Dynamics notes

Dynamics (Meriam and Kraige, Ed. ,2013)

Chapter 1. Introduction

Engineering Mechanics

Statics Dynamics Strength of Materials Vibration

Statics: ,distribution of reaction force from the appliedforce

Dynamics: , 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 deformablebodies

Vibration: x(t) = f(F(t)) on particles and rigid bodies1

a r(F +F )=mx

a r(F +F )=0 rFaF

th7

Newtonian Dynamics‧Kinematics: the relation among

‧Kinetics: the relation between ( ) and ( )x t F t

2

2

( ) ( )( ), ( ), and ( ), and dx t d x tx t x t x t x xdt dt

without reference to applied force

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

Cylindrical Coordinates

Spherical Coordinates

( , , )x y zr

( , , )r zr

( , , )R r

3

Displacement, Velocity, and Acceleration

2 2

1 1

s t

s tds vdt

2 2

1 1

v

v

t

tdv adt

2 2 2 2

1 1 1 1

( )x t t t

x t t tdx xdt xdt dt

4

Velocity and Acceleration Vector

ddt

rr v d

dt

rr a

5

Rectangular Coordinates

x y zx y zx y z

r i j kr i j kr i j k

r

x x xy y yz z z

r r r

6

t td vdt

rr v v e e

: radius of curvature

t t

t n

ddtv v

vv v

rr

e e

e e

Centrifugal and Tangential Acceleration

7

Time derivative of a vector

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

r

r

r rr

r

dddd

d d ddt dt dd dddt dt d

e e

e e

e ee e

e ee e

8

r z

r r z z

r r z

r zr r z z

r r r r r z

r e er e e e e

r e e e e e e

Cylindrical Coordinates (3D)

9

r zr r z e e e

2( ) ( 2 )r zr r r r z e e e

Spherical Coordinates

sin cos

sin cos

cos

cos

r

r r

z

r

r

r r

r

RR R

o

R R R

r er e e

ω e e

e e e

e e e

e ω e

e e

r e e e

10

2 2 2

2

2 2

cos

coscos 2 sin

1 sin cos

R R

R

R R

R

v v v

v Rv Rv R

a a a

a R R Rda R R

R dtda R R

R dt

r v e e e

r a e e e

Velocity and acceleration in Spherical Coordinates

11

Engineering Mechanics Dynamics -- IAA12

Chap.3 Kinetics of Particles

3.1 Force, Mass, and Acceleration3.2 Work and Energy3.3 Impulse and Momentum3.4 Impact and Orbital Mechanics

3.1 Force, Mass, and Acceleration

‧Newton’s Second Law Equation of Motion‧Inertial System: A coordinate system where ‧Free-body diagram

mF rmF r


1 1

2

2 2

3

0

8300 7360 750

1.257 m s

736 75(1.257) 830 N

0 1.257

y y

y

y

y y

F ma

T m g m ya

a

F ma

R m g m yRR

a dt

dt

3.77 m s

Sample 3/1A 75-kg man stands on a springscale in an elevator. The tensionT in the hoisting cable is 8300 N.Find the reading R of the scale innewtons and the velocity υ of theelevator after 3 seconds. The totalmass of the elevator, man, andscale is 750kg.


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

1 1 1

1 1 1

2 2 2

1 2

cos 02 sin

, 4 equation for 4 unknowns

2 constant

N m g m yN T m g m xm g T m yx y

x1

y1y2

x2


2 sin30 sin300

2 cos30 cos30 0

x x

y

F ma

B B mxF

B B mg my

Eliminate B and get3 3gx a

30302A BB

mg

a

x

y

The steel ball is suspended from theaccelerating frame by the two cords Aand B. Determine the acceleration ofthe frame which will cause the tension inA to be twice that in B


cos 0

25(9.81)cos20 0230 N

yF N mg my

NN

2

cos

25(9.81)sin20 0.15(230) 25

1.973 m/s

xF mg N mx

a

a

Frame and sphere as an unit：

Sphere alone：0

( )cos45 10(9.81)cos20 0130.4 N

y

A B

A B

F

T TT T

( )sin45 9.81sin20 10(1.973)19.56 N

x x

B A

B A

F ma

T TT T

75.0 N, 55.4 NA BT T Solution：

45AT

10(9.81) N

x

45BT

20

y

x

y20

25(9.81) N

0.15 NN

problem 03/19The 10-kg sphere is suspended from the 15-kg frame slidingdown the 20° incline. If the coefficient of kinetic frictionbetween the frame and incline is 0.15, compute each tensionof wires A and B


Check for motion. Assume static equilibrium.

From B： 196.2 NT Mass A：

max

0

196.2 (60)(9.81)sin30 09.81N

(0.25)(60)(9.81)cos30 127.4 N ( )

x

s

F

FFF N

a

xy

1m gN

T

AF

2m g

B

T

x

The system is released from rest withthe cable taut. Neglect the smallmass and friction of the pulley andcalculate the acceleration of eachbody and the cable tension T uponrelease if (a) μs = 0.25, μk = 0.2 and (b)μs = 0.15, μk = 0.1

No motion for (a)： 0, 196.2 Na T


20.589 m/s , 208 Na T Solution：

max (0.15)(60)(9.81)cos30 76.5 NF

motion for (b)

(60)(9.81)sin30 (0.1)(60)(9.81)cos30 60T a

A：

B： (20)(9.81) 20y xF ma T a

xy

1m gN

T

AF

2m g

B

T

xx xF ma

The system is released from rest withthe cable taut. Neglect the smallmass and friction of the pulley andcalculate the acceleration of eachbody and the cable tension T uponrelease if (a) μs = 0.25, μk = 0.2 and (b)μs = 0.15, μk = 0.1


0 0

( )0

( ) 0

F N g L bFT g L b T gb

Let = mass / length ：

Solve to obtain：1Lb

The chain is released from rest withthe length b of overhanging links justsufficient to initiate motion. Thecoefficients of static and kinetic fictionbetween the links and the horizontalsurface have essentially the samevalue μ. Determine the velocity υ ofthe chain when the last link leaves theedge. Neglect any friction at thecorner.

