耦合加速度与相对运动
GRE Subject Test · GRE 物理 知识点 12 5:36 英语旁白 · 英中内嵌字幕
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A load released from an aircraft keeps the aircraft's horizontal velocity, so relative to the aircraft its first motion is straight down.
从飞机上释放的货物保持飞机的水平速度，所以相对于飞机，它一开始竖直向下运动。
This topic solves coupled bodies with one shared acceleration, transforms velocities between frames, and checks the answers against force and energy limits.
本主题用同一个加速度求解连接体，在参考系之间变换速度，并用力和能量的极限检验答案。
For a mass m 1 on a frictionless horizontal table connected over an ideal massless pulley to a hanging mass m 2, choose table motion and downward hanging motion as positive.
对于放置在无摩擦水平桌面上、通过理想轻质滑轮悬挂质量为m2的物体、桌面物体质量为m1的系统，规定桌面运动方向和悬挂物体的向下运动方向为正方向。
A taut inextensible string gives one acceleration magnitude.
绷紧且不可伸长的绳子使两物体具有相同大小的加速度。
Write T equals m 1 a, and m 2 g minus T equals m 2 a.
列出方程：T等于m1乘以a，以及m2g减去T等于m2乘以a。
Adding eliminates the internal tension, so a equals m 2 g over m 1 plus m 2, and T equals m 1 m 2 g over m 1 plus m 2.
将两式相加可消去内部张力，因此a等于m2g除以m1加m2，T等于m1乘m2乘g除以m1加m2。
Thus T is less than m 2 g when both masses are positive.
因此，当两个质量均为正值时，T小于m2g。
Setting tension equal to hanging weight silently assumes zero acceleration.
令张力等于悬挂物重，隐含假设了加速度为零。
The common tension assumption requires an ideal massless string and pulley with negligible friction and inertia.
常见的张力假设要求使用理想无质量绳索和滑轮，且摩擦与转动惯量可忽略不计。
A pulley with rotational inertia can have different tensions, with T2 minus T1, times R, equal to I alpha, and a equal to R alpha if the string does not slip.
具有转动惯量的滑轮两端张力可能不同，若绳索不打滑，则满足 T2 减 T1 乘以 R 等于 I alpha，以及 a 等于 R alpha。
Draw each body separately: a force internal to the combined system remains an external force on one chosen body.
请分别绘制每个物体的受力图。 对于整体系统而言是内力的力，对其中任一单独物体而言仍是外力。
Constraint forces can cancel from a system equation while still being needed for individual motion.
约束力在系统方程中可能相互抵消，但仍需用于分析单个物体的运动。
For two inertial frames with constant relative velocity V, Galilean velocity transformation is v sub relative equals v sub ground minus V.
对于两个具有恒定相对速度V的惯性参考系，伽利略速度变换为v下标相对等于v下标地减去V。
A payload released by a level aircraft initially has the aircraft's horizontal speed.
水平飞行的飞机释放的载荷初始时具有飞机的水平速度。
With negligible air resistance, horizontal ground velocity remains constant and vertical velocity becomes minus g t if upward is positive.
在忽略空气阻力的情况下，水平地面速度保持不变，若以向上为正方向，则垂直速度变为负g t。
Relative to that aircraft, horizontal velocity is zero and downward speed is g t.
相对于该飞机，水平速度为零，向下速度为g t。
Position is a different question: vertical displacement is minus one half g t squared.
位置是另一个问题：垂直位移为负二分之一g t平方。
Do not confuse ground speed with relative speed.
不要混淆对地速度与相对速度。
Check limiting cases after solving.
求解后检查极限情况。
As m 1 approaches zero, the ideal coupled acceleration approaches g and tension approaches zero; as m 1 grows very large, acceleration approaches zero and tension approaches m 2 g from below.
