- #1
csmyth3025
- 15
- 1
Homework Statement
The formula for centripetal acceleration (Ac) is $$Ac = \frac {4π^2r} {T^2},$$ where r = radius and T = period of rotation
Homework Equations
The above formula can be rearranged as follows: $$Ac = \frac {2π} {T} × \frac {2π} {T} × \frac {r} {1},$$ $$= \frac {2π} {T} × \frac {2πr} {T},$$ Setting the value of T at 1 (sec) gives the following:$$Ac = \frac {2π} {sec} × \frac {2πr} {sec}$$
The second term is simply the magnitude of the tangential velocity of an object in uniform circular motion. The numeric value of the first term is ##\frac {6.283185} {sec}##. When looking at this problem from the standpoint of vectors, this first term is the factor applied to the magnitude of the tangential velocity to derive the radial (inward) acceleration. Trigonometrically, when breaking the circular motion into very small segments, the sin of the angle formed by the initial and final velocity vectors as the circle segments become smaller and smaller asymptotically approaches 2π.
For example, dividing the circular motion into 100 segments (3.6° each) yields ##\frac {6.279052×10^{-2}} {sec×10^{-2}}##, 1000 segments (0.36° each) yields ##\frac {6.283144×10^{-3}} {sec×10^{-3}}## and 10000 segments (0.036° each) yields ##\frac {6.283185×10^{-4}} {sec×10^{-4}}##.
3. The Attempt at a Solution
What I can't figure out is whether this approximate equality of values between the Sin of the angle formed by the initial and final velocity vectors and 2π is just a happy mathematical coincidence or whether there is some subtle connection there that I'm just not seeing. I don't much believe in happy mathematical coincidences, so I figure I'm missing something.
BTW, this isn't a homework problem. I'm just curious and I've got waaay too much time on my hands.
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