Unit A4 · Space, Time and Motion (HL)Unit A4 · 空间、时间与运动（HL）
Rigid Body Mechanics刚体力学
IB-Style Practice QuestionsIB 风格练习题
MEDIUMHARDPaper 1Paper 1BPaper 2HL ONLY
Syllabus A4.1 to A4.6考纲 A4.1 至 A4.6PHYSICS HL
Name:姓名：Date:日期：
PART I · PAPER 1 STYLE第一部分 · 第一卷风格Short structured · calculator · 30 marks短结构题 · 可用计算器 · 30 分
Short Structured Items短结构题
Show all working in the space below each question. Marks are awarded for correct method as well as final answers. Quote any moment of inertia from the data booklet and state the axis. Keep all angles in radians where rotational kinematics is used. Give numerical answers to an appropriate number of significant figures.在每题下方空白处写出全部解题过程。方法分（method marks）与最终答案同等重要。转动惯量须从数据手册引用并写明转轴（axis）。用到转动运动学时角度一律用弧度。数值答案保留适当的有效数字。
Q1MEDIUMPaper 1HL ONLYtorque and couples力矩与力偶[4 marks]
A mechanic applies a force of $40\ \mathrm{N}$ to a spanner at a point $0.25\ \mathrm{m}$ from the bolt. The force acts at $60^{\circ}$ to the length of the spanner.一名机械师在距螺栓 $0.25\ \mathrm{m}$ 处对扳手施加 $40\ \mathrm{N}$ 的力。该力与扳手长度方向成 $60^{\circ}$。
(a)Calculate the torque about the bolt.计算绕螺栓的力矩。[2]
(b)A second mechanic instead applies a couple to the bolt head, using two parallel $8.0\ \mathrm{N}$ forces acting in opposite directions, separated by a perpendicular distance of $0.30\ \mathrm{m}$. State the net force and calculate the torque of the couple.第二名机械师改用力偶作用于螺栓头：两个方向相反、相距垂直距离 $0.30\ \mathrm{m}$ 的平行 $8.0\ \mathrm{N}$ 力。写出合力并计算力偶矩。[2]
Q2MEDIUMPaper 1HL ONLYmoment of inertia: axis dependence转动惯量：与转轴相关[6 marks]
Two point masses, $1.5\ \mathrm{kg}$ and $2.5\ \mathrm{kg}$, are fixed at the ends of a light (massless) rod of length $0.80\ \mathrm{m}$.两个质点 $1.5\ \mathrm{kg}$ 与 $2.5\ \mathrm{kg}$ 固定在一根长 $0.80\ \mathrm{m}$ 的轻质（无质量）杆两端。
(a)Calculate the moment of inertia of the system about an axis through the centre of the rod, perpendicular to it.计算系统绕过杆中点且垂直于杆的轴的转动惯量。[2]
(b)Calculate the moment of inertia about an axis through the $1.5\ \mathrm{kg}$ mass, perpendicular to the rod.计算绕过 $1.5\ \mathrm{kg}$ 质点且垂直于杆的轴的转动惯量。[2]
(c)A solid disc of mass $2.0\ \mathrm{kg}$ and radius $0.15\ \mathrm{m}$ rotates about its central axis. Using the data booklet, calculate its moment of inertia.一个质量 $2.0\ \mathrm{kg}$、半径 $0.15\ \mathrm{m}$ 的实心圆盘绕中心轴旋转。用数据手册计算其转动惯量。[2]
A turbine rotor starts from rest and accelerates uniformly at a constant angular acceleration of $2.5\ \mathrm{rad\,s^{-2}}$ for $8.0\ \mathrm{s}$.一个涡轮转子从静止开始，以恒定角加速度 $2.5\ \mathrm{rad\,s^{-2}}$ 均匀加速 $8.0\ \mathrm{s}$。
(a)Calculate the angular velocity of the rotor after $8.0\ \mathrm{s}$.计算 $8.0\ \mathrm{s}$ 后转子的角速度。[2]
(b)Calculate the total angle turned through in this time, in radians, and hence the number of complete revolutions made.计算这段时间内转过的总角度（弧度），并由此求完整转过的圈数。[3]
