A compact disc spins at 2.3 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 3.5 cm from the CD's center. What is the coefficient of friction between the ant and the CD?

To find the coefficient of friction between the ant and the CD, we can use a combination of centripetal force and frictional force.

First, we need to find the centripetal force acting on the ant. Since the ant is just beginning to slide off the CD, this force is equal to the maximum static friction force between the ant and the CD. We can express this force as:

F_c = m * a_c

Where F_c is the centripetal force, m is the mass of the ant, and a_c is the acceleration towards the center of the CD.

Next, we need to find the centripetal acceleration. We can use the formula:

a_c = r * ω^2

Where r is the distance of the ant from the center of the CD (3.5 cm or 0.035 m in this case), and ω is the angular velocity of the CD in radians per second. We are given that the CD spins at 2.3 revolutions per second, which is equivalent to 2.3 * 2π radians per second.

Now we have:

a_c = (0.035 m) * (2.3 * 2π rad/s)^2

Next, we substitute this value of a_c into the equation for F_c:

F_c = m * (0.035 m) * (2.3 * 2π rad/s)^2

Finally, we need to equate F_c to the maximum static friction force:

F_c = F_friction

The maximum static friction force can be expressed as:

F_friction = μ * m * g

Where μ is the coefficient of friction between the ant and the CD, m is the mass of the ant, and g is the acceleration due to gravity.

Now we have:

μ * m * g = m * (0.035 m) * (2.3 * 2π rad/s)^2

We can cancel out the mass of the ant on both sides of the equation:

μ * g = (0.035 m) * (2.3 * 2π rad/s)^2

Simplifying further, we have:

μ = [(0.035 m) * (2.3 * 2π rad/s)^2] / g

To find the coefficient of friction, we need to know the value of g, which is approximately 9.8 m/s^2.

Now we can substitute the values into the equation to find the coefficient of friction between the ant and the CD.