# Angular acceleration of a fan

• valeriex0x
In summary, the electric fan's angular velocity decreases uniformly from 500 rev/min to 200 rev/min in 4.00 s, resulting in an angular acceleration of -1.25 rev/s^2. However, to determine the number of revolutions made by the motor in the 4.00 s interval, the initial angular velocity is needed.

#### valeriex0x

angular acceleration of a fan!

## Homework Statement

An electric fan is turned off, and its angular velocity decreases uniformly from 500 rev/min to 200 rm rev/min in 4.00 s.
Find the angular acceleration in rev/s and the number of revolutions made by the motor in the 4.00 interval.

## The Attempt at a Solution

I got -1.25 for alpha in rev/sec^2, which is correct.

How come the number of revolutions isn't 5 revolutions?
(alpha) (time)=
(-1.25)(4 sec)=5 rev (-5) rev

Acceleration x time = velocity, not distance!

ohhhhhhhhhhh... ok thanks. hmmm... I'm not sure what formula i should use. i was going to try x=xo+vot+1/2at^2 but i can;t. .....

valeriex0x said:
ohhhhhhhhhhh... ok thanks. hmmm... I'm not sure what formula i should use. i was going to try x=xo+vot+1/2at^2 but i can;t. .....

Why not? (assuming of course that you'd use the angular form of the equation)

i don't have xo right?
we didn't cover this yet...

valeriex0x said:
i don't have xo right?
we didn't cover this yet...

Assume xo is zero. The count of revolutions (distance) begins, and the clock starts ticking, at the instant the fan is turned off.

okay so :

0+(-1.25 rev/s^2)(4 sec)+ 1/2(-1.25 rev/s^2)(4^2)
x=-5.00+ -12.50
x=-17.50

how do i make this angular?

valeriex0x said:
okay so :

0+(-1.25 rev/s^2)(4 sec)+ 1/2(-1.25 rev/s^2)(4^2)
x=-5.00+ -12.50
x=-17.50

how do i make this angular?

You're using angular values, so it's angular.

But you've got a problem. You're using an acceleration where the initial angular velocity is required.