# Help understanding this Kinetic energy problem

1. Jun 10, 2013

### urbano

1. The problem statement, all variables and given/known data
A 60kg ice skater is spinning around at 300 degrees per second. His radius of gyration is .5m. As he pulls his arms in he does 300 J of work.

i.What is his initial kinetic energy?
ii.What is his final kinetic energy ?
iii.Assuming that his new radius of gyration is .32m, what is his final moment of inertia ?
iv.What is his final angular momentum ?

2. Relevant equations
1/2mv^2 Kinetic energy = 1/2 mass X velocity squared
I= mk^2 Inertia = mass X radius of gyration squared
I x ω Angular momentum = inertia X angular velocity

3. The attempt at a solution
1/2 X 60kg X 5.24 radians squared = 157.2 J

ii. I just added 300J on here as I assumed if the skater done 300J of work as he pulled the arms in I'd just add this on. 457.2 J

iii. Final moment of inertia

inertia = 60kg X .32m^2 = 6.144 kg.m

iV. final angular momentum:

= I X ω
= 6.144kg.m X ω

ω= ..?

or could I go
KE = 1/2m X V^2
so KE = 1/2 I x ω^2
457.2J = 1/2 I x ω^2
√ω = √(457.2/3.07)

2. Jun 10, 2013

### SteamKing

Staff Emeritus
Rotational KE = 1/2*J*ω^2, where ω = angular velocity in rad/s and J = moment of inertial in kg-m^2

3. Jun 10, 2013

### rcgldr

It appears the problem statement may be slightly off. By my math, the skater does about 296.37728 J of work.

4. Jun 10, 2013

### urbano

although I dont understand the question properly , they did say in ithe question the skater does 300J of work, so it looks liek you'd be on the right track rcgld "By my math, the skater does about 296.37728 J of work."

I just have no ideas if my answers are on the right path or not.

5. Jun 11, 2013

### haruspex

As SteamKing points out, you've left out the radius of gyration.
kg.m2.
Can you think of a useful conservation law that would apply?