  dv d  v 0 a= =  dt dt  1 + θ 2 =  a) El vector posición está dado por r (θ ) = r (θ ) ⋅ rˆ , con r (θ ) = a ⋅ θ Calculando el vector velocidad   dr dr ˆ drˆ dr dθ ˆ drˆ dθ v= = ⋅r + r ⋅ = ⋅ ⋅r + r ⋅ ⋅ dt dt dt dθ dt dθ dt = a ⋅ θɺ ⋅ rˆ + a ⋅ θ ⋅ θɺ ⋅ θˆ = a ⋅ θɺ ⋅ rˆ + θ ⋅ θˆ

(

)

Según el enunciado, la magnitud de la velocidad es igual a v0 constante. Así:

 v = v 0 = a ⋅ θɺ ⋅ 1 + θ 2 ⇒ θɺ =

v0

v0 a ⋅ 1+θ 2

(

)

v0

⋅ rˆ + θ ⋅ θˆ =

1+θ 2

(

⋅ rˆ + θ ⋅ θˆ

b) El vector unitario tangente a la trayectoria se define como:  ˆt = v v

(

)

)

(

)

v0 d  v 0  dθ ˆ d ˆ dθ r + θ ⋅θˆ ⋅ ⋅ r + θ ⋅θˆ + ⋅  ⋅ 2 2 dθ  1 + θ  dt dt 1 + θ dθ

(

)

(

 v0 v0  1 = θɺ ⋅ - ⋅ ⋅ 2 ⋅θ ⋅ rˆ + θ ⋅ θˆ + 3 1 +θ 2  2 (1 + θ 2 ) 2 

(

)

 v0 ⋅θ v0  ɺ = θ ⋅ ⋅ rˆ + θ ⋅ θˆ + 3 1 +θ 2  (1 + θ 2 ) 2 

(

)

)

( )

d  drˆ θ ⋅ θˆ + ⋅ dθ  dθ

 drˆ dθˆ ⋅  + θˆ + θ ⋅ dθ  dθ

 v0 ⋅θ v0  ɺ = θ ⋅ ⋅ rˆ + θ ⋅θˆ + ⋅ 2 ⋅θˆ − θ ⋅ rˆ 3 2 2 2 + 1 θ  (1 + θ )    v0 θ ˆ + θ ⋅ θˆ + 2 ⋅ θˆ − θ ⋅ rˆ  = θɺ ⋅ ⋅ ⋅ r 2 1 + θ 2  (1 + θ ) 

(

)

(

a ⋅ 1 +θ 2

Así, el vector velocidad en función de θ está dado por:  v = a⋅

 v0 d ˆ ⋅ r + θ ⋅θˆ  ⋅ rˆ + θ ⋅θˆ + 2 dt 1+θ 

(

     

    

)

  θ   θ2 ˆ θ r + 2 ⋅  +  ⋅   2 a ⋅ (1 + θ 2 )   (1 + θ 2 )   (1 + θ )  v 02 = ⋅ - 2 ⋅θ + θ 3  ⋅ rˆ + 2 + θ 2  ⋅θˆ 2 2 a ⋅ (1 + θ ) v 02

=

)

{

   ⋅ θˆ   

}

d)  Si los vectores velocidad y aceleración son perpendiculares, se cumple que a • v = 0 . Reemplazando los resultados obtenidos en (a) y (c):

Reemplazando por el valor obtenido en a:

(

v0 tˆ =

1 +θ v0

2

1 +θ c) Calculando la aceleración

⋅ rˆ + θ ⋅ θˆ

2

⋅ 1+θ2

    

)

  a•v= =

rˆ + θ ⋅θˆ 1 +θ 2

= =

v 02

a ⋅ (1 + θ

v 02

a ⋅ (1 + θ v 02

a ⋅ (1 + θ

)

2 2

)

2 2

⋅ ⋅

)

2 2

{

}

⋅ - 2 ⋅θ + θ 3  ⋅ rˆ + 2 + θ 2  ⋅θˆ •

v0 1 +θ

2

v0 1 +θ 2

{

}

{

}

⋅ - ( 2 ⋅θ + θ 3 ) + ( 2 + θ 2 ) ⋅θ

⋅ - ( 2 ⋅θ + θ 3 ) + ( 2 ⋅θ + θ 3 ) = 0

v0 1 +θ 2

(

⋅ rˆ + θ ⋅ θˆ

)

