Companion to the IB-Style Practice SetIB 风格练习题的解析配套
Syllabus A4.1 to A4.6考纲 A4.1 至 A4.6PHYSICS HL
Force $40\ \mathrm{N}$ on a spanner at $0.25\ \mathrm{m}$, at $60^{\circ}$ to the arm. (a) torque; (b) a couple of two $8.0\ \mathrm{N}$ forces $0.30\ \mathrm{m}$ apart: net force and torque.扳手上 $0.25\ \mathrm{m}$ 处受 $40\ \mathrm{N}$ 力，与力臂成 $60^{\circ}$。(a) 力矩；(b) 两个相距 $0.30\ \mathrm{m}$ 的 $8.0\ \mathrm{N}$ 力组成力偶：合力与力矩。
Use $\tau = Fr\sin\theta$ with $\theta$ the angle between the force and the arm: $\tau = (40)(0.25)\sin 60^{\circ}$. (M1)
$$ \tau = (40)(0.25)(0.8660) \approx 8.7\ \mathrm{N\,m}. $$(A1)
Two equal, antiparallel forces cancel, so the net force is $0\ \mathrm{N}$. (A1)
The torque of a couple is one force magnitude times the perpendicular separation: $\tau = Fd = (8.0)(0.30) = 2.4\ \mathrm{N\,m}$. (A1)
用 $\tau = Fr\sin\theta$，$\theta$ 为力与力臂的夹角：$\tau = (40)(0.25)\sin 60^{\circ}$。(M1)
$$ \tau = (40)(0.25)(0.8660) \approx 8.7\ \mathrm{N\,m}. $$(A1)
两个大小相等、反向的力相消，故合力为 $0\ \mathrm{N}$。(A1)
力偶矩为单个力的大小乘以垂直间距：$\tau = Fd = (8.0)(0.30) = 2.4\ \mathrm{N\,m}$。(A1)
Masses $1.5\ \mathrm{kg}$ and $2.5\ \mathrm{kg}$ at the ends of a light rod of length $0.80\ \mathrm{m}$. (a) $I$ about the centre; (b) $I$ about the $1.5\ \mathrm{kg}$ end; (c) $I$ of a solid disc, $M = 2.0\ \mathrm{kg}$, $R = 0.15\ \mathrm{m}$.长 $0.80\ \mathrm{m}$ 的轻杆两端有 $1.5\ \mathrm{kg}$ 与 $2.5\ \mathrm{kg}$。(a) 绕中点的 $I$；(b) 绕 $1.5\ \mathrm{kg}$ 端的 $I$；(c) 实心圆盘 $M = 2.0\ \mathrm{kg}$、$R = 0.15\ \mathrm{m}$ 的 $I$。
Use $I = \sum m r^2$. About the centre each mass is $0.40\ \mathrm{m}$ from the axis: (M1)
$$ I = (1.5)(0.40)^2 + (2.5)(0.40)^2 = (0.24) + (0.40) = 0.64\ \mathrm{kg\,m^2}. $$(A1)
The $1.5\ \mathrm{kg}$ mass is now on the axis ($r = 0$); the $2.5\ \mathrm{kg}$ mass is the full $0.80\ \mathrm{m}$ away: (M1)
$$ I = (1.5)(0)^2 + (2.5)(0.80)^2 = 0 + (2.5)(0.64) = 1.6\ \mathrm{kg\,m^2}. $$(A1)
Quote the data booklet value $I = \tfrac12 MR^2$ for a solid disc about its central axis: (M1)
$$ I = \tfrac12 (2.0)(0.15)^2 = \tfrac12 (2.0)(0.0225) = 0.0225\ \mathrm{kg\,m^2}. $$(A1)
用 $I = \sum m r^2$。绕中点时每个质量距轴 $0.40\ \mathrm{m}$：(M1)
$$ I = (1.5)(0.40)^2 + (2.5)(0.40)^2 = (0.24) + (0.40) = 0.64\ \mathrm{kg\,m^2}. $$(A1)
$1.5\ \mathrm{kg}$ 质量现在轴上（$r = 0$）；$2.5\ \mathrm{kg}$ 质量距轴整 $0.80\ \mathrm{m}$：(M1)
