Question

A compact disc spins at 2.8 revolutions per second. An ant is walking on the CD and finds that it just begins to slide off the CD when it reaches a point 3.2 cm from the CD's center.

What is the coefficient of friction between the ant and the CD?

Answer 1

2.8 * 2 pi radians/s = 17.6 rad/s
r = .032 meters

Ac = w^2 r = (17.6)^2(.032) = 9.9 m/s^2

m g (mu) = m Ac
mu g = 9.9
mu (9.8) = 9.9
mu is about 1

Answer 2

Thank you so much!