( )L b g

( )N L b g F

gb

0T

Non-constant Acceleration


1gL

Eliminate T to obtain：

0

22

[ (1 ) ]

1 [ (1 ) ]2 2

L

b

Lb

gd x L dxL

g x LxL

[ (1 ) ]ga x x LL

d xdx

( ) ( )F maT g L x L x agx T xa

Substitute b and simplify：

( )g L x

( )g L x gx

T( )g L x

Non-constant Acceleration


2 2

1 2

1

1

1, 0

1

t

at at

xg L x g Lx

xg g L x Lxgx x LLg

x x gL

gx e a a

LLx c e c e

Non-constant AccelerationAnother approach:


Curvilinear Motion in Polar Coordinates

2

2r

mF r r

mF r r

F r

r z

r r z z

r r z

r zr r z z

r r r r r z

r e er e e e e

r e e e e e e

r zr r z e e e

2( ) ( 2 )r zr r r r z e e e


2 ( )

( 2 )

r rF ma T m r r

F ma F m r r

case (a)case (b)

20

20

2

2

T mr F mb

T mr F mb

Sample 3/10Tube A rotates about the vertical O-axiswith a constant angular rate andcontains a small cylindrical plug B if massm whose radial position is controlled bythe cord wound around the drum ofradius b. Determine the tension T in thecord and the horizontal force Fθ exertedby tube on the plug if the constant angularrate of rotation of the drum is ω0 first inthe direction for case (a) and second inthe direction for case (b). Neglect friction.


22

2

( )

(3.5)2(9.81) 22.4

9.41 N

r

B

B

F m r r mr

N

N

Loss of contact at A：

2

2cos30

2.44.52 m/s

rF mr

mg m

0AN

BN

t

n

mg

BF

0AN

t

n

mg

AF

30

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

r

rr

θ

θ


Equilibrium：

10F T mg

20 sin30 0n nF ma T mg Motion：

2

1

sin30 0.5T mgkT mg

3030

1T

mg

1T

30

mg

2T

n

30 t

problem 03/66

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

w try using x-y coordinates


0

cos 0/cos

yF

N mgN mg

2

2

2

( )

sin ( sin )

( )sin sincos

cos

rF m r r

N m rmg mr

gr

Note that2 g

r

2cos 1gr

is a restriction.

N

y

n

mg

r

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


2 2

0 cos30 0

sin301200

y

r

F N mg

F m N mr

Solve：

For no slipping tendency, set F to zero on

0.9s with

1.155 , 149.4 ft/secN mg

min 0 as1 1

max tan tan (0.9) 42.0 30s

For max , set max sF F N

22max

0 cos30 sin30 0

cos30 sin30

y s

r s

F N mg N

F m N N mr r

max

2.40345 ft/sec

N mg

N

F

mg

n

y

30

problem 03/76Determine the speed υ at which the race car will have no relianceon friction to the banked track. In addition, determine the minimumand maximum speeds, using the coefficient of static friction μs =0.9.

r


0 0

2 20

, sin , sin

, sin ( )

2 (1 cos )

F ma mg ma a g

d a ds d g Rd

gR

When

2

20

20

, cos

cos 2 (1 cos )

(3cos 2 )

r rF ma mg N mR

N mg m mgR

mggR

0,N so2 2

10 023cos 2 cos ( )3 3gR gR

For 10

20, cos ( ) 48.23

R N

n

t

0 mg

A small vehicle enters the top A of the circular path with a horizontalvelocity υ0 and gathers speed as it moves down the path. Determine anexpression for the angle β which locates the point where the vehicleleaves the path and becomes a projectile. Evaluate your expression forυ0 = 0. Neglect friction.

θ


3.2 Work and Energy

12

T

T

T

T

U d

m d

m d

m

F r

r r

r r

r r

Work and Kinetic Energy


Work and Potential Energy

2

1

2

1

2 2

1 2 1 1

2 1

2 2

1 2 21 1

2

2 1

2

2 1

( ) ( )

( )

1 1( )

1 1( )

y

y

er r

r

e r

e

U d mg dx dy

mg dy mg y y

Gm mU d drr

drGm mr

Gm mr r

mgRr r

F r j i j

F r e e


21-2

2 1

2 2 21 2

2 1

2 2 22 1

2 1

1 1

1 1 1 1 12 2 2

1 12

U mgRr r

m mgR mr r

gRr r

2 3 322 32

6 6 6 2

2

30 000 10 102(9.81) (6371)(10)3.6 6371 1200 6371 500

69.44(10 ) 10.72(10 ) 58.73(10 ) ( /s)7663 /s

mm

Sample 3/15A satellite of mass m is put into an elliptical orbit around the earth. At point A, itsdistance 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 = 1200km from the earth.


Potential Energy

TV d

mgdy

mgh

F r 2

1

2

2

2

2

T

r

r

V d

GMm drr

mgR drr

mgRr

F r


Principle of Work and Energy

V W T

TT TdU d dP

dt dt dt

F r rF F r=

‧Kinetic energy

‧Power

0T W V

Conservative Force

x y zV

i j k

F


Constant total energy is A A p pE T V T V

Thus2 2

2 21 12 2A p

A p

mgR mgRm mr r

2 2 2 1 12 ( )A pp A

gRr r

2 2 1 12 ( )A pp A

gRr r

A satellite is put into an elliptical orbit around the earth and has avelocity υP at the perigee position P. Determine the expression forthe 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.


