(a) 1200 rad/s
The angular acceleration of the rotor is given by:
[tex]\alpha = \frac{\omega_f - \omega_i}{t}[/tex]
where we have
[tex]\alpha = -80.0 rad/s^2[/tex] is the angular acceleration (negative since the rotor is slowing down)
[tex]\omega_f [/tex] is the final angular speed
[tex]\omega_i = 2000 rad/s[/tex] is the initial angular speed
t = 10.0 s is the time interval
Solving for [tex]\omega_f[/tex], we find the final angular speed after 10.0 s:
[tex]\omega_f = \omega_i + \alpha t = 2000 rad/s + (-80.0 rad/s^2)(10.0 s)=1200 rad/s[/tex]
(b) 25 s
We can calculate the time needed for the rotor to come to rest, by using again the same formula:
[tex]\alpha = \frac{\omega_f - \omega_i}{t}[/tex]
If we re-arrange it for t, we get:
[tex]t = \frac{\omega_f - \omega_i}{\alpha}[/tex]
where here we have
[tex]\omega_i = 2000 rad/s[/tex] is the initial angular speed
[tex]\omega_f=0[/tex] is the final angular speed
[tex]\alpha = -80.0 rad/s^2[/tex] is the angular acceleration
Solving the equation,
[tex]t=\frac{0-2000 rad/s}{-80.0 rad/s^2}=25 s[/tex]
