# Rotational motion

1. May 23, 2008

### Jason03

heres the problem im working on

http://img55.imageshack.us/img55/8568/urgei5.jpg

I got the first part but I cant get the correct answer to how many revolutions the rotor executes before coming to rest....

I tried using the kinematic equations but cant come up with 27,900 revolutions, which is the answer...

any ideas?

2. May 23, 2008

### Ed Aboud

3. May 23, 2008

### Jason03

4. May 23, 2008

### rock.freak667

So you know $\omega=9300RPM$

which means that in 1 min. there are 9300 revolutions, OR in 60s there are 9300 revs.
How much in 1s now?

and also, you didn't really need to convert the angular velocity to rad/s and time to seconds since they gave you it in RPM and minutes respectively.

5. May 23, 2008

### alphysicist

Hi Jason03,

I think the kinematic equation should give you 27900 revolutions. Can you post what numbers you used in the kinematic equations that gave you a different $(\Delta\theta)$? A common error when solving for theta is to forget to make the angular acceleration negative when it is slowing down (the angular acceleration and initial angular velocity need to have opposite signs) but without seeing the numbers you used there's no way to tell.

6. May 24, 2008

### Jason03

i tried using

theta = 0 + 973.8 -.5(-2.7)(360)^2

Last edited: May 24, 2008
7. May 24, 2008

### Jason03

i figured out the problem....

the equ. is

$$\omega^2 = \omega_{o} + 2\alpha(\theta-\theta_{o})$$

$$973.8^2 = 0 -2(2.7)(\theta)$$

$$\theta = -175.6 e ^3$$

The problem I was having is converting the last part from radians to revolutions.....I realized all I had to do is divide the 175.6e^3 by$$2\pi$$........however when I entered that into my calculator like this $$175.6e3/2\pi$$......I came up with the wrong answer.....I had to enter it as $$175.6e3/(2\pi)$$.....which gives you 27.9e^3....:)