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Chapter 5 · Mechanics第5章 · 力学

Torque & Rotational Dynamics

AP-Style Practice QuestionsAP 风格练习题

EASY MEDIUM HARD

Topics主题 5.1 - 5.6MECH



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PART ITopics 5.1 - 5.6主题 5.1 - 5.6

Multiple Choice Questions选择题

Show all supporting work on scratch paper. Each item is labeled with its Mechanics topic and whether a calculator is permitted on that AP Exam section. Take $g = 9.8~\mathrm{m/s^2}$ unless a problem says otherwise.请在草稿纸上写出所有辅助计算过程。每道题均标注了对应的力学主题以及该 AP 考试部分是否允许使用计算器。除非题目另有说明,取 $g = 9.8~\mathrm{m/s^2}$。

Q1EASY 5.1 Rotational Kinematics5.1 转动运动学No Calculator

A wheel rotates at $60~\mathrm{rev/min}$. Its angular speed in $\mathrm{rad/s}$ is一个轮子以 $60~\mathrm{rev/min}$ 旋转。其角速度(单位 $\mathrm{rad/s}$)为

Q2EASY 5.2 Linear & Rotational5.2 线量与转动量No Calculator

A point on the rim of a disk of radius $0.50~\mathrm{m}$ moves with the disk at angular speed $\omega = 4.0~\mathrm{rad/s}$. The linear speed of that point is半径为 $0.50~\mathrm{m}$ 的圆盘边缘上一点随圆盘以角速度 $\omega = 4.0~\mathrm{rad/s}$ 转动。该点的线速度为

Q3EASY 5.3 Torque5.3 力矩No Calculator

A force of $20~\mathrm{N}$ is applied perpendicular to the handle of a wrench at a distance $0.30~\mathrm{m}$ from the axis of rotation. The magnitude of the torque about the axis is一个 $20~\mathrm{N}$ 的力垂直作用于扳手手柄,作用点距转动轴 $0.30~\mathrm{m}$。关于该轴的力矩大小为

Q4EASY 5.4 Rotational Inertia5.4 转动惯量No Calculator

The moment of inertia of a thin uniform rod of mass $M$ and length $L$ about an axis through its center, perpendicular to the rod, is质量为 $M$、长度为 $L$ 的细均匀杆,绕过其中心且垂直于杆的轴的转动惯量为

Q5MEDIUM 5.1 Constant Angular Acceleration5.1 匀角加速度No Calculator

A wheel initially rotating at $\omega_0 = 2.0~\mathrm{rad/s}$ undergoes a constant angular acceleration $\alpha = 4.0~\mathrm{rad/s^2}$ for $3.0~\mathrm{s}$. Its final angular speed is一个轮子初始角速度为 $\omega_0 = 2.0~\mathrm{rad/s}$,以匀角加速度 $\alpha = 4.0~\mathrm{rad/s^2}$ 转动 $3.0~\mathrm{s}$。其末角速度为

Q6MEDIUM 5.2 Centripetal Acceleration5.2 向心加速度No Calculator

A disk of radius $0.20~\mathrm{m}$ rotates about its central axis at constant angular speed $\omega = 10~\mathrm{rad/s}$. The centripetal acceleration of a point on the rim is半径为 $0.20~\mathrm{m}$ 的圆盘绕中心轴以匀角速度 $\omega = 10~\mathrm{rad/s}$ 转动。圆盘边缘上一点的向心加速度为

Q7MEDIUM 5.3 Torque at an Angle5.3 斜向力矩No Calculator

A force of $30~\mathrm{N}$ is applied at the end of a wrench of length $0.40~\mathrm{m}$, but at an angle of $30^\circ$ to the wrench handle. The magnitude of the torque about the axis is一个 $30~\mathrm{N}$ 的力施加在长 $0.40~\mathrm{m}$ 的扳手末端,方向与扳手柄成 $30^\circ$ 角。关于轴的力矩大小为

