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theta$};
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\begin_layout Standard

\series bold
Exercice : Pendule accroché à un ressort
\end_layout

\begin_layout Standard

\series bold
1.

\series default
 La masse est soumise à son poids 
\begin_inset Formula $\vec{P}$
\end_inset

, à la tension du fil 
\begin_inset Formula $\vec{T}$
\end_inset

 et à la force de rappel élastique 
\begin_inset Formula $\vec{F}$
\end_inset

 exercée par le ressort.
 On les représente sur le schéma.
 La règle de la main droite indique que 
\begin_inset Formula $\mathcal{M}_{z}(\vec{F})<0$
\end_inset

 et 
\begin_inset Formula $\mathcal{M}_{z}(\vec{P})>0$
\end_inset

 (car l'axe 
\begin_inset Formula $(Oz)$
\end_inset

 est 
\begin_inset Formula $\otimes$
\end_inset

).
 La tension est dirigée vers 
\begin_inset Formula $(Oz)$
\end_inset

 donc 
\begin_inset Formula $\boxed{\mathcal{M}_{z}(\vec{T})=0}$
\end_inset

.
\begin_inset VSpace smallskip
\end_inset


\end_layout

\begin_layout Standard
On trace les droites d'action de 
\begin_inset Formula $\vec{P}$
\end_inset

 et 
\begin_inset Formula $\vec{F}$
\end_inset

.
 Le bras de levier du poids est égal à 
\begin_inset Formula $a\cos\theta$
\end_inset

 donc 
\begin_inset Formula $\boxed{M_{z}(\vec{P})=mga\cos\theta}$
\end_inset

.
 
\end_layout

\begin_layout Standard
On définit 
\begin_inset Formula $H$
\end_inset

 le projeté de 
\begin_inset Formula $O$
\end_inset

 sur la droite d'action de 
\begin_inset Formula $\vec{F}$
\end_inset

.
 Par définition le bras de levier de la force de rappel est égal à la distance
 
\begin_inset Formula $OH$
\end_inset

.
 Pour la calculer on remarque que le triangle 
\begin_inset Formula $OAM$
\end_inset

 est isocèle en 
\begin_inset Formula $O$
\end_inset

 donc la droite 
\begin_inset Formula $(OH)$
\end_inset

 est la bissectrice de l'angle 
\begin_inset Formula $\theta$
\end_inset

.
 Ainsi 
\begin_inset Formula $OH=a\cos\frac{\theta}{2}$
\end_inset

 et, puisque la longueur à vide du ressort est nulle : 
\begin_inset Formula $\bigl\Vert\vec{F}\bigr\Vert=kAM=2kAH=2ka\sin\frac{\theta}{2}$
\end_inset

.
 
\end_layout

\begin_layout Standard
On en déduit que : 
\begin_inset Formula $\mathcal{M}_{z}(\vec{F})=-OH\times\bigl\Vert\vec{F}\bigr\Vert=-ka^{2}\times2\cos\frac{\theta}{2}\sin\frac{\theta}{2}$
\end_inset

.
\end_layout

\begin_layout Standard
Cette expression se simplifie sous la forme : 
\begin_inset Formula $\boxed{\mathcal{M}_{z}(\vec{F})=-ka^{2}\sin\theta}$
\end_inset

.
\begin_inset VSpace medskip
\end_inset


\end_layout

\begin_layout Standard

\series bold
2.

\series default
 On applique le théorème du moment cinétique à 
\begin_inset Formula $M$
\end_inset

, par rapport à l'axe orienté fixe 
\begin_inset Formula $(Oz)$
\end_inset

, dans le référentiel terrestre supposé galiléen.
 On se place à l'équilibre donc le moment cinétique est nul : 
\begin_inset Formula $\mathcal{M}_{z}(\vec{P})+\mathcal{M}_{z}(\vec{F})=0\,\,\Longleftrightarrow\,\,mga\cos\theta_{\scriptsize{\mbox{eq}}}-ka^{2}\sin\theta_{\scriptsize{\mbox{eq}}}=0\,\,\Longleftrightarrow\,\,\boxed{\tan\theta_{\scriptsize{\mbox{eq}}}={\textstyle \frac{mg}{ka}}}$
\end_inset

.
\begin_inset VSpace medskip
\end_inset


\end_layout

\begin_layout Standard

\series bold
3.