= mass per unit lengthFor equil. at start ( )kgb g L b

1k

k

Lb

gU T V 2

0

( )2

L b

k kL bU dF x gxdx g

212

T L ( )( )2g

L bV g L b

L b

bdFdx x b

2L b 2

L b

2L b

The chain starts from rest with a sufficient number of links hangingover the edge to barely initiate motion in overcoming frictionbetween the remainder of the chain and the horizontal supportingsurface. Determine the velocity υ of the chain as the last link leavesthe edge. The coefficient of kinetic friction is μk. Neglect and frictionat the edge.

Thus2 2 2

2( ) 12 2 2

k

L b L bg L g

2 (1 )( [ ])kbg L b L bL

Now substitute b

So 2 (1 )( [1 ] [ ])1 1 1

1

k k kk

k k k

k

Lg L L

gL

1 k

gL


3.3 Impulse and Momentum

( )

mmd mdtddt

G rF r

r

G

G

Linear momentum

Impulse

2

1|dt m dt md m rrF r r r G

Conservation of Linear Momentumdt F G if 0 0 F G


1mv

2

1

t

tmgdt

2

1

txtR dt

2

1

tytR dt

15

y

x=+

Sample 3/19

2

1

2

1

1 2

1 2

2 2 2 2

1

( ) ( )

4 /16 4 /16(50) (0.02) (70cos15 )32.2 32.2

( ) ( )

4 /16 4 /16(0) (0.02) (70sin15 )32.2 32.2

45.7 lb, 7.03 lb

45.7 7.03 46.2 lb

tan

t

x x xt

x

t

y y yt

y

x y

x y

y

m F dt m

R

m F dt m

R

R R

R R R

R 1 7.03tan 8.7545.7

xR

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 balland the angle β made by R with the horizontal


R T m

R ma

2 20 2ad

:

:

:

1.62/16(0.001)= (150)32.2

R =472 lbR

1.62/16472=32.2

a

,

2150,000 / sec (4660 )a ft g,

2 2150 0 2(150,000)d , 0.075d ft or 0.900in

25

0mg 150 / secft

problem 03/207

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


Angular Impulse and Momentum

O m H r v

( ) ( ) ( )

( )

( )( )

O

z y x z y x

z yx

O y x z

x y z y xz

mm v y v z m v z v x m v x v y

m v y v zHx y z H m v z v xv v v m v x v yH

H r vi j k

i j kH


O

O O

O O

mm m m m

H r vH r v r v v v r v r F M

M H

Time Derivative of Angular Momemtum

Conservation of Angular Momentum

2

12 1( ) ( )

O Ot

O O O Otdt

M H

M H H H

2

1O 1 2( ) ( )

t

O Otdt H M H

The total angular impulse on a particle of mass mabout a fixed point O equals the correspondingchange in angular momentum about that point.

Principle of Conservation of Angular Momentumif , then 0O H 1 2( ) =( )O OH Hor0O M

0


9

6

( ) ( )

6(10 )74075(10 )

59 200 m/s

O A O B

A A B B

A AB

B

B

H H

mr mr

rr

Sample 3/25 Molynia OrbitA comet is in the highly eccentric orbit shown in the figure. Its speed ate the mostdistant 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.


0 0OM H

so OH constant

min 6371 390 6761r km

max 2(13520) 6761 20279r km

For OH constant6371(33880) 11720 20279B A

11300 /B km h 19540 /A km h

B B

A

33880 /km h11720 /km h

A

ro

maxr minr

2(13520) km

The central attractive force F on an earthsatellite can have no moment about the centerO of the earth. For the particular elliptical orbitwith major and minor axes as shown, a satellitewill have a velocity of 33880 km/h at the perigeealtitude of 390 km. Determine the velocity of thesatellite at point B and at apogee A. The radiusof the earth is 6371 km.


0H ;02 ( ) 2 (2 ) (2 ) 0mr r m r r 0 / 4

2 2 2 200 0

1 12( [ ] ) 2( [2 ] ) (3 / 4)2 2 4

T m r m r mr

2 2 2 20 0

3/ / 3 / 44

n T T mr mr

problem 03/228

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


From

0 0

2 2cos ( )

cos

M Hdmgl ml mldt

gl

so at

d d

2

0 0cos

2 | g dl

290

2sin g g

l l

90 2l gl

By work-energy21 2

2V T mgl m gl

mg

lT

O

The simple pendulum of mass m and length lis released from rest at θ = 0. Using only theprinciple of angular impulse and momentum,determine the expression for in terms of θand the velocity υ of the pendulum at θ = 90°.Compare this approach with a solution by thework-energy principle.


Direct Central Impact


0

0

0

0

' '1 1 0 0 1

2 0 1 1 00

' '2 2 0 2 0

2 0 2 0 20

' '2 1

1 2

[ ( )][ ( )]

( )( )

relative velocity of separationrelative velocity of approach

t

rtt

d

t

rtt

d

F dt memF dt

F dt memF dt

e

Coefficient of Restitution



Oblique Central Impact

' '1 1 2 2 1 1 2 2

'1 1 1 1

'2 2 2 2

' '2 1

1 2

( ) ( ) ( ) ( )

( ) ( )

( ) ( )

( ) ( )( ) ( )

n n n n

t t

t t

n n n

n n n

m m m m

m mm m

VeV


Sample 3/29

' ' '2 1 1

1 2'1

'1 1

'1 1

' ' 2 ' 2 2 21 1

'' 1 11

'1

( ) ( ) 0 ( )0.5( ) ( ) 50sin30 0

( ) 12.5 ft/sec

( ) ( )

( ) ( ) 50cos30 43.3 ft/sec

( ) ( ) 12.5 43.345.1 ft/sec

( ) 12.5tan tan( ) 43.3

n n n

n n

n

t t

t t

n t

n

t

e

m m

16.10

A ball is projected onto the heavy plate with a velocity of 50 ft/sec at the30° angular shown. If the effective coefficient of restitution is 0.5, computethe rebound velocity υ′ and its angle θ′.


