posted by Becky .
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)
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)
dv/dt=-w^2*R coswt i +w^2*R sinwt j
so there it is.
b) a=dv/dt=-w^2 * R(t) where the bold means a vector