Solving Rotational Motion Problem: Get Answer for Revolutions

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
6 replies · 2K views
Jason03
Messages
161
Reaction score
0
heres the problem I am working on

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

I got the first part but I can't get the correct answer to how many revolutions the rotor executes before coming to rest...


I tried using the kinematic equations but can't come up with 27,900 revolutions, which is the answer...

any ideas?
 
Last edited by a moderator:
Physics news on Phys.org
Can you show your attempt at the question please?
 
http://img502.imageshack.us/img502/2429/rolg1.jpg
 
Last edited by a moderator:
Jason03 said:
http://img502.imageshack.us/img502/2429/rolg1.jpg

So you know [itex]\omega=9300RPM[/itex]

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.
 
Last edited by a moderator:
Hi Jason03,

Jason03 said:
heres the problem I am working on

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

I got the first part but I can't get the correct answer to how many revolutions the rotor executes before coming to rest...


I tried using the kinematic equations but can't come up with 27,900 revolutions, which is the answer...

any ideas?

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 [itex](\Delta\theta)[/itex]? 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.
 
Last edited by a moderator:
i tried using


theta = 0 + 973.8 -.5(-2.7)(360)^2
 
Last edited:
i figured out the problem...

the equ. is

[tex]\omega^2 = \omega_{o} + 2\alpha(\theta-\theta_{o})[/tex]

[tex]973.8^2 = 0 -2(2.7)(\theta)[/tex]

[tex]\theta = -175.6 e ^3[/tex]

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[tex]2\pi[/tex]...however when I entered that into my calculator like this [tex]175.6e3/2\pi[/tex]...I came up with the wrong answer...I had to enter it as [tex]175.6e3/(2\pi)[/tex]...which gives you 27.9e^3...:)