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

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Answer:

(a) ωf = 1200 rad/s

(b) t  = 25 s

Explanation:

Kinematic Equation to the rotor

ωf= ω₀ + α*t  Formula (1)

Where:

α : angular acceleration (rad/s²)

t : time interval (s)

ω₀ : initial angular velocity ( rad/s)

ωf : final angular velocity  ( rad/s)

(a) Angular velocity after 10.0 s

Data

α = -80 rad/s²

t = 10 s

ω₀ = 2000 rad/s

We apply the formula (1):

ωf = ω₀ - α*t

ωf = 2000 -  80*(10)

ωf = 2000 - 800

ωf = 1200 rad/s

(b) Time the rotor stops

Data

α = -80 rad/s²

ω₀ = 2000 rad/s

ωf = 0

We apply the formula (1):

ωf = ω₀ - α*t

0 = 2000 -  80*(t)

80*(t) = 2000

t = 2000 / 80

t  = 25 s

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