Fe de erratas: los resultados son correctos si la velocidad aumenta a un ritmo constante de 20 [mm/s2] y no 2 [mm/s2]

En el punto de contacto C de la leva A con el seguidor B, lasvelocidades tangenciales son iguales, es decir v AC = v BC

La aceleración total de este sistema se compone de: Las aceleraciones normales o centrípetas de la leva y el seguidor en el punto C son, respectivamente: aAC =

2 v AC 2 ⇒ v AC = aAC ⋅ RA RA

aBC =

2 v BC 2 ⇒ v BC = aBC ⋅ RB RB

• •

  Una aceleración tangencial at , de magnitud at = 20  mm 2  constante. s     v2 Una aceleración normal o centrípeta acp , de magnitud acp = , donde R R = 90 [mm] es el radio de la ranura y v = v ( t ) = 20 ⋅ t  mm  es su s  velocidad tangencial.

a) En el instante t = 0: como el pasador parte del reposo, v(0) = 0. Donde RA es la distancia del centro de masa de la leva a C, y RB es el radio del seguidor. Igualando las velocidades: v AC = v BC ⇒ aAC ⋅ RA = aBC ⋅ RB ⇒ RB =

aAC ⋅ RA aBC 2

Luego, la aceleración centrípeta en ese instante se hace nula. En consecuencia, sólo hay componente tangencial de aceleración, por lo que la magnitud de la aceleración es igual a la de la aceleración tangencial, es decir   a = at = 20 mm 2  s  

2

Reemplazando los datos RA = 65 [mm], aAC = 0.65 [m/s ] y aBC = 6.68 [m/s ]: RB =

0.65 ⋅ 65 [ mm ] = 6.325 [ mm ] 6.68

b) En el instante t = 2 [s], la velocidad tangencial del pasador tiene magnitud v(2) = 20·2 [mm/s] = 40 [mm/s]. Luego, la magnitud de la aceleración centrípeta es:

Finalmente, el diámetro del seguidor está dado por

(

40 mm   s  acp = 90 [ mm ]

d B = 2 ⋅ RB = 2 ⋅ 6.325 [ mm ] = 12.65 [ mm ]

)

2

=

160 mm  s 2  9 

Finalmente, la magnitud de la aceleración total está dada por:  a =

 at

2

 + acp

2

=

( 20 )

2

2

 160   mm  + = 26.76  mm 2    s 2  s   9  

A partir de la figura adjunta, se puede deducir que x = L ⋅ cos (θ )

De la figura tg (θ ) =

y = L ⋅ sen (θ )

s ⇒ s = s (θ ) = b ⋅ tg (θ ) b

y

Donde s(θ) es la posición de la partícula. La velocidad de la partícula está dada por:

s (θ )

x

Derivando ambas expresiones con respecto al tiempo. v x = xɺ = -L ⋅ sen (θ ) ⋅ θɺ v = yɺ = L ⋅ cos (θ ) ⋅ θɺ

v s = sɺ = b ⋅ sec 2 (θ ) ⋅ θɺ

y

La aceleración de la partícula está dada por: as = vɺs = b ⋅  sec 2 (θ ) ⋅ θɺɺ + θɺ ⋅ 2 ⋅ sec (θ ) ⋅ sec (θ ) ⋅ tg (θ ) = b ⋅ sec 2 (θ ) ⋅ θɺɺ + 2 ⋅ tg (θ ) ⋅ θɺ 

Como, en el instante en que θ = θ0 la velocidad del extremo inferior de la escalera es v: v x = v = -L ⋅ sen (θ 0 ) ⋅ θɺ ⇒ θɺ = −

v L ⋅ sen (θ 0 )

Reemplazando este valor: v y = − L ⋅ cos (θ 0 ) ⋅

v = -v ⋅ ctg (θ 0 ) L ⋅ sen (θ 0 )

Derivando las expresiones de las velocidades con respecto a t para obtener las aceleraciones:

