}{k}},\qquad T_{\text{pendulum}}=2\pi\sqrt{\frac{\ell}{g}}.$$

    Two classic conceptual traps: the period of SHM never depends on the amplitude 振幅, and each system ignores one obvious variable – the pendulum's period does not involve its mass, and the spring's period does not involve $g$ (a spring–block oscillator keeps perfect time in orbit; a pendulum clock does not).

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    Time a pendulum's swing · ⁨진자의 진동 시간 측정⁩

    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. · ⁨진자의 주기는 길이에 중력加速度(중력)에 의존하며 질량이나 (작은) 진폭에는 의하지 않습니다: $T=2\pi\sqrt{L/g}$. 길이를 늘리면 한 번의 진동에 더 많은 시간이 걸립니다.⁩

    Vocabulary · ⁨어휘⁩ Train · ⁨연습하기⁩
    English 한국어
    angular frequency/ˈæŋɡjʊlə ˈfriːkwənsi/ 각주파수
    period/ˈpɪərɪəd/ 주기
    frequency/ˈfriːkwənsi/ 주파수
    amplitude/ˈæmplɪtjuːd/ 진폭
    7.3

    Representing and Analyzing SHM

    Syllabus
    Learning ObjectiveEssential Knowledge

    7.3.A
    Describe the displacement, velocity, and acceleration of an object exhibiting SHM.

    • 7.3.A.1 For an object exhibiting SHM, the displacement of that object measured from its equilibrium position can be represented by the equations $x = A\cos(2\pi ft)$ or $x = A\sin(2\pi ft)$.
      • 7.3.A.1.i Minima, maxima, and zeros of displacement, velocity, and acceleration are features of harmonic motion.
      • 7.3.A.1.ii Recognizing the positions or times at which the displacement, velocity, and acceleration for SHM have extrema or zeros can help in qualitatively describing the behavior of the motion.
    • 7.3.A.2 The position as a function of time for an object exhibiting SHM is a solution of the second-order differential equation derived from the application of Newton's second law.
      • Derived equation: $\dfrac{d^2 x}{dt^2} = -\omega^2 x$
    • 7.3.A.3 Characteristics of SHM, such as velocity and acceleration, can be determined by or derived from the equation $x = A\cos(\omega t + \phi).$
      • 7.3.A.3.i The acceleration of an object exhibiting SHM is related to the object's angular frequency and position.
        • Derived equation: $a = -\omega^2 x$
      • 7.3.A.3.ii It can be shown that the maximum velocity and acceleration of an object exhibiting SHM are related to the angular frequency of the object's motion.
        • Derived equations: $v_{\max} = A\omega$
        • $a_{\max} = A\omega^2$
    • 7.3.A.4 In the presence of a sinusoidal external force, a system may exhibit resonance.
      • 7.3.A.4.i Resonance occurs when an external force is exerted at the natural frequency of an oscillating system.
      • 7.3.A.4.ii Resonance increases the amplitude of oscillating motion.
      • 7.3.A.4.iii The natural frequency of a system is the frequency at which the system will oscillate when it is displaced from its equilibrium position.
    • 7.3.A.5 Changing the amplitude of a system exhibiting SHM will not change its period.
    • 7.3.A.6 Properties of SHM can be determined and analyzed using graphical representations.

    Boundary statement: AP Physics C: Mechanics only expects students to know the solution to the second-order differential equation that describes SHM, as well as be able to identify SHM. AP Physics C: Mechanics does not expect students to mathematically prove that the solution is correct.

    Source: College Board AP Course and Exam Description · ⁨출처: College Board AP Course and Exam Description⁩

    Differentiate $x(t)=A\cos(\omega t+\phi)$ twice:

    $$v=-A\omega\sin(\omega t+\phi),\qquad a=-A\omega^2\cos(\omega t+\phi)=-\omega^2x.$$

    So at the ends of the swing ($x=\pm A$) the speed is zero and the acceleration is largest; passing through equilibrium ($x=0$) the acceleration is zero and the speed is largest, $v_{\max}=A\omega$. The amplitude $A$ and the phase constant 相位常数 $\phi$ come from the initial conditions: where the object starts and how fast it is moving. Eliminating $t$ (or using energy, below) gives speed as a function of position:

    $$v=\pm\,\omega\sqrt{A^2-x^2}.$$
    Displacement varies sinusoidally with time in simple harmonic motion
    Displacement varies sinusoidally with time in simple harmonic motion

    Worked example. A $0.50\ \text{kg}$ mass on a $k=200\ \text{N/m}$ spring: $\omega=\sqrt{k/m}=20\ \text{rad/s}$, $T=2\pi/\omega=0.31\ \text{s}$. With $A=0.10\ \text{m}$: $v_{\max}=A\omega=2.0\ \text{m/s}$ at equilibrium, and at $x=0.050\ \text{m}$ the speed is $v=\omega\sqrt{A^2-x^2}=20\sqrt{0.10^2-0.050^2}=1.7\ \text{m/s}$.

