✔ 最佳答案
In fact, in physics, acceleration, a = dv/dt and v = dx/dt
Bear in mind of these relations since they are important in the following proof.
To begin with, please refer to the following figure:
http://i726.photobucket.com/albums/ww265/physicsworld2010/physicsworld01Feb181935.jpg
Suppose a particle moves around a circle of radius r from P to Q as shown above. By definition of linear velocity, v = dx/dt
For small θ, x = arc PQ = rθ
Therefore, v = d(rθ)/dt
Since r is a constant, v = r dθ/dt
v = rω
Where ω is angular frequency (or angular velocity), which is defined as ω = 2πf, where f is the frequency.
So, by definition, linear acceleration a is defined as a = dv/dt
Please take a look of the following figure
圖片參考：
http://i726.photobucket.com/albums/ww265/physicsworld2010/physicsworld02Feb181935.jpg
Considera particle moving in a circular path from A to B with a uniform speed v(Velocity is not uniform since the direction is changing). The changein velocity of the particle is:
△v = vsinθ - (-vsinθ) = 2vsinθ
If the particle takes time t to travel from A to B, the acceleration of the particle is
a = 2vsinθ / t
Also, by the relation vt = 2rθ
a = 2vsinθ / (2rθ / v) = v^2/r X sinθ/θ
Taking θ→ 0
the instantaneous acceleration of the particle is
lim θ→ 0 a
= v^2 / r lim θ→ 0 (sinθ/θ)
= v^2 / r (Since lim θ→ 0 (sinθ/θ) = 1)
We have: a = v^2 / r
As by above, v / r = ω
So, we have acceleration, a = ωv = 2πfv
And then we have finished the proof.
2009-06-28 16:50:32 補充：
本人不懂用中文證明...
2009-06-28 16:51:22 補充：
因為實在有太多詞彙不懂翻譯，怕譯得怪怪的...
