Problemas resueltos de dinamica

UNIVERSIDAD NACIONAL EXPERIMENTAL EXPERIMENTAL “FRANCISCO DE MIRANDA” ÁREA TECNOLOGÍA DEPARTAMENTO DE ESTRUCTURA

MECANICA RACIONAL GUIA UNIDAD I y II

PROFESORES:

José Contreras Giovanny Galoti 

Joan Gil

ABRIL 2010 EJERCICIOS RESUELTOS

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1. Una partícula se mueve en el espacio de tal forma que: 2

r =2 a t 

;

θ= πt   ;

2

 z =5 t 

Donde r y z están en metros, θ en radianes y t en seg. Determine los vectores velocidad y aceleración en el instante en que la componente radial de la aceleración {a} rsub {r} es 0 v⃗ =? a⃗ =? −−−− →Parat =? cuan cuandoa doa r =0 v´ =v r e r^ + v θ e θ^ + v z e z^ θ= πt  θ´ = π  θ´ = 0

r  !a t 2 ´r   "at   "a ´r  # 

2

z =5 t   z =10 t  z =10

v r =´r = 4 at  v θ =r θ´ =( 2 a t  ) . ( π )=2 πa t  2

2

v z = z´ =10 t  ar =0 −−→ t =?

$ara cuando

ar =´r −θ ´r =0 4 a−( 2 a t  ) . ( π ) = 0 4 a −2 π  a t  =0 2

t =

√

4a 2

2 π  a

t =

2

√

2 2

π 

t =

2

2

2

√ 2 π 

%ustituyendo en las componentes de la velocidad: vr =

4

√ 2 a π  2

vθ

√ 2 =2 πa ( ) =4 a / π  π 

v z=10 √ 2 / π 

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a´ =ar e r^ + a θ eθ^ + a z e z^ ar =0

aθ =( r ´θ + 2 ´r θ´ )=2 a t  .0 + 4 atπ =8 πat  a z= z´ =10 2

&valuando

t =

a ⃗ para

√ 2  seg. π 

ar =0 aθ =8 √ 2 a a z =10 a´ =0 e r^ + 8 √ 2 a eθ^ + 10 e z^

!. 'a ace acellera eración ión de un co(e co(ete te dur dura ante nte un inter interva vallo )rev )reve e la da la ecuac ecuació ión n a =45 −3 t + 2 t 2 . *l princ princip ipio io del del inte interv rval alo, o, la posi posici ción ón y la velo veloci cida dad d del del co(e co(ete te son son !+ !+ pies pies y --0 --0 pies piess; s; resp respec ecti tiva vame ment nte. e. Determine la posición, la velocidad y la aceleración del co(ete cuando t =4  seg.

Solución Datos: 2 a =45 −3 t + 2 t  $ara t  0 seg s,v, a ,  / •



v

  --0 piesseg

o



'a aceleración para el tiempo t  " seg será: at =4= 45− 3 ( 4 ) + 2 ( 4 )

2

•

so  !+ pies y t  " seg

2

at =4= 65 pies / seg

 piesseg!

'a velocidad se o)tiene integrando la ecuación de aceleración tf 

tf 

∫¿

tf 

∫¿

v = a dt 

∫¿ 45− 3 t + 2 t  dt pies / seg 2

v = a dt v f −v o=

v

v

=45 t 

/

2

3 2 t 

+ 2 / 3 t  4 pies / seg 3

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•

'a posición del co(ete para t  " seg puede o)tenerse integrando la ecuación de velocidad tf 

tf 

∫¿ v dt 

tf 

∫¿ v d t s −s =∫¿ 45 t −3 / 2 t  +2 / 3 t  dt pies

s=

2

s=

f 

3

o

2 3 4 s t = 4− st =0=45 / 2 t  −1 / 2 t  + 1 / 6 t  4  pies

0

2

3

4

s t = 4−275 =45 /2 ( 4 ) −1 / 2 ( t ) + 1 / 6 ( t )  pies

s t = 4=275 + 360 −32 + 42.67 pies  pies s t =4 =645,67  pies

. 'a acel aceler erac ació ión n de una una partíc rtícul ula a en movi movim mient iento o recti ectillíneo íneo est esta 2 ! e1presada por la ecuación a =−0.15 v  pulgseg . %i % o  0 y 2 o  3 pulgseg cuando t  0 seg, determine la posición, la velocidad y la aceleración de la partícula cuando t seg.

Solución Datos: 2 a =−0.15 v $ara t  0 seg s,v, a ,  /



so  0 pies y v t   seg

  3 pulgseg

o



a4 5alculo 5alculo de de la veloci velocidad dad de de la partíc partícula ula Si a =

dv  entoncesdv =a dt  dt 

Sustituyendo Sustituyendo y agrupando agrupando dv =( −0.15 v ) dt = ¿ 2

v f 

dv

dv

t f 

v

2

=−0.15 dt 

∫ v  =∫−0.15 dt 

 Integrando amboslados ambos lados de la ecuación

2

vo

t o

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1

1

v f 

− = 0.15 t  + 0.15 t o =¿

v f =

vo

f 

1

v f 

=0.15 t  + 0.15 t o + f 

 1

vo

1 1

0.15 t f  + 0.15 t o+

vo

&valuando en t   seg

1

v t =5=

( )+ 0.15 ( 0 ) +

0.15 5

1

1

 pulg / seg v t =5 =

( )+

0.15 5

(36 )