(

) )

ax = vɺ x = -L ⋅ sen (θ ) ⋅ θɺɺ + cos (θ ) ⋅ θɺ2

(

ay = vɺ y = L ⋅ cos (θ ) ⋅θɺɺ − sen (θ ) ⋅ θɺ2

Como, en el instante en que θ = θ0 la aceleración del extremo inferior de la escalera es a, por lo que:

  v2 ax = a = -L ⋅ sen (θ 0 ) ⋅θɺɺ + cos (θ 0 ) ⋅θɺ2 = -L ⋅  sen (θ 0 ) ⋅θɺɺ + cos (θ 0 ) ⋅ 2  ⇒ 2  θ L ⋅ sen ( ) 0  

yɺɺ = -

 a v 2 ⋅ ctg (θ0 )   v 2 ⋅ ctg (θ0 )  1 − + 2 ⋅ a +  = sen (θ0 ) ⋅ θɺɺ ⇒ θɺɺ = −  L ⋅ sen (θ0 )  L ⋅ sen (θ0 )   L L ⋅ sen (θ0 ) 

=-

v 2  1 ⋅  (1 + ctg 2 (θ 0 ) ) + cos (θ 0 ) ⋅ a  sen (θ0 )  L 

=-

v 2  1 ⋅  csc 2 (θ 0 ) + cos (θ 0 ) ⋅ a  sen (θ0 )  L 

(

)

Finalmente, la aceleración del extremo superior de la escalera está dada por:    v 2 ⋅ ctg (θ 0 )  1 v2 ay = L ⋅  - cos (θ 0 ) ⋅ ⋅ a + − sen (θ 0 ) ⋅ 2   2  θ θ θ L ⋅ sen L ⋅ sen L ⋅ sen ( 0)  ( 0 )  ( 0)  =−

  v 2 ⋅ ctg (θ0 )  v 2 L ⋅  cos (θ 0 ) ⋅ a + + L ⋅ sen (θ0 )  L ⋅ sen (θ0 )  L 

=−

   1 v2 v2  1 v2 ⋅  a ⋅ cos (θ 0 ) + ⋅ ctg 2 (θ 0 ) + ⋅  a ⋅ cos (θ 0 ) + ( ctg 2 (θ 0 ) + 1 )  =− sen (θ 0 )  L L  sen (θ 0 )  L 

=−

  v2 1 ⋅  a ⋅ cos (θ 0 ) + csc 2 (θ 0 )  sen (θ 0 )  L 

   

Otra forma de solucionarlo De la figura, aplicando el Teorema de Pitágoras x 2 + y 2 = L2 Derivando con respecto del tiempo 2 ⋅ x ⋅ xɺ + 2 ⋅ y ⋅ yɺ = 0 ⇒ x ⋅ xɺ + y ⋅ yɺ = 0 ⇒ yɺ = -

x ⋅ xɺ y

Sabiendo que x = L ⋅ cos (θ 0 ) , y = L ⋅ sen (θ0 ) y xɺ = v yɺ = -

L ⋅ cos (θ0 ) ⋅ v = −v ⋅ ctg (θ 0 ) L ⋅ sen (θ 0 )

Derivando nuevamente con respecto del tiempo: 1 xɺ ⋅ xɺ +x ⋅ xɺɺ + yɺ ⋅ yɺ + y ⋅ yɺɺ = 0 ⇒ xɺ 2 +x ⋅ xɺɺ + yɺ 2 + y ⋅ yɺɺ = 0 ⇒ yɺɺ = - ⋅  xɺ 2 +x ⋅ xɺɺ + yɺ 2  y Sabiendo que x = L ⋅ cos (θ 0 ) , y = L ⋅ sen (θ0 ) , xɺ = v , xɺɺ = a e yɺ = − v ⋅ ctg (θ0 )

1 ⋅ v 2 + L ⋅ cos (θ0 ) ⋅ a + v 2 ⋅ ctg 2 (θ 0 ) L ⋅ sen (θ 0 ) 

 El vector posición está dado por r ( t ) = r ( t ) ⋅ rˆ , con r ( t ) = 100 ⋅ t 2 La rapidez angular está dada por θ ( t ) = t 3 Calculando el vector velocidad:   dr dr ˆ drˆ ɺ ˆ drˆ dθ v= = ⋅r + r⋅ = r⋅r + r⋅ ⋅ = 200 ⋅ t ⋅ rˆ + 100 ⋅ t 2 ⋅ θˆ ⋅ 3 ⋅ t 2 = 200 ⋅ t ⋅ rˆ + 300 ⋅ t 4 ⋅ θˆ dt dt dt dθ dt 