Q8MEDIUM 5.4 Parallel-Axis Theorem5.4 平行轴定理No Calculator

A solid uniform disk of mass $M$ and radius $R$ rotates about an axis perpendicular to the disk through a point on its edge. Its moment of inertia about that axis is质量为 $M$、半径为 $R$ 的实心均匀圆盘,绕过其边缘上一点且垂直于圆盘的轴转动。关于该轴的转动惯量为

Q9MEDIUM 5.5 Static Equilibrium5.5 静力平衡Calculator

A uniform horizontal beam of mass $10~\mathrm{kg}$ and length $4.0~\mathrm{m}$ is hinged at the wall at one end. The far end is supported by a vertical cable. The tension in the cable is一根质量为 $10~\mathrm{kg}$、长 $4.0~\mathrm{m}$ 的均匀水平梁,一端用铰链固定在墙上,另一端由竖直绳索支撑。绳索中的张力为

Q10MEDIUM 5.6 Newton's 2nd Law (Rotation)5.6 转动牛顿第二定律No Calculator

A net torque of $6.0~\mathrm{N \cdot m}$ acts on a wheel with moment of inertia $I = 2.0~\mathrm{kg \cdot m^2}$. The wheel's angular acceleration is一个合力矩 $6.0~\mathrm{N \cdot m}$ 作用在转动惯量 $I = 2.0~\mathrm{kg \cdot m^2}$ 的轮子上。该轮子的角加速度为

Q11MEDIUM 5.1 Angular Displacement5.1 角位移No Calculator

A wheel starts from rest and undergoes constant angular acceleration $\alpha = 2.0~\mathrm{rad/s^2}$. Its angular displacement after $5.0~\mathrm{s}$ is一个轮子从静止开始,以匀角加速度 $\alpha = 2.0~\mathrm{rad/s^2}$ 转动。$5.0~\mathrm{s}$ 后的角位移为

Q12MEDIUM 5.4 Comparing Moments of Inertia5.4 转动惯量比较No Calculator

A thin hoop, a solid disk, and a solid sphere all have the same mass $M$ and the same radius $R$, and rotate about axes through their centers. Ranked from greatest to least moment of inertia, they are一个细圆环、一个实心圆盘和一个实心球体,质量均为 $M$,半径均为 $R$,均绕通过各自中心的轴转动。按转动惯量从大到小排列,顺序为

Q13MEDIUM 5.6 Disk Angular Acceleration5.6 圆盘角加速度No Calculator

A net torque $\tau$ acts on a uniform solid disk of mass $M$ and radius $R$ about its central axis. The angular acceleration of the disk is合力矩 $\tau$ 作用于质量为 $M$、半径为 $R$ 的均匀实心圆盘的中心轴上。圆盘的角加速度为

Q14HARD 5.6 Atwood with Massive Pulley5.6 有质量滑轮的阿特伍德机Calculator

Two masses, $m_1 = 4.0~\mathrm{kg}$ and $m_2 = 2.0~\mathrm{kg}$, are connected by a massless string passing over a uniform-disk pulley of mass $M = 4.0~\mathrm{kg}$ and radius $R$. The string does not slip on the pulley. Take $g = 10~\mathrm{m/s^2}$. The acceleration of the masses is closest to两个质量 $m_1 = 4.0~\mathrm{kg}$ 和 $m_2 = 2.0~\mathrm{kg}$ 通过无质量绳索跨过质量为 $M = 4.0~\mathrm{kg}$、半径为 $R$ 的均匀圆盘滑轮相连。绳索在滑轮上不打滑。取 $g = 10~\mathrm{m/s^2}$。两质量的加速度最接近

Q15HARD 5.5 Ladder Equilibrium5.5 梯子平衡No Calculator

A uniform ladder of mass $M$ leans at angle $\theta$ above the horizontal against a frictionless vertical wall. The coefficient of static friction between the ladder and the floor is $\mu$. The minimum value of $\theta$ for which the ladder does not slip is given by质量为 $M$ 的均匀梯子以与水平面成 $\theta$ 角靠在无摩擦的竖直墙壁上。梯子与地面之间的静摩擦系数为 $\mu$。梯子不滑动的最小 $\theta$ 值满足