    Vocabulary · ⁨어휘⁩ Train · ⁨연습하기⁩
    English 한국어
    phase constant/feɪz ˈkɒnstənt/ 위상 상수
    7.4

    Energy of Simple Harmonic Oscillators

    Syllabus
    English

    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.
      • Relevant equation: $E_{\text{total}} = U + K$
    • 7.4.A.2 Conservation of energy indicates that the total energy of a system exhibiting SHM is constant.
    • 7.4.A.3 The kinetic energy of a system exhibiting SHM is at a maximum when the system's potential energy is at a minimum.
    • 7.4.A.4 The potential energy of a system exhibiting SHM is at a maximum when the system's kinetic energy is at a minimum.
      • 7.4.A.4.i The minimum kinetic energy of a system exhibiting SHM is zero.
      • 7.4.A.4.ii Changing the amplitude of a system exhibiting SHM will change the maximum potential energy of the system and, therefore, the total energy of the system.
        • Relevant equation for a spring–object system: $E_{\text{total}} = \dfrac{1}{2}kA^2$
    한국어

    학습 목표 7.4.A: SHM을 나타내는 시스템의 역학적 에너지를 설명하시오.

    • 7.4.A.1 SHM을 나타내는 시스템의 총 에너지는 시스템의 운동에너지와 포텐셜에너지의 합이다.
      • 관련 식: $E_{\text{total}} = U + K$
    • 7.4.A.2 에너지 보존에 따라 SHM을 나타내는 시스템의 총 에너지는 일정하다.
    • 7.4.A.3 SHM을 나타내는 시스템의 운동에너지는 시스템의 포텐셜에너지가 최소일 때 최대가 된다.
    • 7.4.A.4 진동계가 단순조화운동(SHM)을 할 때, 운동 에너지가 최소일 때 시스템의ポテン셜 에너지는 최대가 된다.
      • 7.4.A.4.i 단순조화운동을 하는 시스템의 운동 에너지의 최소값은 0이다.
      • 7.4.A.4.ii 단순조화운동을 하는 시스템의 진폭을 변경하면 시스템의 최대 포텐셜 에너지와 따라서 시스템의 총 에너지가 바뀐다.
        • 스프링-물체 시스템에 대한 관련 방정식: $E_{\text{total}} = \dfrac{1}{2}kA^2$

    Source: College Board AP Course and Exam Description · ⁨출처: College Board AP Course and Exam Description⁩

    The total energy of the oscillator is the sum $E=U+K$, and with no friction it is constant – energy just trades back and forth between the spring's potential energy and the mass's kinetic energy:

    $$E=\tfrac12kA^2=\tfrac12kx^2+\tfrac12mv^2.$$

    All potential at the ends (where $K=0$), all kinetic at equilibrium (where $U$ is minimum). Because $E\propto A^2$, doubling the amplitude quadruples the energy.

    Kinetic and potential energy swap over a cycle while the total energy stays constant
    Kinetic and potential energy swap over a cycle while the total energy stays constant

    Worked example. Where is the energy split evenly? Set $\tfrac12kx^2=\tfrac12E=\tfrac14kA^2$, so $x=A/\sqrt2\approx0.71A$ – much closer to the end than to the middle. For the system above ($A=0.10\ \text{m}$): $x=0.071\ \text{m}$.

    Left alone, a system oscillates at its own natural frequency 固有频率 (set by $k$ and $m$: $f_0=\frac{1}{2\pi}\sqrt{k/m}$). Drive it with a periodic force at that same frequency and the amplitude grows dramatically – resonance 共振. Each well-timed push adds energy faster than damping removes it, so the driven amplitude peaks sharply when the driving frequency matches $f_0$ (a swing pumped in time, a bridge shaken by wind).

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    Trade kinetic and potential energy in SHM · ⁨SHM에서의 운동 에너지와 위치 에너지 교환⁩

    In simple harmonic motion, energy sloshes between kinetic (fastest at the centre) and potential (greatest at the extremes) while the total stays constant. · ⁨**简单简谐运动(SHM)**에서 에너지는 중심부에서 가장 빠르고 극단에서 가장 큰 위치 에너지와 운동 에너지 사이를 오가며, 전체 에너지는 일정하게 유지됩니다.⁩

    Vocabulary · ⁨어휘⁩ Train · ⁨연습하기⁩
    English 한국어
    natural frequency/ˈnætʃərəl ˈfriːkwənsi/ 고유 진동수
    resonance/ˈrezənəns/ 공명
    7.5

    Simple and Physical Pendulums

    Syllabus
    Learning ObjectiveEssential Knowledge

    7.5.A
    Describe the properties of a physical pendulum.