' '1 1 2 2 1 1 2 2

' '1 2

' ' ' '2 1 2 1

1 2' '1 2

' '1 1 1 1 1 1

( ) ( ) ( ) ( )

5.20 0 ( ) ( )

( ) ( ) ( ) ( ) 0.6 ( ) ( ) 5.20 0

( ) 1.039 m/s ( ) 4.16 m/s

( ) ( ) ( ) (

n n n n

n n

n n n n

n n

n n

t t t

m m m m

e

m m

' '

2 2 2 2 2 2

) 3 m/s

( ) ( ) ( ) ( ) 0t

t t t tm m

Sample 3/30

Spherical particle 1 has a velocity υ1 = 6 m/s in thedirection shown and collides with spherical particle2 of equal mass and diameter and initially at rest. Ifthe coefficient of restitution for these conditionsis , determine the resulting motion of eachparticle following impact. Also calculate thepercentage loss of energy due to the impact.

0.6e


2gh , ' '2gh

' ' 1100 0.7242100

heh

'

2100 11002100

47.6%

mgh mghnmgh

As a check of the basketball before the start of a game, the refereereleases the ball from the overhead position shown, and the ballrebounds to about waist level. Determine the coefficient ofrestitution and the percentage n of the original energy lost duringthe impact.


Central-Force Motion

202

2

20

mGmm r rr

r r

mr h

F r


Orbital Mechanics

02

1 cos GmCr h


D’Alembert’s Principle and Inertia Force

inertia forcemr

F ma 0F mr 0

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


fig_04_001

i i

c

mm

F f rF r

Equation of motion

c i im m r rMass center

Chap. 4 Kinetics of Systems of Particles

Principle of motion of the masscenter the resultant of theexternal forces on any system ofmasses equal the total masstimes the mass centeracceleration.


Linear Momentum

( )

( )

i i

i c i

i c i i

c

mm

dm mdt

m

G rr ρ

r ρ

r


fig_04_003

Kinetic Energy

12

1 ( 2 )21 12 2

i c i

i c i

Ti i i

T T Ti c c i c i i i i

T Tc c i i i

T m

m m m

m m

r r ρr r ρ

r r

r r r ρ ρ ρ

r r ρ ρ


Angular Momentum about a Fixed Point

( )

( )O i i i

O i i i i i

i i

O

m

m mO

H r r

H r r r rr F

M

Angular Momentum about c.g.

( )( )( )

G i i i

i i c i

i i c i i i

i i i

mmm mm

H ρ rρ r ρρ r ρ ρρ ρ

( )

G i i C i i i i

i i

G

m mH ρ r ρ ρ rρ FM


p_04_020

1 2 0 1 1 2 2 0(0) ( ) ( ) ( )2 2l lm l m m m s l x m s x m s

ls

2x 1x2m 1m

0mAB

C

With respect to C, i im x constant

Simplify and get

But they meet when

1 1 2 2

0 1 2

m x m xsm m m

2 1x x l so 1 2 1 2

0 1 2

( )m m x m lsm m m

problem 04/22

The man of mass m1 and the woman of mass m2 arestanding on opposite ends of the platform of mass m0 whichmoves with negligible friction and is initially at rest with s = 0.The man and woman begin to approach each other. Derivean expression for the displacement s of the platform whenthe two meet in terms if the displacement x1 of the relative tothe platform.


sp_04_03_01

Sample 4/4

A shell with a mass of 20 kg is fired frompoint O, with a velocity u = 300 m/s in thevertical 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 Ais observed to rise vertically a distance of 500m above P, and fragment B is seen to have ahorizontal velocity vB and eventually lands atpoint Q. When recovered, the masses of thefragments A, B, and C are found to be 5, 9,and 6 kg, respectively. Calculate the velocitywhich fragment C has immediately after theexplosion. Neglect atmosphere resistance.


sp_04_03_01

2 2

/ 300(4/5)/9.81 24.5s

[(300)(4/5)] 2940m2 2(9.81)

2 2(9.81)(500) 99.0m/s/ 4000/ 24.5 163.5m/s

z

z

A A

B

t u g

uhg

ghs t

1 2[ ] A A B B C Cm m m m G G v v v v

2 2 2

320(300)( ) 5(99.0 ) 9(163.5)( cos45 sin45 ) 65

6 2560 1040 495427 173.4 82.5 m/s

(427) (173.4) (82.5) 468m/s

C

C

C

C

i k i j v

v i j kv i j k

Sample 4/4


Steady Mass Flow

1 1 1 2 2 2A A m

2 1 2 1( ) ( ) ( )m m m G v v v v

m F v


( )

106(680 1000/3.6) 4(680)45400 N 45.4 kN

0

45.4 32 4.6(9.81)sin 0

sin 0.296 17.22

a f

x x

T m u m u

F ma

problem 04/35

T

N

4.6(9.81) kNmg

1000 km/h

x32 kNR

The jet aircraft has a mass of 4.6 Mg and a drag (airresistance) of 32 kN at a speed of 1000 km/h at a particularaltitude. The aircraft consumes air at the rate of 106 kg/sthrough 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 tothe exhaust nozzle, determine the maximum angle ofelevation at which the jet can fly with a constant speed of1000 km/h at the particular altitude in question.


problem 04/38

Density of salt water, 31030 kg/m

Resistance R equals net trust T

where ( )T m u Nozzle velocity 20.082/ 41.8 m/s(0.050)

4

u Q A

10001030(0.082) 84.5 kg/s, 70 19.44 m/s3600

84.5(41.8 19.44) 1885 N

m Q

R T

The jet water ski has reached its maximum velocity of 70 km/hwhen operating in salt water. The water intake is in thehorizontal tunnel in the bottom of the hull, so the water entersthe intake at the velocity of 70 km/h relative to the ski. Themotorized pump discharge water from the horizontal exhaustnozzle of 50-mm diameter at the rate of 0.082 m3/s. Calculatethe resistance R of the water to the hull at the operating speed.

