Dynamics full version

Dynamics th (Meriam and Kraige, 7 Ed. ,2013) Chapter 1. Introduction Engineering Mechanics Statics:

Statics Dynamics Strength of Materials Vibration

 (F

+ Fr ) = 0 ,distribution of reaction force Fr from the applied force Fa

Dynamics:

a

 (F

a

+ Fr ) = mx , x(t)= f(F(t)) displacement as a function of

time and applied force Strength of Materials: δ = f(P) deflection and applied force on deformable bodies Vibration: x(t) = f(F(t)) on particles and rigid bodies 1

Newtonian Dynamics ‧Kinematics: the relation among

dx(t ) d 2 x(t ) x(t ), x (t ), and  x(t ), x  and  x dt dt 2 without reference to applied force ‧Kinetics: the relation between

 x(t ) and F (t )

Terms to Know ‧Reference frame: Coordinate system ‧Inertial System: Newton’s 2nd Law of motion ‧Particle and Rigid body ‧Scalar and Vector 2

Chap. 2 Kinematics of Particles

Rectangular Coordinates r ( x, y, z )

3

Cylindrical Coordinates

r (r , , z )

Spherical Coordinates

r ( R,  ,  )

Displacement, Velocity, and Acceleration



s

2

s



1

v

v



x

x

4

ds 

dv 

1

2

1

2

dx 

t



2

t

1



t

t



t

t

vdt

adt

1

2

1

2

  xdt



t

t

2

1

t

 )dt (  2 xdt t

1

Velocity and Acceleration Vector

dr r  v dt

 r 5

dr a dt

Rectangular Coordinates

r

r  xi  yj  zk x x   x      r  xi  yj  zk r   y  r   y    r  y         z   z    r   xi   yj   zk z  6

Centrifugal and Tangential Acceleration dr r   v  v et  vet dt Time derivative of a vector

dr  r dt  vet  ve t v  vet  v en   : radius of curvature 7

Time Derivative of the Unit Vectors in Polar (Cylindrical) Coordinates (2D)

de r  e d de  e r d de r d  de r  e r     e dt dt d de d de e     e r dt dt d 8

Cylindrical Coordinates (3D) r  re r  ze z r  re r  re r  ze z  ze z  rer  re  ze z  r   re r  re r  re  re  re    ze z  ( r  r 2 )e r  ( r  2r)e   ze z

9

Spherical Coordinates r  Re r r  R e  Re r

r

ω  e z  e    sin  e r  cos  e   e   sin e r  e   cos  e e r  ω  e r  o  e   cos  e  r  R e r  R cos  e  Re 10

Velocity and acceleration in Spherical Coordinates r  v  v e  v e  v e v  R R



R







R

v  R cos  v  R 



 r aa e a e a e   R  R cos  a R R



R 2





2

2



R

cos  d   sin  a   R    2 R R dt 1 d a  R    R sin  cos   R dt 2



2



11

2

Chap.3 Kinetics of Particles 3.1 3.2 3.3 3.4

Force, Mass, and Acceleration Work and Energy Impulse and Momentum Impact and Orbital Mechanics

3.1 Force, Mass, and Acceleration ‧Newton’s Second Law Equation of Motion F  mr ‧Inertial System: A coordinate system where F  mr ‧Free-body diagram

12

Engineering Mechanics Dynamics -- IAA

Sample 3/1 A 75-kg man stands on a spring scale in an elevator. The tension T in the hoisting cable is 8300 N. Find the reading R of the scale in newtons and the velocity υ of the elevator after 3 seconds. The total mass of the elevator, man, and scale is 750kg.

 Fy  may  T  m1g  m1  y 8300  7360  750ay ay  1.257 m s

2

 Fy  may  R  m2 g  m2  y R  736  75(1.257) R  830 N   a dt     3

  0   1.257 dt 0

  3.77 m s 13

Engineering Mechanics Dynamics -- IAA

Sample 3/3 The 250-lb concrete block A is released from rest in the position shown and pulls the 400-lb log up the 30° ramp. If the coefficient of kinetic friction between the log and the ramp is 0.5, determine the velocity of the block as it hits the ground at B.

y1  0  N  m1g cos  m1   N  2T  m g sin  m   1 1 x1 , 4 equation for 4 unknowns  y2 m2 g  T  m2  2 x1  y2  constant 14

y2

y1 x1

x2

Engineering Mechanics Dynamics -- IAA

B

30 30

A  2B

a y

x

The steel ball is suspended from the accelerating frame by the two cords A and B. Determine the acceleration of the frame which will cause the tension in A to be twice that in B

mg

 Fx  max 2B sin30  B sin30  mx

 Fy  0 2B cos30  B cos30  mg  my  0 xa Eliminate B and get 

15

g 3 3

Engineering Mechanics Dynamics -- IAA

problem 03/19

The 10-kg sphere is suspended from the 15-kg frame sliding down the 20° incline. If the coefficient of kinetic friction between the frame and incline is 0.15, compute each tension TA of wires A and B 25(9.81) N

y 45

45

TB

y

20

N

x

0.15 N

Sphere alone：  Fy  0

x 20 10(9.81) N

(TA  TB )cos45 10(9.81)cos20  0

Frame and sphere as an unit：  Fy  N  mg cos  my  0

TA  TB  130.4 N

 Fx  max

N  25(9.81)cos20  0 N  230 N  Fx  mg cos   N  mx

(TB  TA )sin 45  9.81sin 20  10(1.973) TB  TA  19.56 N

25(9.81)sin20  0.15(230)  25a

Solution： TA  75.0 N, TB  55.4 N

a  1.973 m/s2 16

Engineering Mechanics Dynamics -- IAA

The system is released from rest with the cable taut. Neglect the small mass and friction of the pulley and calculate the acceleration of each body and the cable tension T upon release if (a) μs = 0.25, μk = 0.2 and (b) μs = 0.15, μk = 0.1

T B

Check for motion. Assume static equilibrium.

From B：T  196.2 N Mass A：  Fx  0

m2 g y

x

T

196.2  F  (60)(9.81)sin30  0 F  9.81 N

A F N m1g

x

Fmax  s N  (0.25)(60)(9.81)cos30  127.4 N (a) No motion for (a)： a  0, T  196.2 N 17

Engineering Mechanics Dynamics -- IAA

The system is released from rest with the cable taut. Neglect the small mass and friction of the pulley and calculate the acceleration of each body and the cable tension T upon release if (a) μs = 0.25, μk = 0.2 and (b) μs = 0.15, μk = 0.1

T B

Fmax  (0.15)(60)(9.81)cos30  76.5 N

m2 g y

motion for (b)

x

T A F N

x

A： Fx  max

T  (60)(9.81)sin30  (0.1)(60)(9.81)cos30  60a B：  Fy  max  (20)(9.81)  T  20a 2 Solution：a  0.589 m/s , T  208 N

m1g 18

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Non-constant Acceleration The chain is released from rest with the length b of overhanging links just sufficient to initiate motion. The coefficients of static and kinetic fiction between the links and the horizontal surface have essentially the same value μ. Determine the velocity υ of the chain when the last link leaves the edge. Neglect any friction at the corner.

