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September 30, 2014

September 30, 2014

Posted by **Becky** on Friday, November 13, 2009 at 8:14pm.

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)

- Physics -
**bobpursley**, Friday, November 13, 2009 at 9:18pmThis 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 ***R(t)**where the bold means a vector

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