LA VELOCIDAD
Enviado por teyshi • 27 de Marzo de 2015 • 212 Palabras (1 Páginas) • 151 Visitas
La velocidad
Vr = r ̇= -4.468 m/s
Vθ = r θ ̇= √2 (2.824)=2.579 m/s
V = √(V_(r^2 )+V_(θ^2 ) ) = √((-4.468)^2+ (2.579)^2 ) =5.16 m/s
La aceleración
ar = r ̈-rθ ̇^2= -32.86- √(2 ) (1.824)^2= -37.57 m/s^2
aθ = rθ ̈ + 2(rθ ̇ ) ̇= √2 (3.812)+2 (-4.468) (1.824)= -10.91 m/s^2
a = √(a_1^2+a_θ^2 )= √((-37.57)^2+(-10.91)^2 )=39.1 m/s^2
15. Solución
Derivando con respecto al tiempo
r = 4 m θ = 0.2 t/t=6 = 1.2 rad
r ̇=0 θ ̇ = 0.2 rad/s
r ̈=0 θ ̈=0
Z = 0.5cos θ Z ̇ = -0.5 sen θ ├ θ ̈ ┤|_(θ=1.2) = -0.0932 m/s
Z ̈= -0.5 ├ [cos〖θθ ̇^2 〗+sen θθ ̈ ]┤|_(θ=1.2)= -0.007247 m/s^2
Vr = r ̇=0
Vθ = rθ ̇=4 (0.2)=0.8m/s
VZ = Z ̇= -0.0932 m/s
ar = r ̈-r θ ̇^2=0-4(0.2)^2=-0.16 m/s^2
aθ = rθ ̈+2 r ̇ θ ̇ = 4(0) + 2(0) (0.2) =
aZ = Z ̈= -0.007247 m/s^2
16. solución
y = 10-6 X3
dy/dx=3 (10 )^(-6) ├ X^2 ┤|_(X= 600)=1.08
(d^2 y)/(dx^2 )=6 (10)^(-6) ├ X┤|_(X=600) = 3.6 (10)-3
Radio de curvatura en x =600
ρ= (1+ (dy/dx)^2 )^(3⁄2)/|(d^2 y)/(dx^2 )| = [1+ (1.08)^2 ]^(3⁄2)/|3.6 (10)^(-3) | =885.7 pies
an = V^2/ρ = 〖40〗^2/885.7=1.81 pies/s^2
a = √(a t^2+a_n^2 ) = √(0+(1.81)^2 )=1.81 pies/s^2
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