Angular Velocity of Ice Skater Around a Pole

  • Thread starter Gwozdzilla
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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


Li = Lf
Iiωi = Ifωf
or maybe...
L = mvr = Iω
or possibly...
mviri = mvfrf

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

I'm mostly confused about which formula to use.

The Attempt at a Solution



mvri = 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?
 

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  • #2
haruspex
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mvri = 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?
 
  • #3
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I = Icm + 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?
 
  • #4
haruspex
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I = Icm + 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.
 

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