MECANIQUE RATIONNELLE

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UMBB Boumerdès, Faculté des sciences, Département de physique Cours exercices, Mécanique Rationnelle : TCT et LMD-ST sem :3 A.KADI 2 ⎡mL ⎤ ⎡ ⎤ ⎢ 0 0 ⎥ ⎡0⎤ → ⎢ 0 3 ⎥ mL mL σ G ( S / R0 ) z 1 ⎢ ⎥ 3 3 ⎢ mL 0 0 ⎥ ⎢⎣ α⎥ mL ⎦ ⎢ ⎥ ⎢⎣ ⎥ R α 1 3 ⎦ R ⎣ 3 ⎦ → 2 • → 2 • → 0 = I ⎢ ⎥ ⎢ ⎥ = ⎢ ⎥ G / R . Ω1 = ⎢ 0 0 0 ⎥ ⎢ 0 = = • ⎥ 0 α z1 α 0 2 2 • → → 0 2 d σ •• → G ( S / R0 ) mL δ G ( S / R0 ) = = α z0 (5) dt 3 1 Pour calculer le moment des forces extérieures on doit déterminer les vecteurs : GC et GA : Nous avons : sinα = x AC ⇔ x ⎛ x ⎞ AC = donc : GC = AC − AG = ⎜ − L⎟ sinα ⎝ sinα ⎠ −−→ −−→ on obtient : ⎧ ⎛ x ⎞ ⎪− ⎜ − L⎟sinα ⎪ ⎝ sinα ⎠ ⎧ Lsinα −−→ ⎛ x −−→ ⎞ ⎪ GC = ⎨ ⎜ − L⎟cosα et GA= ⎨− L cosα ; ⎪ ⎝ sinα ⎠ ⎪ ⎪ 0 R ⎩ 0 0 ⎪ R ⎩ 0 ∑ i −→ → M ( F ext ) / G ⎧ Lsinα ⎪ = ⎨− L cosα∧ ⎪ R ⎩ 0 R 0 0 ⎧ 0 ⎪ ⎨R ⎪ ⎩ 0 A + ⎧ ⎛ x ⎞ ⎪− ⎜ − L⎟sinα ⎪ ⎝ sinα ⎠ ⎛ x ⎞ ⎨ ⎜ − L⎟cosα ∧ ⎪ ⎝ sinα ⎠ ⎪ 0 R ⎪ R ⎩ 0 0 ⎧RC ⎪ ⎨RC ⎪ ⎩ 0 cosα sinα ⎧ ⎪ −→ → → ∑ M ( Fext ) / G = ⎨ 0 = ⎜ LRA sinα − RC ⎜ − L⎟⎟ z0 i sinα ⎪ ⎪ LR R ⎩ 0 A 0 sinα − R C ⎛ x ⎞ ⎜ − L⎟ ⎝ sinα ⎠ ⎛ ⎝ ⎛ ⎝ x ⎞⎞ ⎠⎠ (6) L’égalité des moments dans les équations (5) et (6) donne : 2 •• ⎛ x ⎞ mL LRA sinα − RC ⎜ − L⎟ = α (7) ⎝ sinα ⎠ 3 387
UMBB Boumerdès, Faculté des sciences, Département de physique Cours exercices, Mécanique Rationnelle : TCT et LMD-ST sem :3 A.KADI 4. Théorème de l’énergie cinétique, appliqué à la barre. La variation de l’énergie cinétique de la barre est égale au travail des forces extérieures. dE C = dW ⇔ dE C = dt dW dt L’énergie cinétique de la barre est donnée par : E C E C E C = = = 1 2 1 2 1 2 ⎛ m⎜V ⎝ → 0 • ⎛ 2 m⎜ x ⎝ ⎛ m⎜ x ⎝ ⎞ ( G) ⎟ ⎠ + L • 2 + 2 + 1 2 → 0T Ω1 . I G / R 1 . → 0 Ω1 2 ⎡mL ⎤ ⎢ 0 0 ⎥⎡0⎤ • • • ⎞ • ⎛ ⎞⎢ 3 2 2 1 α − 2L xα cosα ⎟ + ⎥⎢ ⎥ ⎜0,0, α ⎟ ⎝ ⎠⎢ 0 0 0 ⎥⎢ 0 • ⎥ ⎠ 2 2 ⎢ mL 0 0 ⎥⎢⎣ α⎥⎦ ⎢⎣ 3 ⎥⎦ 2 L α • 2 − • • ⎞ 2L xα cosα ⎟ + ⎠ 1 2 2 mL α 3 • 2 dE C dt • •• • •• •• • • •• 1 ⎛ 2 ⎛ = m ⎜2x x+ 2L αα− 2L⎜ xα cosα + xα cosα − 2 ⎝ ⎝ Nous avons aussi : dW dt → → → 0 0 0 = m g • V ( G) + R • V ( A) + R • V ( C) , A → → C → • • 2 ⎞⎞ L xα sinα ⎟ ⎟ + ⎠ ⎠ 1 2 2 • mL 2 α 3 → → → → → → → → R A ⊥ mais : R • 0 A V ( A) = 0 car V 0 ( A) et R • V 0 C ( C) = 0 car ⊥ V 0 ( C) → R C → dW dt = m g → → 0 • V ⎧ 0 ⎪ ( G) = ⎨− mg ⎪ R ⎩ 0 L’égalité entre les deux termes donne : 0 • • ⎧ ⎪ x− Lα cosα • • ⎨ − Lα sinα ⎪ 0 R ⎪⎩ 0 • = mgLα sinα 1 • 2 • ⎛ • •• • •• 2 ⎛ •• • • •• • 2 ⎞⎞ 1 mL 2 2 α α 2 α cosα α cosα α sinα α 2 • m ⎜ ⎟ + = α sinα 2 x x+ L − L⎜ x + x − L x ⎟ mgL ⎝ ⎝ ⎠⎠ 2 3 • •• • •• • 2 • 2 ⎛ •• • • •• • 2 ⎞ L • 2 x x+ L α α − L⎜ xα cosα + xα cosα − L xα sinα ⎟ + α = gLα sinα ⎝ ⎠ 6 388
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CLASSES PREPARATOIRES AUX GRANDES E
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Préface Quand Ali KADI m’a amica
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UMBB Boumerdès, Faculté des scien
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