\series default
 Dans le cas général le TMC donne : 
\begin_inset Formula $ma^{2}\ddot{\theta}=mga\cos\theta-ka^{2}\sin\theta$
\end_inset

.
 On pose le changement de variable
\begin_inset space ~
\end_inset

:
\end_layout

\begin_layout Standard
\begin_inset Formula $\bullet$
\end_inset

 
\begin_inset Formula $\ddot{\theta}=\ddot{\varepsilon}$
\end_inset

,
\end_layout

\begin_layout Standard
\begin_inset Formula $\bullet$
\end_inset

 
\begin_inset Formula $\cos\theta=\cos(\theta_{\scriptsize{\mbox{eq}}}+\varepsilon)\simeq\cos\theta_{\scriptsize{\mbox{eq}}}-\varepsilon\sin\theta_{\scriptsize{\mbox{eq}}}$
\end_inset

,
\end_layout

\begin_layout Standard
\begin_inset Formula $\bullet$
\end_inset

 
\begin_inset Formula $\sin\theta=\sin(\theta_{\scriptsize{\mbox{eq}}}+\varepsilon)\simeq\sin\theta_{\scriptsize{\mbox{eq}}}+\varepsilon\cos\theta_{\scriptsize{\mbox{eq}}}$
\end_inset

,
\end_layout

\begin_layout Standard
ce qui conduit à : 
\begin_inset Formula $ma^{2}\ddot{\varepsilon}=\underset{=0\,\,\text{(cf question 2)}}{\underbrace{mga\cos\theta_{\scriptsize{\mbox{eq}}}-ka^{2}\sin\theta_{\scriptsize{\mbox{eq}}}}}-\left(mga\sin\theta_{\scriptsize{\mbox{eq}}}+ka^{2}\cos\theta_{\scriptsize{\mbox{eq}}}\right)\varepsilon$
\end_inset


\end_layout

\begin_layout Standard
Après simplification on aboutit à : 
\begin_inset Formula $\ddot{\varepsilon}+\left(\frac{g}{a}\sin\theta_{\scriptsize{\mbox{eq}}}+\frac{k}{m}\cos\theta_{\scriptsize{\mbox{eq}}}\right)\varepsilon=0$
\end_inset

.
 
\end_layout

\begin_layout Standard
On conclut que les petits mouvements du système autour de la position 
\begin_inset Formula $\theta_{\scriptsize{\mbox{eq}}}$
\end_inset

 sont des oscillations harmoniques de pulsation : 
\begin_inset Formula $\boxed{\omega_{0}=\sqrt{{\textstyle \frac{g}{a}\sin\theta_{\scriptsize{\mbox{eq}}}+\frac{k}{m}\cos\theta_{\scriptsize{\mbox{eq}}}}}}$
\end_inset

.
\begin_inset Newpage newpage
\end_inset


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\series bold
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has_inner_box 0
inner_pos "c"
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height_special "totalheight"
thickness "0.4pt"
separation "3pt"
shadowsize "4pt"
framecolor "black"
backgroundcolor "none"
status open

\begin_layout Plain Layout

\series bold
\size large
Corrigé DM19
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\begin_layout Standard
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status open

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\backslash
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\backslash
node[above=3pt,fill=white,inner sep=2pt] at (0,0) {$O$};
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\backslash
node[below=3pt] at (0,0) {$z$};
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\backslash
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\backslash
fontsize{13}{10}
\backslash
selectfont,fill=white,inner sep=-0.5pt] at (0,0) {$
\backslash
otimes$};
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\begin_layout Plain Layout

\end_layout

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%droites d'action
\end_layout

\begin_layout Plain Layout


\backslash
draw[dashed] (2.5,-2)--(2.5,0.8);
\end_layout

\begin_layout Plain Layout


\backslash
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\backslash
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\begin_layout Plain Layout

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\backslash
draw[->,line width=1.5pt] (2.5,-2)--(2.5,-2.8) node[right] {$
\backslash
vec{P}$};
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\backslash
draw[->,line width=1.5pt,rotate around={-128.66:(2.5,-2)}] (2.5,-2)--(2.5,-2.8)
 node[above right] {$
\backslash
vec{T}$};
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\begin_layout Plain Layout


