Torque and Rotational Dynamics
AP Physics C: Mechanics Topic 5 7:22 English narration · English + 中文 subtitles burned in
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Push a door at the handle and it swings open easily.
推门把手，门轻松就开了。
Now push just as hard, right next to the hinge. Almost nothing happens. The force was the same both times.
现在用同样大的力，紧贴着合页去推——几乎推不动。
What changed was where you pushed — and that is the whole of this unit in one sentence.
两次的力是一样的，改变的只是你推的位置。 这一句话，就是本单元的全部内容。
This is Unit Five: torque and rotational dynamics.
这是第五单元：力矩与转动动力学。
Every idea from the first four units comes back, rewritten for turning. Angle instead of position, torque instead of force, rotational inertia instead of mass.
前四个单元的每一个概念都会回来， 只不过换成了转动的版本：角度代替位置，力矩代替力，转动惯量代替质量。
Let's begin.
让我们开始吧。
Start with the words.
先从名词开始。
Angular displacement is the angle turned through.
角位移是转过的角度。
Angular velocity is the derivative of that angle with respect to time.
角速度是这个角度对时间的导数。
Angular acceleration is the derivative of the angular velocity.
角加速度是角速度的导数。
The same chain as before — just with angles.
还是同一条链条，只不过换成了角度。
Work in radians, always.
永远用弧度。
Degrees will quietly break every formula in this unit.
用角度制会悄悄地把本单元里的每一个公式都算错。
And when the angular acceleration is constant, the three kinematic equations come straight back, with each letter swapped for its angular partner.
而当角加速度恒定时，那三个运动学公式原封不动地回来，只是每个字母都换成了对应的角量。
Now connect the two worlds.
现在把两个世界连起来。
A point at a distance r from the axis travels along an arc.
距离转轴 r 处的一个点沿着圆弧运动。
Its arc length is r times the angle, its speed is r times the angular velocity, and its tangential acceleration is r times the angular acceleration.
它的弧长等于 r 乘以角度，速率等于 r 乘以角速度，切向加速度等于 r 乘以角加速度。
So points further out move faster.
所以离轴越远的点走得越快。
But a turning point has two accelerations at once. The tangential one speeds it up along the arc. The centripetal one turns it, and points at the axis.
但一个转动的点同时有两个加速度： 切向加速度让它沿弧线加快，向心加速度让它转弯，方向指向转轴。
They are perpendicular, so to get the total you square, add, and take the square root.
两者互相垂直，所以求总加速度要平方、相加、再开平方根。
Torque is the turning effect of a force.
力矩是力的转动效果。
It is to rotation what force is to straight-line motion.
力矩之于转动，就像力之于直线运动。
Watch a beam balance. A small force far from the pivot can hold up a large force close to it, because what matters is not the force alone but the force multiplied by its distance from the axis.
看一根平衡的横梁：离支点远处的小力，可以撑起离支点近处的大力， 因为起作用的不是力本身，而是力乘以它到转轴的距离。
Torque is the size of the force, times the distance from the axis, times the sine of the angle between them.
力矩等于力的大小，乘以到转轴的距离，再乘以两者夹角的正弦。
Equivalently, it is the force times the perpendicular distance from the axis to the line of the force.
等价地说，它等于力乘以从转轴到力的作用线的垂直距离。
That perpendicular distance has a name: the moment arm.
这个垂直距离有个名字：力臂。
A force pointing straight at the axis, or straight away from it, gives no torque at all — its moment arm is zero.
指向转轴或背离转轴的力，力臂为零，因此完全不产生力矩。
In a plane, call one turning direction positive and the other negative, then add.
在平面内，把一个转向定为正，另一个定为负，然后相加。
As a vector product, torque is r cross F. It points perpendicular to the plane, and the right-hand rule fixes which way: out of the page for anticlockwise, into it for clockwise.
作为矢量积，力矩等于 r 叉乘 F，方向垂直于这个平面，由右手定则决定： 逆时针转动时指向纸面外，顺时针时指向纸面内。
Rotational inertia is the rotational version of mass: it measures how hard it is to change the spinning.
转动惯量是质量的转动版本：它衡量改变转动状态有多难。
But unlike mass, it depends on where the mass sits.
但与质量不同，它取决于质量分布在哪里。
For point masses, add each mass times the square of its distance from the axis.
对质点，把每个质量乘以它到转轴距离的平方，再相加。
Notice the r squared. Mass far from the axis matters far more than mass close in — move the mass twice as far out and you quadruple its contribution.
注意那个距离的平方：远离转轴的质量，其影响远大于靠近转轴的质量—— 把质量移到两倍远处，它的贡献就变成四倍。
For a solid body the sum becomes an integral over the mass elements.
对实心物体，求和变成对质量元的积分。
Take a uniform rod of mass M and length L, spinning about its centre.
取一根质量为 M、长度为 L 的均匀细杆，绕它的中心转动。
Its linear density is the mass over the length, so a small piece of length dx has that density times dx of mass.
它的线密度是质量除以长度，所以长度为 dx 的小段，质量就是线密度乘以 dx。
Integrate x squared times that, from minus half L to plus half L, and you get one twelfth M L squared.
