Solucionario de dinámica

solucionario de dinamicaCOSMOS: Complete Online Solutions Manual Organization System

Chapter 11, Solution 1.
x = t 2 − ( t − 3) m
3

v=
a= (a) Time at a = 0.

dx 2 = 2t − 3 ( t − 3) m/s dt
dv = 2 − 6 ( t − 3) m/s 2 dt

0 = 2 − 3 ( t0 − 3) = 10 − 3t0

t0 =

10 3

t0 = 3.33 s

(b)

Corresponding position and velocity.
⎛ 10 ⎞ ⎛ 10 ⎞ x = ⎜ ⎟ − ⎜ − 3 ⎟ = 11.074 m ⎝ 3⎠ ⎝ 3 ⎠ ⎛10 ⎞ ⎛ 10 ⎞ v = ( 2 ) ⎜ ⎟ − ( 3) ⎜ − 3 ⎟ = 6.333 m/s ⎝ 3⎠ ⎝ 3 ⎠
2 2 3

x = 11.07 m

v = 6.33 m/s

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 11,Solution 2.
x = t3 − (t − 2) m
2

v= a= (a) Time at a = 0.

dx = 3t 2 − 2 ( t − 2 ) m/s dt dv = 6t − 2 m/s 2 dt

0 = 6t0 − 2 = 0 t0 = 1 3

t0 = 0.333 s

(b)

Corresponding position and velocity.
⎛1⎞ ⎛1 ⎞ x = ⎜ ⎟ − ⎜ − 2 ⎟ = − 2.741 m ⎝3⎠ ⎝3 ⎠
3 2

x = − 2.74 m

⎛1⎞ ⎛1 ⎞ v = 3 ⎜ ⎟ − 2 ⎜ − 2 ⎟ = 3.666 m/s ⎝3⎠ ⎝3 ⎠

2

v = 3.67 m/s

Vector Mechanics for Engineers: Statics andDynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 11, Solution 3.
Position:
Velocity: x = 5t 4 − 4t 3 + 3t − 2 ft v= a= dx = 20t 3 − 12t 2 + 3 ft/s dt dv = 60t 2 − 24t ft/s 2 dt

Acceleration:
When t= 2 s,

x = ( 5 )( 2 ) − ( 4 )( 2 ) − ( 3)( 2 ) − 2 v = ( 20 )( 2 ) − (12 )( 2 ) + 3 a = ( 60 )( 2 ) − ( 24 )( 2 )
2 3 2

4

3

x = 52 ft v = 115 ft/s a = 192 ft/s 2

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.COSMOS: Complete Online Solutions Manual Organization System

Chapter 11, Solution 4.
Position:
Velocity: x = 6t 4 + 8t 3 − 14t 2 − 10t + 16 in.

v=
a=

dx = 24t 3 + 24t 2 − 28t − 10 in./s dt
dv = 72t 2 + 48t − 28 in./s 2 dt

Acceleration:
When t = 3 s,

x = ( 6 )( 3) + ( 8 )( 3) − (14 )( 3) − (10 )( 3) + 16 v = ( 24 )( 3) + ( 24 )( 3) − ( 28 )( 3) − 10 a = ( 72 )( 3) + ( 48 )( 3) − 282 3 2

4

3

2

x = 562 in. ! v = 770 in./s ! a = 764 in./s 2 !

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 11, Solution 5.
Position:Velocity: x = 500sin kt mm v= a= and k = 10 rad/s kt = (10 )( 0.05 ) = 0.5 rad x = 500sin ( 0.5 ) dx = 500k cos kt mm/s dt dv = − 500k 2 sin kt mm /s 2 dt

Acceleration: When t = 0.05 s,

x = 240 mm ! v = 4390 mm/s ! a = − 24.0 × 103 mm/s 2 !

v = ( 500 )(10 ) cos ( 0.5 )
a = − ( 500 )(10 ) sin ( 0.5 )
2

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. RussellJohnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 11, Solution 6.
Position: Where Let
x = 50sin k1t − k2t 2 mm k1 = 1 rad/s dθ = (1 − t ) rad/s dt x = 50sin θ mm v= and k2 = 0.5 rad/s 2

(

)

θ = k1t − k2t 2 = t − 0.5t 2 rad
and d 2θ = −1rad/s 2 dt 2

Position:
Velocity:

Acceleration:

dx dθ = 50cosθ mm/s dt dt dv a= dt a = 50cosθ d 2θ ⎛ dθ ⎞ 2 − 50sin θ ⎜ ⎟ mm/s dt ⎠ dt 2 ⎝ either cosθ = 0 t =1s
2

When v = 0,

or

dθ =1− t = 0 dt Over 0 ≤ t ≤ 2 s, values of cosθ are: t (s) 0 0 1.0 0.5

1.0 0.5 0.878

1.5 0.375 0.981

2.0 0 1.0

θ ( rad )
cosθ

0.375 0.931

No solutions cosθ = 0 in this range....