Further Mechanics
A-Level Further Mathematics Topic 3 8:04 English narration · English + 中文 subtitles burned in
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Throw a ball, fire a jet of water, launch anything into the air — and it traces the same elegant curve: a parabola.
扔一个球，喷一道水柱，把任何东西抛向空中——它都会划出同一条优美的曲线：抛物线。
The secret is to split the motion in two.
秘诀是把运动一分为二。
The sideways part cruises along at a steady speed, while the up-and-down part is pulled back by gravity.
水平方向以稳定的速度前进，而竖直方向被重力拉回。
Put them together, and out comes the arc.
把两者合起来，就得到这道弧线。
Further Mechanics is the mathematics of motion, force, and impact — and it starts right here.
进阶力学是关于运动、力和碰撞的数学——它就从这里开始。
Further Mechanics takes the mechanics you know and pushes it further.
进阶力学把你已知的力学推向更远。
We will launch projectiles and predict where they land, balance rigid bodies with moments, whirl particles in circles, stretch elastic strings, handle forces that change with position, and finally, watch objects collide and bounce.
我们会抛出抛射体，预测它们的落点；用力矩平衡刚体； 让粒子绕圈旋转；拉伸弹性绳；处理随位置变化的力；最后，看物体碰撞与反弹。
Six powerful ideas about how the physical world moves.
六个关于物质世界如何运动的强大想法。
Let's begin.
让我们开始吧。
Here is the master trick for projectiles: split the velocity into two independent parts.
抛射体的高招是：把速度分成两个互不相干的部分。
The horizontal part, u cosine alpha, never changes — there is no sideways force.
水平部分，u 乘 cosine alpha，永不改变—— 没有水平方向的力。
The vertical part, u sine alpha, is slowed and reversed by gravity.
竖直部分，u 乘 sine alpha，被重力减速并反向。
The two motions share only the clock.
两个运动只共享同一个时钟。
Eliminate the time between them and you get the Cartesian equation of the trajectory — the path — and that path is a parabola.
把它们之间的时间消去，就得到轨迹的直角坐标方程，而这条轨迹是一条抛物线。
Keep the launch speed fixed but vary the angle, and every trajectory you can reach lies underneath one curve, the bounding parabola.
保持抛出速率不变、只改变角度，所有能到达的轨迹都落在同一条曲线之下，这条曲线叫包络抛物线。
From this you get two key results: the range along the ground, and the greatest height.
由此得到两个关键结果：沿地面的射程，以及最大高度。
Let's put numbers in.
我们代入数字。
A ball is thrown at twenty metres per second, at thirty degrees, and gravity is ten.
一个球以每秒二十米、三十度角抛出，重力是十。
First the range: twenty squared, times the sine of sixty degrees, over ten.
先求射程：二十的平方， 乘六十度的正弦，除以十。
That is forty times zero point eight six six, about thirty-four point six metres.
也就是四十乘零点八六六，约三十四点六米。
Now the greatest height: twenty squared, times the sine squared of thirty degrees, over two times ten.
再求最大高度： 二十的平方，乘三十度正弦的平方，除以二乘十。
That works out to five metres.
算出来是五米。
A rigid body under coplanar forces is in equilibrium when two conditions both hold: the vector sum of the forces is zero, and the sum of the moments about any point is zero.
刚体在共面力作用下处于平衡，需要两个条件同时成立：力的矢量和为零， 而且对任意一点的力矩之和也为零。
The turning effect of a force is its moment — force times perpendicular distance.
一个力的转动效果就是它的力矩——力乘以垂直距离。
A body's weight acts at its centre of mass; for a uniform flat shape — a lamina — you can often find it by symmetry, and for a composite body you treat each part as a particle at its own centre of mass.
物体的重量作用在它的质心； 对均匀的平面薄片，常常可以用对称性找到质心；对组合物体， 则把每一部分当作位于自身质心的质点。
Take a uniform beam on a pivot at A, with its weight of one hundred newtons acting at the centre of mass, and a force F holding up the far end.
取一根均匀的横梁，支点在 A， 它一百牛顿的重量作用在质心，一个力 F 撑住远端。
Take moments about A.
对 A 取矩。
F times four equals one hundred times two, so F is fifty newtons.
F 乘四等于一百乘二， 所以 F 是五十牛顿。
Choosing to turn about A made the pivot's own force disappear from the equation.
选择绕 A 转动，让支点自身的力从方程里消失了。
A rigid body in equilibrium can still be on the edge of failing in two ways: toppling, turning over about an edge, or sliding, slipping along the surface — and an exam question will ask which happens first.
处于平衡的刚体仍然可能濒临两种失效：翻倒，绕着一条边翻过去；或者滑动，沿着表面滑走—— 考题会问你哪一种先发生。
When a particle moves in a circle, two vectors tell the whole story.
当一个粒子做圆周运动时，两个向量道尽了全部。
Its velocity always points along the tangent — straight ahead, along the edge of the circle.
它的速度总是沿切线方向——笔直向前， 沿着圆的边缘。
Its speed and the angular speed are linked simply: v equals r omega.
它的速率和角速度有简单的关系：v 等于 r omega。
But even at a steady speed, the particle is accelerating, because its direction keeps changing.
但即使速率不变， 粒子也在加速，因为它的方向一直在改变。
That acceleration points straight in, towards the centre — the centripetal acceleration — equal to r omega squared, or v squared over r.
那个加速度笔直指向内侧，朝向圆心—— 这就是向心加速度——等于 r omega 平方，或 v 平方除以 r。
Quick example.
