Angular Velocity of Ice Skater Around a Pole

[Gwozdzilla]

Homework Statement

An Ice skater of mass m = 70kg is initially traveling at a speed v = 4 m/s along a straight path that brings his center of mass to within a distance b = .6m of a vertical pole fixed in the ice. He reaches out with his hand as he passes the pole and hangs on so that he pulls himself into a circular path of radius r = .8m (measured from the axis of rotation to his center of mass) around the pole. If the skater's moment of inertia about his center of mass is I = 1.40 kgm² , what is his final angular velocity ω about the pole, assuming that no torques act during this maneuver?

Homework Equations

L_i = L_f
I_iω_i = I_fω_f
or maybe...
L = mvr = Iω
or possibly...
mv_ir_i = mv_fr_f

or if energy is conserved...
.5mv² = .5Iω²

I'm mostly confused about which formula to use.

The Attempt at a Solution

mvr_i = Iω
(70)(4)(.6) = (1.4)ω
ω = 120 rad/s
But I thought that this was a relatively unreasonable speed for the skater to be going, so I tried this...

.5(70)(4)² = .5(1.4)ω²
ω = 28.3 rad/s

But I don't think energy is conserved, so I also tried this...

mvr = mvr
(70)(4)(.6) = (70)(v)(.8)
v = 2.14 m/s
v=ωr
3 = ω(.8)
ω = 3.75 rad/s

This seemed reasonable to me, but I didn't use the moment of inertia, and I have nothing to check my work against. How do I know which formula is the correct one to use in this situation?

 

[haruspex]

mvr_i = Iω
(70)(4)(.6) = (1.4)ω

That I is the MoI about the skater's centre of mass, but the rotation is about the pole. How do you correct for that?

 

[Gwozdzilla]

I = I_cm + Md²
I = (1.4) + (70)(.8)²
I = 46.2 kgm²

mvr = Iω

(70)(4)(.6) = (46.2)ω
3.64 rad/s = ω

Would this be the correct answer?

 

[haruspex]

I = I_cm + Md²
I = (1.4) + (70)(.8)²
I = 46.2 kgm²

mvr = Iω

(70)(4)(.6) = (46.2)ω
3.64 rad/s = ω

Would this be the correct answer?

Looks right to me.