01-engineering dynamics - (v 2558)

: ()

(ENGINEERING DYNAMICS)

-

2

(Engineering Dynamics) 2 (Engineering Statics) 2 Engineering Mechanics DYNAMICS SI version Sixth edition J.L. Meriam L.G. Kraige John Wiley & Sons

2 (Particle) (Rigid body) (Kinematics) (Kinetics)

1: (Engineering Mechanics)

2: (1 ) (2 )

3: 3 1 2 3 3 3 section A, section B section C

4: (1 ) 3 3

5:

6: 3 1 2 3 6 section A, section B section C

i

............................................................................................................................................... ........................................................................................................................................... ........................................................................................................................

1 (Introduction to Dynamics) ............................................................. 1-1 1.1 .......................................................................................................... 1-2 1.2 .......................................................................................................... 1-2 1.3 ................................................................................................................. 1-3 1.4 ............................................................................................................ 1-3 1.5 .......................................................................................... 1-5 2 (Kinematics of Particles) ........................................................... 2-1 2.1 00 ......................................................................................... 2-2 2.2 00 ........................................................................ 2-9 2.2.1 00 (x-y) ........................................................ 2-9 2.2.2 00 (n-t) ................................... 2-13 2.2.3 00 (r-) .................................................... 2-14 2.3 00 ....................................................................................... 2-21 2.4 000 ....................................................................... 2-25 3 (Kinetics of Particles) 3A (Force, Mass and Acceleration) .................... 3A-1 3A.1 0 00 ....................................... 3A-1 3A.2 .............................................................. 3A-2

()

3B (Work and Energy) ................................................. 3B-1 3B.1 .................................................................................................. 3B-1 3B.2 ........................................................... 3B-3 3B.3 ............................................................................. 3B-8 3B.4 ............................................ 3B-8 3C (Impulse and Momentum) ............................. 3C-1 3C.1 ....................................................... 3C-1 3C.2 .......................................................... 3C-3 3C.3 ................................................................................................ 3C-12 3C.3.1 ............................................................. 3C-12 3C.3.2 ............................................................ 3C-14 4 (Kinetics of Systems of Particles) ....................................... 4-1 4.1 00 ................................................................................................ 4-1 4.2 .................................................................................... 4-2 4.3 .................................................................................. 4-4 4.3.1 ...................................................................................... 4-4 4.3.2 ........................................................................................ 4-4 4.4 ....................................................... 4-6 4.4.1 ............................................................................. 4-6 4.4.2 .......................................................................... 4-6 5 (Kinetics of Systems of Particles) ....................................... 5-1 5.1 00 ....................................................................... 5-1 5.2 00 ............................................................................................ 5-2 5.3 00 ....................................................................................... 5-7

()

5.4 00 .................................................... 5-11 5.4.1 0 .......................................................... 5-11 5.4.2 0 .............................................................................. 5-13 5.4.3 ....................................................................................... 5-20 5.5 00 ........................................................ 5-26 6 (Kinetics of Rigid bodies) 6A (Force, Mass and Acceleration) .................... 6A-1 6A.1 000 ................................................................. 6A-2 6A.2 000 ....................................................................... 6A-3 6A.2.1 0 ............................................................ 6A-3 6A.2.2 0 ............................................................ 6A-3 6A.3 000 ......................................................... 6A-4 6A.4 000 ............................................................ 6A-5 6B (Work and Energy) ................................................. 6B-1 6B.1 ............................................... 6B-1 6B.2 ............................................................................................. 6B-1 6B.3 0 ......................................................... 6B-3 6C (Impulse and Momentum) ............................. 6C-1 6C.1 ....................................................... 6C-1 6C.2 .......................................................... 6C-2 TABLE D/4 PROPERTIES OF HOMOGENEOUS SOLIDS ........................................... -1 .................................................................................................................................... -1

va = (m/s2) av

= (m/s2)

/A Bav = A B (m/s2)

( )/A B xyav = A B x,y (m/s2)

e = . (Coefficient of restitution) , ,m e te e e =

,n te e = (unit vector) B (n-t) ,re e = (unit vector) (r-) ,x ye e = (unit vector) (x-y) fv

= (internal force) (N) ,k sf f

v v = (N)

Fv

= (external force) (N) g = (m/s2)

G = (constant of gravitation) ( )11 3 26.673 10 /G m kg s= Gv

= (kgm/s)

OHv

= O (kgm2/s) I = (kgm2) k = (springs stiffness) (N/m), (m) m = (kg)

em = 245.976 10 kg

oMv

= O (Nm) P = (W, J/s) Pv

= (N) r = , B 2 (m) rv

= (m) rv

= (m)

/A Brv = A B (m)

()

() R = 66.371 10 m sv

= 1 (m) t = (s) T = (J)

1 2U = B 1 2 (J)

1 2U = B B 1 2 (J)

vv

= (m/s) vv

= (m/s) /A Bv

v = A B (m/s)

( )/A B xyvv = A B x,y (m/s)

,e gV V = (J) Wv = B (N)

v = (rad/s2) v = B (n-t) (rad) v = (rad)

,k s = . . v = (rad/s) = n-t (m)

2

1

t

tdtFv

= (Ns kgm/s)

2

1O

t

tdtM

v

= O (Nms kgm2/s)

( )/2 A B xyv v v = (Coriolis acceleration) (m/s2)

1

-1

1

(Introduction to Dynamics)

Mechanics () (physical science)

(bodies) (rest) (motion) <

D (statics) (dynamics)

(Statics)

(Dynamics)

1

-2

1.1 1 (Basic Concepts) <

1

-3

1.3 (Units) SI (International System of

metric units) <

U.S.

units (United State customary units)

SI U.S. customary units g

1

-5

1.5 (Solving Problem in Dynamics)

2

2-1

2

(Kinematics of Particles)

Kinematics (dynamics)

-

; (displacement) (velocity)

(acceleration) ; (time)

; ;

[1] ; ;

+

- (Rectilinear motion)

- (Plane curvilinear motion)

1) (Rectangular coordinates: x y )

2) ; (Normal and Tangential coordinates: n t )

3) (Polar coordinates: r )

- (Relative motion)

- ; (Constrained Motion of Connected Particles)

(Coordinates)

6.8

W.Y (path)

P 3

;

1) , , :x y z () Rectangular coordinate

2) , , :r z () Cylindrical coordinate

3) , , :R () Spherical coordinate

4) , :n t (;) Normal and Tangential

coordinate (path variables)

2

2-2

2.1 (Rectilinear Motion) (rectilinear motion)

1

6.2

W.W t

P O s

t P

s O

s+ s

;

(Average velocity: avv ) t

av

sv

t

=

(Instantaneous velocity: vv )

t 0

limv

v

t

sv

t

=

vv v&dsv s

dt= = (2.1)

(Average acceleration: aav) t

av

va

t

=

(Instantaneous acceleration: va )

t 0

limv

v

t

va

t

=

vv v&dva v

dt= =

2

2

vv v&&d sa s

dt= = (2.2)

(2.1) (2.2)

v v v vv dv a ds =

v v v v& & & &s ds v dv = (2.3)

(2.1) (2.2) (2.3)

;

;

;;

;

vs (m)

vv (m/s)

va (m/s2)

t (s)

8.8 (Graphical

Interpretations)

1)

() (s-t curve)

2

2-3

() (v-t curve)

() (a-t curve)

6.2

W.W (s-t curve)

t v

v dsv

dt= t ;

W.W (v-t curve)

t v

v dva

dt= t ;

tY t2 ;

2 2

1 1

s t

s t

ds vdt= 2 1s s = . v-t

W.W (a-t curve)

tY t2

;

2 2

1 1

s t

s t

dv adt= 2 1v v = . a-t

2)

() (a-s curve)

() (v-s curve)

6.3

W.3 (a-s curve)

sY s2 ; a-s

2 2

1 1

v s

v s

vdv ads= 2 22 1

2v v

= . a-s

2

2-4

W.3 (v-s curve)

A

A

( )dvvdv ads a v ds= = A Aa v= (Slope)A

8.6 (Constant Acceleration)

; (

W.Y, 2.2 W.g )

, ,v v

v v v v v vds dvv a v dv a ds

dt dt= = =

()

W.Y, 2.2 W.g)