来个小例子。
A particle moves in a horizontal circle of radius two metres, with an angular speed of three radians per second.
一个粒子在半径两米的水平圆上运动，角速度是每秒三弧度。
The speed is r omega — two times three — six metres per second.
速率是 r omega——二乘三——每秒六米。
And the acceleration towards the centre is r omega squared — two times nine — eighteen metres per second squared.
而指向圆心的加速度是 r omega 平方—— 二乘九——每秒每秒十八米。
Two formulas, two clean answers.
两个公式，两个干净的答案。
In a horizontal circle like this the speed stays constant — the standard example is the conical pendulum, a mass swung round on a string that sweeps out a cone.
在这样的水平圆里速率保持不变——最典型的例子是圆锥摆： 用一根绳把重物甩起来，绳子扫出一个圆锥面。
A vertical circle is trickier, because the speed changes with height — so you use energy conservation, not constant speed.
竖直圆更棘手，因为速率随高度改变——所以要用能量守恒，而不是匀速。
Whatever is doing the pushing or pulling — the string tension, or the normal contact force from a track or the inside of a loop — provides the centripetal force.
提供向心力的，是绳的张力，或者轨道、圆环内壁给的法向接触力。
The hardest point is the very top.
最难的点是最顶端。
There, the particle moves slowest, and both gravity and the tension pull straight down, toward the centre.
在那里，粒子速率最慢，重力和张力都笔直向下，指向圆心。
For the string to stay tight at the top, gravity alone must be enough to bend the path — so the minimum speed satisfies m g equals m v squared over r, giving v squared equals g r.
要让绳在顶端保持绷紧， 仅靠重力就必须足以弯折路径——于是最小速率满足 m g 等于 m v 平方除以 r，得到 v 平方等于 g r。
An elastic string obeys Hooke's law: the tension is proportional to how far it is stretched.
弹性绳遵守胡克定律：张力与它被拉伸的长度成正比。
The tension equals lambda x over L, where L is the natural length and lambda is the modulus of elasticity.
张力等于 lambda x 除以 L， 其中 L 是自然长度，lambda 是弹性模量。
Because tension rises in a straight line, the elastic potential energy stored is the triangle underneath it — lambda x squared, over two L.
因为张力沿直线上升，储存的能量就是它下方的三角形—— lambda x 平方，除以二 L。
Try it: a string of natural length two, modulus fifty, stretched by half a metre.
试一试：一根自然长度为二、模量为五十的绳，被拉伸半米。
The tension is twelve point five newtons, and the stored energy is three point one two five joules.
张力是十二点五牛顿，储存的能量是三点一二五焦耳。
Sometimes the force depends on where the object is, not on time.
有时力取决于物体在哪里，而不是取决于时间。
The clever move is to write the acceleration as v times dv by dx.
高招是把加速度写成 v 乘以 dv 比 dx。
Newton's second law then becomes a differential equation linking speed and position.
牛顿第二定律于是变成一个联系速率与位置的微分方程。
Here, eight v dv by dx equals x cubed plus four x.
这里，八 v 乘 dv 比 dx 等于 x 三次方加四 x。
Separate the variables and integrate both sides.
分离变量，两边积分。
Finally, use the initial condition — v is one when x is zero — to fix the constant, and the speed comes out as one quarter x squared, plus one.
最后，用初始条件——x 为零时 v 是一——定出常数， 速率就求得为四分之一 x 平方，加一。
In any collision, the total linear momentum before equals the total linear momentum after — that is the conservation of linear momentum.
在任何碰撞中，碰前的总线动量等于碰后的总线动量——这就是线动量守恒。
How bouncy the collision is comes down to one number, given by Newton's experimental law, which defines the coefficient of restitution e: the speed of separation, divided by the speed of approach, always between zero and one.
碰撞有多“弹”，由牛顿实验定律给出，它定义了恢复系数 e： 分离速率除以接近速率，永远在零和一之间。
e equals one is perfectly elastic, with no energy lost; e equals zero is inelastic, and the bodies stay together.
e 等于一是完全弹性碰撞，没有能量损失；e 等于零是完全非弹性碰撞，两个物体粘在一起。
Take a two-kilogram sphere at five metres per second, striking a three-kilogram sphere at rest, with e of one half.
取一个两千克的球，以每秒五米， 撞上一个静止的三千克的球，e 为二分之一。
Momentum gives one equation; restitution gives another.
动量给出一个方程；恢复系数给出另一个。
Solve the pair, and afterwards A moves at zero point five, and B at three metres per second.
解这一对方程，碰后 A 以零点五、B 以每秒三米运动。
If the collision is an oblique impact — the spheres meeting off-centre — resolve the velocities along and perpendicular to the line of impact, and apply restitution only along that line.
如果是斜碰撞——两球不是正对着相撞——就把速度沿碰撞线和垂直于碰撞线分解， 并且只在碰撞线方向上使用恢复系数。
Before you go, four ways to keep your marks.
结束之前，四个保住分数的办法。
First, for projectiles, split into horizontal and vertical, and remember they share the same time.
第一，抛射体要分解成水平和竖直，并记住它们共享同一个时间。
Second, when you take moments, choose the point that makes an unknown force vanish.
第二，取矩时，选那个能让某个未知力消失的点。
Third, for circular motion, point the resultant force at the centre — and in a vertical circle, check the minimum speed at the top.
第三，圆周运动要把合力指向圆心—— 竖直圆里，检查顶端的最小速率。
Fourth, for a variable force, use v dv by dx and integrate with respect to x.
第四，变力要用 v 乘 dv 比 dx，并对 x 积分。