 ( L  b) g T0 F N   ( L  b) g  gb

Let  = mass / length ： F   N   g  ( L  b)

F  0 T0   g  ( L  b)  0  T0   gb L Solve to obtain： b  1  19

Engineering Mechanics Dynamics -- IAA

Non-constant Acceleration  g ( L  x) T

 g ( L  x)  g ( L  x)  gx

 F  ma T   g  ( L  x)   ( L  x)a

 gx  T   xa Eliminate T to obtain： g a   x  [ x(1   )   L] L  d   xdx

g [ x (1   )   L ]dx 0 b L 1 2 g x2   [ (1   )   Lx]bL 2 L 2 Substitute b and simplify： 

 20

 d  

L

gL 1  Engineering Mechanics Dynamics -- IAA

Non-constant Acceleration Another approach:

 xg    L  x  g    Lx xg   g  L  x   Lx g x 1      L   L g 1     x x  g L  x

t

2

2

x e ,  a 0 a  at

x  c1e  c2e

 at

g 1    L

L  1  21

Engineering Mechanics Dynamics -- IAA

Curvilinear Motion in Polar Coordinates

F  mr    Fr  r  r 2   F   m         r  2r 

r  re r  ze z r  re r  re r  ze z  ze z

 re r  re  ze z

 r   re r  re r  re  re  re   ze z  ( r  r 2 )e r  (r  2r)e   ze z 22

Engineering Mechanics Dynamics -- IAA

Sample 3/10 Tube A rotates about the vertical O-axis with a constant angular rate    and contains a small cylindrical plug B if mass m whose radial position is controlled by the cord wound around the drum of radius b. Determine the tension T in the cord and the horizontal force Fθ exerted by tube on the plug if the constant angular rate of rotation of the drum is ω0 first in the direction for case (a) and second in the direction for case (b). Neglect friction.

23

 Fr  mar 

 T  m(r  r2 )

 F  ma 

F  m(r  2r)

case (a) T  mr 2

F  2mb0

case (b) T  mr 2

F  2mb0

Engineering Mechanics Dynamics -- IAA

If the 2-kg block passes over the top B of the circular portion of the path with a speed of 3.5 m/s, calculate the magnitude NB of the normal force exerted by the path on the block. Determine the maximum speed υ which the block can have at A without losing contact with the path.

2  Fr  m(r  r )  m r (3.5)2  2(9.81)  N B  2 2.4 N B  9.41 N 2

r

Loss of contact at A： N A  0

mg

mg

FB

tθ NB nr

FA

2  Fr  m r

θt

NA  0

30 rn 24

2   mg cos30  m 2.4   4.52 m/s Engineering Mechanics Dynamics -- IAA

problem 03/66

The small sphere of mass m is suspended initially at rest by the two wires. If one wire is suddenly cut, determine the ratio k of the tension in the remaining wire immediately after the other wire is cut to the initial equilibrium tension.

30

30

T1

Equilibrium：  F  0  T1  mg

T1

mg

30

Motion：  Fn  man  0  T2  mg sin 30  0

T2 mg sin 30 k   0.5 T1 mg

T2

 mg 30

w try using x-y coordinates 25

n

t

Engineering Mechanics Dynamics -- IAA

A small bead of mass m is carried by a circular hoop of radius r which rotates about a fixed vertical axis. Show how one might determine the angular speed ω of the hoop by observing the angle θ which locates the bead. Neglect friction in your analysis.

 Fy  0 N cos  mg  0

 y

 n



2  F  m ( r  r  ) r

 N sin  m(r sin ) 2

r N

N  mg / cos

(

mg )sin  mr sin 2 cos



mg

g r cos

g cos   1 Note that 2 r g 2  is a restriction. r 26

Engineering Mechanics Dynamics -- IAA

problem 03/76

Determine the speed υ at which the race car will have no reliance on friction to the banked track. In addition, determine the minimum and maximum speeds, using the coefficient of static friction μs = 0.9.

For no slipping tendency, set F to zero on  F  0 N cos30  mg  0  y

r

2 2   Fr  m r N sin 30  m 1200 Solve： N  1.155 mg ,   149.4 ft/sec

min  0 as  max  tan 1  s  tan 1 (0.9)  42.0  30

y

For max , set F  Fmax   s N

mg

  F  0 N cos30  mg   N sin 30 0  y s

F 30

n

N

2 max 2    Fr  m r   s N cos30  N sin 30  m r with  s  0.9 N  2.40 mg max  345 ft/sec 27

Engineering Mechanics Dynamics -- IAA

A small vehicle enters the top A of the circular path with a horizontal velocity υ0 and gathers speed as it moves down the path. Determine an expression for the angle β which locates the point where the vehicle leaves the path and becomes a projectile. Evaluate your expression for υ0 = 0. Neglect friction.

 F  ma ,

mg sin   ma , a  g sin  



 d  a ds,   d  0 g sin  ( Rd ) 0

 2  02  2 gR (1  cos )

mg

0

 N

R

t

 n

2  Fr  mar ,  mg cos  N  m R 02 N  mg cos   m  2mg (1  cos  ) R 02  mg (3cos   2  ) gR θ 02 02 1 2 When N  0, so 3cos   2    cos (  ) gR 3 3gR 2 For 0  0,   cos 1 ( )  48.2 3 28

Engineering Mechanics Dynamics -- IAA

3.2 Work and Energy Work and Kinetic Energy

U   F T dr   m r T dr   mr T dr 1  mr T r 2

29

Engineering Mechanics Dynamics -- IAA

Work and Potential Energy 2

2

U12   F  dr   (mgj)  (dxi  dyj) 1

1

y2

  mg  dy  mg ( y2  y1 ) y1

Gme m e r  dre r 2 1 1 r r2 dr  Gme m  2 r1 r 1 1  Gme m(  ) r2 r1 2

U12   F  dr  

 mgR 2 (

30

2

1 1  ) r2 r1

Engineering Mechanics Dynamics -- IAA

Sample 3/15 A satellite of mass m is put into an elliptical orbit around the earth. At point A, its distance from the earth is h1 = 500 km and it has a velocity υ1 = 30000 km/h.

Determine the velocity υ2 of the satellite as it reaches point B, a distance h2 = 1200 km from the earth.

 1 1 U1-2  mgR2     r2 r1  1 2 1 1 1 2 1 m1  mgR     m22 2 2  r2 r1  2  1 1 2    2gR     r2 r1  2

2 1

2

2

103 103   30 000  3 2 2      2(9.81) (6371)(10)    6371  1200 6371  500  3.6    2

 69.44(106 ) 10.72(106 )  58.73(106 ) (m/s)2

2  7663 m/s 31

Engineering Mechanics Dynamics -- IAA

Potential Energy

V   F T dr GMm dr 2 r mgR 2    2 dr r  

V   F T dr

mgR 2  r

  mgdy

r2

r1

 mgh

32

Engineering Mechanics Dynamics -- IAA

Conservative Force

Principle of Work and Energy

T  W  V  0

V  W  T ‧Kinetic energy

dU FT dr dr ‧Power P=   FT  FT r dt dt dt

   i  j k x y z F  V



33

Engineering Mechanics Dynamics -- IAA

A satellite is put into an elliptical orbit around the earth and has a velocity υP at the perigee position P. Determine the expression for the velocity υA at the apogee position A. The radii to A and P are, respectively, rA and rP. Note that the total energy remains constant.

Constant total energy is E  TA  VA  Tp  V p Thus

1 2 mgR 2 1 2 mgR 2 m A   m p  2 rA 2 rp

 A2   p2  2 gR 2 (

1 1  ) rp rA

 A   p2  2 gR 2 ( 34

1 1  ) rp rA

Engineering Mechanics Dynamics -- IAA

The chain starts from rest with a sufficient number of links hanging over the edge to barely initiate motion in overcoming friction between the remainder of the chain and the horizontal supporting surface. Determine the velocity υ of the chain as the last link leaves the edge. The coefficient of kinetic friction is μk. Neglect and friction at the edge.