Q16HARD 5.4 Composite Rigid Body5.4 复合刚体No Calculator

A point mass $m$ is rigidly attached to one end of a uniform rod of mass $M$ and length $L$. The system rotates about an axis through the rod's free end (i.e. the end opposite the point mass), perpendicular to the rod. Its moment of inertia about that axis is质量为 $m$ 的质点固定在质量为 $M$、长度为 $L$ 的均匀杆的一端。该系统绕过杆的自由端(即与质点相对的一端)且垂直于杆的轴转动。关于该轴的转动惯量为

Q17HARD 5.3 Vector Torque5.3 矢量力矩No Calculator

A force $\vec F = 3\,\hat\jmath~\mathrm{N}$ is applied at the position $\vec r = 2\,\hat\imath~\mathrm{m}$ relative to a fixed pivot. The torque about that pivot is一个力 $\vec F = 3\,\hat\jmath~\mathrm{N}$ 施加在相对于固定转轴位置为 $\vec r = 2\,\hat\imath~\mathrm{m}$ 处。关于该转轴的力矩为

Q18HARD 5.6 Yo-Yo on Fixed String5.6 固定绳上的悠悠球No Calculator

A uniform solid disk of mass $M$ and radius $R$ has a light cord wrapped around it; the upper end of the cord is fixed. The disk is released from rest and falls vertically as the cord unwinds. The downward acceleration of the disk's center is质量为 $M$、半径为 $R$ 的均匀实心圆盘上绕有一轻绳,绳的上端固定。圆盘从静止开始释放,随绳展开竖直下落。圆盘中心的向下加速度为

PART IIFree-Response · Topics 5.1 - 5.6自由作答 · 主题 5.1 - 5.6

Free-Response Questions自由作答题

Show all work in the space provided. Partial credit is awarded for correct setup, units, and reasoning. Use $g = 9.8~\mathrm{m/s^2}$ unless otherwise stated.请在所提供的空白处写出所有解题过程。正确的建模、单位和推理均可获得部分分数。除非另有说明,使用 $g = 9.8~\mathrm{m/s^2}$。

FRQ 1MEDIUM 5.6 Massive Pulley + Hanging Block5.6 有质量滑轮与悬挂物块Calculator

A uniform-disk pulley of mass $M = 2.0~\mathrm{kg}$ and radius $R = 0.10~\mathrm{m}$ has its axis fixed and frictionless. A light cord is wrapped around the pulley; a block of mass $m = 1.0~\mathrm{kg}$ hangs from the other end. The cord does not slip. The system is released from rest.质量为 $M = 2.0~\mathrm{kg}$、半径为 $R = 0.10~\mathrm{m}$ 的均匀圆盘滑轮,其轴固定且无摩擦。一轻绳绕在滑轮上,另一端悬挂质量为 $m = 1.0~\mathrm{kg}$ 的物块。绳索不打滑。系统从静止开始释放。

(a) Draw a free-body diagram for the block and a separate diagram showing the forces and torques on the pulley.画出物块的受力分析图,以及滑轮所受力和力矩的受力分析图。
(b) Apply Newton's 2nd law to the block (translational form) and to the pulley (rotational form). Use the no-slip condition $a = R\alpha$ to solve for the linear acceleration $a$ of the block and the tension $T$ in the cord.对物块应用牛顿第二定律(平动形式),对滑轮应用牛顿第二定律(转动形式)。利用不打滑条件 $a = R\alpha$ 求物块的线加速度 $a$ 和绳中张力 $T$。
(c) Determine the angular speed $\omega$ of the pulley after the block has fallen a distance $h = 1.0~\mathrm{m}$.求物块下落 $h = 1.0~\mathrm{m}$ 后滑轮的角速度 $\omega$。
(d) Use energy conservation as an independent check on your answer to (c).用能量守恒对 (c) 的答案进行独立验证。
FRQ 2MEDIUM 5.5 Beam-and-Cable Equilibrium5.5 梁与绳索的平衡Calculator