$$ I = (1.5)(0)^2 + (2.5)(0.80)^2 = 0 + (2.5)(0.64) = 1.6\ \mathrm{kg\,m^2}. $$(A1)
引用数据手册中实心圆盘绕中心轴的值 $I = \tfrac12 MR^2$：(M1)
$$ I = \tfrac12 (2.0)(0.15)^2 = \tfrac12 (2.0)(0.0225) = 0.0225\ \mathrm{kg\,m^2}. $$(A1)
Rotor from rest, $\alpha = 2.5\ \mathrm{rad\,s^{-2}}$ for $8.0\ \mathrm{s}$. (a) final $\omega$; (b) angle turned and number of revolutions; (c) tangential speed of a point $0.40\ \mathrm{m}$ from the axis at $8.0\ \mathrm{s}$.转子从静止，$\alpha = 2.5\ \mathrm{rad\,s^{-2}}$ 持续 $8.0\ \mathrm{s}$。(a) 末 $\omega$；(b) 转过角度与圈数；(c) 距轴 $0.40\ \mathrm{m}$ 一点在 $8.0\ \mathrm{s}$ 时的切向速率。
Constant $\alpha$, so use the angular suvat $\omega = \omega_0 + \alpha t$ with $\omega_0 = 0$: (M1)
$$ \omega = 0 + (2.5)(8.0) = 20\ \mathrm{rad\,s^{-1}}. $$(A1)
Use $\theta = \omega_0 t + \tfrac12 \alpha t^2$ with $\omega_0 = 0$: (M1)
$$ \theta = \tfrac12 (2.5)(8.0)^2 = \tfrac12 (2.5)(64) = 80\ \mathrm{rad}. $$(A1)
Number of revolutions $= \dfrac{\theta}{2\pi} = \dfrac{80}{2\pi} \approx 12.7$, so $12$ complete revolutions. (A1)
Use $v = r\omega = (0.40)(20) = 8.0\ \mathrm{m\,s^{-1}}$. (A1)
$\alpha$ 恒定，用角量 suvat $\omega = \omega_0 + \alpha t$，$\omega_0 = 0$：(M1)
$$ \omega = 0 + (2.5)(8.0) = 20\ \mathrm{rad\,s^{-1}}. $$(A1)
用 $\theta = \omega_0 t + \tfrac12 \alpha t^2$，$\omega_0 = 0$：(M1)
$$ \theta = \tfrac12 (2.5)(8.0)^2 = \tfrac12 (2.5)(64) = 80\ \mathrm{rad}. $$(A1)
圈数 $= \dfrac{\theta}{2\pi} = \dfrac{80}{2\pi} \approx 12.7$，故 $12$ 个完整圈。(A1)
用 $v = r\omega = (0.40)(20) = 8.0\ \mathrm{m\,s^{-1}}$。(A1)
Solid-disc flywheel $M = 8.0\ \mathrm{kg}$, $R = 0.20\ \mathrm{m}$; tangential force $12\ \mathrm{N}$ at the rim from rest. (a) $I$ and torque; (b) angular acceleration; (c) time to reach $30\ \mathrm{rad\,s^{-1}}$.实心盘飞轮 $M = 8.0\ \mathrm{kg}$、$R = 0.20\ \mathrm{m}$；轮缘切向力 $12\ \mathrm{N}$，从静止。(a) $I$ 与力矩；(b) 角加速度；(c) 达 $30\ \mathrm{rad\,s^{-1}}$ 的时间。
Solid disc about its axis: $I = \tfrac12 MR^2 = \tfrac12 (8.0)(0.20)^2 = 0.16\ \mathrm{kg\,m^2}$. (A1)
The tangential force at the rim acts at $90^{\circ}$ to the radius: $\tau = FR = (12)(0.20) = 2.4\ \mathrm{N\,m}$. (A1)
Apply Newton's second law for rotation $\tau = I\alpha$, so $\alpha = \tau/I$: (M1)
$$ \alpha = \frac{2.4}{0.16} = 15\ \mathrm{rad\,s^{-2}}. $$(A1)
