Derive an expression for the centripetal acceleration of a body in uniform circular motion.

(I assume familiarity with positions represented by vectors and differentiation of trigonometric functions). Consider the coordinates of a point moving in a circle of radius r around the origin. The equation of the circle is (rsin theta)2 +(rcos theta)2 = r2. So the position vector x is (rcos theta; rsin theta) - this is a column vector. So differentiate with respect to time to get tangential velocity dx/dt: (-rsin thetadtheta/dt; rcos thetadtheta/dt). Differentiate again to get acceleration d2x/dt2: (-rcos theta(dtheta/dt)2-rsin thetad2theta/dt2; -rsin theta(dtheta/dt)2+rcos thetad2theta/dt2). Now dtheta/dt is of course constant since it's constant motion, which means d2theta/dt2 = 0! So acceleration is now simply (-rcos theta(dtheta/dt)2; -rsin theta(dtheta/dt)2). We can relate velocity and position as |v|=r(dtheta/dt), and acceleration as a=|v|2/r.

Answered by Hubert A. Physics tutor

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