CINEMÁTICA PLANA DE UN CUERPO RÍGIDO1.docx

CINEMÁTICA PLANA DE UN CUERPO RÍGIDO 1. Problema 16.3 El gancho se mueve desde el reposo con aceleración de 20 /  . Si se encuentra unido a una cuerda que esta enrollada alrededor del tambor, determine la aceleración angular del tambor y su velocidad angular después que el tambor ha completado 10 revoluciones ¿Cuántas revoluciones más realizará el tambor después que haya completado las primeras 10 10 y el gancho continúe su descenso descenso durante 4 segundos?

Solución Tenemos como datos: at  r   0







20 pies / s

2

2 pies / s

0rev

    f  



10rev

 

0



0rad 

Como la cuerda esta enrollada sobre el tambor, entonces la aceleración será: a

t 



  r 

20     2     10rad

/s

2

Como   es constante, entonces se tiene que:

 (d )



 ( d  ) 

integrando



 f

0



 f 

 d      d      0

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2

 



/

2



2



2

2

2

 



2

 

/ 

0

        2        0  2 10  10 rev   20 10 rev   20  10  2 rad  

 0 

2 

2



0

2

0

 



0

  

0



20  10  2 rad 



   35.45rad / s

2. Problema 16-14 Un disco de 6 pulg de radio gira alrededor de un eje fijo con una velocidad angular de ω= (2t+3) rad/s, donde t esta en segundos. Determine las componentes tangencial y normal de la aceleración de un punto localizado en el borde del disco cuando el desplazamiento angular es θ=40 rad. Solución Sabemos que: d



 dt 

Integrando, se tiene



 

0

t 

d

  /  0 

   dt  0

t 

  2t  3 dt  0

   0   t 2  3t  / t 0    t 2  3t  Cuando θ= 40 rad 40  t t

2

2

 3t 

 3t  

40

0

 t  8  t   5  0 t  5s

Además

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Ahora los valores de los componentes tangencial y normal de la aceleración de un punto en el borde del disco serán:

6     r  2     1rad / s 2  12  6 2 2 2 an     r  13     84.5rad / s  12  at 

3. Problema 16-38 El bloque se mueve hacia la izquierda con una velocidad constante la velocidad y aceleración angular de la barra en función de θ.

Solución De la gráfica, la posición de la barra en función del ángulo θ será: tg

a





 x

x



a



tg  

x



actg  

Derivando respecto al tiempo, se tiene: 

 x



a





2



csc  d   

La velocidad de la barra

v

o



x

 d    vo   a csc2      dt   vo   a csc2      

vo a csc vo a

2

sen

  2

 

1

Ahora calculando la aceleración angular  Derivando respecto al tiempo la ecuación





1



 

v

0

. determine

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  

d   dt 

 vo  2 sen    a   vo  2 sen cos  d     d  



dt







 

vo

a

dt

 

 sen2 

a vo a

 vo  2  a sen   sen2   

v    o  a

2 2

 sen  sen 2 

4. Problema 16-41

La manivela AB gira a una velocidad angular constante de 5 rad/s. determine la velocidad del bloque C y la l a velocidad angular del eslabón BC cuando θ=30°

Solución

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De la figura anterior se obtiene que:  x 

0.6 cos 

0.6 sen





    0.15                   2

0.3 cos                     1

0.3sen

Desarrollando la ecuación (1) 2

0.3



 0.6

 sen  

 x 

0.6 co cos    0.3

 x 

0.6 cos 



0.3

 x 

0.6 cos 



0.09  0.36sen 

 x 

0.6 cos 



0.3sen

0.15



2

0.3 2



 0.6

sen   2



0.15 



2

0.3sen   0.0225

2

0.36sen    0.1125

Derivando dx dt



d 

 0.6 cos

dt  

vC   0.6 sen  



d   dt 

2

0.3 sen

 0.36sen   

0.3 cos 

 0.72 sen  cos  



 

vC  0.6 sen   AB 

vC   0.6 sen30  5 



d   dt 

2

0.3se sen

0.3 cos 

0 .1 1 2 5

 0.36sen    0.1125  0.72 sen  cos   AB

0.3 sen

2

 0.36sen    0.1125

; donde  AB

0.3 co cos 30 30  0.72 sen 30  cos 30 30  5 2

0.3 sen30  0.36sen 30  0.1125

3

 3.708 2 vC   5.208m / s

vC   

Derivando la ecuación (2) d dt

 0.6 sen  

0.6 c co os 

d dt

0.6 cos  AB

d  dt

 0.3sen   0.15  

d  



0.3 cos  



0.3 cos BC ; donde AB

dt

 

0.6cos30 6cos30  5  0.3co 3cos30  BC 

  BC   10rad / s



d dt

,  BC   

d   dt  



d  dt 

; donde    30

 

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5. Problema 16-43

El extremo A de la barra se mueve a la izquierda a una velocidad constante Determine la velocidad angular posición  x .

