Answer:
20.13841 rad/s²
Explanation:
[tex]\omega_i[/tex] = Initial angular velocity = [tex]500\times \frac{2\pi}{60}\ rad/s[/tex]
[tex]\omega_f[/tex] = Final angular velocity = 0
t = Time taken = 2.6 s
[tex]\alpha[/tex] = Angular acceleration
Equation of rotational motion
[tex]\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\frac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\frac{0-500\times \frac{2\pi}{60}}{2.6}\\\Rightarrow \alpha=-20.13841\ rad/s^2[/tex]
The magnitude of the angular acceleration of the CD, as it spins to a stop is 20.13841 rad/s²
