Rotation, flottabilité et équilibre à vitesse terminale
GRE Subject Test · GRE Physics Sujet 11 5:36 Narration en anglais · Sous-titres anglais + 中文 incrustés
Chapitres
Transcript
Turning a floating block upside down can change how much wood is submerged, even though the total weight stays the same.
把漂浮的木块翻转过来，浸没的木头体积可能改变，尽管总重量不变。
This topic uses pivot torque and rotational inertia consistently, splits rolling kinetic energy with the no-slip constraint, and balances buoyancy and drag for composite bodies.
本主题一致地使用绕支点的力矩和转动惯量，用无滑动约束分配滚动动能，并对组合物体平衡浮力和阻力。
Torque about a fixed pivot is r cross F; its magnitude uses the perpendicular lever arm.
绕固定支点的力矩为 r 叉乘 F，其大小使用垂直力臂计算。
For a uniform rod of length L pivoted at one end, gravity acts at its centre, L over 2, and the moment of inertia is M L squared over 3.
对于长度为 L、一端为支点的均匀细杆，重力作用于其中心，即 L 的一半处，转动惯量为 M L 平方除以 3。
If its angle theta is measured from vertical, the gravitational torque magnitude is M g times L over 2 times sine theta, so the angular-acceleration magnitude is 3 g sine theta over 2 L.
若角度 θ 从竖直方向量起，重力矩大小为 M g 乘以 L 的一半再乘以 sin θ，因此角加速度大小为 3 g sin θ 除以 2L。
Define the positive rotation direction before assigning a sign.
确定符号之前，先规定正的转动方向。
Using the centre-of-mass inertia M L squared over 12 without the parallel-axis correction gives a wrong acceleration.
若使用质心转动惯量 M L 平方除以 12 而不做平行轴修正，会得到错误的加速度。
An object rolling without slipping satisfies v equals R omega.
无滑动滚动的物体满足v等于R乘以ω。
Write the total kinetic energy as one half M v squared plus one half I omega squared.
总动能写为二分之一Mv平方加二分之一Iω平方。
With I equal to beta M R squared, the rotational fraction is beta over 1 plus beta; for a solid disk beta is one half and the fraction is one third.
当I等于βMR平方时，转动部分占比为β除以1加β。 对于实心圆盘，β等于二分之一，该占比为三分之一。
A hoop has beta equal to 1 and a fraction of one half.
圆环的β等于1，占比为二分之一。
Static friction can supply the torque required for rolling without dissipating energy at an instantaneously stationary contact on a fixed surface.
静摩擦力可以在固定表面上瞬时静止的接触点处提供滚动所需的力矩，且不耗散能量。
A sliding object does not satisfy the no-slip relation automatically.
滑动物体不会自动满足无滑动关系。
For a floating composite, buoyancy equals the weight of displaced fluid, summed over every submerged part.
对于漂浮的复合体，浮力等于所有浸没部分排开流体的重量之和。
Let a wood block have volume V and a dense stone volume Vs.
设木块体积为V，致密石块体积为Vs。
If the stone is above the water, only submerged wood contributes: ρwater g fV equals total weight.
若石块位于水面之上，仅浸没的木材产生贡献：ρ水g f V等于总重量。
If the stone is attached below and fully submerged, its displaced volume reduces the wood’s required submerged fraction to f−Vs/V.
若石块附着于下方且完全浸没，其排开水的体积使木块所需的浸没分数减小至f减去Vs除以V。
The stone does not cease to weigh anything; the water now supplies some of its support.
石块并未失去重量；此时水提供了部分支撑。
Check whether the assumed orientation and flotation are physically possible.
请检查所假设的姿态与漂浮状态在物理上是否可行。
Terminal speed means zero acceleration, not zero speed or zero force.
终端速度意味着加速度为零，而不是速度为零或受力为零。
With quadratic drag C v squared and a downward weight M g, neglecting buoyancy gives a terminal speed equal to the square root of M g over C.