 

v A .

 y la aceleración angular    en función de su

Solución

De la figura se tiene que la posición x de l a barra en función del ángulo θ es:  sen 

r

r 

 x 

 x  r  csc 

sen 

 x

Derivando respecto al tiempo

dx dt dx



d  r  csc    dt  

d       r   csc  ctg     dt dt    

dx

  r csc  ctg ;   

d   

dt dt   Como la velocidad V esta hacia la izquierda, entonces   v  r csc ctg   

v r 

Ahora de

tag  sen 

 

   derivando respecto al tiempo 

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









d   dt



d v  tag  sen    dx  r   

v d d     sec2   sen  tag   cos    r dt dt    d  

v

 sec2   sen    tag   cos   ;   r dt  v v  v   sec2   sen  tag  sen   tag  cos  tag  sen     r r   r   2

v 2 2 2     sec   tg  sen   tag    cos     sen      r     2 2 2 2 2 2     v x r r r x r r               2 2  2 x x  r    x 2  r 2  x 2  r 2 x x 2  r 2      3 2 2 2 2 3 r x r      x r  v         3 3  2 2 2  r    x 2 x 2  r 2 x x  r       2 2 3 3 2 5  v   x r  r x  r         3   r     x 2 x 2  r 2      2 3 2 5  v   2r x  r          3   r    x 2 x 2  r 2     3  2 2 2 r x r  2     v   

























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6. Problema 16-46

Cuando θ=30°, la manivela AB gira a una velocidad y aceleración angulares de    10rad

/ s  y

2rad / s

2

, respectivamente. Determine la velocidad y aceleración angulares de la barra de conexión BC en este instante. Considere a 0.3m  y b 0.5 0.5m . 

 





Solución

Por ley de senos 0.5 0.3  sen  sen

sen  



  sen



0.6sen 

Cuando θ=30°  sen   0.6 sen30    0.6  0.5  0.3

 sen

arcsin  0.3    arcsin

   17.46



0.3 0.5

1

sen 

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   30    17.46   AB

 10rad

  BC  

/s

0.6cos30 cos17.46

10

  BC   5.45rad / s Derivando la ecuación (2)  sen     cos    0.6 sen    0.6 cos   2

2

 sen   cos    0.6 sen   0.6 cos    sen

2

cos  BC   BC  

 BC

 cos  BC  0.6 sen

 0.6 sen

0.6 sen

2  AB

2 AB

2 AB

 

 0.6 cos  AB  

  0.6 cos  AB  sen

 0.6 cos  AB  sen

2

2

 

BC 

 

BC 

cos   2

  BC  

 

6cos30 s30 (2) (2)  sen17.46 (5.45) 0.6 sen30(10)  0.6co

  BC   21.01rad / s

2

cos17.46 2

7. Problema 16-48

El hombre tira de la cuerda a una razón constante de 0.5 m/s. determine la velocidad y aceleración angulares de la viga AB cuando θ= 60°. La viga gira en torno a A. ignore el espesor de la viga y el tamaño de la polea.

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2 s s  72sen           ( ) si    60 2

72 cos 60 60   s  6 m  s  72  72 reemplazando 2(6) s  72 sen60       AB  0.096 rad

/s

Derivando (*)

 cos     sen     36 cos   2 2    s  s s  36  cos   sen       s s  s s

 s

0

2    s  36  cos cos    sen      2  0.52     cos     sen     36  2 2  1  0.5        cos      sen   36    si   60 2





2  0.52           cos 60    sen60  36  

1

d/

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Solucion

Determinaremos su posición, de la grafica anterior  y B



3sen 

Derivando respecto al tiempo

 y B  3 cos  

()

v B  3cos     si :   30;     10rad / s

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9. Problema 16-52

Si la cuña se mueve hacia la izquierda a una velocidad constante v, determine la velocidad angular de la barra en función de θ.

Solución

Pide la velocidad angular ω De la figura anterior, aplicando la ley de senos, se tiene  x

l 

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10. Problema 16-53

En el instante que se muestra, el disco gira a una velocidad angular ω y una aceleración angular α. Determine la velocidad y aceleración del cilindro B en este instante. Ignore el tamaño de la polea C.

