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?

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

Thank you so much!

To find the coefficient of friction between the ant and the CD, we can use the Centripetal Force equation which relates the friction force to the mass, radius, and speed of the object.

The formula for centripetal force is as follows:

F = (m * v²) / r

Where:
F = centripetal force
m = mass of the object
v = velocity of the object
r = radius of the object

In this case, the ant is just about to slide off the CD, which means the friction force acting on the ant is equal to the maximum static friction force. The maximum static friction force can be calculated using the equation:

Friction Force (maximum) = coefficient of friction * Normal force

Assuming the normal force acting on the ant is equal to its weight (mg), and the weight of the ant is neglected compared to the CD, we can say that the normal force is equal to the ant's weight.

Since the ant is sliding off the CD due to the centripetal force acting on it, we can equate the maximum friction force with the centripetal force:

coefficient of friction * Normal force = (m * v²) / r

Now, we can solve for the coefficient of friction:

coefficient of friction = ((m * v²) / r) / Normal force

To calculate the coefficient of friction, we need information about the ant's mass, velocity, radius, and the acceleration due to gravity.