(c)A point on the rotor lies $0.40\ \mathrm{m}$ from the axis. State the linear (tangential) speed of this point at $t = 8.0\ \mathrm{s}$.转子上一点距轴 $0.40\ \mathrm{m}$。写出该点在 $t = 8.0\ \mathrm{s}$ 时的线（切向）速率。[1]
Q4HARDPaper 1HL ONLYNewton's 2nd law for rotation转动的牛顿第二定律[6 marks]
A flywheel modelled as a solid disc has mass $8.0\ \mathrm{kg}$ and radius $0.20\ \mathrm{m}$, and is free to rotate about its central axis. A constant tangential force of $12\ \mathrm{N}$ is applied at the rim, starting from rest. Friction is negligible.一个建模为实心圆盘的飞轮，质量 $8.0\ \mathrm{kg}$、半径 $0.20\ \mathrm{m}$，可绕中心轴自由转动。在轮缘施加恒定切向力 $12\ \mathrm{N}$，从静止开始。摩擦可忽略。
(a)Calculate the moment of inertia of the flywheel and the torque produced by the applied force.计算飞轮的转动惯量与所施力产生的力矩。[2]
(b)Using Newton's second law for rotation, calculate the angular acceleration of the flywheel.用转动的牛顿第二定律计算飞轮的角加速度。[2]
(c)Calculate the time taken for the flywheel to reach an angular velocity of $30\ \mathrm{rad\,s^{-1}}$.计算飞轮达到 $30\ \mathrm{rad\,s^{-1}}$ 角速度所需的时间。[2]
Q5HARDPaper 1HL ONLYconservation of angular momentum角动量守恒[8 marks]
A figure skater spins about a vertical axis with her arms outstretched, with a moment of inertia of $5.0\ \mathrm{kg\,m^2}$ and an angular velocity of $1.8\ \mathrm{rad\,s^{-1}}$. She then pulls her arms in, reducing her moment of inertia to $2.0\ \mathrm{kg\,m^2}$. Friction at the ice is negligible.一名花样滑冰者绕竖直轴张开双臂旋转，转动惯量为 $5.0\ \mathrm{kg\,m^2}$，角速度为 $1.8\ \mathrm{rad\,s^{-1}}$。她随后收回双臂，使转动惯量减至 $2.0\ \mathrm{kg\,m^2}$。冰面摩擦可忽略。
(a)State the condition under which angular momentum is conserved, and explain why it applies here.写出角动量守恒的条件，并解释为何此处适用。[2]
(b)Calculate the skater's angular velocity after she pulls her arms in.计算滑冰者收臂后的角速度。[2]
(c)Calculate the rotational kinetic energy before and after, and comment on whether kinetic energy is conserved.计算收臂前后的转动动能，并说明动能是否守恒。[3]
(d)State the source of any change in kinetic energy.写出动能变化的来源。[1]
PART II · PAPER 1B / DATA ANALYSIS第二部分 · 第一卷 B / 数据分析Graphs · data · uncertainties · 22 marks图像 · 数据 · 不确定度 · 22 分
Graph and Data Questions图像与数据题
These items reward careful reading of gradients, correct handling of uncertainties, and clean rotational-equilibrium setups. Quote uncertainties to one significant figure and round the value to match.这些题考查对斜率的细致读取、对不确定度的正确处理以及干净的转动平衡列式。不确定度保留 1 位有效数字，并使数值的末位与之对齐。
Q6HARDPaper 1BHL ONLYangular acceleration from a graph + uncertainty由图像求角加速度与不确定度[10 marks]
A torque is applied to a wheel of moment of inertia $0.45\ \mathrm{kg\,m^2}$, which is already turning. A student measures the angular velocity $\omega$ at several times $t$ and tabulates the results:对一个已在转动、转动惯量为 $0.45\ \mathrm{kg\,m^2}$ 的轮子施加力矩。学生测量若干时刻 $t$ 对应的角速度 $\omega$ 并列表：
$t\ /\ \mathrm{s}$
$2.0$
$4.0$
$6.0$
$8.0$
$\omega\ /\ \mathrm{rad\,s^{-1}}$
$5.0$
$9.0$
$13.0$
$17.0$
