How Many Revolutions Does a Slowing Ceiling Fan Make?

[cbrown7]

A ceiling fan is rotating 1.0 rev/s. when turned off, it slows uniformly to a stop in 2.4 min.
How many revolutions does the fan make in this time?
Using the result from the above question, find the number of revolutions the fan must make for its speed to decrease from 1.0 rev/s to 0.50 rev/s.

Help!
thx

 

[Kurdt]

Have a read of this page and then try the question.

http://hyperphysics.phy-astr.gsu.edu/hbase/rotq.html

 

[thomate1]

Based on the given information, the ceiling fan makes 144 revolutions in 2.4 minutes (2.4 min x 60 s/min x 1 rev/s = 144 rev).

To find the number of revolutions needed for the fan to decrease from 1.0 rev/s to 0.50 rev/s, we can use the equation:

(vf)^2 = (vi)^2 + 2ad

Where vf is the final velocity (0.50 rev/s), vi is the initial velocity (1.0 rev/s), a is the acceleration (unknown), and d is the distance (unknown).

We know that the fan slows uniformly, so the acceleration is constant. We also know that the initial velocity is 1.0 rev/s and the final velocity is 0.50 rev/s.

Plugging in these values, we get:

(0.50)^2 = (1.0)^2 + 2a(144)

Simplifying, we get:

0.25 = 1 + 288a

Solving for a, we get:

a = -0.00347 rev/s^2

Now, we can use this value for acceleration in the equation:

vf = vi + at

Where vf is the final velocity (0.50 rev/s), vi is the initial velocity (1.0 rev/s), a is the acceleration (-0.00347 rev/s^2), and t is the time (unknown).

Plugging in these values, we get:

0.50 = 1.0 + (-0.00347)t

Solving for t, we get:

t = 287.99 s

Therefore, the fan would make approximately 17,279.4 revolutions (287.99 s x 60 s/min x 1 rev/s) to decrease from 1.0 rev/s to 0.50 rev/s.

I hope this helps! If you need further clarification or have any other questions, please let me know.