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Rotation problem

  1. May 11, 2008 #1
    1. The problem statement, all variables and given/known data
    [​IMG]

    3. The attempt at a solution
    Could someone see if my solution is correct?

    Part a:
    [itex]I = M R^2[/itex] for a circular hoop.

    [tex]L = \vec{r} \times \vec{p} = I \omega[/tex]

    [tex]m_0 v_0 R \sin(\theta) = M R^2 \omega[/tex]

    [tex]\omega = \frac{m_0 v_0 \sin(\theta)}{M R}[/tex]

    Part b:
    Using conservation of momentum to find the velocity [itex]v[/itex] of the dart+wheel system:
    [tex]m_0 v_0 = (m_0 + M) v[/tex]
    [tex]v = \frac{m_0 v_0}{m_0 + M}[/tex]

    [tex]K_i = \frac{1}{2} m_0 v_0^2[/tex]

    [tex]K_f = K_{translational} + K_{rotation} = \frac{1}{2}(M + m_0) v^2 + \frac{1}{2} (M + m_0) R^2 \omega^2[/tex]

    And then just plug [itex]v[/itex] and [itex]\omega[/itex] in from above and calculate the ratio of final to initial? So, after a bunch of algebra:

    [tex]\frac{K_f}{K_i} = m_0 \left(\frac{\sin^2(\theta)}{M}+\frac{\sin^2(\theta) m_0}{M^2}+\frac{1}{M+m_0}\right)[/tex]
     
    Last edited: May 11, 2008
  2. jcsd
  3. May 11, 2008 #2
    for part a you should have used [tex] M + m_0 [/tex] instead of M as the mass of the complete system. the dart still has some angular momentum after it sticks to the now rotating wheel.

    for part b I think the axle doesn't move, so [tex]K_{translational} = 0 [/tex]
     
  4. May 11, 2008 #3
    Yeah, you're absolutely right for both of them. Thanks.

    The final answer for any future googlers is (oh wait, the problem text was in an image):

    [tex]\frac{m_0 \sin^2(\theta)}{m_0 + M}[/tex]
     
    Last edited: May 11, 2008
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