 = mass per unit length k L b   gb    g ( L  b ) For equil. at start k 1 

k

U  T  Vg (L  b)2 U    dF  x     k  gxdx    k  g 0 2 1 Lb  T   L 2  V g    g ( L  b )( ) 2 2 (L  b)2 1 L2  b 2 2   L   g Thus   k  g L b

Lb dF dx

2

x

b

b

Lb Lb 2 2 Lb 2

2

2

b )( L  b   k [ L  b ]) Now substitute L k k b k L 2 )( L [1  ]  k [L  ]) So   g (1  1  k 1  k 1  k

 2  g (1 



gL 1  k 35

 

gL 1  k Engineering Mechanics Dynamics -- IAA

3.3 Impulse and Momentum Linear momentum G  m r F  m r

d ( m r ) dt d  G dt  G 

Impulse

r2    F dt  m r dt  md r  m r | r1  G   

Conservation of Linear Momentum

 Fdt  G

if F  0  G  0 36

Engineering Mechanics Dynamics -- IAA

Sample 3/19 The horizontal velocities of the ball just before and after impact are separately υ1 = 50 ft/sec and υ2 = 70 ft/sec. If the 4-oz ball is in contact with the racket for 0.02 sec, determine the magnitude of the average force R exerted by the racket on the ball and the angle β made by R with the horizontal

m( )  t2 F dt  m( )  x 2  x 1 t1  x  4 /16 4 /16  (50)  Rx (0.02)  (70cos15 ) 32.2 32.2 m( )  t2 F dt  m( )  y 1 y 2 t1  y  

t2

t

4 /16 4 /16 (0)  Ry (0.02)  (70sin15 ) 32.2 32.2 Rx  45.7 lb, Ry  7.03 lb

y

mgdt

1

mv1

+

t2

t

=

Rx dt

15 x

R  Rx 2  Ry 2  45.72  7.032  46.2 lb

1

t2

t

  tan

Ry dt

1

37

1

Ry Rx

 tan1

7.03  8.75 45.7

Engineering Mechanics Dynamics -- IAA

problem 03/207

The 1.62-oz golf ball is struck by the five-iron and acquires the velocity shown in a time period of 0.001 sec. Determine the magnitude R of the average force exerted by the club on the ball. What acceleration magnitude ɑ does this force cause, and what is the distance d over which the launch velocity is achieved, assuming constant acceleration?

mg  0

150 ft / sec

25  

RT  m : R(0.001)=

1.62/16 (150) , R =472 lb 32.2

1.62/16 a , a  150,000 ft / sec2 (4660 g ) 32.2  2  02  2ad : 1502  02  2(150,000)d , d  0.075 ft or 0.900in R  ma : 472=

38

Engineering Mechanics Dynamics -- IAA

Angular Impulse and Momentum

H O  r  mv

HO  r  mv  m(vz y  vy z)i  m(vx z  vz x) j  m(vy x  vx y)k i HO  x

j y

vx vy 39

k  H x  m(vz y  vy z)      z   H y   m(vx z  vz x)  vz  H z  m(vy x  vx y) Engineering Mechanics Dynamics -- IAA

Time Derivative of Angular Momemtum HO  r  mv 0  HO  r  mv  r  mv  v  mv  r  mv  r F  MO  M  H O

O

Conservation of Angular Momentum   MO  H O



t2

t1

 M O dt  ( H O ) 2  ( H O )1  H O

The total angular impulse on a particle of mass m about a fixed point O equals the corresponding change in angular momentum about that point. t2

(H O )1    M O dt  ( H O ) 2 t1

Principle of Conservation of Angular Momentum if  M O  0 , then H O  0 or (H O )1 =(H O ) 2 40

Engineering Mechanics Dynamics -- IAA

Sample 3/25 Molynia Orbit A comet is in the highly eccentric orbit shown in the figure. Its speed ate the most distant point A, which is at the outer edge of the solar system, is υA = 740 m/s .

Determine its speed at the point B of closest approach to the sun.

(HO ) A  (HO )B mrA A  mrB B rA A 6(109 )740 B   rB 75(106 )

 B  59 200 m/s

41

Engineering Mechanics Dynamics -- IAA

The central attractive force F on an earth satellite can have no moment about the center O of the earth. For the particular elliptical orbit with major and minor axes as shown, a satellite will have a velocity of 33880 km/h at the perigee altitude of 390 km. Determine the velocity of the satellite at point B and at apogee A. The radius of the earth is 6371 km.

B 11720 km / h

M B

r

 O

 H 0  0 so H O  constant

rmin  6371  390  6761 km

33880 km / h

rmax  2(13520)  6761  20279 km

o

For H O constant

A

6371(33880)  11720 B  20279 A  A  19540km / h  B  11300km / h

A rmax 2(13520) km

rmin 42

Engineering Mechanics Dynamics -- IAA

problem 03/228

The two spheres of equal mass m are able to slide along the horizontal rotating rod. If they are initially latched in position a distance r from the rotating axis with the assembly rotating freely with an angular velocity ω0 , determine the new angular velocity ω after the spheres are released and finally assume positions at the ends of the rod at a radial distance of 2r. Also find the fraction n of the initial kinetic energy of the system which is lost. Neglect the small mass of the rod and shaft.

H  0 ; 2mr0 (r )  2m(2r ) (2r )  0   0 / 4  1 1 T  2( m[r0 ]2 )  2( m[2r 0 ]2 )  mr 202 (3 / 4) 2 2 4 3 n  T / T  mr 202 / mr 202  3 / 4 4 43

Engineering Mechanics Dynamics -- IAA

 M 0  H 0 mgl cos   The simple pendulum of mass m and length l is released from rest at θ = 0. Using only the principle of angular impulse and momentum, determine the expression for  in terms of θ and the velocity υ of the pendulum at θ = 90°.

Compare this approach with a solution by the work-energy principle.

O



T

d (ml 2)  ml 2 dt

g   cos  l From  d   d

2   g |  cos d 2 0 0 l g  2  sin  l

2g  90  l

so at   90

  l  2 gl

l By work-energy

V  T

1 mgl  m 2 2

  2 gl

mg 44

Engineering Mechanics Dynamics -- IAA

Direct Central Impact

45

Engineering Mechanics Dynamics -- IAA

Coefficient of Restitution

 e   e 

t

t0 t0

0 t

t0 t0

0

Fr dt

m1[1'  (0 )] 0' 1   Fd dt m2[0  (1 )] 1 0

Fr dt

m2 (2' 0 ) 2' 0   Fd dt m2 (0 2 ) 0 2

2' 1' relative velocity of separation e  1 2 relative velocity of approach 46

Engineering Mechanics Dynamics -- IAA

47

Engineering Mechanics Dynamics -- IAA

Oblique Central Impact

m1 (1 )n  m2 (2 )n  m1 (1' )n  m2 (2' )n m1 (1 )t  m1 (1' )t m2 (2 )t  m2 (2' )t (2' )n  (1' )n Vn e  (1 )n  (2 )n Vn 48

Engineering Mechanics Dynamics -- IAA

Sample 3/29 A ball is projected onto the heavy plate with a velocity of 50 ft/sec at the 30° angular shown. If the effective coefficient of restitution is 0.5, compute the rebound velocity υ′ and its angle θ′.