64

Engineering Mechanics Dynamics -- IAA

With reversers in place

so

F mu

sin30

(50 0.65)(650)sin30 50(55.6 0)16460 2780 19240 N

R g aT m u m

19240 0.63830100

n

problem 04/45

Without reversers

(50 0.65)650 50(55.6)32900 2780 30100 N

g aT m u m

RT x

650 m/s

650 m/s

30

30

200/3.6 55.6 m/s

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

65


Variable Mass system

0( )R m muF R m

F m mu


mu pA mg R m

Rocket Propulsion


Sample 4/12

00 0

0

,

ln

m t

m

T mg m T mu mu mu mg mdm dmd u gdt d u g dtm mmu gtm

0

0max 0

( ) /

ln ( )

b b

bb

t m m mm gu m mm m

let mb：mass of rocket when burnout occurs

Solution I (F=ma solution)

A rocket of initial total mass m0 is fired vertically with constant acceleration until the fuelis exhausted. The relative nozzle velocity of the exhaust gas has a constant value u atatmospheric pressure throughout the flight. If the residual mass of the rocket structureand machinery is mb when burnout occurs, determine the expression for the maximumvelocity reached by the rocket. Neglect atmospheric resistance and the variation ofgravity with altitude.


Sample 4/12

Solution II (Variable-Mass solution)

,F mg F m mu mg m mumu mu T T mg m

same as Solution I

A rocket of initial total mass m0 is fired vertically with constant acceleration until the fuelis exhausted. The relative nozzle velocity of the exhaust gas has a constant value u atatmospheric pressure throughout the flight. If the residual mass of the rocket structureand machinery is mb when burnout occurs, determine the expression for the maximumvelocity reached by the rocket. Neglect atmospheric resistance and the variation ofgravity with altitude.


fig_05_001

Chap.5 Planar Kinematics


fig_05_004

( )

rr r

r r rω

ω ω ω

displacement velocity acceleration

Fixed Axis RotationZ

Y

X


sp_05_03_01

2

2

2

[ ] 2 (0.4 0.3 ) 0.6 0.8 m/ s

[ ( )] 2 (0.6 0.8 ) 1.6 1.2 m/ s

[ ] 4 (0.4 0.3 ) 1.2 1.6 m/ s

[ ] 2.8 0.4 m/ s

n n

t t

n t

v v k i j i j

a a k i j i j

a a k i j i j

a a a a i j

r

r

r

2 rad/s k

24 rad/s k

2 20.6 0.8 1m/s

2 22.8 0.4 2.83 m/sa

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

Write the vector expressions for the velocity and acceleration of point A when


p_05_002

2

2

2

( 6k 45 j)

270i mm/s

4k 45 j 6 (45 j)

180i 1620 j mm/s

A A

A A A

r

a r r

(a)

(b)

2

2

2

6k ( 30i 45 j)

270i 180 j mm/s

4k ( 30i 45 j) 6 ( 30i 45 j)

900i 1740 j mm/s

B B

B B B

r

a r r

24rad/s rad/s, 6

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


sp_05_04_01

,where , ,

, and

O O

O O O

s rr a r

s a s

2

sin ( sin )

(1 cos ) (1 cos )

(1 cos ) sin

(1 cos ) sin

O

O O

O

x s r rx r

x

a r

2

cos (1 cos )

sin sin

sin cos

sin cos

O

O O

O

y r r ry r

y

a r

20x y r and

Sample 5/4

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.

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.


Sample 5/7

3 0 0 10 3 1.7320.1732 0.1 0

4 1.732 m/s

A

i j kv i i j i

i j

2 24 (1.732) 19 4.36 m/sA

/ 0A O A O O v v v v rω

0

-10 rad/ s0.2( cos30 sin30 ) 0.1732 0.1 m3 m/ s

O

kr i j i jv i

ω=

Calculate the velocity of point A on the wheel without slipping for the instant represented.

sp_05_07_01


Sample 5/11

[ / ]

/ 3/ 0.300 10 rad / sO

r

OC

2 2(0.300) (0.200) 2(0.300)(0.200)cos1200.436 m

AC

[ ]

0.436(10) 4.36 m/sA

r

AC

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


sp_05_08_01

Sample 5/8

1 2 3

1 1 2 2 3 3

1 2

1 2

2

1 2

0 100 (175 50 ) 2 75100 50

175 1503 6,7 7

ω ωω ωω

ω ω

r r r rr r r r

k j k i j k iω ω ω

r

r1

r2

r3


sp_05_08_01

Sample 5/14

1 2 3

1 1 2 2 3 3

1 1 1 1 1

2 2 2 2 2

3 3 3 3 3

( )( )( )

r r r rr r r rr r r

r rr r

ω ω ωω ω ωω ω ωω ω ω

r

r1

r2

r3

1 23 3100 ( ) ( 100 ) ( 175 50 )7 7

6 6( [( ) ( 175 50 )] 0 2 (2 [ 75 ])7 7

r k j k k j k i j

)k k i j k k i

21 0.1050 rad/s 2

2 4.34 rad/s


sp_05_09_01

Sample 5/9

/A B A B v v v

1

[ ]5 1500(2 ) 65.4 f / sec

12 605 14

sin sin60sin 0.309 18.02

B

r

t

//

65.4 67.3 ft /secsin78.0 sin72.0

65.4 34.4 ft /secsin30 sin72.0

AA

A BA B

/ 34.4[ / ] 29.5 rad/sec14/12

A BABr

AB

/G B G B v v v

/ /4 (34.4) 9.83 ft / sec

1464.1ft /sec

G B AB A B

G

GBGBAB


sp_05_09_01

Sample 5/9

r

r1r2

y

x1 2

1 1 2 2

1 1 1 1 1

2 2 2 2 2

( )( )

