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Angular momentum - impulse on the axis

  1. Jun 1, 2007 #1
    1. The problem statement, all variables and given/known data

    A uniform square board ABCD had mass 5kg and sides of length 60cm. The board is hinged at A so that it can rotate freely in a vertical plane. A light string of length 1 metre is attached to B and to a point E, 80cm above A. The bards rests in equilibrium with AB horizontal and D vertically below A. It is now raised through a right angle so that B is vertically above A and AD is horizontal, and released from rest in this position. When the board returns to its original position the string becomes taught and the board is brought immediately to rest. Find the impulsive tension in the string when this occurs and the horizontal and vertical components of the impulse on the hinge.

    3. The attempt at a solution

    This is what I did for the first bit.
    [itex]I_A = I_{AB} + I_{AD} = 2\times (\frac{4}{3} \times 5 \times 0.3^2) = \frac{6}{5}[/itex] (perpendicular axes rule)

    Then by energy changes, angular velocity just before it stops is [itex]\omega = \sqrt{24.5}[/itex]

    So the impulse is the change in angular momentum which is [itex]\frac{6}{5} \times \sqrt{24.5} = 5.9396...Ns[/itex] = the wrong answer.
    The right answer is 17.5 Ns. I don't know what I did wrong.

    Also I have no idea about the second part.

    It would be really amazingly immensely helpful if anyone could point me in the direction of some angular momentum/impulse on the axis notes, because I am really not getting it.
  2. jcsd
  3. Jun 1, 2007 #2

    How do you obtain the angular speed [tex]\sqrt{24.5}[/tex] ?
    I have a different result about the angular speed.
    Please check the quantity again.

    Best regards
  4. Jun 1, 2007 #3
    By energy changes:

    At the top, [itex]E_p = mgh = 5 \times 9.8 \times 0.3 = 14.7 J[/itex]
    And rotational kinetic energy = 0

    Just before it stops, all the potential energy is converted to rotational kinetic, so:

    [itex]14.7 = \frac{1}{2} \times \frac{6}{5} \times {\omega}^2[/itex]

    so [itex]\omega = \sqrt{24.5}[/itex]
  5. Jun 1, 2007 #4
    The height that the center of mass drop is 0.3(m).
    Why not 0.6(m)?
  6. Jun 1, 2007 #5
    Oh I see, the change in height is 0.6m, silly me !
    But I still have the wrong answer. I get [itex]\omega = 7[/itex] which gives impulse = 8.4 Ns
  7. Jun 1, 2007 #6
    The impulse is defined by [tex]\vec{F}\delta t[/tex].
    You have calculated the change in angular momentum, of course use the correct one.
    If you like, the change in angular momentum is "the angular impulse".
    [tex]\text{the impulse}\times\text{arm length}=\text{the angular momentum change}[/tex]
  8. Jun 1, 2007 #7
    Or simply,
    [tex]\vec{N}=\frac{\Delta\vec{L}}{\Delta t}=\vec{r}\times\vec{F}\quad\Rightarrow\quad\Delta\vec{L}=\vec{r}\times(\vec{F}\Delta t)=\vec{r}\times\text{the impulse}[/tex]
  9. Jun 2, 2007 #8
    so that would mean impulse = 8.4/0.6 = 14, if I understadnd you correctly, but that's still the wrong answer. Also, what do the arrows above F, r etc mean?
  10. Jun 2, 2007 #9
    The force [tex]\vec{F}[/tex] is along the direction of the string.
    [tex]\vec{r}[/tex] is the positin vector from the pivot A to B.
    They are vectors.
    [tex]\frac{6}{5}\times7=0.6\cdot J\cdot\sin\theta[/tex]
    where [tex]\theta[/tex] is the angle of the two vectors.
    [tex]\frac{6}{5}\times7=0.6\cdot J\cdot\frac{4}{5}\quad\Rightarrow\quad J=17.5\text{(N-s)}[/tex]
  11. Jun 3, 2007 #10
    Ah OK I understand that now.
    Any ideas on the second part?
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