\backslash
draw[->,line width=1.5pt,rotate around={165.96:(2.5,-2)}] (2.5,-2)--(2.5,-2.8)
 node[below right=3pt] {$
\backslash
vec{F}$};
\end_layout

\begin_layout Plain Layout

\end_layout

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\backslash
draw[fill=white] (2.5,-2) circle (0.07) node[below right] {$M$};
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\backslash
draw[->] (0.8,0) arc (0:-39.5:0.8) node[midway,right=1.5pt,yshift=-1mm,fill=white,in
ner sep=2pt] {$
\backslash
theta$};
\end_layout

\begin_layout Plain Layout


\backslash
draw[->] (1.5,0) arc (0:-19.33:1.5) node[midway,right,yshift=-0.5mm] {$
\backslash
tfrac{
\backslash
theta}{2}$};
\end_layout

\begin_layout Plain Layout

\end_layout

\begin_layout Plain Layout


\backslash
fill (2.75,-1) circle (0.07) node[right=3pt,fill=white,inner sep=2pt,yshift=1mm]
 {$H$};
\end_layout

\begin_layout Plain Layout


\backslash
draw[<->] (0,0.5)--(2.5,0.5) node[midway,above] {$a
\backslash
cos(
\backslash
theta)$};
\end_layout

\begin_layout Plain Layout

\end_layout

\begin_layout Plain Layout


\backslash
end{tikzpicture}
\end_layout

\begin_layout Plain Layout


\backslash
end{wrapfigure}
\end_layout

\end_inset


\end_layout

\begin_layout Standard

\series bold
Exercice : Pendule accroché à un ressort
\end_layout

\begin_layout Standard

\series bold
1.

\series default
 La masse est soumise à son poids 
\begin_inset Formula $\vec{P}$
\end_inset

, à la tension du fil 
\begin_inset Formula $\vec{T}$
\end_inset

 et à la force de rappel élastique 
\begin_inset Formula $\vec{F}$
\end_inset

 exercée par le ressort.
 On les représente sur le schéma.
 La règle de la main droite indique que 
\begin_inset Formula $\mathcal{M}_{z}(\vec{F})<0$
\end_inset

 et 
\begin_inset Formula $\mathcal{M}_{z}(\vec{P})>0$
\end_inset

 (car l'axe 
\begin_inset Formula $(Oz)$
\end_inset

 est 
\begin_inset Formula $\otimes$
\end_inset

).
 La tension est dirigée vers 
\begin_inset Formula $(Oz)$
\end_inset

 donc 
\begin_inset Formula $\boxed{\mathcal{M}_{z}(\vec{T})=0}$
\end_inset

.
\begin_inset VSpace smallskip
\end_inset


\end_layout

\begin_layout Standard
On trace les droites d'action de 
\begin_inset Formula $\vec{P}$
\end_inset

 et 
\begin_inset Formula $\vec{F}$
\end_inset

.
 Le bras de levier du poids est égal à 
\begin_inset Formula $a\cos\theta$
\end_inset

 donc 
\begin_inset Formula $\boxed{M_{z}(\vec{P})=mga\cos\theta}$
\end_inset

.
 
\end_layout

\begin_layout Standard
On définit 
\begin_inset Formula $H$
\end_inset

 le projeté de 
\begin_inset Formula $O$
\end_inset

 sur la droite d'action de 
\begin_inset Formula $\vec{F}$
\end_inset

.
 Par définition le bras de levier de la force de rappel est égal à la distance
 
\begin_inset Formula $OH$
\end_inset

.
 Pour la calculer on remarque que le triangle 
\begin_inset Formula $OAM$
\end_inset

 est isocèle en 
\begin_inset Formula $O$
\end_inset

 donc la droite 
\begin_inset Formula $(OH)$
\end_inset

 est la bissectrice de l'angle 
\begin_inset Formula $\theta$
\end_inset

.
 Ainsi 
\begin_inset Formula $OH=a\cos\frac{\theta}{2}$
\end_inset

 et, puisque la longueur à vide du ressort est nulle : 
\begin_inset Formula $\bigl\Vert\vec{F}\bigr\Vert=kAM=2kAH=2ka\sin\frac{\theta}{2}$
\end_inset

.
 