Solucion

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v B

      sen   cos         15 

 2   sen   cos   15 

v B

a

a B

 sen    15 

B

2

   34  30cos 0cos       



34  30 30 co cos

 B

  34  30cos 0cos    2

15  sen  

 

34  30cos 0cos  



2



2

2

3 0 co cos    34  30

a

34  30cos 0cos  

cos   34  30 cos 

 sen     15 

30 sen  

1 34  30 cos    sen     2

  15

2

2

sen 

  

34  30 3 0 co cos  

cos   34  30 cos 



0cos     34  30cos

  15

2

2

sen 

  

3 2

11. Problema 16-68

Si la velocidad angular de la barra AB 4rad/s. determine determine la velocidad del bloque corredizo C en el instante que se muestra.

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v B



 

v B



4(0.15)

v B



0.6m / s

AB



r AB

En el eslabón BC

v C/ B  v B   C/ B  r C / B v C/B  0.6 cos 30i  0.6sen 30 j    CBk   0.2 cos 60i  0.6sen 60 j ˆ

ˆ

ˆ



 0.52i  0.3 j     k   0.1i  0.17 j    0.52i  0.3 j    0.1 j  0.17  i 

v C/B  v C/B



ˆ

ˆ

ˆ

ˆ

ˆ

ˆ

ˆ

ˆ

CB

CB

ˆ

ˆ

CB

ˆ

v C/B   0.52  0.17CB  i   0.3  0.1 CB  j ˆ

ˆ

vCi   0.52  0.17CB  i   0.3  0.1 CB  j ˆ

ˆ

ˆ

vC  0.52 0.52  0.17 0.17 CB

0.3  0.1C/ B  0   CB  3rad / s  vC  0.52  0.17  3  vC  1.04m / s 12. problema 16-73 Si la velocidad angular del eslabón AB es ω=4 rad/s en el instante que se muestra,

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v B  8 cos 60i  8sen 60 j ˆ

ˆ

v D   vD i

ˆ

 BD    BD k  ˆ

r D/ B  1i

ˆ

v D/ B  v B   D/ B  r D/ B v D/ B   8 cos 60i  8sen 60 j    BD k  1i ˆ

ˆ

v D/B  4i  6.93 j   BD j ˆ

ˆ

ˆ

v D i  4i   BD  6.93 j ˆ

ˆ

ˆ

v D  4  vD  4m / s BD  6.93  0   BD  6.93rad / s Por último en la barra DE

ˆ

  ˆ

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13. problema 16-81 si el bloque corredizo A se mueve a la derecha a v A



8 pies /

s , determine la velocidad de

los bloques B y C en el instante que se muestra. El elemento CD esta conectado por medio de un pasador al elemento ADB.

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v A   ADB  r D/ A

vD/ A



vD

8i



ˆ





2.83k  1.41i ˆ

 v D  x i   vD  y

ˆ

 v D  x i   vD  y

ˆ

ˆ

ˆ

ˆ

j  8i

ˆ



vD



2

4.01

 3.99

5.66 pies /

En la barra CD

 3.99

j  4.01i

ˆ

 v D  x  4.01 pies / s  v D  y  3.99 pies / s vD

 1.41

s

2



j

ˆ



j  3.99i

ˆ

ˆ

3.99 j

ˆ

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De la gráfica anterior

rC / IC vC



0.6tag 30  rC / IC

  

0.35 m

r 

  CD  CD

vC    6    0.6  vC   3.6m / s También

vC

  BC  r C / IC   BC  

   BC  

3.6 0.35

vC  r C / IC 

 10.39rad / s

v B   AB  r A/ IC 

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v B



 

v B



 

BC

BC 



r BC 

(0.4)

De la siguiente gráfica: aplicando el Centro Instantáneo (CI)

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Solucion

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2

r 

D/ IC 

30   0.32  0.152  2  0.3  0.15  cos 30

r D/ IC 



rD/ IC 

 0.186m

2

0.3

 0.152  2  0.3  30  0.15  cos 30

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v B

   r   5  2

v B

 10m /

vC

 vB    r C / B

i

ˆ



10

s

45 i

ˆ

10

45 45  j

ˆ



k

ˆ



25

30 i

ˆ

25

30  j

ˆ



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En el instante dado el elemento AB tiene el movimiento angular que se muestra. Determine la velocidad y la aceleración del bloque corredizo C en este instante.

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v B   AB  r AB  3  7 v B  21 pul / s vC  vB    r C / B i

ˆ

21i

ˆ

k

ˆ



5i

ˆ

12 j

ˆ



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 a B n    r  3  7  63in / s    2

2

2

 a B t    r  2  7  14in / s    2

a B  14i  63 j ˆ

ˆ

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