# Centripetal Acceleration derivation help

1. Nov 14, 2009

### qazxsw11111

w=2π /T
v=rw=2πr/T
a=rw2=r(4π2r2/T)

This I know, but why cant I just differentiate v with respect to t to get a? (But the answer is wrong) Can anyone tell me why?

Thanks.

2. Nov 14, 2009

### mg0stisha

Also, $$a_{c}=\frac{v^{2}}{R}$$ in case you didn't have this equation at your disposal.

Are you trying to differentiate v=rw=2πr/T with respect to time?

3. Nov 14, 2009

### Delphi51

Error in the third line - should be a = a=rw²=r(2π /T)² = 4π²r/T²
Differentiating v = 2πr/T does no good because this formula is for the (constant) magnitude of velocity, so the dv/dt = 0. In fact the direction of the velocity is continuously changing so a vector derivative must be done to get the acceleration that way. It would be something like this:
v = 2πr/T[cos(ωt),sin(ωt)]
dv/dt = 2πr/T[-ω*sin(ωt), ω*cos(ωt)]
= 2πr/T*ω[-sin(ωt), cos(ωt)] and since ω = v/r and v = 2πr/T this is
= v²/r[-sin(ωt), cos(ωt)]
showing that the magnitude of acceleration is good old v²/r and that its direction rotates with time.