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Conservation of Angular MomentumIce Skater
Reallife 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.
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
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)
ωf = 6.925 rad/s
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)
Homework Statement
Reallife 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.
Homework 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
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)
ωf = 6.925 rad/s
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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