From rest with constant $\alpha$, use $\omega = \omega_0 + \alpha t$: $30 = 0 + (15)t$. (M1)
$$ t = \frac{30}{15} = 2.0\ \mathrm{s}. $$(A1)
实心盘绕轴：$I = \tfrac12 MR^2 = \tfrac12 (8.0)(0.20)^2 = 0.16\ \mathrm{kg\,m^2}$。(A1)
轮缘切向力与半径成 $90^{\circ}$：$\tau = FR = (12)(0.20) = 2.4\ \mathrm{N\,m}$。(A1)
用转动的牛顿第二定律 $\tau = I\alpha$，故 $\alpha = \tau/I$：(M1)
$$ \alpha = \frac{2.4}{0.16} = 15\ \mathrm{rad\,s^{-2}}. $$(A1)
从静止、$\alpha$ 恒定，用 $\omega = \omega_0 + \alpha t$：$30 = 0 + (15)t$。(M1)
$$ t = \frac{30}{15} = 2.0\ \mathrm{s}. $$(A1)
Skater: $I_1 = 5.0\ \mathrm{kg\,m^2}$, $\omega_1 = 1.8\ \mathrm{rad\,s^{-1}}$, pulls arms in to $I_2 = 2.0\ \mathrm{kg\,m^2}$. (a) condition for conservation; (b) new $\omega$; (c) rotational KE before and after; (d) source of any KE change.滑冰者：$I_1 = 5.0\ \mathrm{kg\,m^2}$、$\omega_1 = 1.8\ \mathrm{rad\,s^{-1}}$，收臂至 $I_2 = 2.0\ \mathrm{kg\,m^2}$。(a) 守恒条件；(b) 新 $\omega$；(c) 收臂前后转动动能；(d) 动能变化来源。
Angular momentum is conserved when no resultant external torque acts on the system. (A1)
Here friction at the ice is negligible and the muscle forces that pull the arms in are internal, so there is no external torque about the spin axis. (R1)
Conservation of angular momentum gives $I_1\omega_1 = I_2\omega_2$: (M1)
$$ \omega_2 = \frac{I_1\omega_1}{I_2} = \frac{(5.0)(1.8)}{2.0} = \frac{9.0}{2.0} = 4.5\ \mathrm{rad\,s^{-1}}. $$(A1)
Use $E_k = \tfrac12 I\omega^2$ for each state: (M1)
$$ E_{k1} = \tfrac12 (5.0)(1.8)^2 = \tfrac12 (5.0)(3.24) = 8.1\ \mathrm{J}. $$(A1)
$$ E_{k2} = \tfrac12 (2.0)(4.5)^2 = \tfrac12 (2.0)(20.25) \approx 20.3\ \mathrm{J}. $$The kinetic energy increases (by a factor of $2.5$), so kinetic energy is NOT conserved even though angular momentum is. (A1)
The extra kinetic energy comes from work done by the skater's muscles in pulling the arms inward against their tendency to move outward. (B1)
系统不受合外力矩时角动量守恒。(A1)
这里冰面摩擦可忽略，收臂的肌肉力是内力，故绕旋转轴无外力矩。(R1)
角动量守恒给出 $I_1\omega_1 = I_2\omega_2$：(M1)
$$ \omega_2 = \frac{I_1\omega_1}{I_2} = \frac{(5.0)(1.8)}{2.0} = \frac{9.0}{2.0} = 4.5\ \mathrm{rad\,s^{-1}}. $$(A1)
对两个状态各用 $E_k = \tfrac12 I\omega^2$：(M1)
$$ E_{k1} = \tfrac12 (5.0)(1.8)^2 = \tfrac12 (5.0)(3.24) = 8.1\ \mathrm{J}. $$(A1)
$$ E_{k2} = \tfrac12 (2.0)(4.5)^2 = \tfrac12 (2.0)(20.25) \approx 20.3\ \mathrm{J}. $$动能增大（约 $2.5$ 倍），故尽管角动量守恒，动能并不守恒。(A1)
