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Chapter 6 · Mechanics第6章 · 力学

Energy & Momentum of Rotating Systems转动系统的能量与动量

AP-Style Practice QuestionsAP 风格练习题

EASY MEDIUM HARD

Topics主题 6.1 - 6.6MECH



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PART ITopics 6.1 - 6.6主题 6.1 - 6.6

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; $G = 6.67 \times 10^{-11}~\mathrm{N \cdot m^2/kg^2}$.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 $g = 9.8~\mathrm{m/s^2}$;$G = 6.67 \times 10^{-11}~\mathrm{N \cdot m^2/kg^2}$。

Q1EASY 6.1 Rotational Kinetic Energy6.1 转动动能No Calculator

A flywheel of moment of inertia $I = 2.0~\mathrm{kg \cdot m^2}$ spins about its axis at $\omega = 3.0~\mathrm{rad/s}$. Its rotational kinetic energy is一个转动惯量为 $I = 2.0~\mathrm{kg \cdot m^2}$ 的飞轮以 $\omega = 3.0~\mathrm{rad/s}$ 绕轴旋转。其转动动能为

Q2EASY 6.3 Angular Momentum6.3 角动量No Calculator

A particle of mass $2.0~\mathrm{kg}$ moves at $3.0~\mathrm{m/s}$ in a circle of radius $0.50~\mathrm{m}$. The magnitude of its angular momentum about the center of the circle is一个质量为 $2.0~\mathrm{kg}$ 的质点以 $3.0~\mathrm{m/s}$ 的速度在半径 $0.50~\mathrm{m}$ 的圆形轨道上运动。其相对于圆心的角动量大小为

Q3EASY 6.4 Conservation of L6.4 角动量守恒No Calculator

A spinning ice skater pulls her arms in toward her body. With no external torques acting on her, her angular speed一位旋转的花样滑冰运动员将手臂向身体收拢。在没有外力矩作用的情况下,她的角速度

Q4EASY 6.6 Orbital Speed6.6 轨道速度No Calculator

A satellite is in a circular orbit around Earth. As the orbital radius is increased, the satellite's orbital speed一颗卫星围绕地球做圆轨道运动。随着轨道半径增大,卫星的轨道速度

Q5MEDIUM 6.1 Disk vs. Hoop KE6.1 圆盘与圆环转动动能比较No Calculator

A solid disk and a thin hoop have the same mass $M$ and the same radius $R$, and rotate about their central axes at the same angular speed $\omega$. The ratio of the disk's rotational kinetic energy to the hoop's rotational kinetic energy is一个实心圆盘和一个细圆环质量均为 $M$,半径均为 $R$,以相同的角速度 $\omega$ 绕各自的中心轴旋转。圆盘的转动动能与圆环的转动动能之比为

Q6MEDIUM 6.2 Rotational Work6.2 转动做功No Calculator

A constant torque of $5.0~\mathrm{N \cdot m}$ rotates a wheel through an angular displacement of $4.0~\mathrm{rad}$. The work done on the wheel by the torque is一个大小为 $5.0~\mathrm{N \cdot m}$ 的恒定力矩使一个轮子转过 $4.0~\mathrm{rad}$ 的角位移。该力矩对轮子做的功为

Q7MEDIUM 6.4 Disk-Drop Collision6.4 圆盘叠落碰撞No Calculator

A disk of moment of inertia $I$ spins about a vertical axis at angular speed $\omega_0$. A second disk of moment of inertia $2I$, initially at rest, is gently dropped onto the first disk and sticks to it. The final angular speed of the combined disks is一个转动惯量为 $I$ 的圆盘以角速度 $\omega_0$ 绕竖直轴旋转。另一个转动惯量为 $2I$ 的圆盘从静止开始轻轻叠落在第一个圆盘上并与之粘合。两圆盘合并后的最终角速度为

Q8MEDIUM 6.5 Sphere Rolling Down Incline6.5 实心球沿斜面滚动No Calculator

A uniform solid sphere is released from rest at the top of an incline of vertical height $h$ and rolls without slipping to the bottom. Its speed at the bottom is一个均匀实心球从高度为 $h$ 的斜面顶端由静止开始释放,无滑动地滚到底部。到达底部时的速度为

