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GRE Subject Test · GRE Physics Chủ đề 27 6:05 Giọng đọc tiếng Anh · phụ đề tiếng Anh + 中文 được ghi trực tiếp
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A person standing still on a turning platform still needs a real inward force, even though their acceleration relative to the platform is zero.
站在转台上静止不动的人仍然需要一个真实的向心力，尽管他相对转台的加速度为零。
This topic transforms acceleration into translating and rotating frames, finds the directions of the centrifugal, Coriolis and Euler terms, and keeps apparent forces apart from real interactions.
本主题把加速度变换到平动和转动参考系中，确定离心力、科里奥利力和欧拉力的方向，并把惯性力与真实相互作用区分开。
In a translating frame with origin acceleration A, Newton's equation becomes m a prime equals F real minus m A.
在具有原点加速度 A 的平移参考系中，牛顿方程变为 m a prime 等于 F real 减去 m A。
An upward-accelerating elevator therefore has N minus m g equals m a in the ground frame, or N minus m g minus m a equals zero for its stationary passenger in the elevator frame.
因此，向上加速的电梯在地面参考系中有 N 减去 m g 等于 m a，或者对于电梯参考系中静止的乘客而言，N 减去 m g 减去 m a 等于零。
These are the same prediction.
这两个预测是一致的。
The additional term is an apparent force caused by the chosen accelerating coordinates; it is not a new contact with another body.
附加项是由所选加速坐标系引起的表观力；它并非与另一物体的新接触。
Uniform translation with A equals zero changes velocity but needs no apparent force.
当 A 等于零时的均匀平动会改变速度，但不需要表观力。
For a rotating basis, differentiating a vector adds Omega cross that vector.
对于旋转基，对向量求导会加上 Omega 叉乘该向量。
Applying this twice gives a equals a sub origin plus a prime plus 2 Omega cross v prime plus Omega cross open parenthesis Omega cross r close parenthesis plus Omega dot cross r. Here r and v prime are measured relative to the moving origin in its rotating axes. All vectors in a calculation must be expressed in the same basis at the same instant.
应用两次后得到 a 等于 a 下标 origin 加 a prime 加 2 Omega 叉乘 v prime 加 Omega 叉乘 (Omega 叉乘 r) 加 Omega dot 叉乘 r。
Move the last three rotational terms to the force side to obtain Coriolis minus 2 m Omega cross v prime, centrifugal minus m Omega cross open parenthesis Omega cross r close parenthesis, and Euler minus m Omega dot cross r.
其中 r 和 v prime 是相对于运动原点在其旋转轴系中测量的；计算中的所有向量必须在同一时刻用同一基表示。 将最后三项旋转项移至力侧，可得科里奥利力 −2mΩ×v′、离心力 −mΩ×(Ω×r) 和欧拉力 −mΩdot×r。
A constant rotation removes the Euler term, not the Coriolis term.
恒定旋转会消除欧拉项，但不会消除科里奥利项。
Take Omega along positive z, an anticlockwise platform viewed from above, and a particle at r along positive x.
取 Ω 沿正 z 方向，即从上方看平台逆时针转动，质点位于正 x 方向的 r 处。
The centrifugal force is along positive x with magnitude m Omega squared r.
离心力沿正 x 方向，大小为 m Ω 平方 r。
If the particle moves outward with v prime along positive x, then Omega cross v prime is along positive y, so the Coriolis force points along negative y.
若质点以相对速度 v 撇沿正 x 方向向外运动，则 Ω 叉乘 v 撇沿正 y 方向，因此科里奥利力指向负 y 方向。
Reversing the relative velocity reverses Coriolis; keeping v prime at zero makes it vanish.
相对速度反向，科里奥利力也反向；v 撇为零时科里奥利力消失。
Coriolis is perpendicular to v prime and does no instantaneous work on that relative motion.
科里奥利力垂直于 v 撇，对这一相对运动不做瞬时功。
Centrifugal force is outward from the rotation axis, not necessarily outward from the chosen origin in an arbitrary three-dimensional position.
离心力背离转轴向外，在任意三维位置上不一定背离所选原点。
A body at rest on the platform has a prime equals v prime equals zero.
