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Classical example of conservation of angular momentum

  1. Dec 6, 2011 #1
    Trying to get the idea behind this, but it's kind of new thinking all this rotational dynamics.
    The classical example of conservation of angular momentum is when a ballerina pulls in her arms as she spins to spin faster. The angular momentum theorem tells us:
    I1*α1 = I2*α2
    So as she decreases her moment of intertia from I1 to I2 the angular acceleration MUST increase from α1 to α2 due to conservation of angular momentum. Now an increase in angular acceleration must mean that a torque has acted on her - or must it? I'm not really sure, since the moment of inertia isn't the same before and after.
    So I wanted to know what role the torques play in this situation. Like in linear dynamics, an example of conservation of linear momentum is often if you throw something heavy away from you. Then according to Newtons 3rd law the force you exert on the heave object it exerts back such that you also get a velocity in the opposite direction. Here a force argument is used, so can't the same be done for the rotational example above?
     
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  3. Dec 6, 2011 #2

    rcgldr

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    A better example would be an ice skater pulling their arms inward to spin faster since there's much less friction with the ice.

    As the arms are drawn inwards, the hands follow an inwards spiral path, so part of the inwards force is in the direction of velocity, which increases the speed. That increase in speed will correspond to angular momentum being conserved. There's internal work done, equal to two times the inwards force times the distance each arm is moved in, and the increase in angular kinetic energy will equal the internal work done. Torque occurs because not all of the mass is located in the skater's hands. As the skater's outer parts (hands and wrists) attemp to increase speed and rate of rotation, the inner parts resist this due to angular inertia, which results in internal torques that keep all parts (eventually) rotating at the same rate.

    As a simpler example, imagine a puck sliding on a frictionless surface, attached to a massless string that goes into a tube at the center of rotation. Angular momentum is also preserved in this case. Again if the string is pulled inwards or the string is allowed to be pulled outwards, the puck goes into a spiral path and part of the tension force is in the direction of velocity, increasing or decreasing the speed of the puck. The work done by the string (and whatever pulls the string through the tube) equals tension times radial distance moved, and equals the change in angular energy of the puck. Here's a link to an image of this situation. The short inward lines are perpendicular to the path, while the long lines represent the string under tension.

    puck_with_string_pulled_through_tube.jpg
     
    Last edited: Dec 6, 2011
  4. Dec 6, 2011 #3
    Hmm wouldnt the velocity always be tangent to the inwards force? I was not thinking about the case where you would pull your arms down in a circle, rather the case where you could pull them in in a straight, horizontal line. We could also just imagine that you had two dumbbels in your hands and out of nowhere chose to drop them. That would certainly decrease the moment of inertia => increase angular acceleration, but would there be any torque in this case?
     
  5. Dec 6, 2011 #4

    BruceW

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    Your equation is not correct. It should be:
    [tex]I_1 \omega_1 = I_2 \omega_2[/tex]
    since the angular momentum is equal to [itex]I \omega[/itex] and the angular momentum is conserved. (Around a vertical axis, of course)
     
  6. Dec 6, 2011 #5

    AlephZero

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    In that case there is no change in angular momentum of the complete system. When you let go of the weights, they don't drop vertially, they also move along the tangent of the circle they were moving around. Ignoring air resistance, the velocity along the tangent stays constant (untill they hit the ground of course).

    You will continue to rotate at the same speed before and after releasing the weights.
     
  7. Dec 6, 2011 #6

    Ken G

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    Also, there is no torque if there is no friction. The skater uses torque to get spinning, but after that the conservation of angular momentum means there is no torque-- no torque is required to pull her arms in, and that's why angular momentum is conserved.
     
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