(2' )n  (1' )n e (1 )n  (2 )n

0  (1' )n 0.5  50sin30  0

(1' )n  12.5 ft/sec m(1 )t  m(1' )t (1' )t  (1 )t  50cos30  43.3 ft/sec

 '  (1' )n2  (1' )t 2  12.52  43.32  45.1 ft/sec  (1' )n   12.5     tan  '   tan1   16.10   43.3   (1 )t  '

49

1

Engineering Mechanics Dynamics -- IAA

Sample 3/30 Spherical particle 1 has a velocity υ1 = 6 m/s in the direction shown and collides with spherical particle 2 of equal mass and diameter and initially at rest. If the coefficient of restitution for these conditions is e  0.6 , determine the resulting motion of each particle following impact.

Also calculate the percentage loss of energy due to the impact.

m1 (1 )n  m2 (2 )n  m1 (1' )n  m2 (2' )n 5.20  0  (1' )n  (2' )n (2' )n  (1' )n e (1 )n  (2 )n

(2' )n  (1' )n 0.6  5.20  0

(1' )n  1.039 m/s

(2' )n  4.16 m/s

m1 (1 )t  m1 (1' )t

(1' )t  (1 )t  3 m/s

m2 (2 )t  m2 (2' )t

(2' )t  (2 )t  0

50

Engineering Mechanics Dynamics -- IAA

As a check of the basketball before the start of a game, the referee releases the ball from the overhead position shown, and the ball rebounds to about waist level. Determine the coefficient of restitution and the percentage n of the original energy lost during the impact.

  2gh ,  '  2gh'

' h' 1100 e    0.724  h 2100 mgh  mgh' n mgh 2100  1100  2100  47.6% 51

Engineering Mechanics Dynamics -- IAA

Central-Force Motion

F  mr  Gmm0  2     r  r   r2     r  2r  0   mr 2  h

52

Engineering Mechanics Dynamics -- IAA

Orbital Mechanics

Gm 1  C cos  2 0 r h

53

Engineering Mechanics Dynamics -- IAA

D’Alembert’s Principle and Inertia Force F  ma  0 F  mr  0 mr inertia force

D’Alembert’s Principle using inertia force to treat dynamics by statics

54

Engineering Mechanics Dynamics -- IAA

Chap. 4 Kinetics of Systems of Particles Mass center mrc   mi ri Equation of motion  F   f   mi ri   F  mrc Principle of motion of the mass center the resultant of the external forces on any system of masses equal the total mass times the mass center acceleration.

fig_04_001

55

Engineering Mechanics Dynamics -- IAA

Linear Momentum

G   miri   mi (rc  ρ i ) d   mirc  ( miρi ) dt  mrc

56

Engineering Mechanics Dynamics -- IAA

Kinetic Energy ri  rc  ρi ri  rc  ρ i 1 T   mi  riT ri 2 1  (mircT rc  2mircT ρ i  miρ Tiρ i ) 2 1 T 1    mrc rc   miρ iT ρ i 2 2

fig_04_003

57

Engineering Mechanics Dynamics -- IAA

Angular Momentum about a Fixed Point   (r  m r ) H O i i i   (r  m r  r  m r ) H O

i i

i

i i

 O   ri  Fi  MO

Angular Momentum about c.g.    ρ  m (r  ρ )   ρ  m r H G i i C i i i i

HG   ρi  miri   ρi  mi (rc  ρ i )

  ρi  Fi

 (ρi  mirc  ρi  mi ρ i )  (ρi  miρ i ) 58

  MG Engineering Mechanics Dynamics -- IAA

problem 04/22

The man of mass m1 and the woman of mass m2 are standing on opposite ends of the platform of mass m0 which moves with negligible friction and is initially at rest with s = 0.

The man and woman begin to approach each other. Derive an expression for the displacement s of the platform when the two meet in terms if the displacement x1 of the relative to s the platform.

l

m1 x 1

x2 m2 m0

B

A

C

p_04_020

With respect to C,mi xi  constant l l m1l  m2 (0)  m0  m1(s  l  x1)  m2 (s  x2 )  m0 (s  ) 2 2 m x  m2 x2 Simplify and get s  1 1 m0  m1  m2 (m  m2 ) x1  m2l But they meet whenx2  x1  l so s  1 m0  m1  m2 59

Engineering Mechanics Dynamics -- IAA

Sample 4/4

sp_04_03_01

A shell with a mass of 20 kg is fired from point O, with a velocity u = 300 m/s in the vertical x-z plane at the inclination shown.

When it reaches the top of its trajectory at P, it explodes into three fragments A, B, and C. Immediately after the explosion, fragment A is observed to rise vertically a distance of 500 m above P, and fragment B is seen to have a horizontal velocity vB and eventually lands at point Q. When recovered, the masses of the fragments A, B, and C are found to be 5, 9, and 6 kg, respectively. Calculate the velocity which fragment C has immediately after the explosion. Neglect atmosphere resistance.

60

Engineering Mechanics Dynamics -- IAA

Sample 4/4 t  uz / g  300(4/5)/9.81  24.5s uz2 [(300)(4/5)]2 h   2940m 2g 2(9.81)

A  2ghA  2(9.81)(500)  99.0m/s B  s / t  4000/ 24.5  163.5m/s [G1  G2 ]

sp_04_03_01

mv  mAv A  mB vB  mC vC

3 20(300)( )i  5(99.0k)  9(163.5)(i cos45  jsin 45 )  6vC 5 6vC  2560i 1040j  495k vC  427i 173.4j  82.5k m/s

C  (427)2  (173.4)2  (82.5)2  468m/s 61

Engineering Mechanics Dynamics -- IAA

Steady Mass Flow

1 A11  2 A22  m G  (m)v2  (m)v1  m(v2  v1 )

F  mv 62

Engineering Mechanics Dynamics -- IAA

problem 04/35

The jet aircraft has a mass of 4.6 Mg and a drag (air resistance) of 32 kN at a speed of 1000 km/h at a particular altitude. The aircraft consumes air at the rate of 106 kg/s through its intake scoop and uses fuel at the rate of 4 kg/s.

If the exhaust has a rearward velocity of 680 m/s relative to the exhaust nozzle, determine the maximum angle of elevation at which the jet can fly with a constant speed of 1000 km/h at the particular altitude in question.

T  ma (u )  mf u  106(680  1000/3.6)  4(680)

mg  4.6(9.81) kN

T

 45400 N  45.4 kN

R  32 kN x

 Fx  max  0 45.4  32  4.6(9.81)sin  0



  1000 km/h N 63

sin  0.296    17.22

Engineering Mechanics Dynamics -- IAA

problem 04/38

The jet water ski has reached its maximum velocity of 70 km/h when operating in salt water. The water intake is in the horizontal tunnel in the bottom of the hull, so the water enters the intake at the velocity of 70 km/h relative to the ski. The motorized pump discharge water from the horizontal exhaust nozzle of 50-mm diameter at the rate of 0.082 m3/s. Calculate the resistance R of the water to the hull at the operating speed.

Resistance R equals net trust T where T  m(u  ) Density of salt water,

Nozzle velocity u  Q / A 

  1030 kg/m3

m  Q  1030(0.082)  84.5 kg/s,   70

0.082  41.8 m/s 2  (0.050) 4

1000  19.44 m/s 3600

R  T  84.5(41.8 19.44)  1885 N 64

Engineering Mechanics Dynamics -- IAA

problem 04/45

The jet-engine thrust reverser to reduce an aircraft speed of 200 km/h after landing employs folding vanes which deflect the exhaust gases in the direction indicated. If the engine is consuming 59 kg of air and 0.65 kg of fuel per second, calculate the braking thrust as a fraction n of the engine thrust without the deflector vanes. The exhaust gases have a velocity of 650 m/s relative to the nozzle.

