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Conservation of angular momentum in a spinning figure skater

  1. Mar 12, 2012 #1
    1. The problem statement, all variables and given/known data
    The outstretched hands and arms of a figure skater preparing for a spin can be considered a slender rod pivoting about an axis through its center. When the skater's hand and arms are brought in and wrapped around his body to execute the spin, the hands and arms can be considered a thin-walled hollow cilinder. His hands and arms have a combine and arms have a combined mass of 8kg. When outstreched, they span 1.8m; when wrapped, they form a cylinder of radius 25cm. The moment of inertia about the axis of rotation of the remainder of his body is constant and equal to .40kg*m^2. If the skater's original angular speed is .40rev/s, what is his final angular speed?


    2. Relevant equations
    moment of inertia for slender rod, axis through center I= 1/12ML^2
    moment of inertia for hollow cylinder 1/2M(R_1^2 + R_2^2)
    conservation of angular momentum I_i*ω_i = I_f*ω_f


    3. The attempt at a solution
    .4rev/s = 2.51 rad/s
    I_i = 1/2*8*1.8^2 = 12.96
    ω_i = 2.51
    I_f = 1/2*8*(.9^2 + .125^2) = 3.3 +.4 = 3.7

    putting those numbers in the conservation of energy forumla and i get ω_f = 8.7 rad/s. the back of my book says the answer is 7.16 rad/s.
     
  2. jcsd
  3. Mar 12, 2012 #2
    First of all you do not use conservation of energy for this problem. You use conservation of angular momentum.

    I_i = 1/2*8*1.8^2 = 12.96
    Above is not correct formula. You miscopied it from your previous.

    Secondly you need the formula for a thin cylindrical shell, not a hollow cylinder. You do not know the inner radius so how can you use it?
     
    Last edited: Mar 12, 2012
  4. Mar 12, 2012 #3
    im still off.
    I_i = 2.16
    I_f = M(R^2) = .5 + .4 = .9 i used thin walled hollow cylinder
    2.16*2.5= 5.42
    5.42 = .9ω_f
    ω_f = 6.02rad/s
    did i do something wrong? am i suppose to use thin walled hollow sphere? it gets me 7.39
     
    Last edited: Mar 12, 2012
  5. Mar 12, 2012 #4
    What about inertia of torso? You forgot it....
     
    Last edited: Mar 12, 2012
  6. Mar 12, 2012 #5
    is there an equation for that or is it already given in the problem. my book says nothing about inertia of torso
     
  7. Mar 12, 2012 #6
    Torso means body!
     
  8. Mar 12, 2012 #7
    Habeas corpus!
     
  9. Mar 12, 2012 #8
    oh well it doesnt matter. i just took my test tonight and probably failed.
     
  10. Mar 13, 2012 #9
    You omitted the rotational inertia of the body when the arms are outstreached from the calculation. The 2.16 should be 0.40+ 2.16 = 2.56.
     
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