Q9MEDIUM 6.6 Kepler's 3rd Law6.6 开普勒第三定律No Calculator

Two satellites orbit the same planet in circular orbits. The orbital radius of satellite 2 is twice that of satellite 1. The ratio $T_2/T_1$ of their orbital periods is两颗卫星在同一行星的圆形轨道上运行。卫星 2 的轨道半径是卫星 1 的两倍。它们的轨道周期之比 $T_2/T_1$ 为

Q10MEDIUM 6.3 Angular Impulse6.3 角冲量No Calculator

A constant net torque of $4.0~\mathrm{N \cdot m}$ acts on a wheel of moment of inertia $2.0~\mathrm{kg \cdot m^2}$, initially at rest, for $3.0~\mathrm{s}$. The wheel's final angular speed is一个大小为 $4.0~\mathrm{N \cdot m}$ 的恒定合力矩作用在转动惯量为 $2.0~\mathrm{kg \cdot m^2}$ 的轮子上,持续时间 $3.0~\mathrm{s}$,轮子初始静止。轮子的最终角速度为

Q11MEDIUM 6.5 Rolling KE Fraction6.5 滚动动能占比No Calculator

A uniform solid disk rolls without slipping along a horizontal surface with center-of-mass speed $v$. The fraction of its total kinetic energy that is rotational is一个均匀实心圆盘以质心速度 $v$ 在水平面上无滑动地滚动。其动能中转动动能所占的比例为

Q12MEDIUM 6.4 Bullet-into-Pivoted-Rod6.4 子弹射入转动杆No Calculator

A uniform rod of mass $M$ and length $L$, pivoted at one end and initially at rest, is struck and embedded by a bullet of mass $m$ moving with speed $v$ at the rod's free end (perpendicular to the rod). The angular speed of the rod-plus-bullet system just after the collision is一根质量为 $M$、长度为 $L$ 的均匀细杆,一端铰支,初始静止,被质量为 $m$、速度为 $v$ 的子弹垂直射入其自由端并嵌入其中。碰撞后,杆与子弹组合体的角速度为

Q13MEDIUM 6.1 Work to Stop Wheel6.1 使轮子停转所做的功No Calculator

A wheel of moment of inertia $0.50~\mathrm{kg \cdot m^2}$ spins at $4.0~\mathrm{rad/s}$. The magnitude of the work that must be done on the wheel to bring it to rest is一个转动惯量为 $0.50~\mathrm{kg \cdot m^2}$ 的轮子以 $4.0~\mathrm{rad/s}$ 旋转。使轮子停止旋转所需做的功的大小为

Q14HARD 6.4 Person on Rotating Platform6.4 人站在旋转台上Calculator

A horizontal turntable of moment of inertia $I_p = 100~\mathrm{kg \cdot m^2}$ rotates freely about a vertical axis at $\omega_0 = 1.0~\mathrm{rad/s}$. A person of mass $50~\mathrm{kg}$ stands at rest on the turntable at $r = 1.0~\mathrm{m}$ from the axis. The person then walks to the axis (treat the person as a point mass throughout). The new angular speed of the turntable is一个转动惯量为 $I_p = 100~\mathrm{kg \cdot m^2}$ 的水平转台以 $\omega_0 = 1.0~\mathrm{rad/s}$ 绕竖直轴自由旋转。一个质量为 $50~\mathrm{kg}$ 的人静止站在距轴 $r = 1.0~\mathrm{m}$ 处。该人随后走向轴心(全程将人视为质点)。转台的新角速度为

Q15HARD 6.6 Geosynchronous Orbit6.6 地球同步轨道Calculator

A satellite is placed in a circular orbit such that its orbital period equals one Earth day. Take $GM_\mathrm{E} = 4.0 \times 10^{14}~\mathrm{m^3/s^2}$ and $T_\mathrm{day} = 8.64 \times 10^4~\mathrm{s}$. The orbital radius (measured from Earth's center) is closest to一颗卫星被置于圆形轨道上,使其轨道周期等于地球自转一天。取 $GM_\mathrm{E} = 4.0 \times 10^{14}~\mathrm{m^3/s^2}$,$T_\mathrm{day} = 8.64 \times 10^4~\mathrm{s}$。轨道半径(从地心算起)最接近

Q16HARD 6.5 Race Down the Incline6.5 斜面竞速No Calculator

A solid sphere, a uniform solid disk, and a thin hoop, all of mass $M$ and radius $R$, are released simultaneously from rest at the top of the same incline and roll without slipping to the bottom. Which arrives first?一个实心球、一个均匀实心圆盘和一个细圆环,质量均为 $M$,半径均为 $R$,同时从同一斜面顶端由静止释放,无滑动地滚到底部。哪个最先到达?