平台上静止的物体，其加速度 a prime 等于速度 v prime 等于零。
With constant rotation and a fixed origin, its real force must be m Omega cross Omega cross r, inward, cancelling the outward apparent term in the rotating equation.
在恒定旋转且原点固定的情况下，其真实力必须为 m Omega 叉乘 Omega 叉乘 r，方向向内，以抵消旋转方程中的向外表观项。
If rotation changes, a tangential real force must also balance the Euler term.
若旋转发生变化，切向真实力也必须平衡欧拉项。
Always check Omega approaches zero and A approaches zero: ordinary inertial Newtonian motion must return.
务必检查 Omega 趋于零和 A 趋于零的情形：此时普通惯性牛顿运动必须回归。
Do not mix the real inward centripetal requirement with an added outward real reaction on the same body; interaction partners belong in separate free-body diagrams.
切勿将向内的真实向心力需求与同一物体上额外的向外真实反作用力混淆；相互作用对象应分别绘制在独立的受力分析图中。
In a rotating frame the Coriolis force is minus 2 m Omega cross v prime, using the velocity relative to the frame, and the centrifugal force is minus m Omega cross, Omega cross r.
在转动参考系中，科里奥利力等于负 2m 乘 Ω 叉乘 v′，用的是相对参考系的速度；离心力等于负 m 乘 Ω 叉乘（Ω 叉乘 r）。
Take a 2 kilogram mass at 1 metre on the x axis, with Omega 2 radians per second about z.
取一个 2 千克的质量，位于 x 轴上 1 米处，Ω 为绕 z 轴 2 弧度每秒。
The centrifugal force is m Omega squared r: 2 times 4 times 1, or 8 newtons along plus x.
离心力等于 m Ω 平方 r：2 乘 4 乘 1，即沿 +x 方向 8 牛。
Moving outward at 3 metres per second, the mass feels a Coriolis force 2 m Omega v prime: 2 times 2 times 2 times 3, or 24 newtons along minus y.
以 3 米每秒向外运动时，该质量受到的科里奥利力为 2m Ω v′：2 乘 2 乘 2 乘 3，即沿 −y 方向 24 牛。
The Euler force is zero only because Omega is constant.
欧拉力为零，只是因为 Ω 恒定。
A particle is instantaneously at rest relative to a platform at radius r along positive x.
一个质点相对于平台瞬时静止，位于沿正x轴方向、半径为r处。
The platform has angular velocity Omega along positive z and angular acceleration alpha along positive z.
平台的角速度Omega沿正z轴方向，角加速度alpha也沿正z轴方向。
Find the real force needed to keep the particle fixed on the platform.
求使该质点固定在平台上所需的真实力。
Four marks.
四分。
With v prime equals a prime equals zero, the inertial acceleration is Omega cross Omega cross r plus dot Omega cross r, which equals negative Omega squared r x hat plus alpha r y hat.
当v'等于a'等于零时，惯性加速度为Ω叉乘(Ω叉乘r)加上点号Ω叉乘r，即负Ω平方r乘以x单位矢量加αr乘以y单位矢量。
Real force is m times negative Omega squared r x hat plus alpha r y hat.
真实力为m乘以负Ω平方r乘以x单位矢量加αr乘以y单位矢量。
Coriolis vanishes, while the Euler apparent force points in the negative y direction and centrifugal points in the positive x direction.
科里奥利力消失，而欧拉表观力指向负y方向，离心力指向正x方向。
Award centripetal term, tangential term, real force and apparent-force distinction.
给分点包括向心项、切向项、真实力和表观力的区别。
The relative velocity v′ belongs in the Coriolis term.
相对速度v'应出现在科里奥利项中。
Using the full inertial velocity counts rotation twice.
若使用完整的惯性速度，则会将旋转计算两次。
Coriolis force: The apparent rotating-frame force −2mΩ×v′ caused by relative motion.
科里奥利力：由相对运动引起的视旋转系力，即负二mΩ叉乘v撇。
centrifugal force: The apparent rotating-frame force −mΩ×(Ω×r), directed away from the rotation axis.
离心力：视旋转系力，即负mΩ叉乘(Ω叉乘r)，方向背离旋转轴。
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。