W.W

0 0

v vv t

vdv a dt=

0 , orv v at v u at= + = +

W.3

0 0

v v v vv s

v sv dv a ds =

( )2 2 2 202 , 2o orv v a s s v u aS= + = +

ov v at= + W .Y

( )0 0 0

v v v vs t t

osds vdt v at dt= = +

2 20

1 1,

2 2oors s v t at S ut at = + = +

t = 0

ov = u

0s

0S s s= t = 0 t

:

4 (

)

8.K

( )va f t= ,

( )va f s=

( )va f v=

;

;

v

v

v v v v

vv

dva

dt

v dv a ds

dsv

dt

=

=

=

2

2-5

8: ([1] Sample Prob. 2/1)

The position coordinate of a particle which is confined to move along a straight line is given by32 24 6s t t= + , where s is measured in meters from a convenient origin and t is in seconds. Determine (a)

the time required for the particle to reach a velocity of 72 m/s from its initial condition at t = 0, (b) the acceleration of the particle when v = 30 m/s, and (c) the net displacement of the particle during the interval from t = 1 s to t = 4 s.

a) t 72 m/s: 72 / ?v m st = =

b) 30 m/s: 30 / ?v m sa = =v

c) t =1 4 : 1 4sec ?S =

d) 1t = 4t =

a) t 72 m/s

: 26 24dsv v tdt

= = (1)

: 12dva a tdt

= = (2)

; t 72 m/s

(1) 272 6 24t= ; 4 sect Ans=

b) 30 m/s

(1) 30 m/s

230 6 24t= 3 sect =

3 sect = (2)

212(3) 36 /a m s Ans= =

c) t =1 4

32 24 6s t t= +

1-4

3 32(4) 24(4) 6 2(1) 24(1) 6s = + +

54s m Ans =

d) t = 1- 4

; 1t = 4t =

(

)

2

/6 24 0 2v om sv t t s== = =

;; 1 2sec 2 4secs s +

10 74 .64Total distance m Ans= =+

2

2-6


A particle moves along the x-axis with an initial velocity vx = 50 m/s at the origin when t = 0. For the first 4 seconds it has no acceleration, and thereafter it is acted on by a retarding force which gives it a constant acceleration ax = -10 m/s2. Calculate the velocity and the x-coordinate of the particle for the conditions of t = 8 s and t = 12 s. and fine the maximum positive x-coordinate reached by the particle.

a) v s t = 8 s t = 12 s

b) x: (xmax)

c) ; t = 0 s t = 12 s

v t

: 0 t 4 s

50

50

/

/ 0

v m s

s v dt t

a dv dt

=

= =

= =

: 4 t 12s

50 4 410

50 10( 4) 90 10 #

x

x

v t t

tt

dv a dt dt

v v

=

= =

=

( )

( )50(4) 4

2

4

2

4

90 10

90 80 #10200 90 52

s t t

t

t

t t t t

ds v dt

s

dt

s

+

= =

= =

a) t = 8 t = 12 s

: 4 t 12 s

;

8

12

8 90 10(8)

12 90 10(12)

10 /

309

/0 10

:

:xm

mt

s

s

t s v

tA s

s vnv

=

=

= =

== =

28

212

2 8 90(8) 5(8 ) 80

12 90(12) 5(12 ) 80

320#

28090 80

:

:5

mt Ans

mt

ts

s

t s

=

=

=

== =

=

b) x

_ max

_ max

_ max

_ max ( )2

max 3

@

@ 90 10

90(9) 5( 29 ) 0 58

9 #

0

s

s

s t s

t m Ans

dstdt

t v t

x

t s

s

=

=

=

=

= =

c) + t = 0s t = 12s

( ) ( ) ( )1 / 20 9 9 12

0 9 12

3 75 3 025 4

v o m s

s s

t s t s t s

total distance s s

m m nsm A

=

= < = < =

= +

= + =

Q

t = 0 s t = 9 s t = 12 s

S0-9s = 325 m

= 45 m S9-12s

2

2-7


The spring-mounted slider moves in the horizontal guide with negligible friction and has a velocity v0 in the s-direction as it crosses the mid-position where s = 0 and t = 0. The two spring together exert a retarding force to the motion of the slider, which gives it an acceleration proportional to the displacement but the oppositely directed and equal to a = -k2s, where k is constant. (The constant is arbitrarily squared for later convenience in the form of the expressions.) Determine the expressions for the displacement s and velocity v as the function of the time t.

(initial condition)

0 0 200

0 ;0

t s

t s

t sv v

a k ss

=

=

== = =

s

v t

(ODE 1st order)

(1)

(2)

(3)

dva

dt

v dv a ds

dsv

dt

=

=

=

vv

v v v v

vv

a) v t

(2)

( )( )

2

2

0

2 2 2 20

2 2 2( ) 0 #

o

v s

v

s

vdv ads k s ds

v dv k s ds

v v k ks v v s

= =

=

=

=

( )sv (3) ( )ts

2 2 20 00

1

1st

dsv dt dsdt v

dt dsv k s

= =

=

( )12 21: sin:

1

xfrom dxaa x

set x ks

dx kds ds dxk

=

= = =

( ) ( )( )

2 2 2 2 20 0

1

0( )

( ) 0

1

0 0

1

0

1 1/

1 1sin sin

1 sin( )

cos( )

sin t

t

kds dxv k s v x

x ksk v k v

kstk vdsvd

vs kt Ans

k

v v kt Ant

s

=

= =

=

=

=

=

Q

2

2-8


A freighter is moving at a speed of 8 knots when its engines are suddenly stopped. If it takes 10 minutes for the freighter to reduce its speed to 4 knots, determine and plot the distance s in nautical miles moved by the ship and its speed v in knots as functions of the time t during this interval. The deceleration of the ship is proportional to the square of its speed, so that a = -kv2.

- (initial condition)

000

08

; (1 1 / ); 1 1.8520

t s

t s

t sv knots

knot mile hour mile kms mile

=

=

== = = =

- 0 10min10min 4tt v knots== = 2a kv=

a) v (knots) t

b) s (nautical miles: ) t

(ODE 1st order)

(1)

(2)

(3)

dva

dt

v dv a ds

dsv

dt

=

=

=

vv

v v v v

vv

a) s t

(1)

8

( )

2

28 0

8

1

1

1 81 1 1 18

1 8

8 8

#t

v t

v v

v

dva kvdt

dv k dtv

k

vk

tv

ktkt ktv

t

v

=

=

= =

=

=

+ = = +

+

=

=

v knots (mi/hour)

t . (hour)

k v10min= 4 knots

( )1

(10min)6

( )

184

1 8 (1/6)

3/4

81 6

hour

t

k m

v knot

v v

ns

i

t

k

s A

=

=

= =

+

=+

b) s t

( )sv (3) ( )ts

(3)

( ) ( )( ))

0 0

0

(

8 1 61 6

4 ln

n

63

8 l6

1

s t

t

ts

dsv ds v dtdt

s dt tt

t mi Ans

++

=

=

+

=

=

=

2 : r-

2-9

2.2 (Plane Curvilinear Motion) (2 )

(x-y) ; (n-t) (r-)

2.4

2.4 (path) O

t A t

A' ; O

rv r r+v v

;

(Average velocity: avv ) t

av

sv

t

=

(Instantaneous velocity: vv )

t sv rv

0 0

lim limt t

s rv

t t

= =

v vv

drv r

dt= =

vv v& (2.4)

vv (speed)

dsv v s

t= = =

v

&

(Average acceleration:aav) t

av

va

t

=

(Instantaneous acceleration: va )

t 0

limv

v

t

va

t

=

vv v&dva v

dt= = (2.5)

;

rv (position vector)

O (m) vv (m/s) va (m/s2)

t (s)

2.2.1 (x-y)

2.5

2 : r-

2-10

2.5 (path)

x-y

;

:

:

:

x y

x y

Position

Velocity

Acceleration

r xi yj

v r xi yj

v i v j

a v r xi yj

a i a j

= +

= = +

= +

= = = +

= +

v vv

v vv v& & &

v v

v vv v v& && && &&

v v

(2.6)

2 2 1; tan yx yx

vv v v

v = + =

v

2 2x ya a a= +v

(Projectile Motion)

2.6

(

;) ;

x-y x

y

0xa = ya g=

0xa = ya g=

(2.1) (2.2) (2.3)

;

( )( )

0

0 0

0: ( )x

x x

x

x direction a

v v

x x v t

=

=

= +

( )

( )

( ) ( )

0

20 0

220 0

: ( )

12

2

y

y y

y

y y

y direction a g

v v gt

y y v t gt

v v g y y

=

=

= +

=

g

; g = 9.81 m/s2

g

2 : r-

2-11


The curvilinear motion of a particle is defined by vx = 50 - 16t and y = 100 4t2, where vx is in the meters per second, y is in meters, and t is in seconds. It is also known that x = 0 when t = 0. Plot the path of particle and determine its velocity and acceleration when the position y = 0 is reached.

