Question

To find the velocity and acceleration vectors for uniform circular motion and to recognize that this acceleration is the centripetal acceleration.
Suppose that a particle's position is given by the following expression:
r(t) = Rcos(omega*t)i + Rsin(omega*t)j

Velocity equals = -omegaR(sin(omega*t)i+omegaR(cos(omega*t)j

A. Now find the acceleration of the particle.
B. Your calculation is actually a derivation of the centripetal acceleration. To see this, express the acceleration of the particle in terms of its position r(t)

Answer 1

This is the obtuse way to find B.

you are given r(t)
velocity is dr/dt
acceleration is dv/dt

Hmmmm. It just got much simpler..prob states uniform circular motion, which means R is a constant. so R' =0, and angular velocity is constant (w'=0)
Then..
dv/dt=-w^2*R coswt i +w^2*R sinwt j
so there it is.
b) a=dv/dt=-w^2 *<b> R(t)</b> where the bold means a vector