\end_layout

\begin_layout Standard
On en déduit que : 
\begin_inset Formula $\mathcal{M}_{z}(\vec{F})=-OH\times\bigl\Vert\vec{F}\bigr\Vert=-ka^{2}\times2\cos\frac{\theta}{2}\sin\frac{\theta}{2}$
\end_inset

.
\end_layout

\begin_layout Standard
Cette expression se simplifie sous la forme : 
\begin_inset Formula $\boxed{\mathcal{M}_{z}(\vec{F})=-ka^{2}\sin\theta}$
\end_inset

.
\begin_inset VSpace medskip
\end_inset


\end_layout

\begin_layout Standard

\series bold
2.

\series default
 On applique le théorème du moment cinétique à 
\begin_inset Formula $M$
\end_inset

, par rapport à l'axe orienté fixe 
\begin_inset Formula $(Oz)$
\end_inset

, dans le référentiel terrestre supposé galiléen.
 On se place à l'équilibre donc le moment cinétique est nul : 
\begin_inset Formula $\mathcal{M}_{z}(\vec{P})+\mathcal{M}_{z}(\vec{F})=0\,\,\Longleftrightarrow\,\,mga\cos\theta_{\scriptsize{\mbox{eq}}}-ka^{2}\sin\theta_{\scriptsize{\mbox{eq}}}=0\,\,\Longleftrightarrow\,\,\boxed{\tan\theta_{\scriptsize{\mbox{eq}}}={\textstyle \frac{mg}{ka}}}$
\end_inset

.
\begin_inset VSpace medskip
\end_inset


\end_layout

\begin_layout Standard

\series bold
3.

\series default
 Dans le cas général le TMC donne : 
\begin_inset Formula $ma^{2}\ddot{\theta}=mga\cos\theta-ka^{2}\sin\theta$
\end_inset

.
 On pose le changement de variable
\begin_inset space ~
\end_inset

:
\end_layout

\begin_layout Standard
\begin_inset Formula $\bullet$
\end_inset

 
\begin_inset Formula $\ddot{\theta}=\ddot{\varepsilon}$
\end_inset

,
\end_layout

\begin_layout Standard
\begin_inset Formula $\bullet$
\end_inset

 
\begin_inset Formula $\cos\theta=\cos(\theta_{\scriptsize{\mbox{eq}}}+\varepsilon)\simeq\cos\theta_{\scriptsize{\mbox{eq}}}-\varepsilon\sin\theta_{\scriptsize{\mbox{eq}}}$
\end_inset

,
\end_layout

\begin_layout Standard
\begin_inset Formula $\bullet$
\end_inset

 
\begin_inset Formula $\sin\theta=\sin(\theta_{\scriptsize{\mbox{eq}}}+\varepsilon)\simeq\sin\theta_{\scriptsize{\mbox{eq}}}+\varepsilon\cos\theta_{\scriptsize{\mbox{eq}}}$
\end_inset

,
\end_layout

\begin_layout Standard
ce qui conduit à : 
\begin_inset Formula $ma^{2}\ddot{\varepsilon}=\underset{=0\,\,\text{(cf question 2)}}{\underbrace{mga\cos\theta_{\scriptsize{\mbox{eq}}}-ka^{2}\sin\theta_{\scriptsize{\mbox{eq}}}}}-\left(mga\sin\theta_{\scriptsize{\mbox{eq}}}+ka^{2}\cos\theta_{\scriptsize{\mbox{eq}}}\right)\varepsilon$
\end_inset


\end_layout

\begin_layout Standard
Après simplification on aboutit à : 
\begin_inset Formula $\ddot{\varepsilon}+\left(\frac{g}{a}\sin\theta_{\scriptsize{\mbox{eq}}}+\frac{k}{m}\cos\theta_{\scriptsize{\mbox{eq}}}\right)\varepsilon=0$
\end_inset

.
 
\end_layout

\begin_layout Standard
On conclut que les petits mouvements du système autour de la position 
\begin_inset Formula $\theta_{\scriptsize{\mbox{eq}}}$
\end_inset

 sont des oscillations harmoniques de pulsation : 
\begin_inset Formula $\boxed{\omega_{0}=\sqrt{{\textstyle \frac{g}{a}\sin\theta_{\scriptsize{\mbox{eq}}}+\frac{k}{m}\cos\theta_{\scriptsize{\mbox{eq}}}}}}$
\end_inset

.
\end_layout

\end_body
\end_document