多出的动能来自滑冰者肌肉克服手臂外移趋势把手臂收向身体所做的功。(B1)
Wheel $I = 0.45\ \mathrm{kg\,m^2}$; $\omega$ vs $t$ data: $(2.0,5.0),(4.0,9.0),(6.0,13.0),(8.0,17.0)$. (a) why $\omega$ vs $t$ is linear and gradient meaning; (b) gradient and $\alpha$; (c) torque from $\tau = I\alpha$; (d) percentage uncertainty at $t = 8.0$ ($\pm 0.5$); (e) meaning of the non-zero intercept.轮 $I = 0.45\ \mathrm{kg\,m^2}$；$\omega$ 对 $t$ 数据：$(2.0,5.0),(4.0,9.0),(6.0,13.0),(8.0,17.0)$。(a) 为何 $\omega$ 对 $t$ 为直线及斜率含义；(b) 斜率与 $\alpha$；(c) 由 $\tau = I\alpha$ 求力矩；(d) $t = 8.0$ 处的百分比不确定度（$\pm 0.5$）；(e) 非零截距的含义。
For a constant torque the angular acceleration is constant, and the angular suvat $\omega = \omega_0 + \alpha t$ has the form $y = c + mx$. (M1)
So a plot of $\omega$ against $t$ is a straight line whose gradient is the angular acceleration $\alpha$. (A1)
Read the gradient from two well-separated points, e.g. $(2.0,\,5.0)$ and $(8.0,\,17.0)$: (M1)
$$ \text{gradient} = \frac{17.0 - 5.0}{8.0 - 2.0} = \frac{12.0}{6.0} = 2.0\ \mathrm{rad\,s^{-2}}. $$(A1)
The gradient is the angular acceleration directly, so $\alpha = 2.0\ \mathrm{rad\,s^{-2}}$. (A1)
Use $\tau = I\alpha$ with $I = 0.45\ \mathrm{kg\,m^2}$: (M1)
$$ \tau = (0.45)(2.0) = 0.90\ \mathrm{N\,m}. $$(A1)
At that point $\omega = 17.0\ \mathrm{rad\,s^{-1}}$ with absolute uncertainty $\pm 0.5\ \mathrm{rad\,s^{-1}}$: (M1)
$$ \frac{0.5}{17.0}\times 100\% \approx 2.9\% \approx 3\%. $$(A1)
Extrapolating the line to $t = 0$ gives $\omega_0 = 1.0\ \mathrm{rad\,s^{-1}}$, a non-zero value, which shows the wheel was already rotating when timing began. (B1)
恒定力矩下角加速度恒定，角量 suvat $\omega = \omega_0 + \alpha t$ 形如 $y = c + mx$。(M1)
故 $\omega$ 对 $t$ 作图为直线，其斜率即角加速度 $\alpha$。(A1)
用相距较远的两点读斜率，如 $(2.0,\,5.0)$ 与 $(8.0,\,17.0)$：(M1)
$$ \text{斜率} = \frac{17.0 - 5.0}{8.0 - 2.0} = \frac{12.0}{6.0} = 2.0\ \mathrm{rad\,s^{-2}}. $$(A1)
斜率即角加速度，故 $\alpha = 2.0\ \mathrm{rad\,s^{-2}}$。(A1)
用 $\tau = I\alpha$，$I = 0.45\ \mathrm{kg\,m^2}$：(M1)
$$ \tau = (0.45)(2.0) = 0.90\ \mathrm{N\,m}. $$(A1)
该点 $\omega = 17.0\ \mathrm{rad\,s^{-1}}$，绝对不确定度 $\pm 0.5\ \mathrm{rad\,s^{-1}}$：(M1)
$$ \frac{0.5}{17.0}\times 100\% \approx 2.9\% \approx 3\%. $$(A1)
把直线外推到 $t = 0$ 得 $\omega_0 = 1.0\ \mathrm{rad\,s^{-1}}$，非零值，说明开始计时时轮已在转动。(B1)
