- #1

- 95

- 0

v=rw=2πr/T

a=rw

^{2}=r(4π

^{2}r

^{2}/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.

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- Thread starter qazxsw11111
- Start date

- #1

- 95

- 0

v=rw=2πr/T

a=rw

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

- 225

- 0

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

- #3

Delphi51

Homework Helper

- 3,407

- 11

Error in the third line - should be a = a=rw²=r(2π /T)² = 4π²r/T²w=2π /T

v=rw=2πr/T

a=rw²=r(4π²r²/T)

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.

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