r r rr r rr r r

r r

ω ωω ω ωω ω ω


sp_05_15_01

/ /( ) ( )A B A B n A B ta a a a 2[ ]na r

2

2

5 1500[2 ]( )12 60

10280 ft / sec

Ba

2[ ]na r

2/

2

14( ) (29.5)12

1015 ft / sec

A B na

/10280cos60 1015cos18.02 ( ) sin18.02A A B ta

/0 10280sin60 1015cos18.02 ( ) cos18.02A B t

2/( ) 9030 ft /secA B t 23310 ft /secA

[ / ]ta r 2 2

/ 9030/(14/12) ft /sec 7740 rad/secA B

23310 ft / secA 2/( ) 9030 ft /secA B t

Sample 5/15


fig_05_005

General Motion: Rotation + Translation

fig_05_006


fig_05_007

Instantaneous Center


fig_05_011

Body-Fixed Coordinates in Rotation

A B

A B

A B

B

r rr rr r

r

ρω ρ+ ρω ω ρ ω ρ ω ρ+ω ρ+ ρ

ω ω ρ ω ρ+ 2ω ρ ρ

Ar

Coriolis Acceleration

2A B

r r ω ω ρ ω ρ ρ+ ω ρ

Coriolis acceleration


sp_05_16_01

A v ρ ρ4 6 5 24 5 in./secA v k i i j i

2 2(24) (5) 24.5 in./secA

( ) 2A a ρ ρ ρ ρ2( ) 4 (4 6 ) 4 24 961 in./sec k k i k j ρ

210 6 60 in./sec k i j ρ22 2(4 ) 5 40 in./sec k i j ρ

281 in./sec iρ2(81 96) (40 60) 15 20 in./secA a i j i j

2 2 2(15) (20) 25 in./secAa

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

r r

y

x

2


( )

rr r

r r rω

ω ω ω

displacement velocity acceleration

With and without Body-Fixed Coordinate

Let

5 , 2 , 310

-20 30

r i k kr r jr r r

= i + j


p_05_161

( sin j cos k)

22 k ( sin j cos k)

2 sin i

cora

(west)

The track provides the necessary westwardacceleration so that the velocity vector is properlyrotated and reduced in magnitude.

B y

z

A

For 500 km/h

(a) Equator,

(b) North pole,

0 0cora

5 250090 2(7.292 10 ) 0.0203 m/s3.6cora

problem 05/163

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


Chap. 6 Dynamics of Planar Rigid BodyEquation 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.

G

G G

m

F r

M H


Equation of Motion in 2D

2

G i i i

i i i

mm

dm

I

H ρ ρρ ω ρ

ω

ω

Angular momentum

c

G

mI

F rM ω


Moment about a Fixed Point

0,if P is fixed

P G C

G C

P P

P

P O

mI mI m

I I

M H ρ rω ρ rω ρ r

rω ω


when tipping impends

F mg mx

(0.4) (0.6) 0 GM mg mg I

23

ANAF

G

50(9.81)N0.8m

1.2m

Determine the value of the force P whichwould cause the cabinet to begin to tip.What coefficient μs of static friction isnecessary to ensure that tipping occurswithout slipping?

AN mg

23

x g g

As a whole :

1( ) (50 10) 60 40 P m m x x x g


Center of Percussion

or

C

G G

O O

mM I

M I

F rω

The resultant force at the center of percussion is zero.

ok

ok Radius of gyration about point O


Center of Percussion

-

012

23

X

Y

O

F R mxmg R my

F h Iy

x

h

XR

YR

hmg

F


2

2

120(9.81)(0.2) 20(0.4)3

36.8rad/s

O OM I

20.2 36.8 7.36m/s

x r

xG0.2m 0.2m

t

20(9.81)N

2F

O 20(9.81) 2 20( 7.36)12 49.04

24.5N

x

A B

F mxF

F mg

F F F

problem 06/34

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


mgy

O

r GO

mg

O

r GO

22

/ 2

( )2

/ 2

O O

y

M I mgr mr

g rgF my mg O mrr

O mg

(a)

(b) 2 21( )2

2 / 32( )3

/ 3

O O

y

M I mgr mr mr

g rgF my mg O mrr

O mg

problem 06/38

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


Sample 6/5

A metal hoop with a radius r = 6 in. isreleased from rest on the 20° incline. If thecoefficients of static and kinetic friction areμs = 0.15 and μk = 0.12, determine theangular acceleration α of the hoop and thetime for the hoop to move a distance of 10 ftdown the incline.


Sample 6/5[ ]x xF ma sin20mg F ma

[ 0]y yF ma cos20 0N mg

[ ]GM I 2Fr mr

0.1710F mg cos20 0.940N mg mg

max[ ]sF N max 0.15(0.940 ) 0.1410F mg mg

max[ ]kF N 0.12(0.940 ) 0.1128F mg mg

x[ ]xF ma sin20 0.1128mg mg ma 20.229(32.2) 7.38 ft/seca

[ ]GM I 20.1128 ( )mg r mr 20.1128(32.2) 7.26 rad/sec6 /12

21[ ]2

x at2 2(10) 1.646 sec

7.38xt

Assume pure rolling 4 equations for 4 unknowns a r

and

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

So it is slipping Solve again the 4 unknowns . kf N


Sample 6/7

98

The slender bar AB weighs 60 lb and moves in the vertical plane, with its endsconstrained to follow the smooth horizontal and vertical guides. If the 30-lb force isapplied at A with the bar initially at rest in the position for which θ = 30°, calculatethe resulting angular acceleration of the bar and the forces in the small end rollersat A and B.