Q17HARD 6.3 Direction of Angular Momentum6.3 角动量的方向No Calculator

A particle moves in a circular path in the $xy$-plane, traveling counter-clockwise as viewed from the $+z$ axis. Its angular momentum about the center of the circle points in the一个质点在 $xy$ 平面内沿圆形轨道运动,从 $+z$ 轴方向俯视为逆时针。其相对于圆心的角动量指向

Q18HARD 6.6 Total Orbital Energy6.6 轨道总机械能No Calculator

A satellite of mass $m$ orbits a planet of mass $M$ in a circular orbit of radius $r$. The total mechanical energy of the satellite is质量为 $m$ 的卫星在质量为 $M$ 的行星的半径为 $r$ 的圆形轨道上运行。卫星的总机械能为

PART IIFree-Response · Topics 6.1 - 6.6自由回答题 · 主题 6.1 - 6.6

Free-Response Questions自由回答题

Show all work in the space provided. Partial credit is awarded for correct setup, units, and reasoning. Use $g = 9.8~\mathrm{m/s^2}$ and $G = 6.67 \times 10^{-11}~\mathrm{N \cdot m^2/kg^2}$ unless otherwise stated.请在所提供的空白处写出完整的解题过程。正确的建模、单位和推理均可获得部分分数。除非另有说明,取 $g = 9.8~\mathrm{m/s^2}$,$G = 6.67 \times 10^{-11}~\mathrm{N \cdot m^2/kg^2}$。

FRQ 1MEDIUM 6.4 Disk-on-Disk Collision6.4 圆盘叠落碰撞Calculator

A horizontal disk of mass $M = 2.0~\mathrm{kg}$ and radius $R = 0.30~\mathrm{m}$ rotates about a fixed, frictionless vertical axis at angular speed $\omega_0 = 6.0~\mathrm{rad/s}$. A second, identical disk (same $M$, same $R$, axis aligned with the first) is dropped onto the first disk from rest and sticks to it.一个质量为 $M = 2.0~\mathrm{kg}$、半径为 $R = 0.30~\mathrm{m}$ 的水平圆盘以角速度 $\omega_0 = 6.0~\mathrm{rad/s}$ 绕固定的无摩擦竖直轴旋转。另一个完全相同的圆盘(质量、半径相同,轴与第一个圆盘对齐)从静止开始叠落在第一个圆盘上并粘合。

(a) Determine the moment of inertia of one disk about the axis.求一个圆盘相对于该轴的转动惯量。
(b) Apply conservation of angular momentum to find the final angular speed $\omega_f$ of the combined disks.应用角动量守恒定律,求两圆盘合并后的最终角速度 $\omega_f$。
(c) Compute the total rotational kinetic energy before and after the disks join.计算两圆盘合并前后的总转动动能。
(d) Determine the fraction of the original rotational kinetic energy that was lost, and explain physically where the missing energy went.求损失的转动动能占原来转动动能的比例,并从物理角度解释缺失的能量去哪里了。
FRQ 2MEDIUM 6.5 Sphere Rolling Down Incline6.5 实心球沿斜面滚动Calculator

A uniform solid sphere of mass $M = 0.50~\mathrm{kg}$ and radius $R = 0.10~\mathrm{m}$ is released from rest at the top of an incline of angle $\theta = 30^\circ$ and vertical height $h = 0.80~\mathrm{m}$. The sphere rolls without slipping.一个质量为 $M = 0.50~\mathrm{kg}$、半径为 $R = 0.10~\mathrm{m}$ 的均匀实心球从倾角 $\theta = 30^\circ$、竖直高度 $h = 0.80~\mathrm{m}$ 的斜面顶端由静止释放,无滑动地滚动。