 F  mu With reversers in place

TR  mg u sin30  ma  (50  0.65)(650)sin30  50(55.6  0)  16460  2780  19240 N

30

Without reversers

650 m/s

T  mg u  ma

TR

 (50  0.65)650  50(55.6)

x

 32900  2780  30100 N

 

30

650 m/s

so n 

  200/3.6  55.6 m/s 65

19240  0.638 30100

Engineering Mechanics Dynamics -- IAA

Variable Mass system  R  m( 0 )  mu

F  R  m    F  m  mu

66

Engineering Mechanics Dynamics -- IAA

Rocket Propulsion mu  pA  mg  R  m

67

Engineering Mechanics Dynamics -- IAA

Sample 4/12 A rocket of initial total mass m0 is fired vertically with constant acceleration until the fuel is exhausted. The relative nozzle velocity of the exhaust gas has a constant value u at atmospheric pressure throughout the flight. If the residual mass of the rocket structure and machinery is mb when burnout occurs, determine the expression for the maximum velocity reached by the rocket. Neglect atmospheric resistance and the variation of gravity with altitude.

Solution I (F=ma solution)    mu  mg  m T  mg  m, T  mu  mu

 m dm t dm  gdt   d  u   g  dt 0 m0 m 0 m m   u ln 0  gt m let mb：mass of rocket when burnout occurs

d  u

tb  (mb  m0 )/ m

max  u ln

m0 g  (m0  mb )  mb m 68

Engineering Mechanics Dynamics -- IAA

Sample 4/12 A rocket of initial total mass m0 is fired vertically with constant acceleration until the fuel is exhausted. The relative nozzle velocity of the exhaust gas has a constant value u at atmospheric pressure throughout the flight. If the residual mass of the rocket structure and machinery is mb when burnout occurs, determine the expression for the maximum velocity reached by the rocket. Neglect atmospheric resistance and the variation of gravity with altitude.

Solution II (Variable-Mass solution)

   mg  m  mu   F  mg,  F  m  mu   mu  T  T  mg  m mu

69

same as Solution I

Engineering Mechanics Dynamics -- IAA

Chap.5 Planar Kinematics

fig_05_001

70

Engineering Mechanics Dynamics -- IAA

Z

Fixed Axis Rotation Y

X

displacement r velocity r  ω  r

fig_05_00 4

acceleration r  ω  (ω  r)  ω  r 71

Engineering Mechanics Dynamics -- IAA

Sample 5/3 The right-angle bar rotates clockwise with an angular acceleration

  4k rad/s2 Write the vector expressions for the velocity and acceleration of point A when

  2k rad/s sp_05_03_01

[v    r ] [an    (  r )] [at    r ] [a  an  at ]

v  2k  (0.4i  0.3j)  0.6i  0.8j m / s an  2k  (0.6i  0.8j)  1.6i  1.2j m / s2 at  4k  (0.4i  0.3j)  1.2i  1.6j m / s a  2.8i  0.4j m / s2

2

  0.62  0.82  1m/s a  2.82  0.42  2.83 m/s

72

Engineering Mechanics Dynamics -- IAA

Determine the velocity and acceleration of (a) point A and (b) point B with

  6 rad/s,   4rad/s2 (a) A    r A  (6k  45 j)

 270i mm/s aA    r A   2 r A  4k  45 j  62 (45 j)  180i 1620 j mm/s2 (b) B    r B  6k  (30i  45 j)

 270i 180 j mm/s aB    r B   2 r B  4k  (30i  45 j)  62 (30i  45 j)  900i 1740 j mm/s2

p_05_002

73

Engineering Mechanics Dynamics -- IAA

Sample 5/4 A wheel of radius r rolls on a flat surface without slipping. Determine the angular motion of the wheel in terms of the linear motion of its center O. s  r

O  r, aO  r where O  s, aO  O   s,   , and      sp_05_04_01

Determine the acceleration of a point on the rim of the wheel as the point comes into contact with the surface on which the wheel rolls.

y  r  r cos  r (1  cos ) x  s  r sin  r (  sin ) x  r(1  cos )  O (1  cos ) y  r sin  O sin   y   sin    cos x   (1  cos )   sin O

O

 aO (1  cos )  r 2 sin

O

O

 aO sin  r 2 cos 74

 x  0 and  y  r2

Engineering Mechanics Dynamics -- IAA

Sample 5/7 Calculate the velocity of point A on the wheel without slipping for the instant represented.

v A  vO  v A/O  vO  ω r0 ω = -10k rad/ s r0  0.2(i cos30  jsin30 )  0.1732i  0.1j m vO  3i m/ s

v A  3i 

i

j

0

0

0.1732 0.1 sp_05_07_0 1

k 10  3i  1.732j  i 0

 4i  1.732j m/s

A  42  (1.732)2  19  4.36 m/s 75

Engineering Mechanics Dynamics -- IAA

Sample 5/11 Locate the instantaneous center of zero velocity and use it to find the velocity of point A for the position indicated.

[   / r]

  O / OC  3 / 0.300  10 rad / s AC  (0.300)2  (0.200)2  2(0.300)(0.200)cos120  0.436 m

[  r]

 A  AC  0.436(10)  4.36 m/s 76

Engineering Mechanics Dynamics -- IAA

Sample 5/8 r2 r3

r1

r  r1  r2  r3 r  ω1  r1  ω2  r2  ω3  r3

r

0  ω1k 100j  ω2k  (175i  50j)  2k  75i 100ω1  50ω2     175 ω  150  2 

sp_05_08_0 1

3 6 ω1   , ω2   7 7

77

Engineering Mechanics Dynamics -- IAA

Sample 5/14 r2

r  r1  r2  r3 r  ω1  r1  ω2  r2  ω3  r3

r3

r1

r  ω 1  r1  ω1  (ω1  r1 ) ω 2  r2  ω2  (ω2  r2 )

r

ω 3  r3  ω3  (ω3  r3 )

sp_05_08_0 1

3 3 r  1k 100j  ( k )  ( k 100j)   2k  (175i  50j) 7 7 6 6 ( )k  [( k)  (175i  50j)]  0  2k  (2k [75i]) 7 7

1  0.1050 rad/s2

2  4.34 rad/s2 78

Engineering Mechanics Dynamics -- IAA

Sample 5/9

sp_05_09_01

v A  vB  v A/ B [  r] 5 1500(2 ) 65.4 ft / sec 12 60 5 14  sin  sin60

A 65.4  sin78.0 sin72.0 A/ B 65.4  sin30 sin72.0 [   / r ]

AB 

A  67.3 ft /sec A/ B  34.4 ft /sec A/ B 34.4   29.5 rad /sec AB 14/12

vG  v B  vG / B

B 

  sin1 0.309  18.02

G / B  GBAB 

GB 4  A/ B  (34.4)  9.83 ft / sec 14 AB

G  64.1ft /sec 79

Engineering Mechanics Dynamics -- IAA

y

Sample 5/9

r1

r2 r

x

r  r1  r2 r  ω1  r1  ω2  r2

sp_05_09_01

r  ω 1  r1  ω1  (ω1  r1 ) ω 2  r2  ω2  (ω2  r2 )