Sample 6/72cos30 2 cos30 1.732 ft/secxa a 2sin30 2 sin30 1.0 ft/secya a

[ ]GM I 21 6030(2cos30 ) (2sin30 ) (2cos30 ) (4 )12 32.2

A B

[ ]x xF ma[ ]y yF ma

6030 (1.732 )32.2

B

6060 (1.0 )32.2

A

68.2 lbA 15.74 lbB 24.42 rad/sec

[ ]CM I m d 21 6030(4cos30 ) 60(2sin30 ) (4 )12 32.2

60 60(1.732 )(2cos30 ) (1.0 )(2sin30 )32.2 32.2

4.39 9.9424.42 rad/sec

[ ]y yF ma[ ]x xF ma

6060 (1.0)(4.42)32.2

A

6030 (1.732)(4.42)32.2

B

68.2 lbA

15.74 lbB


mg

r

x

G

A

N

0 0OM I I M

0s Hence no friction force and

sinx x AF ma mg mr sinAgr

problem 06/82

Determine the angular acceleration of each of the two wheelsas they roll without slopping down the inclines. For wheel Ainvestigate the case where the mass of the rim and spokes isnegligible and the mass of the bar is concentrated along itscenterline. For wheel B assume that the thickness of the rim isnegligible compared with its radius so that all of the mass isconcentrated in the rim. Also specify the minimum coefficient ofstatic friction μs required to prevent each wheel from slopping.


NC

F

mg

r

x

G

B

2sin 2

sin2

C C B

B

M I mgr mrgr

1 sin / cos2

1 tan2

s

s

F mg mgN

2 sin2gM I Fr mrr

1 sin2

F mg

problem 06/82

Determine the angular acceleration of each of the two wheelsas they roll without slopping down the inclines. For wheel Ainvestigate the case where the mass of the rim and spokes isnegligible and the mass of the bar is concentrated along itscenterline. For wheel B assume that the thickness of the rim isnegligible compared with its radius so that all of the mass isconcentrated in the rim. Also specify the minimum coefficient ofstatic friction μs required to prevent each wheel from slopping.


b

mg

C

45

Ar I

Ama/( )G A tm ar

b

G

T

2

2

12 6 2 2

34

1 3( )6 42

2 2 (12)(9.81) 20.8 N8 8

A

G

M I madmgb b bmb m

gb

M Ib gT mb

b

T mg

problem 06/84

The uniform 12-kg square panel is suspended from point Cby the two wires at A and B. If the wire at B suddenlybreaks, calculate the tension T in the wire at A an instantafter the break occurs.


Kinetic Energy

212

T m

212 OT I

2 21 12 2 CT m I


mg

A

x

mg

B

x

2 2

sin1 12 2

U TU mgx

T m I

2

2 2 2 2

1 021 1 ( )2 2

T m

T m mr mr

2

2

1sin 2 sin2

sin sin

A

B

mgx m gx

mgx m gx

case A：

case A：

case B：

case B：

problem 06/116


1 1 2 2T V T

2 2 21 1 ( )2 2 12 2l lmg x ml m x

22 2

( )2

6 2 2

lg x

l lx x

max 0.211 2 2 2

( 0.211 )2 1.861

0.211 (0.211 )6 2 2

x

lg l gll l l

20d

dx

0.789x l 0.211x lset obtain or

0.789x lThe solutionwould yield the sameonly then the motion is CCW.

max

For the pivoted slender rod of length l, determine the distance xfor which the angular velocity will be a maximum as the barpasses the vertical position after being released in thehorizontal position shown. State the corresponding angularvelocity.


2

11 2

t

tdt

F G

G F G

Linear Momentum Angular Momentum

O OH I

2

11 2( ) ( )

O Ot

O O Ot

M H

H M dt H


Sample 6/16The uniform rectangular block of dimensions shown is sliding to the lefton the horizontal surface with a velocity v1 when it strikes the small stepat O. Assume negligible rebound at the step and compute the minimumvalue of v1 which will permit the block to pivot freely about O and justreach the standing position A with no velocity. Compute the percentageenergy loss n for b = c.


Sample 6/16[ ]O OH I

2 2 2 2 2 22 2 2

1( ) { ( ) [( ) ( ) ]} ( )12 2 2 3O

c b mH m b c m b c

1 2[( ) ( ) ]O OH H 2 21 2( )

2 3b mm b c 1

2 2 23

2( )b

b c

2 2 3 3[ ]T V T V 2 2

22

1 0 0 [ ]2 2 2 2O

b c bI mg

2 2 2 2 212 231 ( )[ ] ( )

2 3 22( )bm mgb c b c b

b c

22 2

1 22( (1 )( )3g c b c b

b

2 2 22 2 2 21 2 2 222 2 2 22 11 2

1 13 32 2 1 11 3 2( )

4 12

OO

m IE k b c bn b cE b c cm

b

62.5%n b c


problem 06/188

The homogeneous sphere of mass m and radius r is projected along theincline 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 tof the period of slopping. In addition, state the velocity v of the masscenter G and the angular velocity ω at the end of the period of slipping.


problem 06/188

0

0

0

0

0

( ) 0 cos

( )

( cos sin ) ( )

ty y y

tx x x

k

F dt m N mg

F dt m

mg mg t m

00

2

( )

2( cos )5

tG

k

M dt I

mg r t mr

r

7 cos 2sink

2 tan7k

0

0

0

2(7 cos 2sin )

57 2tan

57 2 tan

k

k

k

k

k

tg

r r

mgG

r

N kN

y

x(1)

(2)

We desire the time t when (3)Solution of Eqs. (1)-(3)：

For slipping to cease,

or


Chap.7 3D Kinematics and KineticsTranslation Rotation

v r

( ) a r r


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

4 30

112


Sample 7/14 6.28 rad/sx z i k

2

2

6.28 4 25.1 rad/s

25.1 0 25.1 rad/s

x z x z

z x

k i j

j j

(a)

(b)

(c)

(d)

4 0 6.28 4.35 1.60 2.77 m/s0 0.693 0.4

A i j k

v r i j k

2

( )