Uniform beam, weight $300\ \mathrm{N}$, length $6.0\ \mathrm{m}$, on supports at the left end ($R_L$) and at $5.0\ \mathrm{m}$ ($R_R$); load $500\ \mathrm{N}$ at $4.0\ \mathrm{m}$. (a) equilibrium conditions; (b) $R_R$ by moments about the left end; (c) $R_L$; (d) verify $R_L$ about the right support; (e) percentage uncertainty in the beam weight ($\pm 10\ \mathrm{N}$).均匀梁，重 $300\ \mathrm{N}$、长 $6.0\ \mathrm{m}$，支点在左端（$R_L$）与 $5.0\ \mathrm{m}$ 处（$R_R$）；载荷 $500\ \mathrm{N}$ 在 $4.0\ \mathrm{m}$ 处。(a) 平衡条件；(b) 对左端取矩求 $R_R$；(c) $R_L$；(d) 对右支点取矩验证 $R_L$；(e) 梁重的百分比不确定度（$\pm 10\ \mathrm{N}$）。
For a rigid body in equilibrium the resultant force is zero: $\Sigma F = 0$. (A1)
And the resultant torque about any point is zero: $\Sigma\tau = 0$. (A1)
Take moments about the left end; this eliminates $R_L$, whose line of action passes through the pivot so it has zero moment. (M1 for stating the pivot choice)
Anticlockwise positive, the beam weight acts at its centre ($3.0\ \mathrm{m}$), the load at $4.0\ \mathrm{m}$, and $R_R$ at $5.0\ \mathrm{m}$: (M1)
$$ R_R(5.0) - (300)(3.0) - (500)(4.0) = 0. $$ $$ 5.0\,R_R = 900 + 2000 = 2900. $$(A1)
$$ R_R = \frac{2900}{5.0} = 580\ \mathrm{N}. $$(A1)
Vertical equilibrium $\Sigma F = 0$: $R_L + R_R = 300 + 500 = 800\ \mathrm{N}$. (M1)
$$ R_L = 800 - 580 = 220\ \mathrm{N}. $$(A1)
Take moments about the right support at $5.0\ \mathrm{m}$. The weight is $2.0\ \mathrm{m}$ to its left and the load $1.0\ \mathrm{m}$ to its left, while $R_L$ acts $5.0\ \mathrm{m}$ to its left: (M1)
$$ R_L(5.0) = (300)(2.0) + (500)(1.0) = 600 + 500 = 1100. $$ $$ R_L = \frac{1100}{5.0} = 220\ \mathrm{N}. $$This matches part (c), confirming the result. (A1)
The beam weight is $300\ \mathrm{N}$ with absolute uncertainty $\pm 10\ \mathrm{N}$: (M1)
$$ \frac{10}{300}\times 100\% \approx 3.3\% \approx 3\%. $$(A1)
刚体平衡时合力为零：$\Sigma F = 0$。(A1)
且对任意点的合力矩为零：$\Sigma\tau = 0$。(A1)
对左端取矩；这消去 $R_L$，其作用线过转轴故力矩为零。(说明转轴选择得 M1)
取逆时针为正，梁重作用在中心（$3.0\ \mathrm{m}$），载荷在 $4.0\ \mathrm{m}$，$R_R$ 在 $5.0\ \mathrm{m}$：(M1)
$$ R_R(5.0) - (300)(3.0) - (500)(4.0) = 0. $$ $$ 5.0\,R_R = 900 + 2000 = 2900. $$(A1)
$$ R_R = \frac{2900}{5.0} = 580\ \mathrm{N}. $$(A1)
竖直方向平衡 $\Sigma F = 0$：$R_L + R_R = 300 + 500 = 800\ \mathrm{N}$。(M1)
$$ R_L = 800 - 580 = 220\ \mathrm{N}. $$(A1)