当 m1 趋近于零时，理想耦合加速度趋近于 g，张力趋近于零；当 m1 非常大时，加速度趋近于零，张力从下方趋近于 m2 g。
An energy derivation gives m 2 g s equals one half of m 1 plus m 2, times v squared, for release from rest and an ideal pulley, agreeing with v squared equals 2 a s.
对于从静止释放且滑轮理想的情况，能量推导给出 m2 g s 等于二分之一乘以 m1 加 m2 的和，再乘以 v 的平方，这与 v 的平方等于 2as 一致。
This agreement checks the equations; it does not permit energy conservation if friction or other unaccounted work is present.
这种一致性检验了方程；但如果存在摩擦或其他未计入的功，就不能使用能量守恒。
A 3 kilogram block on the table and a 2 kilogram hanging mass share one acceleration through a taut string.
桌上 3 千克的物块和悬挂的 2 千克物体通过绷紧的绳子共享一个加速度。
The table block gives T equals m 1 a; the hanging mass gives m 2 g minus T equals m 2 a.
桌上物块给出 T 等于 m1 乘 a；悬挂物体给出 m2 g 减 T 等于 m2 乘 a。
Adding the two equations removes the tension: a equals m 2 g over m 1 plus m 2, which is 2 times 10 over 5, or 4 metres per second squared.
把两个方程相加消去张力：a 等于 m2 g 除以 m1 加 m2，即 2 乘 10 除以 5，等于 4 米每二次方秒。
The tension is m 1 a: 3 kilograms times 4, which is 12 newtons.
张力等于 m1 乘 a：3 千克乘 4，等于 12 牛。
That is less than the 20 newton hanging weight, because a net downward force is needed to accelerate the hanging mass.
它小于 20 牛的悬挂重力，因为悬挂物体需要一个向下的合力才能加速。
Replace the ideal pulley by a solid disk of mass M and radius R, rotating without axle friction.
将理想滑轮替换为质量为M、半径为R的实心圆盘，且转轴无摩擦。
The string does not slip.
绳子不打滑。
Derive the acceleration for table mass m1 and hanging mass m2.
推导桌面上质量m1和悬挂质量m2的加速度。
Find the two tensions and check the limit M approaching zero.
求出两个张力，并检验M趋近于零时的极限情况。
T sub 1 equals m sub 1 a. m sub 2 g minus T sub 2 equals m sub 2 a. T sub 2 minus T sub 1, times R, equals I alpha.
T₁ = m₁a；m₂g − T₂ = m₂a；(T₂ − T₁)R = Iα。
With I equals M R squared over 2 and alpha equals a over R, T sub 2 minus T sub 1 equals M a over 2.
由 I = MR²/2 及 α = a/R，得 T₂ − T₁ = Ma/2。
Thus a equals m sub 2 g over m sub 1 plus m sub 2 plus M over 2.
因此 a = m₂g / (m₁ + m₂ + M/2)。
T sub 1 equals m sub 1 m sub 2 g over m sub 1 plus m sub 2 plus M over 2. And T sub 2 equals m sub 2 times g minus a.
T₁ = m₁m₂g / (m₁ + m₂ + M/2)，且 T₂ = m₂(g − a)。
At M equals 0, tensions become equal and the ideal result returns.
当 M = 0 时，张力相等，理想结果回归。
Award body equations, torque or constraint, acceleration, tensions, and limit.
评分要点包括：受力方程、力矩或约束条件、加速度、张力以及极限情况。
A massless ideal pulley equates tension, not tension and weight.
无质量理想滑轮使张力相等，而非张力与重力相等。
Subtract frame velocities component by component before taking a speed magnitude.
在计算速率大小之前，需逐分量相减参考系速度。
constraint: A condition linking allowed positions or motions of connected bodies.
约束：连接物体之间允许位置或运动之间的关联条件。
relative velocity: Velocity of one object measured in another moving frame.
相对速度：在一个运动参考系中测量另一个物体的速度。
Retrieve the meanings without reading the definitions, then change a parameter in the original transfer and test the limiting case.
在不看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。