- x y

2: 50 16 ( / ): 100 4 ( )

xx velocity v t m sy position y t m = =

- (initial condition): 0 00 0t st s x m== =

a) (path) x-y t

b) y = 0 m

: ( )

: ( )

: ( )

x y

x y

x y

position r xi yj a

velosity v v i v j b

xi yj

acceleration a a i a j c

v i v j

= +

= +

= +

= +

= +

v vv

v vv

v v& &

v vv

v v& &

2 2 1

2 2 1

; tan

; tan

yx y v

x

yx y a

x

vv v v

v

aa a a

a

= + =

= + =

v

v

x

0

2

0

: 50 16

16

50 8

:

:

x

x

x t

x

x

velocity

acceleration

x position dx v

v t

a v

x t t

dt

=

= =

=

=

+

&

y 2:

:

8:

100 4

8y

y y

y t

v y

y position

velocity

acceleratio

t

a vn

=

= =

= =

&

&

a) (path) x-y

t

250 8x t t= + 2100 4y t=

b) y = 0 m

t y = 0 m 2

2

100 4

0 1 4 5

:

00 #

y ty pos

s

i

t

it on

t

=

= =

; 5 sec ;

( )

( )

2 2

1

50 16 50 /30 /

4

5

(

8 0 / 53.

5

15

)

t n

:

a

x yx

yy

ov

x

Velocity

mv v vv i sm s

s

v mv

jsv

t

= += = = = =

=

=

=

vv

v

2

2

2

2 2

( 5 ):

16 /17.89 /

8 /

16

8

x

x y

y

A

m s

ecelera

a ia a

tion t s

aa

sj

mm s

= = = +

=

=

=

=

v

vv

2 : r-

2-12


A rocket has expended all its fuel when it reaches position A, where it has a velocity of uv at angle with respect to the horizontal. It then begins unpowered flight and attains a maximum added height h at position B after traveling a horizontal distance s form A. Determine expressions for h and s, the time t of flight from A to B and the equation of the path. For the interval concerned, assume a flat earth with a constant gravitational acceleration g and neglect any atmospheric resistance.

- (initial condition: t0 = 0 s) A

,velocity u angle = =v

-4 (unpowered) ;

(projectile motion: 0xa = ya g= )

h, s t A-B

Projectile

( )( )

0

0 0

0: ( )

(1)

(2)

x

x x

x

x direction a

v v

x x v t

=

=

= +

( )

( )

( ) ( )

0

20 0

220 0

: ( )

(3)

1 (4)2

2 (5)

y

y y

y

y y

y direction a g

v v gt

y y v t gt

v v g y y

=

=

= +

=

1) t B ( B 0 /yv m s= )

(3) tB ( )0:

0 s

i

n

s n

i

y yy velocity v v gt

ut Ag

gt

n

u

s

=

=

=

2) h B ( B sinB ut g= )

tB (4) h

( )

( )

2

2

2 2

0 0

sin2

1:2

sin 1 sin0 sin2

yy position

uh

y y v t gt

u uh u gg g

Ansg

= +

= +

=

3) s B ( B sinB ut g= )

tB (2) s ( )

( )

( ) ( )

( )

0

max

2

2

0

45( )

sin cos

sin

:

sin0 cos

22

x

o

u

x position x x v t

u

s whe

s u

s

g

us n

g

Ang

= +

= +

=

=

=

2 : r-

2-13

2.2.2 + (n-t)

2.7

2.7

(path)

; (normal and tangential coordinate:

n-t) A, B, C

n-t

n-t

; ; (unit

vector)

n-t ; ; ;

te

ner

( )t

tV v e=v

iv

jv

2.8

( )( )

: cos sin

: sin cos

: sin cos

:

: cos s

:

in

nn t

t

n t

t n

Unit vector

Derivative o

normal e i je e

tangential e i j

normal e i j e

tangential e

f Unit vect

i

or

j e

=

= +

= =

= =

& &

& &

v v

v v

v v&

v v&

n-t

2.8

;

( )( ) ( )

( )

( )

( ):

:

t t n n

n

t t

t

n

t n

Position e

Velocit V e e

V e e

V e

y

= =

=

=

=

&

v v && &

v&

v&

v

( )

:

t t

t

t

t

e

where

V v e

v

+

=

=

&

&

v

( )

( ) ( ){ }( ){ }

( )

( ) ( )

( )

2

2

2

(

2

)

:

:

tt t

t t t

t n

t n

n

n

n t

t

t t

t

n

d ea V

dt

a e e e e

e e

e e

Aeceleration

a a e a e

va v

whena v

= =

= + = +

= +

= +

= = = = =

= +

&

&& & && & &

&& &

& &&

vv &

v &

& &

&&

v

&

n-t

22

2 2

:

:

:

:

:

n

t t

n n t t

n

t

n t

Position

Velocity

Acceleration

where v

vwhere a

e

V v e

a

v

a

a e a e

v

a a a

=

= = =

=

=

= +

=

=

+

=

&

v

v

& &

&&&

v

v (2.7)

tv

v

path line ; path line

(tangential)

; ;

( tvv

) ;

2 : r-

2-14

n-t

ne$

te$

v

ta

na

&

2.9

2.9 P path

line r

; n-t

r = =

n-t

r

22

2 2

:

:

:

:

:

n

t

t

tn

t

n n t

t

t

n t

Position

Velocity

Acceleration

r re

V v e

a

where v r

vwhere a r v

r

a e a

a a aa v r

e

=

= = = = +

= =

=

=

= +

v

v

&

&

&&&

v

& v

(2.8)

%: (r )

: r t (rad)

& : (Angular velocity: rad/s)

&& : (Angular acceleration: rad/s2)

2.2.3 (r-)

Path

ere

rr

( )t

x iv

y

jv

A

2.10

2.10

(path)

(polar coordinate: r-)

;

;

( )( )

cos sin

sin cos

sin cos

cos sin

:

:

rr

r

r

Unit vector

Derivative of unit vecto

e i je e

e i j

e i j e

e i j e

r

=

=

= +

= +

= +

=

& &

& &

v v

v v

v v&

v v&

r-

2.10

;

( ) ( )

( ) :

:

t r

t t r

r

r

Position

Velocity

r r

v r r e

e

r e r e

r e

=

=

= +

= +

&

v v && &

v

&

( ) ( )( ) ( )

( ) ( )

( ) ( )

2 2

:

t t r r

r r

r

Acceleration

a v r e r e r e r e r e

r r e r r e

r e r e r e r e r e

= = + + + +

= + + +

= + +

& && &

& & &&

&

&

& & &

&

v v & && && & &

&

&& &

& & &

2 : r-

2-15

r-

( )

( )

( )

2 2( )

22 2

(

(

2

)

)2

:

:

.

.

1.