对 $5.0\ \mathrm{m}$ 处的右支点取矩。梁重在其左 $2.0\ \mathrm{m}$，载荷在其左 $1.0\ \mathrm{m}$，$R_L$ 在其左 $5.0\ \mathrm{m}$：(M1)
$$ R_L(5.0) = (300)(2.0) + (500)(1.0) = 600 + 500 = 1100. $$ $$ R_L = \frac{1100}{5.0} = 220\ \mathrm{N}. $$与 (c) 一致，结果得证。(A1)
梁重为 $300\ \mathrm{N}$，绝对不确定度 $\pm 10\ \mathrm{N}$：(M1)
$$ \frac{10}{300}\times 100\% \approx 3.3\% \approx 3\%. $$(A1)
Solid-disc pulley $M = 2.0\ \mathrm{kg}$, $R = 0.10\ \mathrm{m}$; block $m = 1.5\ \mathrm{kg}$ on a string round the rim, released from rest, no slip. (a) the two Newton equations; (b) show $a = \dfrac{mg}{m + \tfrac12 M}$ and find it; (c) tension and angular acceleration; (d) speed after falling $1.2\ \mathrm{m}$.实心盘滑轮 $M = 2.0\ \mathrm{kg}$、$R = 0.10\ \mathrm{m}$；物块 $m = 1.5\ \mathrm{kg}$ 系绳绕轮缘，从静止释放，不打滑。(a) 两个牛顿方程；(b) 证明 $a = \dfrac{mg}{m + \tfrac12 M}$ 并求之；(c) 张力与角加速度；(d) 下落 $1.2\ \mathrm{m}$ 后的速率。
For the falling block, downward positive, with tension $T$ acting upward: $mg - T = ma$. (M1·A1)
For the pulley, the string tension acts at the rim, giving a torque $TR$; Newton's second law for rotation is $TR = I\alpha$, with $I = \tfrac12 MR^2$. (A1)
The string does not slip, so $a = \alpha R$, i.e. $\alpha = a/R$. Substitute into $TR = I\alpha$ with $I = \tfrac12 MR^2$: (M1)
$$ TR = \tfrac12 MR^2 \cdot \frac{a}{R} \;\Rightarrow\; T = \tfrac12 M a. $$(M1)
Put this into $mg - T = ma$: $mg - \tfrac12 M a = ma$, so $mg = a\left(m + \tfrac12 M\right)$, giving (A1)
$$ a = \frac{mg}{m + \tfrac12 M} = \frac{(1.5)(9.81)}{1.5 + \tfrac12(2.0)} = \frac{14.715}{2.5} \approx 5.9\ \mathrm{m\,s^{-2}}. $$(A1)
From $T = \tfrac12 M a = \tfrac12 (2.0)(5.886) \approx 5.9\ \mathrm{N}$. (M1·A1)
Angular acceleration $\alpha = a/R = 5.886/0.10 \approx 59\ \mathrm{rad\,s^{-2}}$. (A1)
Constant acceleration from rest, use $v^2 = u^2 + 2as$ with $u = 0$: (M1)
$$ v = \sqrt{2(5.886)(1.2)} = \sqrt{14.13} \approx 3.8\ \mathrm{m\,s^{-1}}. $$(A1)
对下落物块，取向下为正，张力 $T$ 向上：$mg - T = ma$。(M1·A1)
对滑轮，绳张力作用在轮缘，产生力矩 $TR$；转动的牛顿第二定律为 $TR = I\alpha$，$I = \tfrac12 MR^2$。(A1)
绳不打滑，故 $a = \alpha R$，即 $\alpha = a/R$。代入 $TR = I\alpha$，$I = \tfrac12 MR^2$：(M1)
$$ TR = \tfrac12 MR^2 \cdot \frac{a}{R} \;\Rightarrow\; T = \tfrac12 M a. $$(M1)
代入 $mg - T = ma$：$mg - \tfrac12 M a = ma$，故 $mg = a\left(m + \tfrac12 M\right)$，得 (A1)
