Moment of inertia and Kinetic Energy (Rotational Motion)

miamirulz29

1. The problem statement, all variables and given/known data
A skater spins with an angular speed of 8.3 rad/s with her arms outstretched. She lowers her arms, decreasing her moment of inertia by a factor of 8.9. Ignoring the friction on the skates, determine the percent of change in her kinetic energy. Answer in percent.


2. Relevant equations
KE = (1/2)Iw^2
Li = Lf

[b]3. The attempt at a solution[/b
wi = omega initial

wf = omega final

If Li = Lf,

Iwi = (I/8.9)wf

8.9wi = wf

So KEi = (1/2)I(wi^2)

KEf = (1/2)(I/8.9)(wf^2)

So KEf = (1/2)(I/8.9)(8.9wi^2)

So I(wi^2)/2 = I(8.9wi^2)/17.8

Does that mean there is no change in kinetic energy? Also, if there is a change in kinetic energy how do I convert that to a percent?

miamirulz29

So I just realized I could simplify the last part to:

So is the percent change 4.45 /2 = 2.225 = 222.5 %?

miamirulz29

Anybody?

nrqed

You forgot to square the factor of 8.9 that comes with omega_f.
It should be (8.9 omega_i)^2 , not 8.9 omega_i^2

miamirulz29

I didn't correct it in my second post?

miamirulz29

wait I am so lost in this problem, can somebody help me please?

miamirulz29

Oh wait I just figured it out. If wf = 8.9wi, then (8.9wi - wi)/wi = 7.9 *100, 790%