:

t r

t r r

rt r

rt

t r r

r

Position

Velocity

Acceleration

where v rv v v

v r

where a r ra a a

a r

r r e

v

r

dor

v e v e

a

a rr dt

a e a e

= = +

=

= = +

=

= +

=

= +

=

+

&

&

&& &

&

& v

&

v

v

&

v

&

v (2.9)

r-

re$e$ v

a

P

r

&

ra

2.11

2.11 P path

line r

r-

( ) .tr r const= =

( )t ;

; r

; ;

;

2 2( )

22 2

(

( )

)

( )

( )

:

:

:

. 0

.

t r

t r r

t r

t r

r

r

rt r

Position

Velocity

Acceleration

where vv v v

v r

where a ra a a

a r

r r e

v v e v e

a a e a e

=

= = +

=

= = +

=

= +

= +

&

&

&&

v

v

v

v

v (2.10)

;

4

r-

( 2.9 ) path line

r n-t r-

; 2 ;

r

nt r

Position nr re=v ( ) t rr r e=

v

Velocity

( ) t t tv v e=v

: twhere v r= & ( ) t r rv v e v e = +v

. 0rwhere v

v r

=

= &

tv v= te e=

Acceleration

n n t ta a e a e= +v

2

2tn

t t

va r v

r

a v r

= = =

= =

& &

&&&

( ) t r ra a e a e = +v

2

2r

va r v

r

a v r

= = =

= =

& &

&&&

ta a= , n ra a= te e=

: n ra a=

4

2 : n-t

2-16


To anticipate the dip and hump in the road, the driver of a car applies her brakes to produce a uniform deceleration. Her speed is 100 km/h at the bottom A of the dip and 50 km/h at the top C of the hump, which is 120 m along the road from A. If the passengers experience a total acceleration of 3 m/s2 at A and if the radius of curvature of the hump at C is 150 m, calculate (a) the radius of curvature at A, (b) the acceleration at the inflection point B, and (c) the total acceleration at C.

- Brake , , ,t A t B t C ta a a a= = =

- A-C = 120 m

- , 100 /t Av km h= , 50 /t Cv km h=

- ?A = 150C m =

- A 2 2 2, ,3 / ;A A t A n Aa m s a a a= = +v v

a) A , ?A = , b) B ?Ba =v

c) C ?Ca =v

a) A , ?A =

A 2,

,

2,

, (1)t A

At

n

An A

A A

va

va = =

(1) ,100

/3.6t A

v m s= , ?n Aa = 2 2 2

, , (2)A n A t Aa a a= +

(2) 23 /Aa m s= , ?t Aa = ,

,

120

0

50/3.6 120

100/3.6 0(3)

t C

t A

v m

t t tv

m

t t t

v dv a ds

v dv a ds

=

=

(3)

( ) ( )( )

22

250/3.6 100/3.62 120

2.41 /t m sa

= =

; 2, 2.41 /t A ta a m s= = (2)

2,

2 2 2, ,

2 2 2,3 1.785 /( 2.41)

A n A t A

n A n A

a a a

a a m s

= +

= + =

; 2, 1.785 /n Aa m s= (1)

( )2 2,,

100/3.61.785

432

t AA

n A

A A

va

m ns

=

= =

b) B , ?Ba =v

, ,

2 2 2, ,

2 2

2,

2,

, 0

2.41

2.41

/

.

/

t B t B

n B

t

B

B t

t B t

B n B t B

B

a a a

v

a a a m s A

vwhere a

a a m

ns

s

= =

= =

= = =

=

+

=

c) C , ?Ca =v

( )

( )

,

2 2, ,

22

,

2

2

,

2

2

2

2

1.286 2.41 /

1.286 2.41 2.73

50/3.6.

11.286 /

2.41

0

/

5

/

t C

n C

t

C n C t C

t

C n t

C

C

C

a

a a a

vwher m s

a

a e e m s

a m s An

e

a m

s

s

= =

= =

=

=

=

=

+ =

+

v

v

v

2 : n-t

2-17

: 4

4 4

4

1)

2) +

+ +

3) +

** Problem 2/7

2 : n-t

2-18


A certain rocket maintains a horizontal attitude of its axis during the powered phase of its flight at high altitude. The thrust imparts a horizontal component of acceleration of 6 m/s2, and the downward acceleration component is the acceleration due to gravity at that altitude, which is g = 9 m/s2. At the instant represented, the velocity of the mass center G of the rocket along the 15o direction of its trajectory is 20(103) km/h. For this position determine (a) the radius of curvature of the fight trajectory, (b) the rate at which the speed v is increasing, (c) the angular rate & of the radial line from G to the center of curvature C, and (d) the vector expression for the acceleration av of the rocket.

- n-t

a) C, ? =

b) (Rate speed v is increasing) ?t tv a= =v v&

c) (Angular rate) &

d) av

a) C , ? =

2 2

(1)ttn

n

va

va

= =

(1) 3

,

20(10 )/

3.6t Av m s= ?na =

n-t 2

2

7.14 /

8.

:

1

9cos15 6sin15

: 9sin15 2os 5 /6c 1

n

t

n

t

normal a

tangetial a

a m s

a m s

=

==

= +

; 27.14 /na m s= (1)

( )3 62

20 107.14

4.32(10 ) .m Ans = =

b) Rate speed v is increasing : ?t tv a= =v v&

28.12 /t tv a m s Ans= =&

c) Angular rate : &

&

( )( )( )

3

4

6

:

20 10 /3.6

4.32

12.85 10 .

10

/

tt

vfro

rad s Ans

m v

= =

=

=

& &

&

d) Total acceleration : av

7.14

8.12

n n t t

n te ee

Aa

na e

sa =

++

=

v

2 : r-

2-19


Rotation of the radially slotted arm is governed by = 0.2t + 0.02t3, where is in radians and t is in seconds. Simultaneously, the power screw in the arm engages the slider B and controls its distance from O according to r = 0.2 + 0.04t2, where r is in meters and t is in seconds. Calculate the magnitudes of the velocity and acceleration of the slider for the instant when t = 3 s.

3: 0.2 0.02Angular t t = + 2: 0.2 0.04Radial r t= +

t =3 sec (3sec) (3sec),v av v

(3sec) (3sec),v av v

(3sec) (3sec),v av v

2 2( )

22 2

)

( )

(

( )

2

.

.

:

:

r

t r r

r

r

t

r

t r

t

r

Velocity

Acceleration

where v rv v v

v r

where

v v e v e

a a e a

a r ra

a r r

e

a a

= = +

=

=

= +

= +

= +

= +

&

&

&& &

& v

&& v

&

v

v

&, , , ,r r r & &&& && t =3 sec 2 2

(3sec)

(3sec)

2(3sec)

0.2 0.04 0.2 0.04(3) 0.56

0.08 0.08(3) 0.24 /

0.08 0.08 0.08 /

r t r m

r t r m s

r r m s

= + = + =

= = =

= = =

&& &&

& &&

3 3(3sec)

2 2(3sec)

2(3sec)

0.2 0.02 0.2(3) 0.02(3) 1.14 (65.3 )

0.2 0.06 0.2 0.06(3) 0.74 /

0.12 0.12(3) 0.36 /

t t rad

t rad s

t rad s

= + = + =

= + = + =

= = =

& &

&& &&

; t =3 sec

( )( )2 2 2 2

( )2

0.56 0

0.24 /

0.414 /

0

.74

0 .479.24 0 /.414

r

t r

v r

v r

v m s Ansv v

m s

m s

= =

= = =

= + = + =

&

&

v

t =3 sec

( )( )

( ) ( )

2

2

2 2

2 22 2( )

2

2

0.24 0.56 0.74

0.56 0.36 2(0.24)(0.74)

0.22

0.227 /

0.557 /

0.601 /7 0.557

r

t r

m s

m

a r r

a r r

a a a m s An

s

s

= = =

= + = + =

= + = + =

&

&& &

&&

&

v

Path line: 0 t 5sec

cossin

x r

y r

=

=

2 : r-

2-20

4: ([1] Sample Prob. 2/10)

A tracking radar lies in the vertical plane of the path of a rocket which is coasting in unpowered flight above the atmosphere. For the instant when = 30o, the tracking data give r = 8(104) m, r& = 1200 m/s, and & = 0.80 deg/s. The acceleration of the rocket is due only to gravitational attraction and for its particular attitude is 9.20 m/s2 vertically down. For these conditions determine the velocity v of the rocket and values of r&& and &&.

(unpowered) 30 = ;

48(10 ) , 1200 / , 0.80deg/r m r m s s= = =& & 9.20 m/s2

30 = , ,v r v &&&&

1) vv

( ) ( )42 2

:

1200 /

.

8 10 0.8180

1200 1631

16

93

3

6

/

9 /

9

r r

r

Velocity

m s

v v e v e

where v r

sr

m s

v m

v

= +

= =

= = =

= + =

&

v

&

v

2) ,r &&&&

( )

( )

22

2

:

.

