Answer:
8.37758 rad/s²
Explanation:
[tex]\omega_i[/tex] = Initial angular velocity = [tex]20\times \frac{2\pi}{5}[/tex]
[tex]\omega_f[/tex] = Final angular velocity = 0
[tex]\theta[/tex] = Angle of rotation = [tex]6\times 2\pi\ rad[/tex]
Equation of rotational motion
[tex]\omega_f^2-\omega_i^2=2\alpha \theta\\\Rightarrow \alpha=\frac{\omega_f^2-\omega_i^2}{2\theta}\\\Rightarrow \alpha=\frac{0^2-(20\times \frac{2\pi}{5})^2}{2\times 6\times 2\pi}\\\Rightarrow \alpha=-8.37758\ rad/s^2[/tex]
The the magnitude of the angular acceleration of the salad spinner as it slows down is 8.37758 rad/s²