En el instante t = tn, v ( tn ) = 200 ⋅ tn ⋅ rˆ + 300 ⋅ tn4 ⋅θˆ

 El vector posición está dado por r (θ ) = r (θ ) ⋅ rˆ , con r (θ ) = R ⋅ (1-cos (θ ) ) , con R = 0.5 [m] La rapidez angular está dada por θ ( t ) = t 3 Calculando el vector velocidad:

Calculando el vector aceleración:   dv drˆ dθˆ a= = 200 ⋅ rˆ + 200 ⋅ t ⋅ + 1200 ⋅ t 3 ⋅ θˆ + 300 ⋅ t 4 ⋅ dt dt dt ˆ drˆ dθ dθ dθ = 200 ⋅ rˆ + 200 ⋅ t ⋅ ⋅ + 1200 ⋅ t 3 ⋅ θˆ + 300 ⋅ t 4 ⋅ ⋅ dθ dt dθ dt = 200 ⋅ rˆ + 200 ⋅ t ⋅ θˆ ⋅ 3 ⋅ t 2 + 1200 ⋅ t 3 ⋅ θˆ − 300 ⋅ t 4 ⋅ rˆ ⋅ 3 ⋅ t 2 = ( 200 - 900 ⋅ t 6 ) ⋅ rˆ + 1800 ⋅ t 3 ⋅ θˆ

  dr dr ˆ drˆ dr dθ ˆ drˆ dθ v= = ⋅r + r ⋅ = ⋅ ⋅r + r ⋅ ⋅ dt dt dt dθ dt dθ dt = R ⋅ sen (θ ) ⋅θɺ ⋅ rˆ + R ⋅ (1-cos (θ ) ) ⋅θɺ ⋅θˆ = R ⋅θɺ ⋅ sen (θ ) ⋅ rˆ + (1-cos (θ ) ) ⋅θˆ   Despejando θɺ en función de θ, R y la magnitud de v :  2 v = R ⋅θɺ ⋅ sen 2 (θ ) + (1-cos (θ ) ) ⇒ θɺ =

 En el instante t = tn, a ( tn ) = ( 200 - 900 ⋅ tn6 ) ⋅ rˆ + 1800 ⋅ t3n ⋅ θˆ

 v R ⋅ sen2 (θ ) + (1-cos (θ ) )

2

En el instante en que θ = 180º = π [rad], la velocidad de la espiga tiene magnitud v0. Así:

θɺ (θ = π ) =

v0 R ⋅ sen 2 ( π ) + (1-cos ( π ) )

Calculando el vector aceleración:

2

=

v0 R ⋅ 0 +2 2

2

=

v0 v = 0 = v0 2 ⋅R 2 ⋅1 2

 d  dv d drˆ  dθˆ  a= = R⋅ sen (θ ) ⋅ θɺ ⋅ rˆ + sen (θ ) ⋅θɺ ⋅  + R ⋅  (1-cos (θ ) ) ⋅θɺ ⋅θˆ + (1-cos (θ ) ) ⋅θɺ ⋅ dt  dt dt dt dt    

(

(

)

)

ˆ d  d dθ ˆ drˆ dθ  = R⋅ sen (θ ) ⋅ θɺ ⋅ ⋅ r + sen (θ ) ⋅ θɺ ⋅ ⋅ + R⋅ (1-cos (θ ) ) ⋅ θɺ ⋅ ddtθ ⋅θˆ + (1-cos (θ ) ) ⋅θɺ ⋅ ddθθ ⋅ ddtθ  dt dθ dt  d θ  dθ  

(

(

)

( ( (

)

)

(

)

