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anonymous
 one year ago
can anyone help me understand the angular momentum questions
anonymous
 one year ago
can anyone help me understand the angular momentum questions

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rvc
 one year ago
Best ResponseYou've already chosen the best response.0The 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) = \(\Large\sf\frac{ Kg~ m^2}{s}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0excuse 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

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0an 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

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0of 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

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0dw:1435075254161:dw

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

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0substituting 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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