$$ a = \frac{mg}{m + \tfrac12 M} = \frac{(1.5)(9.81)}{1.5 + \tfrac12(2.0)} = \frac{14.715}{2.5} \approx 5.9\ \mathrm{m\,s^{-2}}. $$(A1)
由 $T = \tfrac12 M a = \tfrac12 (2.0)(5.886) \approx 5.9\ \mathrm{N}$。(M1·A1)
角加速度 $\alpha = a/R = 5.886/0.10 \approx 59\ \mathrm{rad\,s^{-2}}$。(A1)
从静止匀加速，用 $v^2 = u^2 + 2as$，$u = 0$：(M1)
$$ v = \sqrt{2(5.886)(1.2)} = \sqrt{14.13} \approx 3.8\ \mathrm{m\,s^{-1}}. $$(A1)
Solid cylinder ($I = \tfrac12 mR^2$) rolls without slipping from rest down a slope, descending $h = 2.0\ \mathrm{m}$. (a) rolling constraint; (b) energy equation, show $mgh = \tfrac34 mv^2$; (c) speed at the bottom; (d) fraction of KE that is rotational and why a sliding block is faster.实心圆柱（$I = \tfrac12 mR^2$）从静止无滑滚下斜坡，下降 $h = 2.0\ \mathrm{m}$。(a) 滚动约束；(b) 能量方程，证明 $mgh = \tfrac34 mv^2$；(c) 坡底速率；(d) 转动动能占比及为何滑块更快。
For rolling without slipping the contact point is instantaneously at rest, which gives $v = \omega R$. (A1)
The lost gravitational potential energy becomes both translational and rotational kinetic energy: (M1)
$$ mgh = \tfrac12 mv^2 + \tfrac12 I\omega^2. $$Substitute $I = \tfrac12 mR^2$ and $\omega = v/R$: (M1)
$$ mgh = \tfrac12 mv^2 + \tfrac12\left(\tfrac12 mR^2\right)\frac{v^2}{R^2} = \tfrac12 mv^2 + \tfrac14 mv^2. $$(A1)
$$ mgh = \tfrac34 mv^2. $$(A1)
Cancel $m$ and solve: $gh = \tfrac34 v^2$, so $v = \sqrt{\tfrac43 gh}$: (M1)
$$ v = \sqrt{\tfrac43 (9.81)(2.0)} = \sqrt{26.16} \approx 5.1\ \mathrm{m\,s^{-1}}. $$(A1)
The rotational share is $\dfrac{\tfrac14 mv^2}{\tfrac34 mv^2} = \dfrac13$ of the total kinetic energy. (A1)
A frictionless sliding block converts all of $mgh$ into translation, $mgh = \tfrac12 mv^2$, giving $v = \sqrt{2gh} \approx 6.3\ \mathrm{m\,s^{-1}}$. (M1)
Since the rolling cylinder must also spin, only two thirds of the energy is available for translation, so it arrives more slowly than the sliding block. (A1)
无滑滚动时接触点瞬时静止，给出 $v = \omega R$。(A1)
损失的重力势能变为平动与转动两部分动能：(M1)
$$ mgh = \tfrac12 mv^2 + \tfrac12 I\omega^2. $$代入 $I = \tfrac12 mR^2$ 与 $\omega = v/R$：(M1)
$$ mgh = \tfrac12 mv^2 + \tfrac12\left(\tfrac12 mR^2\right)\frac{v^2}{R^2} = \tfrac12 mv^2 + \tfrac14 mv^2. $$(A1)
$$ mgh = \tfrac34 mv^2. $$(A1)
约去 $m$ 求解：$gh = \tfrac34 v^2$，故 $v = \sqrt{\tfrac43 gh}$：(M1)
$$ v = \sqrt{\tfrac43 (9.81)(2.0)} = \sqrt{26.16} \approx 5.1\ \mathrm{m\,s^{-1}}. $$(A1)