0 25.1 0 4 0 6.280 0.693 0.4 4.35 1.60 2.77(10.05 ) (10.05 38.4 6.40 )20.1 38.4 6.40 m/s

A

a r r r vi j k i j k

i i j ki j k

113


Body-Fixed Coordinate Translation


Rotation

i i j j k k

/ rel

/ / rel rel( ) 2A B A B

A B A B A B

v v r v

a a r r v a

Inertial coordinates X-Y-ZBody fixed coordinates x-y-zA: point of interestB: origin of body-fixed coordinates, often

the mass center AΩ: angular velocity of the rigid angular

velocity of x-y-z about X-Y-Z

or 2

A B

A B

r rr r


Sample 7/3,7/4

Crank CB rotates about the horizontal axiswith an angular velocity ω1 = 6 rad/s which isconstant for a short interval of motion whichincludes the position shown. The link AB hasa ball-and-socket fitting on each end andconnects crank DA with CB. For the instantshown, determine the angular velocity ω2 ofcrank DA and the angular velocity ωn of linkAB.


Sample 7/3,7/4/

2[ ] 50 , 100(6) 600 mm/sA B n A B

A Br

v v rv j v i i

250 600

50 100 100x y zn n n

i j kj i

2

6

2

0 2

y z

x z

x y

n n

n n

n n

2 6 rad/s

/[ 0] 50 100 100 0x y zn A B n n n r

2 2 22 2 4 5 2 5 rad/s3n

4 8 10rad/s rad/s rad/s3 3 3x y zn n n

2( 2 4 5 ) rad/s3n i j k


Angular velocity

1 2 3

1 1 2 2 3 3T

2 2

2 3

1 2 3

1T

2 2T

3 3

2 3

0

0

, , 0

0

0

r = r r rr = r r r

r

r r rr

r

r

4 equations for 4 unknows; andif and are general vectors,

then

6 equations for 6 unknows; and

1r

2r

r3r

118


Angular Acceleration

1 2 3

1 1 2 2 3 3

1 1 1 1

2 2 2 2

3 3 3 3

T2 2

2 3

0

0

0

r = r r rr = r r r

r = r

r

r

r

4 equations for 4 unknows; and

1r

2r

r3r

119


Angular Momentum

[ ( )]G dm H

[ ( )]O dm H r r

(a)

(b)

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

3 3 3 1

G G

xx xy xz

yx yy yz

zx zy zz

xx x xy y xz z

yx x yy y yz z

zx x zy y zz z

I I II I II I I

I I II I I

I I I

H I


Inertia Matrix

xx x yy y zz zI I I H i j k

Principal Axes

0 00 00 0

xx

yy

zz

II

I

2 2

2 2

2 2

( )

( )

( )

xx xy

yy xz

zz yz

I y z dm I xydm

I z x dm I xzdm

I x y dm I yzdm

xx xy xz

yx yy yz

zx zy zz

I I II I II I I

121


problem 07/64

b

b

x

y

a ay

z

z

Introduce axes0, ,

2 2x y z

2 21 10, (2 )12 3x y y yI I m a ma

0, 0y z x zI I 2 2 2 21 1[(2 ) (2 ) ] ( )

12 3z zI m a b m a b

x y z Eq. 7/11 applied to gives

2 2 21 1( ) [ ( )]3 32 2

y y y z z zI I

ma m a b

H j k

j k1cos45 sin45 ( )2

j j k j k

1sin45 cos45 ( )2

k j k j k

2 2 21 3[ (2 ) ] 20 ( 0.04 0.06 )6 6

( 0.4 0.6 ) Nms

m b a b

H j k j k

j k21 1 (20 ) ( 0.4 0.6 ) 6.0 59.2 J

2 2TT H k j k

But

The rectangular plate, with amass of 3 kg and a uniform smallthickness, is welded at the 45° angle tothe vertical shaft, which rotates with theangular velocity of 20π rad/s. Determinethe angular momentum H of the plateabout O and find the kinetic energy of theplate.


problem 07/64

b

b

x

y

a ay

z

z

z

45 y

20 rad/sz

xM

2

2

22 2

1 2 0 0120 0

10 0 0 2 012

0 0 10 0 4 412

G G

G G

x x'G y y

z z

m aI

I m bI

m a b

H I

H I

I

about x-y-z

about x′ -y′ -z′

1 0 0 0 00 cos sin 0 sin0 sin cos cos

'

G G G G

Α

H A H A I A I

Note that G G I A I A

Kinetic energy T T1 1T2 2G G

I I


Euler’s Equation

F G

M H

( )

( )

xyz

x y z

ddtH H H

HM H

i j k H

( )

( )

( )

x y z z y

y z x x z

z x y y x

H H H

H H H

H H H

M i

j

k

x x y z z y

y y z x x z

z z x y y x

M H H H

M H H H

M H H H

( )

( )

( )

x xx x yy zz y z

y yy y zz xx z x

z zz z xx yy x y

M I I I

M I I I

M I I I

124


problem 07/82

z

45 y

20 rad/sz

xM

Eq. 7/23 2x yz zM I

/ 22 2

/ 2

3 3 2 22

2

2 3(0.2)( ) 0.02 kgm3 6 62 2 2 2

b

yz bI yzdm y dl y dy

b b mb

20.02(20 ) 79.0 NmxM

on plate, 79.0 Nmx M i

but acting on shaft, 79.0 Nm M i

The plate has a mass of 3 kg and iswelded to the fixed vertical shaft, whichrotates at the constant speed of 20πrad/s. Compute the moment M appliedto the shaft by the plate due to dynamicsimbalance.

125


problem 07/82

z

45 y

20 rad/sz

xM

0

00

0

0 00 0

0 0 0

G G G

G G

G

z x

z x

y z

d ddt dt

M H H

I I

I

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