12

t r

rr

rAcceleration

where a r r

a a e a e

r a r

rr aar

r

=

=

= +

=

= +

+ &

&&

&

& & &&

v

&&

&

&&

&

,ra a 2

2

cos (9.20)cos30 7.97 /

sin (9.20)si 4.6 /0 0n 3r m sa g

a sg m

= = =

= = =

; ,r &&&&

( )

( ) ( )4 2

2

24

4

8(10 )

8(10 )

7.97 0.80180

12

14.60 2 1200 0.80

180

7.63 /

3.61(10 ) /

r

m s Ans

rad s Ans

r a r

a rr

= +

= + =

=

=

=

&

&& &

&&

&

2 $

2-21

2.3 (Relative motion)

( )

(absolute) 2.11

1m/s 3m/s

()

1m/s 3m/s

2 m/s (3m/s - 1m/s)

; (Translating

frames of reference)

(Relative

velocity)

-3m/s -2m/s (

;

)

2.11

(Frames of reference)

3

1) Fixed frames of reference: (

)

;

(absolute)

2) Translating frames of reference: (

)

; (relative)

3) Rotating frames of reference: (

)

;

(relative)

;

(Translating frames of reference)

; (Rotating frames

of reference)

(rigid bodies) 5

2.12 (B observe to A)

2 $

2-22

2.12

InitAxes ialX Y axes ()

TranslatAxes x y ing axes ()

A B

/

/

/

:

:

:

A B A B

A B A B

A B A B

Position

Velocity

Acceleration

r r r

v v v

a a a

= +

= +

= +

v v v

v v v

v v v

(2.11)

/ / /A B A B A Br v av v v

(

) A B A

B A relative to B

B A

( 2.13)

/

/

/

:

:

:

B A B A

B A B A

B A B A

Position

Velocity

Acceleration

r r r

v v v

a a a

= +

= +

= +

v v v

v v v

v v v

2.13 (A observe to B)

: Relative motion

1) (Vectors Algebra)

( ) ( )1; tan

x x y yx y

yx yx y

x

i ji j

i ji j

c a b a b a ba a ac

c c cb b b c

++

+ =+

= + = + += ==

v vv v

v vv v

vv vv

v v

2) Law of sines / Law of cosines

sin

:

sin sin

La

a b

wof sine

c

s

= =

2 2 2

2 2 2

2 2 2

2 cos( )

2 cos( )

2 cos( )

:Lawof cosine

a b c bc

b a

s

c ac

c a b ab

= +

= +

= +

2 $

2-23

8: ([1] Sample Prob. 2/1K)

Passengers in the jet transport A flying east at a speed of 800 km/h observe a second jet plane B that passes under the transport in horizontal flight. Although the nose of B is pointed in the 45o northeast direction, plane B appears to the passengers in A to moving away from the transport at 60o angle as show. Determine the true velocity of B.

B Bvv

;

2 (Law of sines/cosines Vectors

Algebra) ;

1. Law of sine

A B

/B A B Av v v= +v v v

Bv

v Sine

( )800

60 75sinsin sinB A

o o

v v= =

60800

75sinsin

717 /o

oB kv m s Ans ==

A

B /B Av

( )/ 80045 75sinsin sin

B A Ao o

v v= =

/ 800 75sin 4 5865 /sin

o

oB A m hv k= =

2. (Vectors Algebra)

A


(a)

/ / /

800

cos 45 sin 45

cos60 sin 60

A

o oB B B

o oB A B A B A

v i

v v i v j

v v i v j

=

= +

= +

vv

v vv

v vv

/A B B Av v vv v v

(a)

[ ]/

/

cos 45 sin 45 800 0

cos 60 sin 60

o o

o o

B A B A

B

B A

v

i j i j

i j

v v

v

v

+ = +

+ +

= +vv v v v

v v

v v

,i jv v

( )( )

/

/

cos45 800 cos60 (1)

sin 45 sin 60 (: 2)

: o oB B Ao o

B B A

i terms

j term

v

v vs

v=

=

v

v

matrix

/

800cos45 cos60

0sin 45 sin60

o oB

o oB A

v

v

=

Cramers Rule 800 cos60

0 sin 60 800sin 60 0cos 45 sin 60 sin 45 cos60cos 45 cos60

sin 45 sin 60

o

o o

o o o oo oB

o o

v += =

717 /Bv km h Ans=

/

cos 45 800

sin 45 0 0 800sin 45cos 45 sin 60 sin 45 cos60cos 45 cos60

sin 45 sin 60

o

o o

o o o oo oB A

o o

v= =

/ 586 /B Av km h=

:

Cramers rule, Gauss elimination, Inverse matrix

2 $

2-24

6: ([1] Sample Prob. 2/1t)

Car A is accelerating in the direction of its motion at the rate of 1.2 m/s2. Car B is rounding a curve of 150 m radius a constant speed of 54 km/h. Determine the velocity and acceleration which car B appears to have to an observer in car A if car A has reached a speed of 72 km/h for the positions represented.

A 2: 1.2 /

observe 30 , 72 /A BAconstant acceleration a m s

at v km h=

= =

v

v

B : 54 /

150B

B

constant velocity v km h

m=

=

v

/ B A /B Avv

/B Aav

a) / ?B Av =v


te

ne

Sine /B Avv

2 2 2/ 2 cos(60 )o

B A A B A Bv v v v v= +

( )( )2 2 2/ 72 54 2 72 54 cos(60 )o

B Av = +

( )/ 64.9 / 18.02 /B Av km h m s Ans =

/

sin sin 60 46.54

1o

B

o

AvAns = =

b) / ?B Aa =v

/B A B Aa a a= +v v v ?Ba =

v

te

ne

( )22, 2,

54/3.61.5150 /

B t

BB nv

m sa = = =

, 0B ta =

; A

, , 1

1.5

. 0

/

5B B n B t t

B n

na a a e e

a e m s

= + = +

=

v v v

v

/B A B Aa a a= +v v v

Sine /B Aav

2 2 2/ 2 cos(30 )o

B A A B A Ba a a a a= +

( )( )2 2 2/ 1.2 1.5 2 1.2 1.5 cos(60 )o

B Aa = + 2

/ 0.757 /B Aa m s Ans =

2 $

2-25

2.4 +

(Constrained Motion of Particles)

1) One Degree of Freedom (1DOF)

(coordinate) ;

; 4

2.14

2.14 (Pulley-

Cable system) A B

A +x 1

B ; 0.5

A - x 1 B

0.5

B ; A ;

;; ;

( 1 DOF)

;

2 1

2 1

2.

2

2 (*)

:2

.

:

L r r b

L co

x

n

y

x y

Total length of cable

when r

st

co r bnst

= + + + +

= + +

= + +

(*) y

. (**)2xy Const= +

2 11. 2 2Const L r r b =

; x y

y -1/2

x + , - ;

(+ - )

(*)

;

2 . ( )

1 : 0 2

:

2 ( )

2 :

(1)

0 2 2 ( )

thA B

ndA B

L x y const a

time x y or v

Total lengthof cab

v b

le

De

time x y or a

rivati

c

e of

a

v

= + +

= + +

= + +

& &

&& &&

2) One Degree of Freedom (2DOF)

(coordinate) ;

; 4

2.15 2DOF

A ; B C

( )

(*), (**), (***)

2.15

2 $

2-26

( )1

2

2 ( )

:

): (

: A A D

B B C C D

L y y C

Total

cable A

cab

length

a

L y y y y C ble

of ca le

B

b

= + +

= + + +

: ( ) 2( )D aeliminate y b+

3 44 (2 "2 " )A B CC y y y C DOF= + + +

1 2 3 4

3 4 1 2

: , , ,

2 , 2A B

when C C C C isconstant

and C L L C C C= + = +

( )Derivative of

0 2 4 2 4 ( )

0 2 4 2 4 (

1

: )

:

2

th

ndA B C A B C

A B C A B C

y y y or v v v

y y y or a a a

= + + + +

= + + + +

& & &

&& && &&

;

(degree of freedom: DOF>2)

;

8: ([1] Sample Prob. 2/1u)

In the pulley configuration shown, cylinder A has a downward velocity of 0.3 m/s. Determine the velocity of B.

