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.

HA
Answered by Hubert A. Physics tutor

3717 Views

See similar Physics A Level tutors

Related Physics A Level answers

All answers ▸

How could I calculate the internal resistance of a cell?


Why do all objects fall at the same rate in a vacuum, independent of mass?


A ball is thrown at speed u = 10.0 m/s at an angle of 30.0 degrees to the ground at height, s = 0. How far does the ball travel horizontally from its starting position? (Ignore air resistance and taking g = 9.81 m/s^2)


A transmitter from a researcher's boat sends a signal to the seabed of waves speed 300m/s and it takes 5 seconds for the signal to return back to the boat. Calculate the depth of the sea there.


We're here to help

contact us iconContact ustelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

MyTutor is part of the IXL family of brands:

© 2025 by IXL Learning