把 x 的平方乘以它，从负二分之一 L 积到正二分之一 L，得到十二分之一 M L 平方。
A solid disk works the same way, built from rings: a ring at radius r and width dr carries a mass M over pi R squared, times two pi r dr.
实心圆盘也一样，把它看成一圈圈的环：半径 r、宽 dr 的环，质量是 M 除以 pi R 平方，再乘以二 pi r dr。
Integrate r squared against that from zero to R and you get one half M R squared.
把 r 的平方乘上去，从零积到 R，就得到二分之一 M R 平方。
Now move the axis to one end.
现在把转轴移到一端。
You could redo the integral — or use the parallel-axis theorem: add M times the square of the shift.
你可以重新积分一遍——也可以用平行轴定理： 加上 M 乘以位移的平方。
Half of L, squared, times M, is one quarter M L squared.
二分之一 L 的平方乘以 M，等于四分之一 M L 平方。
Add it and you get one third M L squared.
加起来就得到三分之一 M L 平方。
Newton's first law has a rotational twin: with zero net torque, the angular velocity stays constant.
牛顿第一定律有一个转动版的孪生兄弟：合力矩为零时，角速度保持不变。
The object either does not turn at all, or turns steadily.
物体要么完全不转，要么匀速转动。
Full static equilibrium needs both conditions — the net force is zero, and the net torque is zero about any axis you like.
完整的静平衡需要两个条件——合力为零， 并且对任意一个你选定的轴，合力矩为零。
And here is the trick that solves nearly every beam and ladder problem: put your axis right where an unknown force acts.
这里有一个几乎能解决所有横梁和梯子问题的技巧： 把转轴取在某个未知力的作用点上。
Its moment arm is then zero, so it drops out of the torque equation completely, and one unknown disappears.
它的力臂就是零，于是它彻底从力矩方程里消失， 未知量少了一个。
A uniform twenty kilogram beam, four metres long, is pivoted at its left end and held horizontal by a vertical rope at its right end.
一根均匀的二十千克横梁，长四米，左端用铰链支住，右端由一根竖直的绳保持水平。
A forty kilogram child sits one metre from the pivot.
一个四十千克的孩子坐在离支点一米处。
Find the tension.
求绳的张力。
Pause here and try it.
先暂停，自己试一试。
Take torques about the pivot, so the pivot force drops out.
对支点取矩，这样支点的力就消失了。
The rope's torque is the tension times four.
绳的力矩是张力乘以四。
Against it: the beam's weight acting at its centre, two metres out, and the child's weight one metre out.
与它对抗的是：横梁的重力作用在中点，也就是两米处；以及孩子的重力，在一米处。
Both give three hundred and ninety-two newton metres.
两者都给出三百九十二牛顿米。
Add them, divide by four, so the tension is one hundred and ninety-six newtons.
相加后除以四，所以张力是一百九十六牛顿。
A net torque produces an angular acceleration, in proportion to the rotational inertia.
合力矩产生角加速度，其大小与转动惯量成反比。 角加速度等于合力矩除以转动惯量。
Angular acceleration equals the net torque divided by the rotational inertia. Every symbol has a partner. Force becomes torque.
每一个符号都有它的对应者： 力变成力矩，质量变成转动惯量，加速度变成角加速度。
Mass becomes rotational inertia. Acceleration becomes angular acceleration. So the method is identical to what you already do: draw the diagram, write one equation per body, and solve.
所以方法和你已经会的完全一样：画图、每个物体写一条方程、然后求解。
Now the classic.
现在来看经典题。
Two masses hang over a pulley that actually has rotational inertia. Four kilograms on one side, two on the other.
两个物体挂在一个真正具有转动惯量的滑轮两侧：一侧四千克，另一侧两千克。
Write three equations, one per body: the heavy mass, the light mass, and the pulley itself.
写三条方程，每个物体一条：重的物体、轻的物体，以及滑轮本身。
For the pulley, the net torque is the difference of the two tensions, times the radius.
对滑轮来说，合力矩等于两个张力之差乘以半径。
The rope does not slip, so the linear acceleration is the radius times the angular acceleration.
绳不打滑， 所以线加速度等于半径乘以角加速度。
Adding them all together gives one point one metres per second squared, and the two tensions come out at thirty-five and twenty-two newtons.
把它们全部相加，得到一点一米每二次方秒， 两个张力分别是三十五牛顿和二十二牛顿。
<slow>They must be different.</slow> If they were equal, nothing would spin the pulley at all.
它们必须不相等—— 如果相等，就没有任何东西能让滑轮转起来。
Three habits that save marks.
三个能保住分数的习惯。
First, keep everything in radians and fix a positive sense of rotation before you write anything.
第一，一切都用弧度，并在动笔之前先定下正的转动方向。
Second, draw the axis on your diagram — the moment arm is the perpendicular distance to it, not the distance to the force.
第二，在图上把转轴画出来——力臂是到转轴的垂直距离，而不是到力的距离。
Third, over a massive pulley the two rope tensions are never equal, and you must state the constraint that links the linear and angular accelerations.
第三，跨过有质量的滑轮时，绳两侧的张力永远不相等， 而且你必须写出把线加速度和角加速度联系起来的约束条件。
Get those three, and this unit is yours.
做到这三点，这个单元就是你的了。