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Rotational Dynamics on a stool

  1. Mar 13, 2006 #1
    A skinny student (40 kg) sits on a stool that can rotate, holding two 8 kg bowling balls with his arms out stretched. The stool weighs 20 kg. The student and stool are spun such that they have an angular velocity of 2 rad.s-1.
    When the student pulls his arms in towards him, what happens to his angular velocity and what is his final angular velocity when they rest in his lap?
    I know I have to find [tex]\omega[\tex] and I think using [tex]L=I\omega[\tex] should be the key. Only L and I are unknown..any ideas?
     
  2. jcsd
  3. Mar 13, 2006 #2
    Well, there are no net external torques on the system so it sounds like angular momentum should be conserved.

    -Dan
     
  4. Mar 13, 2006 #3
    So L(initial)=L(final), so we find the kinertic energy (Iw^2)/2, find I and and substitute it to L=Iw and find w?
     
  5. Mar 13, 2006 #4

    Hootenanny

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    That's about it. Just a quick note on latex, at the end you need to use [/tex] not [\tex], apart from that, your code is correct. If you want to put code on the same line use the [itex] tags instead of [tex] :smile:
     
  6. Mar 13, 2006 #5
    When trying to find I from the k.e equation there was a prob that I ran into. The kinetic energy of the system is not given. K.E=(Iw^2). I donot see how (mv^2)/2 can be used as we do not know v, only w is known.
     
  7. Mar 13, 2006 #6

    BobG

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    It looks like you have to make some assumptions about the persons dimensions and the length of his arms.

    Rotational kinetic energy is [tex]\frac{1}{2}I \omega^2[/tex]

    Moment of Inertia (I) is [tex]I = mr^2[/tex]

    Linear velocity (v) is [tex]v = r \omega[/tex]
     
  8. Mar 13, 2006 #7
    where r=2(pi) right?. Then we find v and sbstitute to (mv^2)/2=(Iw^2)/2 and find I which we sunstitute into L=Iw and find L right?. so whats [tex]I = mr^2[/tex] for?
     
  9. Mar 13, 2006 #8

    BobG

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    r is the radius.

    [tex]KE = \frac{1}{2}m v^2[/tex]
    [tex]= \frac{1}{2}m (r \omega)^2[/tex]
    [tex]= \frac{1}{2}m r^2 \omega^2[/tex]
    [tex]= \frac{1}{2}I \omega^2[/tex]

    The [tex]I = m r^2[/tex] only applies for a point-mass or ring (since all the units of mass have the same I). The moment of inertia for anything else is a sum of the moments of inertia for all of the point-masses that make up the object.

    For the original problem, you need to model the person-chair as a cylinder and find the moment of inertia for a cylinder and model the weights as point masses. You can derive the equation for the moment of inertia for a cylinder yourself ([tex]I = \Sigma m_i r_i^2[/tex]) or look up the formula in a book or on-line.
     
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