Torque and Rotational Dynamics
AP Physics 1 Topic 5 7:41 English narration · English + 中文 subtitles burned in
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Push a heavy door right next to its hinge, as hard as you can. It barely moves.
用尽全力，紧贴门轴推一扇沉重的门，它几乎纹丝不动。
Now push with the same force, out at the handle — and it swings wide open.
现在用同样的力去推门把手—— 门就轻松地大开。
Same force, same door, different result.
同样的力，同样的门，结果却不同。
What changed was the distance from the hinge, and that idea runs through this whole unit.
改变的只是到门轴的距离，而这个想法贯穿整个单元。
This is Unit Five: torque and rotational dynamics.
这是第五单元：力矩与转动动力学。
Everything you know about forces comes back — but now for objects that spin.
你学过的关于力的一切都会回来—— 只不过现在是针对转动的物体。
New words, the same physics, and one equation you will use again and again.
新的词汇，同样的物理， 以及一个你会反复使用的公式。
Start with the angle.
先从角说起。
In rotation we do not use degrees.
在转动中我们不用度。
Take the radius, bend it around the rim, and the angle it covers is one radian.
把半径弯到圆周上，它所对的角就是一弧度。
Now count how many of those arcs fit around a full circle.
现在数一数，一个整圆能放下多少段这样的弧。
Six, and a small piece left over — a full turn is about six point two eight radians.
六段，还剩下一小段—— 一整圈约为六点二八弧度。
Every equation here assumes radians.
这里的每一个公式都默认使用弧度。
Now three new quantities, each mirroring one you already know.
现在有三个新的物理量，每一个都对应你已经熟悉的一个量。
Angular displacement is the angle turned, in radians.
角位移是转过的角， 用弧度度量。
Angular velocity is how fast that angle changes — the spin rate.
角速度是这个角变化的快慢——也就是转动的快慢。
Angular acceleration is how fast the spin rate itself changes.
角加速度是转动快慢本身变化的快慢。
And choose one turning direction as positive before you write anything.
另外，动笔之前先选定一个转向为正。
Because the words match, the equations match.
因为词汇一一对应，公式也一一对应。
Swap every straight-line symbol for its turning twin, and the three constant-acceleration equations come back with the same shape.
把每一个直线运动的符号换成它的转动对应量， 那三个匀加速公式就以相同的形式回来了。
A wheel starts from rest and reaches thirty radians per second in six seconds.
一个轮子从静止开始， 在六秒内加速到每秒三十弧度。
The angular acceleration is thirty divided by six, so five radians per second squared.
角加速度是三十除以六，即每二次方秒五弧度。
And the angle turned is a half, times five, times six squared — ninety radians, a little over fourteen full turns.
转过的角是二分之一，乘五，乘六的平方——九十弧度，也就是十四圈多一点。
A rigid object turns as one piece, so every point shares the same angular velocity — but not the same speed.
刚体作为一个整体转动，所以每一点的角速度都相同，但速率并不相同。
Multiply the angle by the distance from the axis and you get the arc length; multiply the angular velocity by that distance and you get the linear speed.
把角乘以到转轴的距离，就得到弧长；把角速度乘以这个距离，就得到线速度。
The same distance links angular and tangential acceleration.
同一个距离把角加速度和切向加速度联系起来。
Picture a spinning platform with two riders, one near the middle and one at the rim.
想象一个旋转的平台上有两位乘客，一位靠近中心，一位在边缘。
They go round together, so their angular velocity is identical — but a rider twice as far out moves twice as fast.
他们一起转， 所以角速度完全相同，但离中心远一倍的人，速度也快一倍。
A bicycle wheel of radius zero point three five metres turns at twelve radians per second.
一个半径零点三五米的自行车轮以每秒十二弧度转动。
The rim moves at zero point three five times twelve — four point two metres per second.
轮缘的速度是零点三五乘十二—— 四点二米每秒。
Halfway in, that drops to two point one.
到一半的位置，就降到二点一。
Now the cause of the turning: torque, the turning effect of a force.
现在来看转动的原因：力矩，也就是力的转动效果。
Push at a right angle and the torque is just the force times the distance from the axis.
垂直施力时， 力矩等于力乘以到转轴的距离。
Twenty newtons at the end of a wrench zero point three metres long gives six newton metres.
在一把零点三米长的扳手末端用二十牛顿的力， 就得到六牛顿米。
Where you push decides the result.
你在哪里施力决定了结果。
But a real push is rarely at a right angle.
但真实的推力很少恰好垂直。
Only the part of the force that is perpendicular does any turning.
只有力中垂直的那一部分才产生转动效果。
There is a second way to see it: the moment arm, or lever arm — the perpendicular distance from the axis to the line of the force.
还有第二种看法：力臂，也就是从转轴到力的作用线的垂直距离。
Either way, torque is force times distance times the sine of the angle between them.
两种看法都给出：力矩等于力乘以距离，再乘以两者夹角的正弦。
The same twenty newtons at sixty degrees gives five point two newton metres, not six.
同样的二十牛顿，在六十度时给出五点二牛顿米，而不是六。
And push straight along the wrench, and the torque falls to zero — the line of the force now passes through the axis.
而沿着扳手方向推，力矩就降为零——因为力的作用线此时穿过转轴。
Torque spins things up.
力矩让物体转起来。
What resists?
是什么在抵抗呢？
Not mass alone.
不只是质量。
Rotational inertia measures how hard it is to change an object's spin, and it depends on where the mass sits.