(a)Explain why a graph of $\omega$ against $t$ should be a straight line, and state what its gradient represents.解释为何 $\omega$ 对 $t$ 的图应为直线，并说明其斜率代表什么。[2]
(b)Calculate the gradient of the line and hence determine the angular acceleration of the wheel.计算该直线的斜率，由此求轮的角加速度。[3]
(c)Using $\tau = I\alpha$, calculate the magnitude of the applied torque.用 $\tau = I\alpha$ 计算所施力矩的大小。[2]
(d)The reading at $t = 8.0\ \mathrm{s}$ has an absolute uncertainty of $\pm 0.5\ \mathrm{rad\,s^{-1}}$. Calculate the percentage uncertainty in this value of $\omega$.$t = 8.0\ \mathrm{s}$ 处的读数有绝对不确定度 $\pm 0.5\ \mathrm{rad\,s^{-1}}$。计算该 $\omega$ 值的百分比不确定度。[2]
(e)State what the non-zero vertical intercept of the line tells you about the motion of the wheel.说明该直线的非零纵截距反映轮的运动状态。[1]
Q7HARDPaper 1BHL ONLYrotational equilibrium of a loaded beam受载梁的转动平衡[12 marks]
A uniform beam of weight $300\ \mathrm{N}$ and length $6.0\ \mathrm{m}$ rests horizontally on two supports: one at the left end (force $R_L$) and one at $5.0\ \mathrm{m}$ from the left end (force $R_R$). A load of weight $500\ \mathrm{N}$ is placed $4.0\ \mathrm{m}$ from the left end.一根均匀梁，重 $300\ \mathrm{N}$、长 $6.0\ \mathrm{m}$，水平搁在两个支点上：一个在左端（受力 $R_L$），另一个距左端 $5.0\ \mathrm{m}$（受力 $R_R$）。一个重 $500\ \mathrm{N}$ 的载荷放在距左端 $4.0\ \mathrm{m}$ 处。
(a)State the two conditions for the beam to be in equilibrium.写出梁处于平衡的两个条件。[2]
(b)By taking moments about the left end, calculate the force $R_R$ at the right support. State why the left end is a convenient pivot.对左端取矩，计算右支点的力 $R_R$。说明为何左端是方便的转轴。[4]
(c)Hence calculate the force $R_L$ at the left support.由此计算左支点的力 $R_L$。[2]
(d)Verify your value of $R_L$ by taking moments about the right support.通过对右支点取矩验证你的 $R_L$ 值。[2]
(e)The weight of the beam is known to $\pm 10\ \mathrm{N}$. Calculate the percentage uncertainty in the beam's weight.梁的重量已知到 $\pm 10\ \mathrm{N}$。计算梁重的百分比不确定度。[2]
PART III · PAPER 2 STYLE第三部分 · 第二卷风格Extended structured · calculator · 30 marks长结构题 · 可用计算器 · 30 分
Extended Structured Problems长结构问题
Set up each problem with a clear diagram and labelled axis. In coupled problems, write the translational equation ($F = ma$) and the rotational equation ($\tau = I\alpha$) separately, then link them with the constraint $a = \alpha R$. Carry intermediate values to extra figures and round only the final answer.每题先画清晰的示意图并标注转轴。耦合题中，分别写出平动方程（$F = ma$）与转动方程（$\tau = I\alpha$），再用约束 $a = \alpha R$ 把它们联系起来。中间值多保留几位，仅在最终答案处取舍有效数字。
A light, inextensible string is wrapped around the rim of a pulley modelled as a solid disc of mass $2.0\ \mathrm{kg}$ and radius $0.10\ \mathrm{m}$, free to rotate about a fixed horizontal axis. A block of mass $1.5\ \mathrm{kg}$ hangs from the free end of the string and is released from rest. The string does not slip on the pulley and friction at the axle is negligible.一根轻而不可伸长的绳缠绕在滑轮的轮缘上，滑轮建模为质量 $2.0\ \mathrm{kg}$、半径 $0.10\ \mathrm{m}$ 的实心圆盘，可绕固定水平轴自由转动。一个质量 $1.5\ \mathrm{kg}$ 的物块挂在绳的自由端，从静止释放。绳在滑轮上不打滑，轴处摩擦可忽略。
