Question

An electric motor is turned off, and its angular velocity decreases uniformly from 900 rev/min to 400 rev/min in 6.00 s. (a) Find the angular acceleration and the number of revolutions made by the motor in the 6.00-s interval. (b) How many more seconds are required for the motor to come to rest if the angular acceleration remains constant at the value calculated in part (a)?

Answer 1

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Initial angular velocity ω1 = 900 rev/min

                                        = 15 rev/s

Final angular velocity ω2 = 400 rev/min

                                       = 6.667 rev/s

Time taken t = 6 s

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(a)

Angular acceleration of the fan

α = (ω2 - ω1) / t

α = (6.667-15)/6

α = - 1.3889 rev/s^2

Total displacement

θ = ω1*t + (1/2)αt^2

= 15*6 + 0.5 * (-1.3889) * 6^2

= 64.9999 rev

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(b)

Time taken to comes to rest is

t = (ω2 - 0) / α = (6.667-0)/1.3889

    = 4.80 s

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