contestada

Blades of an airplaneâs engine rotate with an initial speed is 40 rev/min. Assuming constant angular acceleration of magnitude 8 rad/s^2 . (a) how long does it take for the blades to reach the angular speed Ïf = 120 rev/ min?
(b)Through how many radians does a blade turn during the time found in (a).
Note : Write down the detail process

Respuesta :

Answer:

Part a)

t = 1.05 s

Part b)

[tex]\theta = 8.78 rad[/tex]

Explanation:

Initial angular speed is given as

[tex]\omega_i = 40 rev/min = 0.66 rev/s[/tex]

[tex]\omega_i = 2\pi (0.66) = 4.19 rad/s[/tex]

angular acceleration is given as

[tex]\alpha = 8 rad/s^2[/tex]

now we have

part a)

final angular speed = 120 rev/min

[tex]\omega_f = 2\pi(\frac{120}{60} rev/s)[/tex]

[tex]\omega_f = 12.57 rad/s[/tex]

now by kinematics we have

[tex]\omega_f = \omega_i + \alpha t[/tex]

[tex]12.57 = 4.19 + 8 t[/tex]

[tex] t = 1.05 s[/tex]

Part b)

Angle turned by the blades is given by

[tex]\theta = \omega_i t + \frac{1}{2}\alpha t^2[/tex]

[tex]\theta = 4.19(1.05) + \frac{1}{2}(8)(1.05)^2[/tex]

[tex]\theta = 8.78 rad[/tex]