80

Engineering Mechanics Dynamics -- IAA

Sample 5/15 aA  10280cos60 1015cos18.02  ( A/ B )t sin18.02 0  10280sin60 1015cos18.02  ( A/ B )t cos18.02 sp_05_15_01

aA  aB  (aA/ B )n  (aA/ B )t

aB 

5 1500[2 ] 2 ( ) 12 60 2

 10280 ft / sec

 A  3310 ft /sec2

[an  r2 ]

[  at / r] [an  r2 ]

14 (29.5)2 12  1015 ft / sec2

(aA/ B )n 

( A/ B )t  9030 ft /sec2

 A/ B  9030/(14/12) ft /sec2  7740 rad/sec2  A  3310 ft / sec2

81

( A/ B )t  9030 ft /sec2

Engineering Mechanics Dynamics -- IAA

General Motion: Rotation + Translation

fig_05_005

fig_05_006

82

Engineering Mechanics Dynamics -- IAA

Instantaneous Center

fig_05_007

83

Engineering Mechanics Dynamics -- IAA

Body-Fixed Coordinates in Rotation Coriolis Acceleration

rA

fig_05_011

rA  rB  ρ rA  rB  ω  ρ+ ρ

rA  rB  ω   ω  ρ  ω  ρ  ρ+  2ω ρ

rA  rB  ω   ω  ρ  ω  ρ  ω ρ+  ω ρ+  ρ    rB  ω   ω  ρ  ω  ρ+ 2ω ρ  ρ

 Coriolis acceleration

84

Engineering Mechanics Dynamics -- IAA

Sample 5/16 The motion of slider A is separately controlled, and at this 2 instant, r = 6 in., r =5 in./sec, and r =81 in./sec . Determine the absolute velocity and acceleration of A for this position. y

v A    ρ  ρ v A  4k  6i  5i  24j  5i in./sec

A  (24)2  (5)2  24.5 in./sec a A    (  ρ)    ρ  2  ρ   ρ x sp_05_16_01

  (  ρ)  4k  (4k  6i)  4k  24j  961 in./sec2   ρ  10k  6i  60j in./sec2 2  ρ  2(4k)  5i  40j in./sec2  ρ  81i in./sec2

a A  (81  96)i  (40  60) j  15i  20j in./sec2

aA  (15)2  (20)2  25 in./sec2 85

Engineering Mechanics Dynamics -- IAA

With and without Body-Fixed Coordinate

Let r  5i,   2k,   3k

r    r  10j  r      r     r = -20i + 30j

displacement r velocity r  ω  r acceleration r  ω  (ω  r)  ω  r 86

Engineering Mechanics Dynamics -- IAA

problem 05/163

A vehicle A travels with constant speed v along a north-south track. Determine the Coriolis acceleration aCor as a function of the latitude θ at (a) the equator and (b) the north pole.

z

B



   ( sin j  cos k) acor  2 

A

 

 2k ( sin j  cos k) y

 2 sin i (west)

For   500 km/h (a) Equator,   0  acor  0 (b) North pole,

500  0.0203 m/s2 3.6 The track provides the necessary westward acceleration so that the velocity vector is properly rotated and reduced in magnitude.

  90  acor  2(7.292 105 )

p_05_161

87

Engineering Mechanics Dynamics -- IAA

Chap. 6 Dynamics of Planar Rigid Body Equation of Motion ‧The resultant of the external forces equals to the inertia of the mass center ‧The resultant moment about C.G. of the external forces equals to the rate change of the angular moment about C.G.

F  mrG  M  H G

88

G

Engineering Mechanics Dynamics -- IAA

Equation of Motion in 2D Angular momentum

HG   ρi  miρ i   ρi   miω  ρi  

  ω dm 2

 Iω

F  mrc  MG  Iω

89

Engineering Mechanics Dynamics -- IAA

Moment about a Fixed Point

  ρ  mr MP  H G C   ρ  mrC  I Gω   ρ  mrP  I Pω if rP  0, P is fixed   I Oω   I Pω

90

Engineering Mechanics Dynamics -- IAA

Determine the value of the force P which would cause the cabinet to begin to tip.

What coefficient μs of static friction is necessary to ensure that tipping occurs without slipping?

0.8m 50(9.81)N 1.2m G 

FA NA

when tipping impends N A  mg



MG  mg(0.4)  mg(0.6)  I  0 

2 3

 F  mg  mx

2  x  g  g 3

As a whole :

P  (m  m1) x  (50  10) x  60 x  40g 91

Engineering Mechanics Dynamics -- IAA

Center of Percussion

F  mrC  MG  IGω or MO  IO

ko

ko

Radius of gyration about point O

The resultant force at the center of percussion is zero.

92

Engineering Mechanics Dynamics -- IAA

Center of Percussion -F  RX  mx mg  RY  my F  h  I 

F

O

y  0 1 x   2 2 h  3

h

mg RX

RY 93

Engineering Mechanics Dynamics -- IAA

problem 06/34

The 20-kg uniform steel plate is freely hinged about the z-axis as shown. Calculate the force supported by each of the bearings at A and B an instant after the plate is released from rest in the horizontal y-z plane.

MO  IO

1  20(9.81)(0.2)  20(0.4)2 3   36.8rad/s2

 x  r

2F

0.2m

G

 x  0.2  36.8  7.36m/s2  Fx  mx

0.2m O

20(9.81)  2F  20(7.36)

20(9.81) N

1 2F  49.0  mg 4 FA  FB  F  24.5N

t 94

Engineering Mechanics Dynamics -- IAA

problem 06/38

Determine the angular acceleration and the force on the bearing at O for (a) the narrow ring of mass m and (b) the flat circular disk of mass m immediately after each is released from rest in the vertical plane with OC horizontal.

(a)

  mgr  2mr 2 M  I   O O

  g / 2r

 Fy  my  mg  O  mr(

g ) 2r

O  mg / 2 O

O

O

r

G

O

r

1 2  M  I   mgr  ( mr  mr 2 ) (b)  O O 2   2g / 3r

G

 Fy  my  mg  O  mr(

y mg

mg

2g ) 3r

O  mg / 3 95

Engineering Mechanics Dynamics -- IAA

Sample 6/5 A metal hoop with a radius r = 6 in. is released from rest on the 20° incline. If the coefficients of static and kinetic friction are μs = 0.15 and μk = 0.12, determine the angular acceleration α of the hoop and the time for the hoop to move a distance of 10 ft down the incline.

96

Engineering Mechanics Dynamics -- IAA

Sample 6/5

[ Fx  max ]

mg sin20  F  ma

[ Fy  may  0] N  mg cos20  0 Fr  mr 2 [ MG  I  ] Assume pure rolling a  r 4 equations for 4 unknowns

F  0.1710mg

and

N  mg cos20  0.940mg

Check if the assumption valid. The friction force be bounded by  N

[ Fmax  s N ]

Fmax  0.15(0.940mg )  0.1410mg

[ Fmax  k N ]

F  0.12(0.940mg )  0.1128mg

So it is slipping f  k N.

[ Fx  max ] [ MG  I  ] 1 [ x  at 2 ] 2

Solve again the 4 unknowns

mg sin20  0.1128mg  ma a  0.229(32.2)  7.38 ft/sec2 0.1128(32.2) 0.1128mg (r )  mr 2    7.26 rad/sec2 6 /12 2x 2(10) t   1.646 sec 7.38  97

Engineering Mechanics Dynamics -- IAA

Sample 6/7 The slender bar AB weighs 60 lb and moves in the vertical plane, with its ends constrained to follow the smooth horizontal and vertical guides. If the 30-lb force is applied at A with the bar initially at rest in the position for which θ = 30°, calculate the resulting angular acceleration of the bar and the forces in the small end rollers at A and B.

