Question

As you finish listening to your favorite compact disc (cd), the cd in the player slows down to a stop. assume that the cd spins down with a constant angular acceleration. part a if the cd rotates clockwise at 500 rpm (revolutions per minute) while the last song is playing, and then spins down to zero angular speed in 2.60 s with constant angular acceleration, what is α, the magnitude of the angular acceleration of the cd, as it spins to a stop?

Answer 1

Remember that the angular acceleration is the rate at which the angular velocity changes. Given that the angular acceleration is constant then we have tho following formula relating the angular acceleration $\alpha$ and the angular velocity $\omega$:

$\alpha =\frac{\omega_f-\omega_i}{\Delta t}$

Where $\omega_f$ is the final angular velocity and $\omega_i$ is the initial angular velocity. $\Delta t$ is the time the rotation took from the initial angular velocity to reach the final angular velocity.
Let's convert $\omega_i$ to rotations per second.
If we have 500 rotations per minute and a minute has 60 seconds we can apply a simple rule of three to get the answer:

$500 rotations\to60sec\\?rotation\to1sec$
We get the following for $\omega_i$.
$\omega_i=\frac{50}{6}$
Finally the angular acceleration is:
$\alpha=\frac{0-\frac{50}{6}}{2.60}=-\frac{50} {6\times2.60}=-3.21rotations/sec^2$
The result is a negative value but this is expected as the rotating motion is deceletating. In the case of an increasing angular velocity we would have a positive angular acceleration.