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Angular Momentum: Puck Spinning

  1. Jan 31, 2015 #1
    1. The problem statement, all variables and given/known data
    The puck in the figure shown below has a mass of 0.120 kg. Its original distance from the center of rotation is 40.0 cm, and it moves with a speed of 80.0 cm/s. The string is pulled downward 15.0 cm through the hole in the frictionless table. Determine the work done on the puck.

    4gWOb2a.png
    2. Relevant equations

    ## L = I \omega##
    ##KE = 1/2 \cdot I \cdot \omega^2 ##
    3. The attempt at a solution

    Consider angular momentum. Note it's conserved.

    Therefore, ## \frac{1}{2}mr_i^2 \omega_i = \frac{1}{2} mr_f^2 \omega_f \Rightarrow r_i^2 \omega_i =r_f^2 \omega_f ##

    Note that ##\omega_i = 80 cm/s = 2 rad/s## (because 2pi radians is one revolution, which is 80pi cm, so 1 radian is 40 cm)

    Threrefore, ##0.40^2 \cdot 2 = 0.25^2 \cdot \omega_f \Rightarrow \omega_f = 5.12##

    Note that Work Done is the change in kinetic energy. KE initial is equal to ## 1/2 (1/2 \cdot 0.12 \cdot 0.4*0.4) \cdot 2^2 ## (1/2 Iw^2 = 1/2 (1/2mr^2)w^2). KE final is simialrly equal to ## 1/2 (1/2 \cdot 0.12 \cdot 0.25 \cdot 0.25 \cdot 5.12^2##.

    We can do the calculation, to get the change as 0.029952. But the book says answer is 5.99*10^-2.

    What's wrong?
     
  2. jcsd
  3. Feb 1, 2015 #2

    Nathanael

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    What's the (extra) factor of 1/2 for? (I see an extra one in both initial and final KE's)

    You don't even need to bother with angular velocities. (You can solve it that way if you'd like, but it's redundant.) You can just use the angular momentum of a particle = mvr
     
  4. Feb 1, 2015 #3

    tms

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    Do you see the discrepancy between the two quoted lines?
     
    Last edited: Feb 1, 2015
  5. Feb 1, 2015 #4
    How do you get that angular momentum of a particle = mvr

    @tms, EDIT: No, the kinetic energy isn't being calculated there. That's the conservation of angular momentum, which is Iw.

    FINAL EDIT: I see the problem now. Moment of inertia is mr^2, not 1/2 mr^2.
     
  6. Feb 1, 2015 #5

    Nathanael

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    Angular momentum of a particle equals Iω, but I=mr2, so the angular momentum is mωr2
    Since, ωr=v, the angular velocity is mvr
     
  7. Feb 1, 2015 #6

    haruspex

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    You mean, the angular momentum is mvr, yes?
     
  8. Feb 1, 2015 #7

    Nathanael

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    Oops! Mindless typo, thanks.
     
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