Deriving Centripetal Acceleration

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SUMMARY

The discussion focuses on deriving the formula for centripetal acceleration, specifically a = (v^2)/r, from the position of a particle moving in a circular path defined by the equation x(t) = r(isin(wt) + jcos(wt)). The user, Dan, successfully computes the acceleration as a(t) = -rw^2(icos(wt) + jsin(wt)) but struggles with eliminating the trigonometric functions. The solution involves finding the magnitude of the acceleration vector and applying the Pythagorean identity sin² + cos² = 1 to simplify the expression.

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  • Understanding of circular motion and centripetal acceleration
  • Familiarity with vector calculus and derivatives
  • Knowledge of trigonometric identities, specifically sin² + cos² = 1
  • Basic proficiency in physics concepts related to motion
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Hi. I'm trying to derive a = (v^2)/r from the position of a particle moving in a circle. Since the equation for a particle moving in a circle is x(t) = r(isin(wt) + jcos(wt)) where w is constant. I took two derivatives to find it's acceleration. I got a(t) = -rw^2(icos(wt)+jsin(wt)). Now I'm stuck. How the heck and I getting rid of those trig functions? Thanks.

-Dan
 
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Find the magnitude of the vector a. You will get i.icos^2+j.jsin^2. Then i.i is unit vector dotted with itself equals 1. Then sin^2+cos^2 is some trig identity.
 

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