Question

An electric fan is turned off, and its angular velocity decreases uniformly from 720 rev/min to 200 rev/min in a time interval of length 3.25
a.find the angular acceleration in rev/s
b.find the number of revolutions made by the motor in the time interval of length 3.25
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?

Answer 1

(a) first, we need to convert the initial and final angular speed into rev/s. Keeping in mind that 1 min=60 s, we should divide the speeds by 60:
$\omega_i = 720 rev/min = 12.0 rev/s$
$\omega_f = 200 rev/min = 3.3 rev/s$

So, the angular acceleration is
$\alpha = \frac{\omega_f-\omega_i}{\Delta t} = \frac{3.3 rev/s - 12.0 rev/s}{3.25 s}=-2.68 rev/s^2$
where the negative sign means it is a deceleration.

(b) The angle covered by the fan (in revolutions) is given by the following equation
$\theta(t) = \omega_i t + \frac{1}{2} \alpha t^2$
where t is the time and $\alpha$ the angular acceleration we find at point (a). Substituting numbers, we get
$\theta (3.25 s)=(12.0 rev/s)(3.25 s)+ \frac{1}{2} (-2.68 rev/s^2)(3.25 s)^2=24.85 rev$

(c) In order for the fan to come to rest, its final angular speed must go to zero:
$\omega(t)=0$

The expression for the angular speed as a function of time is 
$\omega(t) = \omega _i + \alpha t$
By requiring this is equal to zero, we find the time t after which the fan comes to rest:
$\omega_i + \alpha t = 0$
$t=- \frac{\omega_i}{\alpha}= -\frac{12.0 rev/s}{-2.68 rev/s^2} =4.48 s$