If the CD rotates clockwise (let's take clockwise rotation as positive) at 500 (revolutions per minute) while the last song is playing, and then spins down to zero angular speed in 2.60 with constant angular acceleration, what is , the angular acceleration of the CD, as it spins to a stop?

Units????

To find the angular acceleration, we can use the following equation of rotational motion:

ωf = ωi + αt

Where:
ωf is the final angular velocity (0 in this case as the CD comes to a stop).
ωi is the initial angular velocity (500 revolutions per minute converted to radians per second).
α is the angular acceleration (the value we want to find).
t is the time taken for the CD to come to a stop (2.60 seconds).

First, let's convert the initial angular velocity from revolutions per minute to radians per second. Since 1 revolution is equal to 2π radians, we have:

ωi = (500 revolutions/minute) * (2π radians/1 revolution) * (1 minute/60 seconds)

Simplifying, we get:

ωi = (500 * 2π) / 60 radians/second

Now, substituting the given values into the equation, we have:

0 = (500 * 2π) / 60 + α * 2.60

Simplifying further:

0 = (500 * 2π) / 60 + 2.60α

To solve for α, we can rearrange the equation as follows:

2.60α = - (500 * 2π) / 60

Now, divide both sides of the equation by 2.60 to isolate α:

α = - (500 * 2π) / (60 * 2.60)

Calculating with a calculator, we get:

α ≈ -15.707 radians/second²

Therefore, the angular acceleration of the CD as it spins to a stop is approximately -15.707 radians/second² (negative because the CD is slowing down).