DYNAMIQUE_POINT_MATERIEL_correction

# DYNAMIQUE_POINT_MATERIEL_correction - Correction du T.D 5...

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Correction du T.D 5 Exercice 5.1 ef´ erentiel Terrestre suppos´ e galil´ een. m -→ a = -→ F ` a 1 D : ma x = F x donc v = a x t + v 0 = F x m t + v 0 et ` a l’arrˆ et v ( t = t a ) = 0 donc : F x = - mv 0 t a A.N : F x = 4166 , 6 N . Exercice 5.2 ef´ erentiel Terrestre suppos´ e galil´ een. On prend (Oxy) comme rep` ere avec Ox horizontal et Oy suivant la verticale. Le point mat´ eriel est plac´ e ` a t = 0 en O.Le vecteur vitesse initiale est pris dans le plan Oxy. 1 - PFD : m -→ a = m -→ g ¨ x = 0 ¨ y = - g ¨ z = 0 ˙ x = v 0 x = v 0 cos( α ) ˙ y = v 0 y - gt = v 0 sin( α ) ˙ z = 0 x = v 0 cos( α ) t y = - 1 2 gt 2 + v 0 sin( α ) t z = 0 , car x 0 = y 0 = z 0 = 0. On extrait t = x v 0 cos( α ) de x ( t ) et on trouve finalement en rempla¸ cant dans y ( t ): y = - gx 2 2 v 2 0 cos 2 ( α ) + x tan( α ). Port´ ee : y = 0 x p = tan( α )2 v 2 0 cos 2 ( α ) g = v 2 0 g sin(2 α ). 2 - l p = v 2 0 g sin(2 α ) donc 2 α = arcsin( l p g v 2 0 ) 3 - Les coordonn´ ees du point P doivent v´ erifier l’´ equation de la trajectoire soit : Y = - gX 2 2 v 2 0 cos 2 ( α ) + X tan( α ) = - gX 2 (1+tan 2 ( α )) 2 v 2 0 + X tan( α ) soit : gX 2 2 v 2 0 tan 2 ( α ) - X

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