Answer:
[tex]723.2 rad/s^2[/tex]
Explanation:
The angular acceleration of an object in rotation is equal to the rate of change of its angular velocity.
Mathematically:
[tex]\alpha = \frac{\omega_f - \omega_i}{t}[/tex]
where
[tex]\alpha[/tex] is the angular acceleration
[tex]\omega_i[/tex] is the initial angular velocity
[tex]\omega_f[/tex] is the final angular velocity
t is the time elapsed
For the optical disk in this problem, we have:
[tex]\omega_i = 411.5 rad/s[/tex] is the initial angular velocity
[tex]\omega_f=0[/tex] is the final angular velocity (because the disk comes to rest)
t = 0.569 s is the time elapsed
So, the angular acceleration is
[tex]\alpha=\frac{0-411.5}{0.569}=-723.2rad/s^2[/tex]
And since we are only interested in the magnitude, then the magnitude is [tex]723.2 rad/s^2[/tex]