98

Engineering Mechanics Dynamics -- IAA

Sample 6/7 ax  a cos30  2 cos30  1.732 ft/sec2

ay  a sin30  2 sin30  1.0 ft/sec2 1 60 2 [ MG  I  ] 30(2cos30 )  A(2sin30 )  B(2cos30 )  (4 ) 12 32.2 60 30  B  (1.732 ) [ Fx  max ] 32.2 60 A  60  (1.0 ) [ Fy  may ] 32.2 2 A  68.2 lb B  15.74 lb   4.42 rad/sec 1 60 2 4.39  9.94 [ MC  I   m d ] 30(4cos30 )  60(2sin30 )  (4 ) 12 32.2 2   4.42 rad/sec 60 60  (1.732 )(2cos30 )  (1.0 )(2sin30 ) 32.2 32.2

[ Fy  may ]

[ Fx  max ]

60 A 60  (1.0)(4.42) 32.2 60 30  B  (1.732)(4.42) 32.2 99

A  68.2 lb B  15.74 lb Engineering Mechanics Dynamics -- IAA

problem 06/82

Determine the angular acceleration of each of the two wheels as they roll without slopping down the inclines. For wheel A investigate the case where the mass of the rim and spokes is negligible and the mass of the bar is concentrated along its centerline. For wheel B assume that the thickness of the rim is negligible compared with its radius so that all of the mass is concentrated in the rim. Also specify the minimum coefficient of static friction μs required to prevent each wheel from slopping.

MO  I   I  0 M  0

A mg G 

r x

Hence no friction force and s  0

g F  ma  mg sin   mr  x x A   A  sin r

 N 100

Engineering Mechanics Dynamics -- IAA

problem 06/82

Determine the angular acceleration of each of the two wheels as they roll without slopping down the inclines. For wheel A investigate the case where the mass of the rim and spokes is negligible and the mass of the bar is concentrated along its centerline. For wheel B assume that the thickness of the rim is negligible compared with its radius so that all of the mass is concentrated in the rim. Also specify the minimum coefficient of static friction μs required to prevent each wheel from slopping.

MC  IC  mgr sin  2mr2B B  

M  I   Fr  mr 2

B mg G  r

x

C

F

g sin 2r

N

g sin 2r

1  F  mg sin 2 F 1 s   mg sin / mg cos N 2 1 s  tan 2 101

Engineering Mechanics Dynamics -- IAA

problem 06/84

The uniform 12-kg square panel is suspended from point C by the two wires at A and B. If the wire at B suddenly breaks, calculate the tension T in the wire at A an instant after the break occurs.

M A  I   mad

C 

T A

mgb 1 2 b b  mb   m  2 6 2 2 3g  4b  MG  I 

45

I

r

G

m(aG / A )t  r

T b

b 1 2 3g  mb ( ) 4b 2 6

T

mg maA

2 2 mg  (12)(9.81)  20.8 N 8 8

b 102

Engineering Mechanics Dynamics -- IAA

Kinetic Energy 1 T  m 2 2

1 T  IO2 2

1 1 T  m 2  IC 2 2 2

103

Engineering Mechanics Dynamics -- IAA

problem 06/116

A mg

x

 B mg

x

U  T U  mgx sin 1 1 T  m 2  I 2 2 2 1 case A： T  m 2  0 2 1 1  case B： T  m 2  mr 2 ( )2  m 2 2 2 r 1 case A： mgx sin  m 2   A  2gx sin 2 case B： mgx sin  m 2  B  gx sin



104

Engineering Mechanics Dynamics -- IAA

For the pivoted slender rod of length l, determine the distance x for which the angular velocity will be a maximum as the bar passes the vertical position after being released in the horizontal position shown. State the corresponding angular velocity.

T1  V12  T2

l g(  x) 2  2 2 l lx x2   6 2 2

l  l  1 1  mg   x    ml 2  m(  x)2 2 2  2  2 12 

d2 set  0 obtain dx

l g(  0.211l ) g 2 max  x0.211  2  1.861 l l 0.211l 2 (0.211l)2   6 2 2 105

x  0.789l

or

x  0.211l

The solution x  0.789l would yield the same max only then the motion is CCW. Engineering Mechanics Dynamics -- IAA

Linear Momentum

Angular Momentum

HO  IO

MO  H O t2

(HO )1   MOdt  (HO )2

 F  G

t1

t2

G1   Fdt  G2 t1

106

Engineering Mechanics Dynamics -- IAA

Sample 6/16 The uniform rectangular block of dimensions shown is sliding to the left on the horizontal surface with a velocity v1 when it strikes the small step at O. Assume negligible rebound at the step and compute the minimum value of v1 which will permit the block to pivot freely about O and just reach the standing position A with no velocity. Compute the percentage energy loss n for b = c.

107

Engineering Mechanics Dynamics -- IAA

Sample 6/16

[HO  IO] [(HO )1  (HO )2 ]

1 c b m (HO )2  { m(b2  c2 )  m[( )2  ( )2 ]}2  (b2  c2 )2 12 2 2 3 b m 31b m1  (b2  c2 )2 2  2 3 2(b2  c2 ) 2

2

1 b c b [T2  V2  T3  V3 ] IO22  0  0  mg[       ] 2  2  2 2 31b 2 mg 1m 2 2 (b  c )[ 2 2 ]  ( b2  c2  b) 23 2 2(b  c ) g c2 1  2( (1  2 )( b2  c2  b) 3 b 1 2 1 2 2 m   I  2 2 2 2  1 O 2   E 2 k  b c 3b 3 2 n   1  O 2 2  b2  c2   1   2 2  1 2 E 3  c2   1    2(b  c )  m1 41  2  2  b  n  62.5% b  c 108

Engineering Mechanics Dynamics -- IAA

problem 06/188

The homogeneous sphere of mass m and radius r is projected along the incline of angle θ with an initial speed v0 and no angular velocity (ω0 = 0). If the coefficient of kinetic friction is μk, determine the time duration t of the period of slopping. In addition, state the velocity v of the mass center G and the angular velocity ω at the end of the period of slipping.

109

Engineering Mechanics Dynamics -- IAA

problem 06/188

t

0  Fydt  m(y  y )  0  N  mg cos t 0  Fxdt  m(x x )

y

0

x

0

(k mg cos  mg sin )t  m( 0 )

(1)

t

mg

0 MGdt  I (  0 )

G

2 (k mg cos r )t  mr 2 (2) 5 We desire the time t when   r (3)

r

N

Solution of Eqs. (1)-(3)： 20 t g (7k cos  2sin )

k N

For slipping to cease,

50 k  7k  2tan

7k cos  2sin

2 or k  tan 7

50 k  7k r  2r tan 110

Engineering Mechanics Dynamics -- IAA

Chap.7 3D Kinematics and Kinetics Translation

Rotation

v   r a    r    (  r) 111

Engineering Mechanics Dynamics -- IAA

Sample 7/1 The 0.8-m arm OA for a remote-control mechanism is pivoted about the horizontal x-axis of the clevis, and the entire assembly rotates about the z-axis with a constant speed N = 60 rev/min. Simultaneously, the arm is being raised at the constant rate  4 rad/s/ for  the position where   30 , determine (a) the angular velocity of OA, (b) the angular acceleration of OA, (c) the velocity of point A, and (d) the acceleration of point A. If, in addition to the motion described, the vertical shaft and point O had a linear motion, say, in the z-direction, would that motion change the angular velocity or angular acceleration of OA?

