Question

An electric fan is turned off, and its angular velocity decreases uniformly from 460 rev/min to 210 rev/min in a time interval of length 4.15 s . a)Find the angular acceleration in rev/s2. b)Find the number of revolutions made by the motor in the time interval of length 4.15 s . c)How many more seconds are required for the fan to come to rest if the angular acceleration remains constant at the value calculated in part A?

An airplane propeller is 1.83 m in length (from tip to tip) with mass 128 kg and is rotating at 2800 rpm (rev/min) about an axis through its center. You can model the propeller as a slender rod.

a)What is its rotational kinetic energy?

b)Suppose that, due to weight constraints, you had to reduce the propeller's mass to 75.0% of its original mass, but you still needed to keep the same size and kinetic energy. What would its angular speed have to be, in rpm?

Answer 1

Here,

for the electric fan ,

initial angular speed, wi = 460 rev/min

wi = 7.67 rev/s

final angular speed , wf = 210 rev/min

wf = 3.5 rev/s

a) Using first equation of motion

angular acceleration ,a = (wf - wi)/t

a = (3.5 - 7.67)/4.15

a = -1.005 rev/s^2

the angular acceleration is -1.005 rev/s^2

b)

Using seocnd equation of motion

theta = wi * t + 0.5 * a * t^2

theta = 7.67 * 4.15 - 0.5 * 1.005 * 4.15^2

solvnig

theta = 23.17 revs

the number of revolutions made are 23.17 .

c)

let the time taken to stop is t

Using first equation of motion

wf = wi + a * t

0 = 3.5 - 1.005 * t

t = 3.48 s

it will take 3.48 secnds more for the fan to stop

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mass ,m = 128 kg

length , L = 1.83 m

angular speed , w = 2800 rpm

w = 293.16 rad/s

rotational kinetic energy = 0.5 * I * w^2

rotational kinetic energy = 0.5 * m * L^2/12 * w^2

rotational kinetic energy = (0.5 * 128 * 1.83^2/12) * 293.16^2

rotational kinetic energy = 1535006.9 J

the rotational kinetic energy of propeller is 1535006.9 J