(a) Apply conservation of mechanical energy to derive an expression for the speed of the sphere at the bottom of the incline in terms of $g$ and $h$.应用机械能守恒定律,推导球到达斜面底部时的速度表达式(用 $g$ 和 $h$ 表示)。
(b) Compute the numerical value of the bottom-of-incline speed.计算斜面底部速度的数值。
(c) Apply Newton's 2nd law (translational + rotational) to derive the linear acceleration of the sphere along the incline in terms of $g$ and $\theta$.应用牛顿第二定律(平动 + 转动),推导球沿斜面的线加速度表达式(用 $g$ 和 $\theta$ 表示)。
(d) Determine the minimum coefficient of static friction $\mu_s$ between the sphere and the incline that is required to prevent slipping.求球与斜面之间防止滑动所需的最小静摩擦系数 $\mu_s$。
FRQ 3HARD 6.3 / 6.4 Bullet-into-Rod (Ballistic)6.3 / 6.4 子弹射入杆(弹道实验)Calculator

A uniform rod of mass $M = 0.50~\mathrm{kg}$ and length $L = 0.80~\mathrm{m}$ is pivoted at one end about a frictionless horizontal axis and hangs vertically at rest. A bullet of mass $m = 0.020~\mathrm{kg}$ moving horizontally at $v_0 = 200~\mathrm{m/s}$ strikes the rod at its free (lower) end and embeds in it.一根质量为 $M = 0.50~\mathrm{kg}$、长度为 $L = 0.80~\mathrm{m}$ 的均匀细杆,一端铰支于无摩擦的水平轴上,竖直悬挂静止。一颗质量为 $m = 0.020~\mathrm{kg}$、以水平速度 $v_0 = 200~\mathrm{m/s}$ 运动的子弹射入杆的自由端(下端)并嵌入其中。

(a) Determine the moment of inertia of the rod-plus-bullet system about the pivot.求杆与子弹组合体相对于铰支点的转动惯量。
(b) Apply conservation of angular momentum about the pivot to determine the angular speed of the system immediately after the collision.应用相对于铰支点的角动量守恒,求碰撞后系统的角速度。
(c) Apply conservation of mechanical energy to determine the maximum angle $\theta$ from the vertical that the rod-plus-bullet swings up before momentarily stopping.应用机械能守恒定律,求杆与子弹组合体向上摆动后瞬间静止时偏离竖直方向的最大角度 $\theta$。
(d) Determine the fraction of the bullet's original kinetic energy that was lost in the collision.求碰撞中子弹原始动能的损失比例。
FRQ 4HARD 6.6 Orbital Mechanics6.6 轨道力学Calculator

A satellite of mass $m = 500~\mathrm{kg}$ is in a circular orbit of radius $r_1 = 1.5\,R_\mathrm{E}$ around Earth, where $R_\mathrm{E} = 6.4 \times 10^6~\mathrm{m}$ and $M_\mathrm{E} = 6.0 \times 10^{24}~\mathrm{kg}$.质量为 $m = 500~\mathrm{kg}$ 的卫星在半径 $r_1 = 1.5\,R_\mathrm{E}$ 的圆形轨道上围绕地球运行,其中 $R_\mathrm{E} = 6.4 \times 10^6~\mathrm{m}$,$M_\mathrm{E} = 6.0 \times 10^{24}~\mathrm{kg}$。

(a) Apply Newton's 2nd law (centripetal form) to derive an expression for the orbital speed $v_1$ in terms of $G$, $M_\mathrm{E}$, and $r_1$, and compute its numerical value.应用牛顿第二定律(向心力形式),推导轨道速度 $v_1$ 关于 $G$、$M_\mathrm{E}$ 和 $r_1$ 的表达式,并计算数值。
(b) Determine the orbital period $T_1$ of the satellite.求卫星的轨道周期 $T_1$。
(c) Determine the total mechanical energy $E_1$ of the satellite in this orbit.求卫星在该轨道上的总机械能 $E_1$。
(d) The satellite is to be moved to a higher circular orbit of radius $r_2 = 2.0\,R_\mathrm{E}$. Determine the work that an external agent must do on the satellite (i.e. the change in mechanical energy required).卫星将被转移到半径为 $r_2 = 2.0\,R_\mathrm{E}$ 的更高圆形轨道。求外力对卫星所做的功(即所需的机械能变化量)。