If a fan is disengaged from the motor, how long does it take for it to slow down to 1/2 of its original speed. Assuming the only work done during coasting is to accelerate the air the fan pulls in to its exit velocity. Friction loss due to bearings, etc will be ignored.
Energy of Rotor = E_k=1/2 Iω^2 where I=Inertia of rotor, ω=angular velocity
Work done by fan = E_d= 1/2 mv^2 where m = mass of air being blown out, v = air exit velocity.
Each revolution of the fan rotor moves a fixed volume of air, so v is a fuction of ω.
m is a fuction of ω and time. ∆m = kω∆t, k is a constant
The Attempt at a Solution
Rate change of energy of rotor = ∆(Work done by fan) =1/2 ∆mv^2
How do I proceed?