Answer:
22 revolutions
Explanation:
2 rev/s = 2*(2π rad/rev) = 12.57 rad/s
The angular acceleration when it starting
[tex]\alpha_a = \frac{\Delta \omega}{\Delta t} = \frac{12.57}{10} = 1.257 rad/s^2[/tex]
The angular acceleration when it stopping:
[tex]\alpha_o = \frac{\Delta \omega}{\Delta t} = \frac{-12.57}{12} = -1.05 rad/s^2[/tex]
The angular distance it covers when starting from rest:
[tex]\omega^2 - 0^2 = 2\alpha_a\theta_a[/tex]
[tex]\theta_a = \frac{\omega^2}{2\alpha_a} = \frac{12.57^2}{2*1.257} = 62.8 rad[/tex]
The angular distance it covers when coming to complete stop:
[tex]0 - \omega^2 = 2\alpha_o\theta_o[/tex]
[tex]\theta_o = \frac{-\omega^2}{2\alpha_o} = \frac{-12.57^2}{2*(-1.05)} = 75.4 rad[/tex]
So the total angular distance it covers within 22 s is 62.8 + 75.4 = 138.23 rad or 138.23 / (2π) = 22 revolutions