A 0.3 /Av m s=

B ?Bv =

Step 1: ()

3 2 . (

:

1)B AL

Total lengthof ca

y y co

bl

nst

e

= + +

Step 2: 1 (1)

0 3 2

0 3 2 (2

( )

)

1 :

B A

B A

y

derivati

y

v

e o

v

v f

= +

= +

& &

0.3 /Av m s=v (2)

( )

2:32 0.3 0.

0.2

/3/ ( )

2

B

B

B

Av v

v m s

v m s upward Ans

+ =

= =

=

2 $

2-27

6: ([1] Sample Prob. 2/1v)

The tractor A is used to hoist the bale B with the pulley arrangement shown. If A has a forward velocity vA, determine an expression for the upward velocity vB of the bale in terms of x.

A Av +x

B x

( ),B Av f v x=

Step 1:

( ) ( ) 2 22

:

2 (1)L h y l

Total length of cabl

h y x

e

h= + = + +

Step 2: 1 (1)

2 2

2 2

210 22

(1)

2

:

0 AB

x xyh x

derivative o

x vv

f

h x

= ++

= ++

&&

2 2

12

AB

xvv Ans

h x =

+

A-1

(Kinetics of Particles)

Kinetics: ( )0RF v

= 3

1) Force, Mass and Acceleration

2) Work and Energy

3) Impulse and momentum

= =

[1] ==

Section A: (Force, Mass, and Acceleration)

Kinetics

A.7

Fv

m

av

1Fv

1av

m2Fv

2av

m

3Fv

3av

m

3A.1

m

1 2 3

1 2 3

.F F F

const ma a a

= = = =

v v v

v v v

A-2

m

F ma=v v (3A.1)

3A.2

Fv

m

1Fv

2Fv

3Fv

RF F= v v

av

RF F= v v

avm

3A.2

3A.2

(Equation of Motion)

F ma =v v

(3A.2)

(Coordinate) 2

( =- =)

Fv

(Free-Body Diagram: FBD) =

= (W = mg) (T) ( Fs= sN

Fk= kN) - (F = kx)

A.@

(Friction)

= ==

2 Fluid friction Dry friction Coulomb friction

= ( ) =

= Dry friction

A-3

Dry friction 2 (Static friction: fs)

() (Kinetic

friction: fk)

m Pv

m Pv

fv

W mg=

N

f

P45o

s sf N=

k kf N=

f P=

3A.3

3A.3 f P

m (Equilibrium) f P

P (Maximum friction force)

= N (3A.3)

max s sf f N= = (3A.3)

maxf N == s

e = P =

(Motion increasing velocity) 3A.3 (c)

= maxf kf (Kinetic friction)

N (3A.4)

k kf N= (3A.4)

k e =

e k s 20% [2]

<

(Tangent) =

() maxP f max s sf f N= =

maxP f> ==

kf ( )k kf N=

e = k

s 20%

A-4

1) Rectilinear Motion: 1D ()

x ax y - z

=

: 1D

0

0

x x

y

z

F ma

F

F

=

=

=

v v

v

v (3A.5)

2) Curvilinear Motion: 2D ()

=

2.1) (x-y)

2.2) =- (n-t)

2.3) = (r-)

1.

Coordinates Equation of Motion Acceleration Velocity

1. Rectangular (x - y) x xF ma =v v

y yF ma =v v

x xa = &&

y ya = && x xv = &

yv y= &

2. Normal and Tangential (n - t) n nF ma =v v

t tF ma =v v

22t

tnv

va

= = =& &

tt va = &

0nv =

tv = &

3. Polar (r-) r rF ma =v v

F ma =v v

2r r ra = &&&

2r ra = +&& && r rv = &

rv = &

Kinematics 2 2

2. Kinematics

Type of Motion Equations for the Relationship

General Case

( ).a constv

dv

dta =

ds

dtv =

dsv dv a=

Type of Motion Equations for the Relationship

Special Case

( ).a const=v

0v v at+= ( )2 20 02v v a s s= +

20

12o

s s v t at= + +

A-5

: (Rectilinear Motion: 1D)

7: ([1] Sample Prob. /1)

A 75-kg man stands on a spring scale in an elevator. During the first 3 seconds of motion from rest, the tension T in the hoisting cable is 8,300 N. Find the reading R of the scale in newtons during this interval and the upward velocity v of the elevator at the end of 3 seconds. The total mass of the elevator, mass, and scale is 750 kg.

75 ; 750m tm kg m kg= =

= 0-3 s cable 8,300T N=

a) R =

b) 3 s; = 3s ?v =

a) R c

1) FBD R =

R

: y m y

m m y

F m a

R m g m a

+ =

=

( ) (1)y mR a g m = + (1) 275 , 9.81 /mm kg g m s= = ?ya =

2) FBD ?ya =

ya

: y t y

t t y

F m a

T m g m a

+ =

=

( )

( ) 28,300 750(9.81)750

1.257 /

ty

t

T m

m s

ga

m

=

= =

= 21.257 /ya m s= (1)

( ) ( )1.257 9.81 75 830 N nsR A= = +

b) 3s ?v =

ov v at= +

3s 0 1.1257(3 3.77 /) m Anv s s= + =

A-6

@: ([1] Sample Prob. /@)

A small inspection car with a mass of 200 kg runs along a fixed overhead cable and is controlled by the attached cable at A. Determine the acceleration of the car when the control cable is horizontal and under a tension T = 2.4 kN. Also find the total force P exerted by the supporting cable on the wheels.

200m kg=

2.4T kN= a

a) a cable

b) P cable

a) a cable

FBD a P

a

( ) ( ):

12 513 13

x x

x

x F m a

T mg ma

+ =

=

( )( ) ( )( ) ( )2

12 52, 400 200 9.81 200

13 13

7.304 /

x

x

a

a m s Ans

=

=

b) P cable

FBD

( ) ( ) ( ):

12 50 0

13 13

y yy F m a

P mg T m

+ =

= =

( )( ) ( ) ( )12 5200 9.81 2,4002.734

013 13

P kN Ans

P =

=

A-7

: ([1] Sample Prob. /)

The 125-kg concrete block A is released from rest in the position show and pulls the 200-kg log up the 30o 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.

125 ; 200 ; 0.5A l km mkg kg = = =

A =: ?Bv =

c ?Bv =

Bv =

( )2 2 2B A A B Av v a S S= +

+

( )2 20 2 6 0 (1)B Av a= +

(1) ?Aa =

(equation of motion) =

FBD

125Am kg= T

A =

: y yF m a+ =

125(9.81) 125 (2)AT a =

(2) T =

200lm kg=

: x xx F m a+ =

2 200(9.81) sin(30 ) 0.5 200 (3)CT N a =

(3) , CN a =

: y yy F m a+ =

200(9.81) cos(30 ) 0N =

1,699.14 ( )N N y = +

2 C AL y y= +

0 2 (4)2

AC A C

aa a a = + =

1,699.14N N= 2

AC

aa = (3)

( ) ( )2 200(9.81)sin(30 ) 0.5 1,699.14 200 (3 )2AaT = (2) (3)

2

1004.2

1.776 /A

T N

a m s

=

=

= 21.776 /Aa m s= (1)

( )( )2 2 1.667 6 0Bv =

4.62 /Bv m s Ans=

: = A

Equilibrium check

A-8

g: ([1] Sample Prob. /g)

The design model for a new ship has a mass of 10 kg and is tested in an experimental towing tank to determine its resistance to motion through the water at various speeds. The test results are plotted on the accompanying graph, and the resistance R may be closely approximated by the dashed parabolic curve shown. If the model is released when it has a speed of 2 m/s, determine the time required for it to reduce its speed to 1 m/s and the corresponding travel distance x.

10m kg=

R - v relation: 2R kv=

( k = 2 N s2/m2) = 22R v=

2 m/s 1 m/s a) : 2 1 / ?m st = ; b) : 2 1 / ?m sS =

a) 2 1 / ?m st =

Bv =

(ODE 1st order)

( )

( )

( )

dva a

dt

v dv a ds b

dsv c

dt

=

=

=

vv

v v v v

vv

2 1 /m st (a)

1 /

2 /

1(1)

t m s

o m sdt dv

a=

(1) ?a =

(equation of motion) =

FBD

10m kg= a

: x xF m a+ =

2

100

2 100

x

x

R a

v a

=

=

20.2xa v =

= 20.2xa v= (1)

( ) ( )

1

22

1

2 1 /2

2 1 /

10.2

1 15

2.

