Re: [閒聊] a = v^2/r

[Wolfram (Infinite)]

給阿猴..(因為我不確定你的ID是哪個)

我找到一個更好的網頁

http://www.physics.upenn.edu/courses/gladney/mathphys/java/sect4/subsubsection4_1_2_2.html

看起來很長,但其實只要搞清楚幾個部分就可以了...

the position of the particle must be

x = r cos(theta) + x0
y = r sin(theta) + y0

一階微分,找到在x,y分量上的速率

vx = dx/dt = -r sin(theta) d(theta)/dt
vy = dy/dt = r cos(theta) d(theta)/dt

二階微分,找到在x,y分量上的加速度

ax = d(vx)/dt = -r cos(theta) [d(theta)/dt]2
ay = d(vy)/dt = -r sin(theta) [d(theta)/dt]2

很容易算出

a = sqrt(ax2 + ay2) = sqrt[r2(cos(theta)2 + sin(theta)2)* omega4] = r*omega2

然後omega = (theta)/t = (r*theta)rt (上下乘r)

          = v/t

所以

a=r*(v^2/r^2) = v^2/r

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推 ophlia13:這...是去松羅湖時說的那個笑話嗎??
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