112

Engineering Mechanics Dynamics -- IAA

Sample 7/1 (a)   x  z  4i  6.28k rad/s (b)

     x  z  x   z z  x  6.28k  4i  25.1j rad/s2

  25.1j  0  25.1j rad/s2 i j k (c) v A    r  4 0 6.28  4.35i  1.60j  2.77k m/s 0 0.693 0.4 (d) a A    r    (  r)    r    v

i 0

j 25.1

k

i

j

k

0 

4

0

6.28

0 0.693 0.4

4.35 1.60 2.77

 (10.05i)  (10.05i  38.4j  6.40k )  20.1i  38.4j  6.40k m/s2 113

Engineering Mechanics Dynamics -- IAA

Body-Fixed Coordinate Translation

114

Engineering Mechanics Dynamics -- IAA

Rotation Inertial coordinates X-Y-Z Body fixed coordinates x-y-z A: point of interest B: origin of body-fixed coordinates, often the mass center A Ω: angular velocity of the rigid angular velocity of x-y-z about X-Y-Z i    i j    j k    k

v A  vB    rA / B  vrel   r    (  r a a  A

B

A/ B

A/ B

)  2  vrel  arel

or rA  rB                   2      rA   rB   115

Engineering Mechanics Dynamics -- IAA

Sample 7/3,7/4 Crank CB rotates about the horizontal axis with an angular velocity ω1 = 6 rad/s which is constant for a short interval of motion which includes the position shown. The link AB has a ball-and-socket fitting on each end and connects crank DA with CB. For the instant shown, determine the angular velocity ω2 of crank DA and the angular velocity ωn of link AB.

116

Engineering Mechanics Dynamics -- IAA

Sample 7/3,7/4 v A  v B  n  rA/ B [  r] v A  502 j, vB  100(6)i  600i mm/s

i

j

k

502 j  600i  nx ny

nz

6 

50 100 100  ny  nz

2  2nx

 nz

2  6 rad/s

0  2nx  ny [n rA / B  0] 50nx 100ny 100nz  0 4 8 10 nx   rad/s ny   rad/s nz  rad/s 3 3 3 2 2 n  (2i  4j  5k) rad/s n  22  42  52  2 5 rad/s 3 3 117

Engineering Mechanics Dynamics -- IAA

Angular velocity r3

r = r1  r2  r3

r

r2

r = 1  r1  2  r2  3  r3  0  T 2 r2  0 4 equations for 4 unknows; 2 and 3

r1

if r1, r2 , and r3 are general vectors, r1  0  then 2Tr2  0  T 3 r3  0 6 equations for 6 unknows; 2 and 3 118

Engineering Mechanics Dynamics -- IAA

Angular Acceleration r3

r

r2

r = r1  r2  r3 r = 1  r1  2  r2  3  r3  0

r1

 r = 1  1  1  r1      2  2  2  r2     3  3  3  r3   0  T 2 r2  0 4 equations for 4 unknows; 2 and 3

119

Engineering Mechanics Dynamics -- IAA

Angular Momentum x-y-z Body-fixed coordinates at C.G. ω: angular velocity of the rigid body

HG   [  (   )]dm HO   [r  (  r)]dm HG  IG 

(a)

33

31

 I xx I xy I xz     I yx I yy I yz     I zx I zy I zz     I xxx  I xyy  I xzz     I yxx  I yyy  I yzz  I   I   I    zx x zy y zz z 

(b) 120

Engineering Mechanics Dynamics -- IAA

Inertia Matrix  I xx I xy I xz     I I  I yy yz   yx  I zx I zy I zz   

I xx   ( y 2  z 2 )dm

I xy   xydm

I yy   ( z 2  x2 )dm

I xz   xzdm

I zz   ( x2  y2 )dm

I yz   yzdm

Principal Axes  I xx 0 0 I yy   0 0

0 0  I zz 

H  I xxx i  I yyy j  I zzzk

121

Engineering Mechanics Dynamics -- IAA

problem 07/64

Introduce axes The rectangular plate, with a     0,   ,   mass of 3 kg and a uniform small x y z 2 2 thickness, is welded at the 45° angle to 1 1 2 2 I xy  0, I yy  m(2a)  ma the vertical shaft, which rotates with the 12 3 angular velocity of 20π rad/s. Determine I yz  0, I xz  0 the angular momentum H of the plate 1 1 about O and find the kinetic energy of the I zz  m[(2a)2  (2b)2 ]  m(a2  b2 ) 12 3 plate.

Eq. 7/11 applied tox  y  z gives

H  jI yyy  kI zzz

z

1  1   j( ma2 )  k[ m(a2  b2 )] 3 3 2 2 z 1   y But j  jcos45  k sin45  ( j  k) a 2 a 1 k  jsin 45  k cos45  (j  k) y 2 1 3 b H  m[b2 j  (2a2  b2 )k]  20 (0.04j  0.06k) 6 6  x b   (0.4j  0.6k) Nms 1 1 T  T H  (20 k) (0.4j  0.6k )  6.0 2  59.2 J 2 2 122

Engineering Mechanics Dynamics -- IAA

problem 07/64

HG  IG

about x-y-z

HG  IG about x′ -y′ -z′ z z

y a a

 I xx 0 I'G   0 I yy  y 0  0

b x

b

2 1  m 2 a 0 0   12  0   1 2  0  0 m 2b 0    12 I zz   2 1 2 2  0 0 m 4a  4b   12  

0 0  0  0  1 '  Α  0 cos sin    0    sin        0  sin  cos     cos 

z



45 Mx

HG  AHG  AIGA  IG y  Note that IG  A IG A 1 2

1 2

Kinetic energy T  TIG   TIG

z  20 rad/s 123

Engineering Mechanics Dynamics -- IAA

Euler’s Equation

F  G

M M M M M M

M  H dH )xyz    H dt  ( H x i  H y j  H z k)    H

M  (

M  (H x  H yz  Hzy )i

x

 H x  H yz  H zy

y

 H y  H zx  H xz

z

 H z  H xy  H yx

x

 I xx x  (I yy  I zz )yz

y

 I yy y  (I zz  I xx )zx

z

 I zz z  (I xx  I yy )xy

 (H y  H zx  H xz ) j  (H z  H xy  H yx )k

124

Engineering Mechanics Dynamics -- IAA

problem 07/82

The plate has a mass of 3 kg and is welded to the fixed vertical shaft, which rotates at the constant speed of 20π rad/s. Compute the moment M applied to the shaft by the plate due to dynamics imbalance.

Eq. 7/23

2 M  I   x yz z 2

I yz   yzdm   y dl  

b/ 2

b / 2

z

y 2  2dy

 2 b3 b3 mb2 3(0.2)2  (  )   0.02 kgm2 3 2 2 2 2 6 6 45

y

Mx

z  20 rad/s

M x  0.02(20 )2  79.0 Nm

on plate, Mx  79.0i Nm but acting on shaft, M  79.0i Nm 125

Engineering Mechanics Dynamics -- IAA

problem 07/82

d d H   G   HG  dt dt  IG     IG   

MG 

 0     IG     0 z x       z 0 x    y z 0      0  0    0 0      0 0 0 

z

45

y

Mx

z  20 rad/s 126

Engineering Mechanics Dynamics -- IAA