(a)Write Newton's second law for the falling block (taking downward as positive) and Newton's second law for rotation of the pulley.写出下落物块的牛顿第二定律（取向下为正）以及滑轮转动的牛顿第二定律。[3]
(b)Using the rolling/string constraint $a = \alpha R$, show that the acceleration of the block is $a = \dfrac{mg}{m + \tfrac12 M}$, and calculate its value.用绳约束 $a = \alpha R$，证明物块的加速度为 $a = \dfrac{mg}{m + \tfrac12 M}$，并计算其值。[4]
(c)Calculate the tension in the string and the angular acceleration of the pulley.计算绳中的张力与滑轮的角加速度。[3]
(d)Calculate the speed of the block after it has fallen $1.2\ \mathrm{m}$.计算物块下落 $1.2\ \mathrm{m}$ 后的速率。[2]
Q9HARDPaper 2HL ONLYrolling without slipping + energy无滑滚动与能量[10 marks]
A solid uniform cylinder is released from rest and rolls without slipping down a slope, descending a vertical height of $2.0\ \mathrm{m}$. For a solid cylinder about its central axis, $I = \tfrac12 mR^2$.一个实心均匀圆柱从静止释放，无滑滚下斜坡，竖直下降 $2.0\ \mathrm{m}$。实心圆柱绕中心轴的 $I = \tfrac12 mR^2$。
(a)State the rolling constraint that links the linear speed $v$ to the angular speed $\omega$.写出把线速率 $v$ 与角速率 $\omega$ 联系起来的滚动约束。[1]
(b)Write the energy-conservation equation for the cylinder, including both kinetic energy terms, and show that $mgh = \tfrac34 mv^2$.写出圆柱的能量守恒方程（包含两项动能），并证明 $mgh = \tfrac34 mv^2$。[4]
(c)Calculate the linear speed of the cylinder at the bottom of the slope.计算圆柱到达坡底时的线速率。[2]
(d)Calculate the fraction of the total kinetic energy that is rotational, and explain why a frictionless sliding block of the same mass would reach the bottom moving faster.计算总动能中转动部分所占的比例，并解释为何同质量的无摩擦滑块到达坡底时运动更快。[3]
Q10HARDPaper 2HL ONLYangular momentum in a rotational collision转动碰撞中的角动量[8 marks]
A horizontal turntable, modelled as a disc, has a moment of inertia of $0.40\ \mathrm{kg\,m^2}$ and rotates freely about a vertical axis at $4.0\ \mathrm{rad\,s^{-1}}$. A lump of modelling clay of mass $0.50\ \mathrm{kg}$ is dropped vertically and sticks to the turntable at a point $0.30\ \mathrm{m}$ from the axis.一个水平转盘建模为圆盘，转动惯量为 $0.40\ \mathrm{kg\,m^2}$，绕竖直轴以 $4.0\ \mathrm{rad\,s^{-1}}$ 自由旋转。一块质量 $0.50\ \mathrm{kg}$ 的橡皮泥竖直落下，粘在转盘上距轴 $0.30\ \mathrm{m}$ 处。
(a)Explain why angular momentum is conserved during the collision but linear momentum of the clay is not a useful conserved quantity here.解释为何碰撞过程中角动量守恒，而此处橡皮泥的线动量不是有用的守恒量。[2]
(b)Calculate the moment of inertia of the clay about the axis, and hence the total moment of inertia after the clay sticks.计算橡皮泥绕轴的转动惯量，由此求橡皮泥粘上后的总转动惯量。[2]
(c)Calculate the angular velocity of the turntable immediately after the clay sticks.计算橡皮泥粘上后转盘的角速度。[2]
(d)Calculate the rotational kinetic energy before and after the collision, and state whether the collision is elastic.计算碰撞前后的转动动能，并说明该碰撞是否为弹性碰撞。[2]