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anonymous

  • one year ago

can anyone help me understand the angular momentum questions

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  1. rvc
    • one year ago
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    The formula for angular momentum (L) is \(\Large\rm \color{red}L=\color{darkblue}{I~\omega }\) Where I : Moment of Inertia \(\omega\) : Angular velocity The SI unit of A.M (L) = \(\Large\sf\frac{ Kg~ m^2}{s}\)

  2. anonymous
    • one year ago
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    excuse but particularly i can give some example like their is a bat a ball ht the base of bat now the angular momentum of the bat

  3. anonymous
    • one year ago
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    @rvc

  4. rvc
    • one year ago
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    ?

  5. rvc
    • one year ago
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    @Michele_Laino :)

  6. Michele_Laino
    • one year ago
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    an example og the angular momentum can be this: please think about a metallic rod which is oscillating around to its one end, like in the drawing below: |dw:1435074583207:dw|

  7. Michele_Laino
    • one year ago
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    of course our rod is oscillating as effect of gravity on it. Now The Moment of Inertia of our rod, with respect to it center of oscillation, is: \[\Large I = \frac{1}{3}M{L^2}\] where M is the mass of our rod Now the moment of the external force which is acting on our rod, namely its weight, is: \[\Large {\mathbf{M}}_{\mathbf{O}}^{\mathbf{E}} = - \frac{{MgL}}{2}\sin \theta {\mathbf{\hat z}}\] |dw:1435075148006:dw|

  8. Michele_Laino
    • one year ago
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    |dw:1435075254161:dw|

  9. Michele_Laino
    • one year ago
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    so, applying the second cardinal equation of Mechanics, we can write: \[\Large \begin{gathered} \frac{d}{{dt}}\left( {I\omega } \right) = - \frac{{MgL}}{2}\sin \theta \hfill \\ \hfill \\ I\dot \omega = - \frac{{MgL}}{2}\sin \theta \hfill \\ \hfill \\ I\ddot \theta = - \frac{{MgL}}{2}\sin \theta \hfill \\ \hfill \\ \ddot \theta + \frac{{MgL}}{{2I}}\sin \theta = 0 \hfill \\ \end{gathered} \]

  10. Michele_Laino
    • one year ago
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    substituting I=(1/3) M*L^2, we get: \[\Large \ddot \theta + \frac{{3g}}{{2L}}\sin \theta = 0\] finally, considering the "little oscillation" case, namely: \[\Large \sin \theta \cong \theta \] we get: \[\Large \ddot \theta + \frac{{3g}}{{2L}}\theta = 0\] That above, is a simple application of angular momentum of a rigid body Can you compute the period of little oscillation, starting from the last equation?

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