    • 7.5.A.1 A physical pendulum is a rigid body that undergoes oscillation about a fixed axis.
    • 7.5.A.2 For small amplitudes of motion, the period of a physical pendulum is derived from the application of Newton's second law in rotational form.
      • Relevant equation: $T_{\text{phys}} = 2\pi\sqrt{\dfrac{I}{mgd}}$
      • 7.5.A.2.i When displaced from equilibrium, the gravitational force exerted on a physical pendulum's center of mass provides a restoring torque.
        • Derived equation: $\tau = -mgd\sin\theta$
      • 7.5.A.2.ii For small amplitudes of motion, the small-angle approximation can be applied to the restoring torque.
        • Derived equation: $\sin\theta \approx \theta$
        • $\tau = -mgd\theta = I\alpha$
      • 7.5.A.2.iii The small-angle approximation and Newton's second law in rotational form yield a second-order differential equation that describes SHM:
        • Equation: $\dfrac{d^2\theta}{dt^2} = -\omega^2\theta$
    • 7.5.A.3 A simple pendulum is a special case of physical pendulums in which the hanging object can be modeled as a point mass at a distance, $l$, from the pivot point.
      • Relevant equation: $T_p = 2\pi\sqrt{\dfrac{\ell}{g}}$
    • 7.5.A.4 A torsion pendulum is a case of SHM where the restoring torque is proportional to the angular displacement of a rotating system. For example, a horizontal disk that is suspended from a wire attached to its center of mass may undergo rotational oscillations about the wire in the horizontal plane.
      • Derived equation: $I\alpha = -k\Delta\theta$

    Source: College Board AP Course and Exam Description · ⁨출처: College Board AP Course and Exam Description⁩

    Energy conservation: KE ⇄ PE

    A physical pendulum 物理摆 is any rigid body swinging about a fixed pivot. Displace it by an angle $\theta$ and gravity, acting at the center of mass a distance $d$ from the pivot, supplies a restoring torque

    $$\tau=-mgd\sin\theta.$$

    For small angles, apply the small-angle approximation 小角度近似 $\sin\theta\approx\theta$ and Newton's second law in rotational form ($\tau=I\alpha$):

    $$\frac{d^2\theta}{dt^2}=-\frac{mgd}{I}\,\theta=-\omega^2\theta\quad\Rightarrow\quad T_{\text{phys}}=2\pi\sqrt{\frac{I}{mgd}}.$$

    This is the same "$\ddot\theta=-\omega^2\theta$" pattern as before – recognising it is the derivation. A simple pendulum 单摆 is the special case of a point mass on a light string: $I=m\ell^2$ and $d=\ell$ give $T=2\pi\sqrt{\ell/g}$.

    Gravity acting at the center of mass provides a physical pendulum's restoring torque
    Gravity acting at the center of mass provides a physical pendulum's restoring torque

    Worked example. A uniform rod (mass $M$, length $L$) swings from one end: $I=\tfrac13ML^2$, $d=\tfrac{L}{2}$, so

    $$T=2\pi\sqrt{\frac{\tfrac13ML^2}{Mg\,\tfrac{L}{2}}}=2\pi\sqrt{\frac{2L}{3g}}$$

    – shorter than a simple pendulum of length $L$, because the rod's mass sits nearer the pivot.

    A torsion pendulum 扭摆 – a disk hanging from a wire that twists – is SHM one more time: the wire's restoring torque is proportional to the twist angle, $I\alpha=-\kappa\,\Delta\theta$, giving $T=2\pi\sqrt{I/\kappa}$.

    Exam skill. Every pendulum FRQ wants the same three steps: write the restoring torque about the pivot, apply the small-angle approximation, and match the result to $\ddot\theta=-\omega^2\theta$ to read off $\omega$. State the small-angle step explicitly – it is a scored point, and it is why large-amplitude swings are not simple harmonic.

    A large Foucault pendulum swinging over a marked floor
    A Foucault pendulum: a long, heavy pendulum whose slow swing reveals the Earth turning beneath it
    Vocabulary · ⁨어휘⁩ Train · ⁨연습하기⁩
    English 한국어
    physical pendulum/ˈfɪzɪkl ˈpendjʊləm/ 물리적 진자
    small-angle approximation/smɔːl ˈæŋɡl əˌprɒksɪˈmeɪʃn/ 소각도 근사
    simple pendulum/ˈsɪmpl ˈpendjʊləm/ 단진자
    torsion pendulum/ˈtɔːʃn ˈpendjʊləm/ 나선 진자
    7.5

    Exam tips

    • Identify SHM from a linear restoring force $F=-kx$, which gives $\tfrac{d^2x}{dt^2}=-\omega^2 x$ with $\omega=\sqrt{k/m}$.
    • Write the solution $x=A\cos(\omega t+\phi)$ and get $v,a$ by differentiating; period $T=\tfrac{2\pi}{\omega}$ is amplitude-independent.
    • Energy trades between $\tfrac12 kx^2$ and $\tfrac12 mv^2$, with total $\tfrac12 kA^2$.
    • For a pendulum, use the small-angle approximation $\sin\theta\approx\theta$ to reach SHM.
    • Match phase to the start: released from rest at maximum displacement uses cosine.

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