= R ⋅θɺ ⋅  sen (θ ) ⋅ θɺɺ + cos (θ ) ⋅ θɺ ⋅ rˆ + sen (θ ) ⋅θɺ ⋅θˆ  + R ⋅θɺ ⋅  (1-cos (θ ) ) ⋅θɺɺ + sen (θ ) ⋅θɺ ⋅θˆ - (1-cos (θ ) ) ⋅θɺ ⋅ rˆ    ˆ (1-cos (θ ) ) ⋅θɺɺ + sen (θ ) ⋅θɺ + sen (θ ) ⋅θɺ ⋅θˆ  = R ⋅ θɺ ⋅  sen (θ ) ⋅ θɺɺ + cos (θ ) ⋅ θɺ- (1-cos (θ ) ) ⋅θɺ ⋅ r+   ɺ ɺɺ ɺ ɺɺ ɺ ˆ   ˆ = R ⋅ θ ⋅ sen (θ ) ⋅θ + θ ⋅ ( 2 ⋅ cos (θ ) -1) ⋅ r+ (1-cos (θ ) ) ⋅θ + 2 ⋅ sen (θ ) ⋅θ ⋅θ  

) (

) (

)

)

En el instante en que θ = 180º = π [rad], la aceleración de la espiga tiene v magnitud a0 y θɺ = 0 Así: 2 ⋅R 2 2   v v v     a = a0 = R ⋅ 0 ⋅  sen (π ) ⋅θɺɺ + 0 ⋅ ( 2 ⋅ cos (π ) -1)  +  (1-cos (π ) ) ⋅ θɺɺ + 2 ⋅ sen ( π ) ⋅ 0   2 ⋅ R  2 ⋅R 2 ⋅ R    

=

2 v 0  3 ⋅ v 0  ɺɺ ⋅  −  + 2 ⋅θ 2  2 ⋅ R 

⇒ θɺɺ =

(

)

2

a02 9 ⋅ v 02 a2 = 02 − − 2 2 v 0 16 ⋅ R v0

En la figura, se puede calcular el ángulo γ’ tg ( γ ' ) =

Así, γ = 180º - γ ' = 112.62º

1  16 ⋅   2 

2

=

a02 9 ⋅ v 02 − v 02 4 ⋅ R 2

α

γ

γ'

Por otra parte, por teorema del coseno:

 4 ⋅ a02 9 ⋅ v 02 1  4 ⋅ a 2 9 ⋅ v 02  + 4 ⋅θɺɺ2 ⇒ θɺɺ2 = ⋅  2 0 − ⇒ 2 =  2 v0 4 ⋅R 4  v0 4 ⋅ R2   9 ⋅ v 02

β

12 ⇒ γ ' = 67.38º 5

20 2 = x 2 + q 2 - 2 ⋅ x ⋅ q ⋅ cos (γ ) Derivando con respecto a t: 0 = 2 ⋅ x ⋅ xɺ + 2 ⋅ q ⋅ qɺ - 2 ⋅ cos (γ ) ⋅ ( xɺ ⋅ q + x ⋅ qɺ ) ⇒ cos ( γ ) ⋅ ( xɺ ⋅ q + x ⋅ qɺ ) = x ⋅ xɺ + q ⋅ qɺ ⇒ qɺ ⋅ ( x ⋅ cos ( γ ) - q ) = xɺ ⋅ ( x - q ⋅ cos ( γ ) ) ⇒ qɺ = xɺ ⋅

x - q ⋅ cos (γ )

x ⋅ cos ( γ ) - q

Se sabe que para x = 10 [m], xɺ = 2 m  . Basta calcular q en ese instante  s ɺ para poder determinar la rapidez q del punto B. Aplicando Teorema del Seno en el triángulo ABC 20 x x 10 = ⇒ sen ( β ) = ⋅ sen ( γ ) = ⋅ sen (112.62º ) = 0.4615 sen ( γ ) sen ( β ) 20 20 ⇒ β = 27.49º Así, α = 180º - ( β + γ ) = 39.89º . Aplicando nuevamente el Teorema del Seno: sen (α ) sen ( 39.89º ) 20 q = ⇒ q = 20 ⋅ = 20 ⋅ = 13.90 [ m ] sen ( γ ) sen (α ) sen ( γ ) sen (112.62º ) Finalmente, reemplazando datos qɺ = 2 ⋅

10 - 13.90 ⋅ cos (112.62º ) m   = -1.73  m   s 10 ⋅ cos (112.62º ) - 13.90  s 

 El vector posición está dado por r (θ ) = r (θ ) ⋅ rˆ , con r (θ ) = a ⋅ e k ⋅θ Si la rapidez angular de rotación es constante θɺ , la aceleración angular es igual a cero

