How Do You Calculate Kinetic Energy Changes in Rotational Motion?

[Noctix]

Homework Statement

[PLAIN]http://img576.imageshack.us/img576/6328/physics2.jpg

Homework Equations

KE(final)/Ke(initial)

The Attempt at a Solution

1/2mv²/1/2mv²

v(final)²/v(initial)²

v= wr

r(initial)=? and r(final)=?

I originally thought the r's would cancel, but they don't because they're different. what to i do with them?
the corect answer is simply w(final)/w(initial), which is 3.3ish, but I want to understand how they got to that.
thanks.

 

[Denken]

ok, so my understanding of this is kind of like a lever. You have a set amount of torque that is the same on both sides of the focal point. So Torque=Newtons x Distance. If one side is longer than the other (the professor with his arms out) the Newtons are smaller because the distance is great. while the shorter arm (professor with his arms in) has a shorter distance, meaning it will have greater force.
...
so 1.5rad/s x D(initial)=5rad/s x D(final) ... that's what i got out of it

 

[tiny-tim]

Welcome to PF!

Hi Noctix! Welcome to PF!

(have an omega: ω )

I originally thought the r's would cancel, but they don't because they're different. what to i do with them?

ah, you haven't used conservation of https://www.physicsforums.com/library.php?do=view_item&itemid=313" (rv, = r²ω) …

use that, and you should get the correct result ​

(and to account for the change in energy … ie, where did the energy come from or go to? … think about the https://www.physicsforums.com/library.php?do=view_item&itemid=75" )