Question

An electric fan is turned off, and its angular velocity decreases uniformly from 550 rev/min to 180 rev/min in a time interval of length 4.30 s.
A.) Find the angular acceleration in revolutions per second per second.
= -1.43 rev/s^2
B.) Find the number of revolutions made by the fan blades during the time that they are slowing down in Part A.
= _________? rev
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?
= ____________? s

Answer 1

Answer:

(a) $\alpha =-1.43rev/sec^2$

(b) 4.17 rev

(c) 6.40 sec

Explanation:

We have given the angular velocity of fan decreases from 550 rev/min to 180 rev/min

So initial angular speed $\omega _0=550rev/min=\frac{550}{60}=9.166rev/sec$

Final angular speed $\omega =180rev/min=\frac{180}{60}=3rev/sec$

Time is given as t = 4.3 sec

(a) We know that angular acceleration is given by $\alpha =\frac{\omega -\omega _0}{t}=\frac{3-9.166}{4.3}=-1.43rev/sec^2$

(b) We know the relation \Theta =\omega _0t+\frac{1}{2}\alpha t^2=9.166\times 4.3-\frac{1}{2}\times 1.43\times 4.3^2=26.1935rad=\frac{26.1935}{2\pi }=4.17rev

(c) Now final angular velocity $\omega =0rev/sec$

So time $t=\frac{0-9.166}{-1.43}=6.40sec$