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Conservation of Angular Momentum bicycle problem

  1. Feb 12, 2013 #1
    1. The problem statement, all variables and given/known data
    In a demonstration, a bicycle wheel with moment of inertia = .37 kg*m^2 is spun up to 14 rad/s, rotating about a vertical axis. A student hold the wheel while sitting on a rotatable stool. The student and the stool are initially stationary and have a moment of inertia equal to 3.6 kg*m^2. If the student turns the bicycle wheel over so its axis point in the opposite direction, with what angular velocity will the student and stool rotate? Assume the wheel, student, and stool all have the same axis of rotation.

    2. Relevant equations
    Im not entirely sure if you need the equation for angular momentum, or just to know that it is conserved. Ill give the equation anyways.
    Angular Momentum Equation: L=I*ω

    Moment of Inertia Equation: I = m*r^2

    3. The attempt at a solution

    I figured that this question is all about the conservation of angular momentum. If I add the moment of inertia of the stool and student to the negative of the inertia for the bicycle wheel (because its now upside-down, right? ) I would get the total moment of inertia for the system. I have no idea how to proceed from here, becase the equation for moment of inertia is I = m*r^2, and I dont have an r to use. Maybe I need to look into the angular momentum equation some more (derivative of that equation to find the change of ω perhaps?)

    I feel like this should be an easy problem (this is a reading review problem supposedly) but its just not clicking for me. Any help is greatly appreciated.
  2. jcsd
  3. Feb 12, 2013 #2

    Doc Al

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    Staff: Mentor

    Treat the system as having two parts: (1) The wheel, and (2) The student and stool.

    Initially, only the wheel is rotating. So what's the total angular momentum of the system?

    What happens to the angular momentum of the wheel when it is inverted?
  4. Feb 12, 2013 #3
    If I treat this system as two parts, then the angular momentum is just that of the wheel in the beginning, .37 kg*m^2. So does the angular momentum become negative if we flip the wheel on its z axis, becoming -.37kg*m^2?

    this would mean that the angular velocity of the system would flip, becoming -14 rad/s right?

    Does the angular momentum of the student and the stool have any use in this case?
  5. Feb 12, 2013 #4

    Doc Al

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    Staff: Mentor

    The moment of inertia of the wheel doesn't change, but its angular momentum does.

    Consider the total angular momentum of the system. Can that change?
  6. Feb 12, 2013 #5
    Sorry, I was a bit confused while writing my previous post. I mean does the moment of inertia for the student and the stool matter?

    So if I flip over the bicycle wheel, its angular velocity should become negative, right? (relative to the system, its now spinning in the other direction). Should I then use this new angular velocity and the sum of the moments of inertia (which dont change) to calculate the new angular momentum?

    I feel like this is going in circles, since once I calculate that new angular momentum I would have to use all the same values again to calculate the angular velocity.
  7. Feb 12, 2013 #6

    Doc Al

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    Staff: Mentor

    Of course it matters.

    Treat the two parts of the system as separate (but loosely connected) entities.

    Try this:

    Ang Mom (of stool & student) + Ang Mom (of wheel) = constant.

    If Ang Mom (of wheel) changes, how must Ang Mom (of stool & student) change?
  8. Feb 12, 2013 #7
    The Angular Momentum of the stool and student must change to make the equation still valid, since Angular Momentum is conserved.

    So to calculate the Angular Momentum of the wheel:

    AM(wheel) = .37*(-14) = -5.18

    so then:

    AM(wheel) + AM(Student) = constant.

    -5.18 + 3.6*ω(student) = constant.

    How should I proceed from here? I am not sure how to calculate the constant before the switching of direction of the wheel...

    Im off to class for a bit but I will check back after.
  9. Feb 12, 2013 #8

    Doc Al

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    Staff: Mentor

    Good. That's for after the wheel was flipped.

    Do the same calculation for before the wheel is flipped.

    (You're almost there.)
  10. Feb 12, 2013 #9
    So should i do something like:

    AM(student-initial) + AM(wheel-initial)=AM(student-final)+AM(wheel-final)

    and then solve it for the AM of the student-final.

    so it would be:

    3.6*0 +.37*14 = 3.6*x+.37*-14

    and solve for the final angular velocity.

    x = 2.89 rad/s

    Is this correct?

    EDIT: I just checked my work over and typed the answer into the online homework system for my class - i got it right. This was such a simple problem in hindsight!
    Thank you Doc Al for all your help and patience as i struggled to grasp this concept. Your assistance has been invaluable!
  11. Feb 12, 2013 #10

    Doc Al

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    Staff: Mentor

    Excellent! :approve:
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