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 → → • d v •• → • • •• → 2 γ = = ( ρ− ρθ ) u ρ + (2 ρθ − ρθ ) uθ dt avec •• θ = 0 car • θ = ω = Cte •• • 1 2 2 1 2 1 γ cos (1 cos ) 2 ρ = ρ− ρθ = − ρ 0ω ω t − ρ 0ω + ω t = − ρ 0ω (1 + 2cosω t) 2 2 2 • γ θ = • ⎛ 1 ⎞ 2 ρθ = 2 2⎜ − ρ 0ω sinω t ⎟ω = −ρ 0ω sinω t ⎝ 2 ⎠ L’accélération γ du mobile se calcul par : γ = 1 2 2 2 2 2 2 1 γ ρ + γ θ = ρ 0ω (1 + 2cosω t) + (2sinω t) = ρ 0ω 1+ 4(1 + 2 2 cosω t) 1 2 2 γ = ρ 0ω 1+ 8cos 2 ω t 2 c. Détermination de l’accélération normale γ N à partir des accélérations γ ρ et γ θ Nous avons : α = ( , Ox) et β = ( N, OM ) , on sait que : T → → −−→ → → π → → π ( N , T ) = et ( u ρ , u θ ) = 2 2 → On projette les deux accélérations sur l’axe portant la normale N , on obtient : γ N = −γ cos β − γ sin β ρ θ Exprimons l’angle α et β en fonction de θ : En coordonnées polaires nous avons : x = ρ cosθ et y = ρ sinθ alors : tg α = dy dx 2 d( ρ sinθ ) (1 + cosθ )cosθ − sin θ cosθ + cos 2θ ⎛ 3θ ⎞ = = = − = −cot g⎜ ⎟ d( ρ cosθ ) − (1 + cosθ )sinθ − sinθ cosθ sinθ + sin 2θ ⎝ 2 ⎠ ⎛ 3θ ⎞ ⎛ π 3θ ⎞ tg α = − cot g⎜ ⎟ = tg⎜ + ⎟ ⇔ ⎝ 2 ⎠ ⎝ 2 2 ⎠ π 3θ α = + 2 2 π θ nous avons aussi géométriquement : α = θ + β + on déduit : β = 2 2 On remplace β dans l’expression de γ N = −γ cos β − γ sin β = γ N , ce qui donne : 1 2 θ ρ 0ω (1 + 2cosθ )cos 2 2 ρ θ + 2 θ ρ 0ω sinθ sin 2 3 2 θ 3 ω t γ N = ρ cos 2 0ω = ρ 0ω cos 2 2 2 2 205
UMBB Boumerdès, Faculté des sciences, Département de physique Cours exercices, Mécanique Rationnelle : TCT et LMD-ST sem :3 A.KADI c. Rayon de courbure On sait que dans les mouvements curvilignes : Ce qui donne : 2 v R = γ N 2 2 2 ω t ρ 0ω cos = 2 3 2 ω t ρ 0ω cos 2 2 2 v γ N = ⇒ R 2 θ = ρ 0 cos 3 2 2 θ R = ρ 0 cos 3 2 2 v R = γ On peut aussi déduire le rayon de courbure d’une autre manière en sachant que : N θ S = 2ρ 0 sin et 2 α = θ + β + π 2 dS R = = dα θ ρ cos dθ 2 2 = ρ 0 cos 3 3 2 dθ 2 0 θ 206
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