在二次阻力 C v 平方和向下重力 M g 的作用下，忽略浮力可得终端速度等于 M g 除以 C 的平方根。
Two otherwise identical balls share the drag coefficient C; doubling the mass then multiplies terminal speed by the square root of 2.
两个其他条件相同的球具有相同的阻力系数 C；质量加倍时，终端速度变为原来的根号 2 倍。
If buoyancy matters, replace M g by M minus rho fluid V, all times g.
若需考虑浮力，则把 M g 换成 M 减去流体密度乘以体积 V，再乘以 g。
A linear-drag model instead gives a different mass scaling.
线性阻力模型则给出不同的质量缩放关系。
State the specified drag law and compare force balances, rather than transferring a formula between models.
应明确所给的阻力定律并比较受力平衡，而不是在不同模型之间直接套用公式。
For a body rolling without slipping, v equals R omega.
对于无滑动滚动的物体，v 等于 R 乘 ω。
With I equal to beta M R squared, the kinetic energy is one half M v squared plus one half I omega squared, which is one half M v squared times one plus beta.
若 I 等于 β M R 的平方，动能等于二分之一 M v 的平方加二分之一 I ω 的平方，即二分之一 M v 的平方乘一加 β。
The rotational share is beta over one plus beta.
转动动能所占比例为 β 除以一加 β。
For a solid disk, beta is one half, so the share is one half over three halves: one third.
对于实心圆盘，β 等于二分之一，所以比例是二分之一除以二分之三，即三分之一。
For a uniform rod pivoted at one end, use I equals M L squared over 3, not its centre-of-mass inertia.
对于一端固定的均匀杆，使用 I 等于 M L 的平方除以 3，而不是绕质心的转动惯量。
The torque is minus M g times L over 2 times sine theta, so the angular acceleration is minus 3 g sine theta over 2 L.
力矩等于负 M g 乘 L 除以 2 乘 sin θ，所以角加速度等于负 3 g sin θ 除以 2 L。
A solid sphere rolls from rest down a vertical height h without slipping or energy loss.
一个实心球体从静止开始沿竖直高度 h 滚下，且无滑动、无能量损失。
Use I equals two M R squared over five to find its bottom speed.
利用转动惯量 I 等于二分之五乘以 M 乘 R 平方，求其到达底部的速度。
Compare with a frictionless sliding point mass.
将其与无摩擦滑动的质点进行比较。
State which assumption would fail if the sphere slipped.
指出如果球体发生滑动，哪一个假设会失效。
Energy gives M g h equals one half M v squared times one plus two fifths, hence v equals the square root of ten g h over seven.
能量关系给出 Mgh = (1/2)Mv²(1 + 2/5)，因此 v = √(10gh/7)。
The slider has v sub s equals the square root of two g h, so v divided by v sub s equals the square root of five sevenths, which is less than one.
滑块的速度为 v_s = √(2gh)，所以 v/v_s = √(5/7) < 1。
Slipping breaks v equals R omega and can dissipate energy through kinetic friction.
滑动会破坏 v = Rω 的关系，并可能通过动摩擦耗散能量。
Award energy relation, speed, comparison, and the constraint loss distinction.
评分要点包括：能量关系、速度、比较以及约束与损耗的区别。
Use the inertia about the chosen pivot.
使用所选支点处的转动惯量。
In flotation, include the stone’s displaced volume as well as its weight.
在浮力计算中，需同时考虑石块排开液体的体积及其自身重量。
Terminal speed requires a specified drag law.
终端速度需要明确指定的阻力定律。
moment of inertia: Mass-weighted squared perpendicular distance from a specified rotation axis.
转动惯量：质量加权的、距指定转轴的垂直距离的平方。
terminal speed: Steady speed at which opposing forces balance.
终端速度：阻力与驱动力平衡时的恒定速度。
Retrieve the meanings without reading the definitions, then change a parameter in the original transfer and test the limiting case.
在不阅读定义的情况下回忆含义，然后改变原始传递中的一个参数并检验极限情况。