Question

An airliner arrives at the terminal, and its engines are shut off. The rotor of one of the engines has an initial clockwise angular speed of 2000 rad/s. The engine's rotation slows with an angular acceleration of magnitude 80.0 rad/s2. (a) Determine the angular speed after 10.0 s. (b) How long does it take for the rotor to come to rest?

Answer 1

(a) 1200 rad/s

The angular acceleration of the rotor is given by:

`\alpha = (\omega_f - \omega_i)/(t)`

where we have

`\alpha = -80.0 rad/s^2` is the angular acceleration (negative since the rotor is slowing down)

`\omega_f` is the final angular speed

`\omega_i = 2000 rad/s` is the initial angular speed

t = 10.0 s is the time interval

Solving for
`\omega_f`, we find the final angular speed after 10.0 s:

`\omega_f = \omega_i + \alpha t = 2000 rad/s + (-80.0 rad/s^2)(10.0 s)=1200 rad/s`

(b) 25 s

We can calculate the time needed for the rotor to come to rest, by using again the same formula:

`\alpha = (\omega_f - \omega_i)/(t)`

If we re-arrange it for t, we get:

`t = (\omega_f - \omega_i)/(\alpha)`

where here we have

`\omega_i = 2000 rad/s` is the initial angular speed

`\omega_f=0` is the final angular speed

`\alpha = -80.0 rad/s^2` is the angular acceleration

Solving the equation,

`t=(0-2000 rad/s)/(-80.0 rad/s^2)=25 s`