 El vector posición está dado por r (θ ) = r (θ ) ⋅ rˆ , con r (θ ) = 2 ⋅ a ⋅ cos (θ ) . Además, θ =

Calculando el vector velocidad   dr dr ˆ drˆ dr dθ ˆ drˆ dθ v= = ⋅r + r ⋅ = ⋅ ⋅r + r ⋅ ⋅ = k ⋅ a ⋅ e k ⋅θ ⋅θɺ ⋅ rˆ + a ⋅ e k ⋅θ ⋅ θɺ ⋅ θˆ dt dt dt dθ dt dθ dt Así, v r = k ⋅ a ⋅ ek ⋅θ ⋅ θɺ y v θ = a ⋅ e k ⋅θ ⋅θɺ .

a) Calculando el vector velocidad   dr dr drˆ dr d θ ˆ drˆ dθ v= = ⋅ rˆ + r ⋅ = ⋅ ⋅r + r ⋅ ⋅ = −2 ⋅ a ⋅ sen (θ ) ⋅ b ⋅ t ⋅ rˆ + 2 ⋅ a ⋅ cos (θ ) ⋅ b ⋅ t ⋅ θˆ dt dt dt dθ dt dθ dt = 2 ⋅ a ⋅ b ⋅ t ⋅ - sen (θ ) ⋅ rˆ + cos (θ ) ⋅ θˆ

(

Calculando el vector aceleración a partir del vector velocidad:    dv  drˆ ɺ  dθˆ ɺ  a= = k ⋅ a ⋅θɺ ⋅  k ⋅ e k ⋅θ ⋅θɺ ⋅ rˆ + ek ⋅θ ⋅ ⋅θ  + a ⋅θɺ ⋅  k ⋅ e k ⋅θ ⋅ θɺ ⋅θˆ + e k ⋅θ ⋅θ  dt dθ  dθ    = k ⋅ a ⋅ θɺ 2 ⋅ ek ⋅θ ⋅ k ⋅ rˆ + θˆ + a ⋅ θɺ 2 ⋅ e k ⋅θ ⋅ k ⋅ θˆ − rˆ

(

)

(

= a ⋅ θɺ2 ⋅ e k ⋅θ ⋅ ( k 2 -1 ) ⋅ rˆ + 2 ⋅ k ⋅ a ⋅ θɺ2 ⋅ e k ⋅θ ⋅ θˆ

)

Así, ar = a ⋅θɺ2 ⋅ ek ⋅θ ⋅ ( k 2 -1) y aθ = 2 ⋅ k ⋅ a ⋅ θɺ2 ⋅ e k ⋅θ .

b⋅t2 dθ , de donde θɺ = = b⋅t 2 dt

)

La magnitud de la velocidad v está dado por:  v = 2 ⋅ a ⋅ b ⋅ t ⋅ sen 2 (θ ) + cos 2 (θ ) = 2 ⋅ a ⋅ b ⋅ t

Calculando el vector aceleración:   dv d d a= = [2 ⋅ a ⋅ b ⋅ t ] ⋅ - sen (θ ) ⋅ rˆ + cos(θ ) ⋅ θˆ + 2 ⋅ a ⋅ b ⋅ t ⋅ - sen (θ ) ⋅ rˆ + cos(θ ) ⋅ θˆ dt dt dt d dθ = 2 ⋅ a ⋅ b ⋅ - sen(θ ) ⋅ rˆ + cos(θ ) ⋅ θˆ + 2 ⋅ a ⋅ b ⋅ t ⋅ - sen (θ ) ⋅ rˆ + cos(θ ) ⋅ θˆ ⋅ dθ dt  ˆ dr dθˆ  = 2 ⋅ a ⋅ b ⋅ - sen(θ ) ⋅ rˆ + cos(θ ) ⋅ θˆ + 2 ⋅ a ⋅ b 2 ⋅ t 2 ⋅ - cos (θ ) ⋅ rˆ - sen(θ ) ⋅ − sen (θ ) ⋅ θˆ + cos(θ ) ⋅  dθ dt   = 2 ⋅ a ⋅ b ⋅ - sen(θ ) ⋅ rˆ + cos(θ ) ⋅ θˆ + 2 ⋅ a ⋅ b 2 ⋅ t 2 ⋅ - cos (θ ) ⋅ rˆ - sen(θ ) ⋅ θˆ − sen(θ ) ⋅ θˆ − cos(θ ) ⋅ rˆ

