# Conservation of Angular Momentum-Ice Skater

1. Nov 5, 2013

### Olp1217

Conservation of Angular Momentum--Ice Skater

1. The problem statement, all variables and given/known data
Real-life situations are generally complicated, but some can be approximately analyzed by using simple models. Such a model for a skater’s spin is shown, with a cylinder and rods representing the skater. In (a), the skater goes into the spin with the “arms” out, and in (b) the “arms” are over the head to achieve a faster spin by conservation of angular momentum. If the initial spin rate is 1 revolution per 1.5 s, what is the angular speed when the arms are tucked in?

The cylinder represents the skater's torso and has radius r = 20 cm, mass M = 75 kg. The two rods represent the skater's arms and have length L = 80 cm, mass m = 5 kg.

2. Relevant equations
Conservation of angular momentum (L) = I(1)ω(1) = I(2)ω(2)
Moment of Inertia of cylinder = (1/2)MR^2
Angular velocity = 2∏/T
(where T = period)
Parallel axis theorem = I = Icm + Md^2

3. The attempt at a solution
Where i indicates before her arms have been pulled in and f indicates after her arms have been pulled in:

ωi = 2∏/T = 2∏/1.5 = 4.189 rad/s.
Ii(cylinder) = .5Mr^2 = .5(75)(.2)^2 = 1.5 kg m^2
Ii(rod) = I(cylinder) + md^2 = 1.5 + 5 ((.8/2)+.2)^2 = 3.3 kg m^2
Ii(system) = 1.5 + 2(3.3) = 8.1 kg m ^2
La = Iiωi = (4.189)(8.1) = 33.931 kg m^2 / s

If(rod) = I(cylinder) + md^2 = 1.5 + 5(.2)^2 = 1.7 kg m^2
If(system) = 1.5 + 2(1.7) = 4.9 kg m^2

Iiωi = Ifωf => 33.931 = 4.9 (ωf)

I feel like I'm missing something huge and obvious, but for the life of me I can't see it. Please help!
According to my professor, the correct answer is somewhere between 12.5 and 13 radians/s (accounting for rounding)

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Last edited: Nov 5, 2013
2. Nov 5, 2013

### voko

Explain why you compute MoI of the rods that way, both cases.

3. Nov 5, 2013

### Olp1217

I managed to get the right answer, actually. I had been calculating the moment of inertia incorrectly. Here's the right solution:

Ii = I(cylinder) + 2( (1/12) mL^2 + md^2) where d = (L/2) + r
= 1.5 + 2 ( (1/12)(5)(.8)^2 + (5)(.6)^2)
= 5.633 kg m^2
La = Ii ωi = (5.633)(4.189) = 23.598 rad/s

If = I(cylinder) + 2(md^2) where d = r
= 1.5 + 2((5)(.2)^2)
= 1.9 kg m^2

Ii ωi = If ωf => 23.598 = 1.9 ωf