# Circluar motion, omega

1. Oct 31, 2009

### emyt

circluar motion, "omega"

hi, I erased the default format by accident, but it's just a quick question:
i and j are unit vectors, w= omega dtheta/dt ,
theta= angle dependent on time

when we have i(-rwsin(theta)) and j(rwcos(theta)) as our velocity vector components,
and we wish to find the acceleration

i(-rwsin(theta)) and j(rwcos(theta)) is differentiated to i(-rw^2cos(theta)) and
j(-rw^2sin(theta))

the factor of omega remains constant because dtheta/dt will be the same everywhere since it is in uniform motion right? consistent speed?

thanks

edit: and if you did not have uniform circular motion, you would get

i(-r(a/r)wcos(theta)) and j(-rw(a/r)sin(theta)) as acceleration components:

using
w = dtheta/dt = v/r
d(v/r) / dt = 1/r(dv/dt) = a/r
when differentiating
-rwsin(theta), getting -r(a/r)wcos(theta) , remembering that theta is a function of time

thanks

Last edited: Oct 31, 2009
2. Nov 1, 2009

### kuruman

Re: circluar motion, "omega"

You are correct for the uniform circular motion part. For non-uniform (but still circular) motion, ω is not constant so you will need dω/dt terms.

3. Nov 1, 2009

### emyt

Re: circluar motion, "omega"

thanks, dω/dt is d/dt(v/r) right? which is 1/r dv/dt, isn't that a/r?

4. Nov 2, 2009

### kuruman

Re: circluar motion, "omega"

You can say that, but I prefer to think of dω/dt as the angular acceleration alpha and leave it at that.