转动惯量衡量改变一个物体转动状态的难易程度，它取决于质量分布在哪里。
A small mass far from the axis can be harder to spin than a large mass close in.
远离转轴的小质量，可能比靠近转轴的大质量更难转动。
Distance is squared — that is why far out wins.
距离要平方，所以离得远的更难转。
For a point mass it is mass times distance squared, and for several masses you add them together.
对一个质点，它等于质量乘以距离的平方；多个质点则相加。
That squared is the part students forget.
学生最容易忘的就是那个平方。
Move the same mass twice as far from the axis and the rotational inertia grows four times, not two.
把同样的质量移到离转轴两倍远的地方， 转动惯量会变成四倍，而不是两倍。
Here are a hoop and a solid disc with the same mass and radius.
这里有一个圆环和一个实心圆盘， 质量和半径都相同。
The hoop keeps all its mass at the rim, so it is twice as hard to spin.
圆环把全部质量都放在边缘，所以难转一倍。
And inertia is smallest when the axis passes through the centre of mass — move the axis away from there and it only grows.
而且，当转轴穿过质心时，转动惯量最小——把转轴移开，它只会变大。
Now balance.
现在来看平衡。
An object is in rotational equilibrium when the net torque is zero, so its angular velocity never changes.
当净力矩为零时，物体处于转动平衡，角速度不会改变。
For an object that is completely still, the net force must be zero as well.
对于完全静止的物体，合力也必须为零。
Watch the beam: whenever one side's torque is bigger, it tips.
看这根横梁：只要一侧的力矩更大，它就倾斜。
Slide the weight out until the two turning effects are equal, and it holds still.
把重物向外滑动， 直到两侧的转动效果相等，它就静止不动。
Equal torques does not mean equal forces.
力矩相等并不意味着力相等。
Let's use that.
我们来用一用。
A thirty kilogram child sits two metres from the pivot of a seesaw.
一个三十千克的孩子坐在跷跷板支点外两米处。
Where must a forty kilogram child sit to balance it?
一个四十千克的孩子应该坐在哪里才能平衡？
Pause here and try it.
先暂停，自己试一试。
Set the two torques equal.
让两侧力矩相等。
Gravity is on both sides, so it cancels.
重力在两边都有，可以约掉。
Thirty times two equals forty times the distance, so the distance is one point five metres.
三十乘二等于四十乘那个距离，所以距离是一点五米。
The heavier child sits closer in.
较重的孩子坐得更靠近支点。
If the net torque is not zero, the spin has to change.
如果净力矩不为零，转动就必须改变。
The net torque equals the rotational inertia times the angular acceleration.
净力矩等于转动惯量乘以角加速度。
Beside Newton's second law the pattern is complete: torque replaces force, rotational inertia replaces mass, angular acceleration replaces acceleration.
和牛顿第二定律并排来看，这个对应就完整了：力矩代替力， 转动惯量代替质量，角加速度代替加速度。
The same law, dressed for spinning.
同一条定律，只是换上了转动的外衣。
That one equation answers a favourite exam question.
这一个公式就能回答一道考试常客。
Two wheels get exactly the same net torque, but the second has half the rotational inertia of the first.
两个轮子受到完全相同的净力矩， 但第二个的转动惯量只有第一个的一半。
Which spins up faster?
哪一个转得更快？
Angular acceleration is torque divided by rotational inertia, so halving the inertia doubles the angular acceleration.
角加速度等于力矩除以转动惯量，所以惯量减半，角加速度加倍。
In the same time, the second wheel turns through twice the angle.
在相同的时间里，第二个轮子转过的角是第一个的两倍。
Same torque, twice the result.
同样的力矩，两倍的效果。
Now a full exam question.
现在来做一道完整的考题。
A string is wrapped around a solid disc pulley of mass four kilograms and radius zero point five metres, and pulled with a steady six newtons.
一根绳绕在一个实心圆盘滑轮上，滑轮质量四千克， 半径零点五米，绳以恒定的六牛顿被拉动。
Find the angular acceleration, and the acceleration of a point on the rim.
求角加速度，以及轮缘上一点的加速度。
Pause here and try it.
先暂停，自己试一试。
First the rotational inertia: a half, times four, times zero point five squared — zero point five.
首先求转动惯量：二分之一，乘四，乘零点五的平方——零点五。
Then the torque: six times zero point five — three newton metres.
然后求力矩：六乘零点五——三牛顿米。
Now divide: three over zero point five gives six radians per second squared.
现在相除：三除以零点五， 得到每二次方秒六弧度。
And the rim of the pulley accelerates at the radius times that — three metres per second squared.
而轮缘的加速度等于半径乘以它——三米每二次方秒。
Four habits that save marks.
四个能保住分数的习惯。
First, torque uses the perpendicular distance only — a force whose line passes through the axis gives no torque.
第一，力矩只用垂直距离——作用线穿过转轴的力， 力矩为零。
Second, when a problem hides an unknown force, put your axis right at that force and it drops out.
第二，当题目里藏着一个未知的力时，就把转轴取在那个力上， 它就会从方程中消失。
Third, rotational inertia depends on the distance squared, so moving mass outwards costs more than it looks.
第三，转动惯量与距离的平方成正比， 所以把质量往外移，代价比看上去大得多。
Fourth, work in radians, and remember linear speed is radius times angular velocity.
第四，用弧度计算， 并记住线速度等于半径乘以角速度。