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Mastering Physics 10.85 - Dynamics of Rotational Motion

  1. Apr 3, 2012 #1
    This is my first post on here; I hope it will be worth it. I haven't been able to find an adequate solution to this elsewhere.

    1. The problem statement, all variables and given/known data

    A 5.30 kg ball is dropped from a height of 11.5 m above one end of a uniform bar that pivots at its center. The bar has mass 8.50 kg and is 9.00 m in length. At the other end of the bar sits another 5.40 kg ball, unattached to the bar. The dropped ball sticks to the bar after the collision.

    2. Relevant equations

    [itex]W_t=K_t[/itex]
    [itex]L_i=L_f[/itex]
    [itex]L=mvr[/itex]
    [itex]v=r\omega[/itex]

    3. The attempt at a solution

    First, I defined the data given as following:
    [itex]m_1 = 5.3kg, h_1 = 11.5 m, m_b = 8.5 kg, \ell = 9.00m, m_2 = 5.4kg[/itex]

    Then, I looked up the moments of inertia for the bar and balls relative to the center of the bar, then found the total of these:

    [itex]I_b=\frac{1}{12}m_b(\frac{\ell}{2})^2=8.94 kg \, m^2[/itex]
    [itex]I_1=m_1(\frac{\ell}{2})^2=107 kg \, m^2[/itex]
    [itex]I_2=m_2(\frac{\ell}{2})^2=109 kg \, m^2[/itex]
    [itex]I_t=\sum I = 225.6 kg \, m^2[/itex]

    Then, I found the resulting velocity of the 1st ball when it hits the bar:
    [itex]m_1h_1g = I_2\omega_2^2 \implies v = \sqrt{2h_1g} \implies v = 15 \frac{m}{s}[/itex]

    Then I used the conservation of angular momentum to find the final angular velocity after the bar has rotated:
    [itex]L_1 = m_1v_1\frac{\ell}{2}=357.75Nms[/itex]
    [itex]L_2=I_t\omega[/itex]
    [itex]L_1=L_2 \implies I_t\omega = L_1 \implies \omega = \frac{L_1}{I_t} = 1.59 \frac{rad}{sec}[/itex]

    Then I found the initial velocity of the ball moving upwards from the angular velocity and used the work energy theorem to find the final height of the ball.
    [itex]v=\frac{\ell}{2} \omega \implies v=7.14 \frac{m}{s}[/itex]
    [itex]\frac{1}{2}m_2v^2=m_2gh_2 \implies h_2=\frac{v^2}{2g} \implies h_2 = 2.60 m[/itex]

    However, this is the incorrect value for [itex]h_2[/itex]. Any ideas?
     
    Last edited: Apr 3, 2012
  2. jcsd
  3. Apr 3, 2012 #2

    Doc Al

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    Staff: Mentor

    Double check that value.
     
  4. Apr 3, 2012 #3
    That's embarrassing; nice catch! I tried it with this updated value and still got the wrong answer:


    [itex]I_b=\frac{1}{12}m_b(\frac{\ell}{2})^2=14.3 kg \, m^2[/itex]
    [itex]I_1=m_1(\frac{\ell}{2})^2=107 kg \, m^2[/itex]
    [itex]I_2=m_2(\frac{\ell}{2})^2=109 kg \, m^2[/itex]
    [itex]I_t=\sum I = 230 kg \, m^2[/itex]

    Then, I found the resulting velocity of the 1st ball when it hits the bar:
    [itex]m_1h_1g = I_2\omega_2^2 \implies v = \sqrt{2h_1g} \implies v = 15 \frac{m}{s}[/itex]

    Then I used the conservation of angular momentum to find the final angular velocity after the bar has rotated:
    [itex]L_1 = m_1v_1\frac{\ell}{2}=357.75Nms[/itex]
    [itex]L_2=I_t\omega[/itex]
    [itex]L_1=L_2 \implies I_t\omega = L_1 \implies \omega = \frac{L_1}{I_t} = 1.55 \frac{rad}{sec}[/itex]

    Then I found the initial velocity of the ball moving upwards from the angular velocity and used the work energy theorem to find the final height of the ball.
    [itex]v=\frac{\ell}{2} \omega \implies v=7.00 \frac{m}{s}[/itex]
    [itex]\frac{1}{2}m_2v^2=m_2gh_2 \implies h_2=\frac{v^2}{2g} \implies h_2 = 2.50 m[/itex]
     
  5. Apr 3, 2012 #4

    Doc Al

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    Staff: Mentor

    Just curious: What's the initial angle of the bar? Horizontal?
     
  6. Apr 3, 2012 #5
    I assumed it was; the question statement that I posted is all of the information that I have.
     
  7. Apr 3, 2012 #6

    Doc Al

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    Staff: Mentor

    Your work looks OK to me. My only suggestion, if it's an online system that may be fussy about numeric values, is to redo the calculation without round off intermediate steps. (In case that's what you did.) That may affect the answer a bit.

    What book is this from?
     
  8. Apr 3, 2012 #7
    Hm, I just tried the calculation again without any rounding and I got 2.48 m and the online system still rejected it. Thanks for taking a look at this, though!

    It's from Sears & Zemansky's University Physics 13e.

    Edit:
    Somebody pointed out (on Facebook) that I shoudn't be using [itex]\frac{\ell}{2}[/itex] for the calculation of the moment of inertia of the bar and I realized that I didn't notice that when you pointed it out. This made my answer come out right.
     
    Last edited: Apr 3, 2012
  9. Apr 3, 2012 #8

    Doc Al

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    Staff: Mentor

    Good!

    (I should have checked to make sure that you corrected it.)
     
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