1

 pulg / seg

36

v t =5=1.29 pulg / seg )4 5alculo 5alculo de la posic posición ión de de la partícula partícula para t =  seg Si a =v

 dv v dv  entonces  entoncesdx dx = dx a

Sustituye Sustituyendo ndo dx=

v dv

−0.15 v

2

 x f 

∫

v f 

−1  dv ∫ 0.15 v

 Integrando amboslados ambos lados de la ecuación dx =  x o

vo

 − 1 v  ln ( v ) f   x x f =  x o 0.15 vo

 x f − x i=

−1 0.15

 ln ( v f ) +

1 1  −1  ln ( v i )=¿ x t =5 =  ln ( v 5 ) +  ln ( v 0 ) pulg 0.15 0.15 0.15

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&valuando en t   seg

 x t =5=

 − 1 0.15

 ln ( 1.29 )+

1  ln ( 36 ) pulg  pulg v t = 4=22.19  pulg 0.15

c4 'a acel aceler erac ació ión n de la part partíc ícul ula a a los los  seg seg se o)ti o)tien ene e al sust sustit itui uirr el valo valorr o)tenido de velocidad a los  seg en la ecuación a =−0.15 v 2

2

2

a =−0.15 ( 1.29 )  pulg / seg a =−0.25  pulg / seg

2

". &l movimiento nto curvilíneo de una par partícula se descr scri)e por las ecuaciones:  x =2−7 t 2  y =−4 t + 5 t 3  en las cuelas 1 e y están en pies y t en segundos. Determine las magnitudes y direcciones de los vectores de posición, velocidad y aceleración cuando t =  "seg.

Solución Datos: 2

 x =2−7 t  3  y =−4 t + 5 t  6agnitudes y direcciones de

´r , v´ y a´ =?

t = 4 seg

==7

Se !"#$e%e% &'s (e)$*'('s (e &'s e+,'+$!%es-

•

2

3

 x =2−7 t 

 y =−4 t + 5 t 

 x´ =−14 t 

 y´ =−4 + 15 t 

2

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•

r = √  x + y r =√ ( 2−7 t  ) +(−4 t + 5 t  )  pies

'a magnitud de r´ será:

2

2

2 2

3 2

 Parat   Para t = 4 segr = √ ( 2−7 ( 4 ) ) +(−4 ( 4 )+5 ( 4 ) ) r =323.28 pies 2 2

•

'a dirección de r´ y θ= tan

−1

θ= tan

−1

puede darse a trav8s del ángulo que forman 1 e

(

()

3 2

3

 y − 1 − 4 t + 5 t  θ= tan 2  x 2 −7 t 

(

)

)

−4 ( 4 )+ 5 ( 4 )3 θ=−70,10 2 2− 7 ( 4 )

´r =323,28 pies 70,10

•

'a magnitud de v´ será: v =√  x´ 2 + y´ 2 v = √ (− (−14 t )2+(−4 + 15 t 2 )2 pies / seg  Parat = 4 segr = √ (− (−14 ( 4 )) +(−4 + 15 ( 4 ) ) v =242,55 pies / seg 2

•

'a dirección de v´ e  y´ θ= tan

−1

()

( ( ) )

2 2

puede darse a trav8s del ángulo que forman  x´

 ´  y −1 −4 + 15 t  θ= tan  x´ −14 t 

(

4 + 15 4

2

2

)

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θ= tan

−1

θ= tan

−1

( ´´ ) = ( − ) ( ) ( −  ) =−  y −1  30 t  θ tan  x 14 30 4 14

θ

83.35

a´ =120,81 pies / seg 83,35 2

. 'a rot rotac ació ión n de la la )arr )arra a 9* con con resp respec ecto to de de 9 está está de deni nida da por por la la 2 relación relación θ= 2 t  , donde θ se e1presa en radianes y t en segundos. &l collarín  res)ala por la )arra de tal forma que su distancia desde 9 es 2 3 donde e r  se e1presa en pulgadas y t en segundos. r =60 t  −20 t  , dond 5uando t =¿  - s determínense a4 su velocidad, )4 su aceleración total. Utilice sistema de coordenadas tangenciales y normales

Solución Datos 2

3

r =60 t  −20 t  2 θ= 2 t  v´ y ´a=? parat =1 seg

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a´ =√ ( ´s et  ) +( ´s / # e n) 2

2

2

a4 'a velo velocid cidad ad ser será: á: v´ =´s et  ´v =r ´θ e t  ´v =( 60 t  −20 t  ) . ( 4 t ) e t  2

3

$ara t =1 seg v´ =( 60 ( 1 ) −20 ( 1 ) ) . ( 4 ( 1 ) ) e t  ´v =160 e t  pulg / seg 2

3

)4 a´t =´s e t  ´at =r ´θ et  ´at =( 60 t  − 20 t  ) ( 4 ) et  ´ a t =160 e t  pulg / seg 2

3

a´n =´s / # en 2

2

; donde

¿  #= r a´n= ¿

4 t  2

3

θ´ r¿

(60 t  −20 t  )(¿ )( ¿ ) ¿ ¿2 ¿ 4 ( 1)

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'a *celeración total será:

a =√ ( 160 et  ) +( 640 e n)  pulg / seg a =659.69 pulg / seg 2

2

2

2

3. 'a tra traye yect ctor oria ia de de una una part partíc ícul ula a $ es un un cara caraco col. l. &l &l movi movimi mien ento to de de la partícula está denido por las relaciones r  )