1 1/20.

5sec

2

t

o

v

m sv

m s

dt dvv

tv

t Ans

=

=

=

=

=

=

b) 2 1 / ?m sS =

ODE 1st order (b)

( )

( )[ ]

( )

1 / 1 /

22 / 2 /

1 / 1

22 /

2 1 /

0.2

1 15 ln

0.2

5ln 1 3.2 47/

S m s m s

o m s m s

S m s v

vo m s

m s m

v vds dv dv

a v

dt d

s

v vv

S An

=

=

= =

= =

= =

A-9

h: ([1] Sample Prob. /h)

The collar of mass m slides up the vertical shaft under the action of a force F of constant magnitude but variable direction. If = kt where k is a constant and if the collar starts form rest = 0, determine the magnitude F of the force which will result in the collar coming to rest as reaches /2. The coefficient of kinetic friction between the collar and the shaft is .k

m ; .F const= ; kt = .k const= . k

: t = 0 sec 0v = 0 =

= 0v= /2 =

?F=

F

FBD

F

: y yF m a+ =

cos (1)ykNF mg am =

(1) , yN a =

1) N x =

: 0x xF m a+ = =

sin 0NF =

sinN F =

2) ya ODE =

(2)ydvdt

a =

kt = 1

d k dt dt dk

= = (2)

(2 )ydv

kd

a =

c , yN a (1)

( ) ( )cos sink dvm mF dF g k = = F

( )[ ] [ ]

( )[ ]

0 0

0

cos sin

sin cos

v

k

k

F

F

mg d mk dv

mg mkv

=

=

+ =

( )[ ]sin cos 1 (3)kF mg mkv + =

(3) ( )v f =

= F 0v= /2 =

( )[ ] ( )

[ ]2

1 0 1 0

1

2

k

kF mg mk

mgF Ans

=

+ =

: v

(3)

[ ]2 1 kmg

F=

( )v f =

[ ] ( )[ ]sin cos 1 (*)2 1 kkg g

vkk

= +

A-10

: (Curvilinear Motion: 2D)


Determine the maximum speed v which the sliding block may have as it passes point A without losing contact with the surface.

m ;

maxv () A

( 0 0f k NF N= = = )

maxv

FBD A

v

: n nF m a+ = 2

(1)nv

mg N ma m = =

(1) N = 0 maxv = 2ma

ax

x

m

0v

mg

v A s

m

g n

=

=

1 ( )maxv v g< = = N =

= A

2 ( )maxv v g> = = N ()

= =

==

= A (

= (1)

0N= 2v

g = 2v

g =

v = =

)

=

v g>

A-11


Small objects are released from rest at A and slide down the smooth circular surface of radius R to a conveyor B. Determine the expression for the normal contact force N between the guide and each object in terms of and specify the correct angular velocity of the conveyor pulley of radius r to prevent any sliding on the belt as the objects transfer to the conveyor.

small objects m ; R ; smooth surface 0k =

A: t = 0 sec 0v = 0 =

a) ( )N f =

b) ( ) Pulley r

m

a) ( )N f =

FBD m

( )N f =

: n nn F ma+ = 2

sin (1)nmg ma mNvR

= =

(1) v = =

ODE

(2)tv adv ds=

(2) ds Rd= ta =

=

: t tt F ma+ =

cos tamg m =

cost ga =

= ta (2) ( )v f =

( )0 0

cosv

dv dv g R

= 2

2si 2 sinn2

Rg vv

gR = =

2v (1) ( )N f =

( )2 sinsin Rgmg mN R = 3 sinN mg Ans =

b)

m =

m B =

m B : ( )2 sin /2Bv Rg =

: bv r=

=

( )2 sin /2

B bv v

Rg r

=

=

2RgAns

r =

A-12

: ([1] Sample Prob. /m)

A 1,500-kg car centers a section of curved road in the horizontal plane and slows down at a uniform rate from a speed of 100 km/h at A to a speed of 50 km/h as it passes C. The radius of curvature of the road at A is 400 m and at C is 80 m. Determine the total horizontal force exerted by the road on the tires at position A, B and C. Point B is the inflection point where the curvature changes direction.

1,500m kg= ;

400 ; 80A Cm m = =

A-C : - =

, , , .t A t B t C ta a a a const= = = =

- A : , 100 /t Av km h= C : , 50 /t Cv km h=

- 200A CS m =

A, B, C = ?; ?; ?A B CF F F= = =v v v

?; ?; ?A B CF F F= = =v v v

FBD

2 2, ,

,

2 2, ,

(1

:

):

:

A A n A t

B B t

C C n C t

Fat A

a

F F

F F

F F F

t B

at C

= +

=

= +

v

v

v

(1) ,n tF F

equation of motions

2

(2):

:

tn n

t t

vcomponent n

componen

F ma m

t t F ma

= =

=

(2) ,t Av ,t Cv = 2

: tncomv

Fponen n mt =

( )2 2,,

100/3.6(1,500 2,8)

490

00AA

nt

A

Fv

Nm = = =

( )

2 2, ,

,

,

2 2, 50/3.6(1,500)

80

0

3,620

B t B t

B

C

A

B

nt

n

C

v vm m

v

F N

F m N

=

=

= =

= =

component: t (2) at kinematics a =

const.

( )

0

2 20 0

20 0

( )

2 ( )

1( )

2

v v at a

v v a s s b

s s v t at c

= +

= +

= + +

(b) at

( )2 2 2C A t C Av v a s s= +

( ) ( ) ( )2

2 250/3.6 100/3.6 2 20

1 7 /

0

4

0

.4

t

t m

a

a s

= +

=

= : t tcomponent t F ma =

( ), , , (1,500) 1.447 2,170A t B tt A t maF F F N= = = ==

A-13

,n tF F (1) 2 2

2 2

3,620

2

2,890 2,170

3,620 2,1

:

,170

4, 2: 0

:

270

A

B

C

N Ans

N

F

F

at A

at B

at C

Ans

N AnF s

= + =

=

= + =

v

v

v

:

3

(

)

: cc [2]

(Banking angle: ) =-

m v

FBD

2

: n nv

n F m a m + = =

( )2

cos sin (1)s c cv

N N m + =

(1) cN

equation of motions b

: 0b bb F ma+ = =

cos 0c gN m =

coscmg

N =

coscmg

N = (1)

( ) ( ) 2cos sincos coss mg mg vm + = [ ]

2

tan (2)sv

g + =

(2)

(= =) =

1) = =

2) = s

3) = =

c:

= 0s = (2)

[ ]2

0 tanv

g + =

21tan

vAns

g

=

:

1) tancv v g < =

=

A-14


Compute the magnitude v of the velocity required for the spacecraft S to maintain a circular orbit of altitude 320 km above the surface of the earth.

(spacecraft) 320 km

= ?v=

FBD spacecraft

v 2

: n nv

n F m a m + = =

( ) ( )

2

2emm vG m

R hR h=

++

( ) (1)eGmv

R h =

+

1

22 9.825 / ( )

eGmg m s not rotating effectR

= =

= 2eGm gR= (1)

( ) ( )

2

(2)gR g

v RR h R h

= =+ +

2 3 39.825 / ; 6.371 10 ; 320 10g m s R m h m= = =

( )3

3 3

9.8256.37 10

6.37 10 320 1

7 220 /

0

, m s An

v

s=

= +

spacecraft

7,220 m/s

( )g

v RR h

=+

=

A-15


Tube A rotates about the vertical O-axis with a constant angular rate =& and contains a small cylindrical plug B of the mass m whose radial position is controlled by the cord which passes freely through the tube and shaft and is wound around the drum of radius b. Determine the tension T in the cord and the horizontal component F of the force exerted by the tube on the plug if the constant angular rate of rotation of the drum is 0 first in the direction of case (a) and second in the direction of case (b). Neglect friction.

m B:

- O .const = =&

- r , 0f kF N= =

: - .b const= Case (a): 0 .( )const CCW = Case (b): 0 .( )const CW =

?T = A B m: ?F = = 2

FBD m

FT

raa =&

r

0+b

( ) ( )