A uniform horizontal beam of mass $M = 20~\mathrm{kg}$ and length $L = 4.0~\mathrm{m}$ is attached to a wall by a frictionless hinge at one end. The far end of the beam is supported by a light cable that runs from the far end of the beam upward to the wall, making an angle of $30^\circ$ above the horizontal beam. A point load of mass $m = 10~\mathrm{kg}$ hangs from the far end of the beam.质量为 $M = 20~\mathrm{kg}$、长 $L = 4.0~\mathrm{m}$ 的均匀水平梁,一端通过无摩擦铰链固定在墙上。梁的远端由一根轻绳支撑,绳从梁的远端向上连接到墙壁,与水平梁成 $30^\circ$ 角。质量为 $m = 10~\mathrm{kg}$ 的集中载荷悬挂在梁的远端。

(a) Draw a free-body diagram for the beam, identifying all forces (gravity on beam, gravity on load, hinge force components, cable tension).画出梁的受力分析图,标注所有力(梁的重力、载荷的重力、铰链力各分量、绳索张力)。
(b) Take torques about the hinge to find the tension $T$ in the cable.以铰链为支点取力矩,求绳索中的张力 $T$。
(c) Apply $\sum F_x = 0$ and $\sum F_y = 0$ to find the horizontal and vertical components of the hinge force on the beam.应用 $\sum F_x = 0$ 和 $\sum F_y = 0$,求铰链对梁的水平和竖直分力。
(d) Determine the magnitude and direction of the resultant hinge force.求合铰链力的大小和方向。
FRQ 3HARD 5.4 / 5.6 Pivoted Composite Rod5.4 / 5.6 有枢轴的复合杆Calculator

A uniform rod of mass $M = 0.40~\mathrm{kg}$ and length $L = 1.20~\mathrm{m}$ has a small point mass $m = 0.10~\mathrm{kg}$ rigidly attached to one end. The rod is free to rotate about a frictionless pivot at the opposite end (so the heavy end is far from the pivot). The rod is held in a horizontal position and released from rest.质量为 $M = 0.40~\mathrm{kg}$、长 $L = 1.20~\mathrm{m}$ 的均匀杆,在一端固定连接一个质量为 $m = 0.10~\mathrm{kg}$ 的小质点。杆可绕另一端的无摩擦枢轴自由转动(即重端远离枢轴)。杆保持水平位置,从静止释放。

(a) Determine the moment of inertia of the rod-plus-point-mass system about the pivot.求杆与质点系统关于枢轴的转动惯量。
(b) Locate the position of the system's center of mass measured from the pivot.求系统质心相对于枢轴的位置。
(c) Determine the initial angular acceleration of the system at the instant of release.求系统在释放瞬间的初始角加速度。
(d) Use energy conservation to determine the angular speed of the system when the rod has rotated to the vertical (point mass straight down).用能量守恒求杆转到竖直位置(质点朝正下方)时系统的角速度。
FRQ 4HARD 5.6 Rolling Cylinder on Incline5.6 斜面上滚动的圆柱体Calculator

A uniform solid cylinder of mass $M$ and radius $R$ rolls without slipping down an incline of angle $\theta$.质量为 $M$、半径为 $R$ 的均匀实心圆柱体,在倾角为 $\theta$ 的斜面上做无滑动滚动。

(a) Draw a free-body diagram for the cylinder. Identify each force.画出圆柱体的受力分析图,标注每个力。
(b) Apply Newton's 2nd law (translational form, along the incline) and the rotational form (torque about the cylinder's center). Use the no-slip condition $a = R\alpha$.应用牛顿第二定律(平动形式,沿斜面方向)和转动形式(关于圆柱中心的力矩)。利用不打滑条件 $a = R\alpha$。
(c) Solve for the linear acceleration $a$ of the cylinder's center and the static-friction force $f_s$ on the cylinder, in terms of $M$, $g$, and $\theta$.用 $M$、$g$ 和 $\theta$ 表示,求圆柱中心的线加速度 $a$ 和作用在圆柱上的静摩擦力 $f_s$。
(d) Determine the minimum coefficient of static friction $\mu_s$ between the cylinder and the incline that is required for rolling without slipping.求圆柱与斜面之间维持无滑动滚动所需的最小静摩擦系数 $\mu_s$。