(

[

)

(

)

(

)

( ( {

) )

]

[

]

[ [

] ] }

= 2 ⋅ a ⋅ b ⋅ - sen(θ ) ⋅ rˆ + cos(θ ) ⋅ θˆ − 4 ⋅ a ⋅ b 2 ⋅ t 2 ⋅ cos(θ ) ⋅ rˆ + ⋅sen(θ ) ⋅ θˆ = 2 ⋅ a ⋅ b ⋅ (- sen(θ ) − 2 ⋅ b ⋅ t 2 ⋅ cos(θ )) ⋅ rˆ + cos(θ ) − 2 ⋅ b ⋅ t 2 ⋅ sen(θ ) ⋅ θˆ

[

]

La magnitud de la aceleración a está dada por:  a = 2 ⋅a ⋅ b ⋅

[sen(θ ) + 2 ⋅ b ⋅ t

2

] [

]

⋅ cos(θ ) + cos (θ ) − 2 ⋅ b ⋅ t 2 ⋅ sen(θ ) 2

2

= 2 ⋅ a ⋅ b ⋅ sen 2 (θ ) + 4 ⋅ b ⋅ t 2 ⋅ cos (θ ) ⋅ sen (θ ) + 4 ⋅ b 2 ⋅ t 4 ⋅ cos 2 (θ ) + cos 2 (θ ) − 4 ⋅ b ⋅ t 2 ⋅ cos(θ ) ⋅ sen (θ ) + 4 ⋅ b 2 ⋅ t 4 ⋅ sen 2 (θ )

[

]

= 2 ⋅ a ⋅ b ⋅ sen 2 (θ ) + cos 2 (θ ) + 4 ⋅ b 2 ⋅ t 4 ⋅ cos 2 (θ ) + sen 2 (θ ) = 2 ⋅ a ⋅ b ⋅ 1 + 4 ⋅ b2 ⋅ t 4

v3 b) El radio de curvatura de la curva está dado por ρ =   v ×a Haciendo el producto cruz:

  v ×a =

= zˆ ⋅

θˆ

- 2 ⋅ a ⋅ b ⋅ t ⋅ sen (θ )

(

2 ⋅ a ⋅ b ⋅ t ⋅ cos (θ )

)

(

- 2 ⋅ a ⋅ b ⋅ sen (θ ) + 2 ⋅ b ⋅ t 2 ⋅ cos (θ ) - 2 ⋅ a ⋅ b ⋅ t ⋅ sen (θ )

(

)

- 2 ⋅ a ⋅ b ⋅ sen (θ ) + 2 ⋅ b ⋅ t 2 ⋅ cos (θ )

[

2

(

[

= zˆ ⋅ 4 ⋅ a ⋅ b ⋅ t ⋅ - sen (θ ) ⋅ cos (θ ) + 2 ⋅ b ⋅ t 2

2

)

2 ⋅ a ⋅ b ⋅ cos (θ ) − 2 ⋅ b ⋅ t 2 ⋅ sen (θ )

0

2 ⋅ a ⋅ b ⋅ t ⋅ cos (θ )

(

) ⋅ sen (θ )) + 4 ⋅ a ⋅ b ⋅ t ⋅ cos (θ )⋅ (sen (θ ) + 2 ⋅ b ⋅ t ⋅ sen (θ ) + ⋅sen (θ ) ⋅ cos (θ ) + 2 ⋅ b ⋅ t ⋅ cos (θ )]

2 ⋅ a ⋅ b ⋅ cos (θ ) − 2 ⋅ b ⋅ t 2 ⋅ sen (θ )

= zˆ ⋅ - 4 ⋅ a ⋅ b ⋅ t ⋅ sen (θ ) ⋅ cos (θ ) − 2 ⋅ b ⋅ t 2

zˆ 0

2

2

2

2

2

2

2

= 8 ⋅ a 2 ⋅ b 3 ⋅ t 3 ⋅ zˆ



)]

⋅ cos (θ )

2

 a) El vector posición está dado por r (θ ) = r (θ ) ⋅ rˆ , con r (θ ) = A ⋅ ek ⋅θ



