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Homework Help: The effect of a spinning wheel on a turntable

  1. Aug 3, 2008 #1
    1. The problem statement, all variables and given/known data
    A person stands at rest on a 5.0kg, 50cm diameter frictionless turntable.
    He receives spinning bicycle wheel, 180rpm counterclock wise,
    diameter 64cm, and the rim weighs 4.0kg.
    The wheel was held in a horzontal plane, the handles extend outward from the center (rotation axis is vertical).

    This person's legs and torso are 70kg, have an average diameter of 25cm and a height of 180cm. His arms are 2.5kg each, and he hold the handles of the wheel 45cm from the center of his body.

    This person turns the spinning wheel over 180degrees so that the handle that had been pointing up now points down.
    Describe this person's motion and calculate any relevant numerical quantities.

    2. Relevant equations
    Inertia, axis is r distance away: I = m r^2
    Inertia of loop: I= m r^2
    Inertia of disk: I= (1/2) m r^2
    Inertia of rod, axis about center: I = (1/12) m L^2
    Inertia of rod, axis about edge: I = (1/3) m L^2

    Conservation of angular momentum:
    initial L = final L
    change in L would be 0 if L is conserved.
    angular momentum: L = I W

    3. The attempt at a solution
    Bicycle wheel by itself:
    - axis is the handles at the center of wheel

    -inertial of wheel loop = I 'loop' = m'loop' r'loop'^2
    -initial angular momentum of wheel loop = Li = I'loop' W = m'loop' r'loop'^2

    -After turnning the wheel 180degrees, the spinning direction is inverted.
    counter clock wise --> clock wise
    final L = negative initial L
    Lf = - Li
    - change in L = 2Li = 2 I'loop' W = 2 m'loop' r'loop'^2 W

    The person standing on turntable holding the wheel:
    - axis is now the body of the person standing on the turntable disk.

    -inertial total = I disk + I body + I arms + I loop
    = (1/2) m'disk' r'disk'^2 + (1/12) m'body' r'body'^2
    + (1/3) m'arms' r'arms'^2 + m'loop' r'loop'^2

    note: r'body' = the body's average diameter 25cm, NOT the height of the person.

    - angular momentum of this whole system is the change in L:
    L = 2Li = I 'total' W'final'
    We can now calculate the final angular velocity of the system.

    I'm just confused of why my calculations didn't include the person's height?
    Is that just an extra information not necessary to use? or did I do something wrong with my calculations?~~~
    Thanks alot! =)
  2. jcsd
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