转动部分为 $\dfrac{\tfrac14 mv^2}{\tfrac34 mv^2} = \dfrac13$ 的总动能。(A1)
无摩擦滑块把全部 $mgh$ 转为平动，$mgh = \tfrac12 mv^2$，得 $v = \sqrt{2gh} \approx 6.3\ \mathrm{m\,s^{-1}}$。(M1)
由于滚动圆柱还须自转，仅三分之二能量可用于平动，故它比滑块到达更慢。(A1)
Turntable $I = 0.40\ \mathrm{kg\,m^2}$ at $4.0\ \mathrm{rad\,s^{-1}}$; clay $0.50\ \mathrm{kg}$ sticks at $0.30\ \mathrm{m}$ from the axis. (a) why angular momentum is conserved; (b) clay's $I$ and the total $I$; (c) new angular velocity; (d) rotational KE before and after, elastic or not.转盘 $I = 0.40\ \mathrm{kg\,m^2}$、$4.0\ \mathrm{rad\,s^{-1}}$；橡皮泥 $0.50\ \mathrm{kg}$ 粘在距轴 $0.30\ \mathrm{m}$ 处。(a) 为何角动量守恒；(b) 橡皮泥的 $I$ 与总 $I$；(c) 新角速度；(d) 碰撞前后转动动能及是否弹性。
The clay falls vertically and sticks; the impact forces between clay and turntable are internal to the system, and there is no external torque about the vertical axis, so the total angular momentum about that axis is conserved. (R1)
The clay's linear momentum is downward and is absorbed by the axle, so it is not a useful conserved quantity for the spin; angular momentum about the axis is the right tool. (A1)
Treat the clay as a point mass at $r = 0.30\ \mathrm{m}$: $I_\text{clay} = mr^2 = (0.50)(0.30)^2 = 0.045\ \mathrm{kg\,m^2}$. (M1)
$$ I_\text{total} = 0.40 + 0.045 = 0.445\ \mathrm{kg\,m^2}. $$(A1)
Conserve angular momentum: $I_1\omega_1 = I_\text{total}\,\omega_2$: (M1)
$$ \omega_2 = \frac{(0.40)(4.0)}{0.445} = \frac{1.6}{0.445} \approx 3.6\ \mathrm{rad\,s^{-1}}. $$(A1)
$E_{k1} = \tfrac12 (0.40)(4.0)^2 = 3.2\ \mathrm{J}$; $E_{k2} = \tfrac12 (0.445)(3.596)^2 \approx 2.9\ \mathrm{J}$. (M1)
Kinetic energy falls from $3.2\ \mathrm{J}$ to about $2.9\ \mathrm{J}$, so the collision is inelastic. (A1)
橡皮泥竖直落下并粘住；橡皮泥与转盘间的冲击力是系统内力，绕竖直轴无外力矩，故绕该轴的总角动量守恒。(R1)
橡皮泥的线动量向下，被轴吸收，故对转动而言不是有用的守恒量；绕轴的角动量才是正确工具。(A1)
把橡皮泥视为 $r = 0.30\ \mathrm{m}$ 处的质点：$I_\text{clay} = mr^2 = (0.50)(0.30)^2 = 0.045\ \mathrm{kg\,m^2}$。(M1)
$$ I_\text{total} = 0.40 + 0.045 = 0.445\ \mathrm{kg\,m^2}. $$(A1)
角动量守恒：$I_1\omega_1 = I_\text{total}\,\omega_2$：(M1)
$$ \omega_2 = \frac{(0.40)(4.0)}{0.445} = \frac{1.6}{0.445} \approx 3.6\ \mathrm{rad\,s^{-1}}. $$(A1)
$E_{k1} = \tfrac12 (0.40)(4.0)^2 = 3.2\ \mathrm{J}$；$E_{k2} = \tfrac12 (0.445)(3.596)^2 \approx 2.9\ \mathrm{J}$。(M1)
动能从 $3.2\ \mathrm{J}$ 降到约 $2.9\ \mathrm{J}$，故碰撞为非弹性。(A1)