Luego, v × a = 8 ⋅ a 2 ⋅ b 3 ⋅ t 3

 Sabiendo que v = v = 2 ⋅ a ⋅ b ⋅ t , se puede calcular el radio de curvatura

  dr dr ˆ drˆ dr dθ ˆ drˆ dθ v= = ⋅r + r ⋅ = ⋅ ⋅r + r ⋅ ⋅ = k ⋅ A ⋅ e k ⋅θ ⋅ θɺ ⋅ rˆ + A ⋅ ek ⋅θ ⋅ θɺ ⋅ θˆ dt dt dt dθ dt dθ dt = A ⋅ e k ⋅θ ⋅ θɺ ⋅ k ⋅ rˆ + θˆ = r (θ ) ⋅θɺ ⋅ k ⋅ rˆ + θˆ

haciendo:

ρ=

(2 ⋅ a ⋅ b ⋅ t )3 8 ⋅ a2 ⋅ b3 ⋅ t 3

=

8 ⋅ a3 ⋅ b3 ⋅ t 3 =a 8 ⋅ a2 ⋅ b3 ⋅ t 3

(

)

(

)

Si la magnitud de la velocidad es v0 constante:  v = v0 = r (θ ) ⋅ θɺ ⋅ k 2 +1 ⇒ θɺ =

v0 r (θ ) ⋅ k 2 +1

Reemplazando en la expresión para la velocidad:  v = r (θ ) ⋅

v0 r (θ ) ⋅ k +1 2

(

)

⋅ k ⋅ rˆ + θˆ =

v0 k 2 +1

(

⋅ k ⋅ rˆ + θˆ

)

Calculando el vector aceleración:   dv a= = dt

 drˆ dθˆ  d v0 k ⋅ rˆ + θˆ = ⋅ k ⋅ + = k 2 +1 dt k 2 +1  dt dt 

v0

v 02 ⋅  k ⋅ θˆ − rˆ  r (θ ) ⋅ ( k 2 +1 ) 

=

k 2 +1

v0

(

⋅ θɺ  k ⋅ θˆ − rˆ  =

)

 drˆ dθ dθˆ dθ  ⋅ k ⋅ ⋅ + ⋅  k 2 +1  dθ dt dθ dt  v0

b)  Si los vectores velocidad y aceleración son perpendiculares, se cumple que a • v = 0 . Reemplazando los resultados obtenidos en (a):       v 02 v 03  k ⋅ θˆ − rˆ   •  v 0 ⋅ k ⋅ rˆ + θˆ  = a•v=  ⋅ ⋅ [k - k ] = 0 3 2  2  r (θ ) ⋅ ( k +1 )   r (θ ) ⋅ ( k 2 +1) 2   k +1

(

c) De la ecuación para la velocidad angular

)

θɺ =

v0 v0 dθ = ⇒ ek ⋅θ ⋅ dθ = ⋅ dt θ k ⋅ 2 dt A ⋅ e ⋅ k +1 A ⋅ k 2 +1

Integrando: θ

t

v0

k ⋅θ ∫ e ⋅ dθ = ∫

⋅ dt ⇒

1 k ⋅θ ⋅e k

θ

=

v0 ⋅ t

t

0 A ⋅ k 2 +1 A ⋅ k 2 +1 0 v0 ⋅ t v0 ⋅ k ⋅ t 1 ⇒ ⋅ ( e k ⋅θ -1) = ⇒ e k ⋅θ -1 = 2 k A ⋅ k +1 A ⋅ k 2 +1 0

0

⇒ e k ⋅θ =

 v ⋅k ⋅t  +1 ⇒ k ⋅θ = ln  0 +1 2 A ⋅ k +1  A ⋅ k +1  v0 ⋅ k ⋅ t

⇒ θ = θ (t ) =

2

 v ⋅k ⋅t  1 ⋅ ln  0 +1 2 k  A ⋅ k +1 

Derivando se obtiene:

θɺ =

v ⋅k ⋅t v0 ⋅ t 1 1 ⋅ ⋅ 0 = k v 0 ⋅ k ⋅ t +1 A ⋅ k 2 +1 v 0 ⋅ k ⋅ t + A ⋅ k 2 +1 A ⋅ k 2 +1

Probs resueltos 1