0

r m b

r

v

b

v

=

=

&

(T, F)

( m)

( )

:

(1)r

r rr F ma

amT

+ =

=

:

(2)

F m a

F am

+ =

=

,ra a

m r

O-axis .const = =& = 2 2

:2 0 2

r r r rAccelerationr r

r

r

aa

= =

= + = +

&&&

&

& &

&& && (3)

(3) ,r r& &&

B

r B

( .b const= )

0

2 2 20

0

0

0 0

0:

:2 2 0 0

r

r

r bVelocity

r b

r r b b bAcceleration

r r b b

r b

r

vv

aa

= = =

= =

= = =

= + = + =

=

=

&&

&

&&&&&

&&& & &&&

&

&

0r b=& 0r =&& (3)

( )

2 2 2

0 02 0 2 2

0r rr r r r

r r bb

a a

a a

= =

= + +

=

==

&&&

&& &&

= (1) (2)

Case (a): 0+ 2

0, 2T mr F mb Ans = = Case (b): 0

20, 2T mr F mb Ans = =

B-1

Section B: (Work and Energy)

(Work and Energy) ( (Force Mass and Acceleration, Work

and Energy, Impulse and momentum) ((( Kinetics (((

(((((

G (G

[1] GG (

3B.1 (Work)

(Work) s ((

m (( (path) ds (G((

G(( ds ( tF

rv

rv

drv

O

Fv

ds dr=vF

v

costF F

=

v

nF

m

m

2 3B.1

( 3B.1 Fv

m (( ((

ds G(( costF F = rv

rv drv drv ( ds G dU ( Fv

( )cosdU F ds=

( (scalar dot product)

dU F dr= v v

B-2

SI ( )N ( )m G

N m

(( (joule, J )

1 G( 1 ((G

1 m G 1 1 .J N m=

1) 2

2 3B.2

( 3B.2 ((

1s 2s ( Fv

( )2 21 1 x y z

F dx F dy F dzU F dr + += = v v

2

1

S

tSF dsU = . tF s (3B.1)

2) 222

2 3B.3

( 3B.3 ((

1 2 L (

Pv

( ) ( )

( )21

2

1 2 1

2

1

2 1

cos sin

cos cosx

x

P i P j dx i

P dx P x x

U F dr

+

= =

=

=

vv v

v v

cosPL = (3B.2)

B-3

3B.2 (Work and Energy Relation) (Energy) 2

1) (Kinetic Energy: T) ((( ()

21

2mvT = (3B.3)

T ( )J ( )N m (

2) (Potential Energy: V) (( (

( (fixed datum) (Gravitational potential energy)

(Elastic potential energy)

2 3B.4

( 3B.4 m (((

(path) 1 2

Fv dU ( F

v

(( drv

dU F dr= v v (3B.4)

( F ma=v v

(3B.4) ( ) tma dr ma ds=v v

Kinematic tv dv a ds=

G( (3B.4)

( )t ds v dv

dU F dr ma dr

ma m=

= = =

v v v v

vdvdU m=

(((

1 2 G

2

1

2

1 2 1

v

vF dr v dvU m = = v v

( )2 21 2 2 112 v vU m = (3B.5)

( (3B.5)

( )2 2 2 21 2 2 1 2 11 1 12 2 2v v v vU m m m = =

1 2 2 1U T T T = = (3B.6) ( (3B.6)

(work-energy equation) ,

0 0

0

( (3B.6)

1 1 2 2T U T+ = (3B.6a)

B-4

(3B.6a) 00 ,00

0000 0

1) (Gravitational Potential Energy: Vg)

2 3B.5

Case(a): g constant= ()

2M 3B.5a m (( 1-2 (G

W mg j= v v

( dr dx i dy j dz k= + +vv vv G(G(

( ) ( )2 21 2 1 1 mg j i dy j dz kU F dr dx + += = vv v vv v

( )2

12 1

y

ymg dy mg y y = =

: + ( )2 1y y 0, Wv (: y ) 0

0 , () 00

( 3B.5b m (( g

( (datum) 0gV = (

h

gV mgh=

G (( 1h 2h 0

( )2 1gV mg h h mg h = = (3B.7)

B-5

0,00 1h 2h

( )1 2 2 1 gU mg h h mg h V = = = (3B.8)

( (3B.7) (3B.8) 0,0

0

2 3B.6

Case(b): g constant ()

2M 3B.6b m (( 1-2 (

1r 2r ) ( ()

2e rer

Gm mF =v

G((

( )2 2 2

1 2 21 12 1 1 2

1 1 1 1 e r r ee dr e mgRr r r rr

Gm mU F dr Gm m

= =

= = v v

( 3B.6b (

( (( (

0gV = ) ( r (()

2

gmgRV

r=

G ((G 1r 2r 0

2

1 2

1 1gV mgR r r

= (3B.9)

B-6

000 (G 1r 2r

1 22

1 2

1 1gVU mgR r r

=

= (3B.10)

( (3B.9) (3B.10) 0,0

0

2) (Elastic Potential Energy: Ve)

2 3B.7

22 ( 3B.7 ((

(linear elastic spring) (( (spring stiffness) k (

x 3B.7b x (

0 ( (

F kx i= v v (( (

(G)) G00 (negative work)

G

( ) ( )2 21 1

2 2 21 2 2 11 2

1x xx x

kx i i kx k x xU F dr dx dx = = = = v vv v (3B.11)

( 3B.7b ( x

( ( )eV (((

( GG

212e

kxV = +

B-7

( )eV (( (

(G (G)

G (( 1x 2x 0

( )2 22 112eV k x x = (3B.12)

00 ( (( 1x 2x

( )1 2 2 22 112 eVU k x x = = (3B.13)

( (3B.12) (3B.13) 00

0

( (3B.6) (G (

( (3B.6)

1 2

1 2 g e

U T

U V V T

=

=

1 2 g eU T V V = + + (3B.14)

( 1 2U (( (G (

( (3B.14) (( 1 2

1 2

1 2

1 2

2 2

1 2

1 ,1 ,1 2 ,2 ,2

2 2 2 21 1 2 2

1 1 1 12 2 2 2

g e g e

o

mgh mgh

r or

mgR m

r

gRr r

o

T V V U T V V

mv kx U mv kx

+ =

+ + + = + +

+ + + +

(3B.15)

( (3B.15) (Conservation of Energy Equation)

(( 1 ( ) (

) (( 2

B-8

3B.3 (Power: P ) G

dU F dr drP F

dt dt dt

= = =

v v vv

P F V= v v

(3B.16)

SI (Watt, W ) ( 1 1 / 1 . /W J s N m s= =

2 (Mechanical Efficiency: me ) 0

0 00 G

outputm

input

Ppower outpute

power input P= = (3B.17)

(Overall Efficiency: e) ( ( (Mechanical loss)

(Electrical loss) (thermal loss) (

, ,m e te e e G (

m e te e e e= (3B.18)

3B.4 - (datum) (Free Body Diagram: FBD) ()

- 1 1 2 2T U T+ =

1 21 ,1 ,1 2 ,2 ,2g e g eT V V U T V V+ =+ + + +

- ((( 212

mvT = ( 2v )

- ( 12

1 2

S

tSF dsU = G( tF s ( 1 2 1 2F SU =

vv

- ( (

- ( )2 21 2 2 121k x xU = ( (

) 221

eV k x=

- ( ) 21 2 2 1 1 2 2 21 2

1 1,U mg y y U mgR

r r

= =

(

gV mgh= () 2

gmgRV

r= ( ) ( r

R = 6.371(106) m

B-9

: T1 + U1-2 = T2

2 _: ([1] Sample Prob. e/1_)

Calculate the velocity v of the 50-kg crate when it reaches the bottom of the chute at B if it is given an initial velocity of 4 m/s down the chute at A. The coefficient of kinetic friction is 0.30.

1) FBD ( 50 kg

2) Bv (( A-B

[ ]1 1 2 2T U T+ = 2 21 1 2 21 12 2v U vm m+ = (1)

( [ ]1 2U Fs =

( ) ( ) ( )1 2 sin15 50(9.81)sin15 0.3(50)(9.81)cos15 10 151.9 .o o okU mg s R s J = = =

1 2 151.9U J = (1)

22 2

21 1151.92 2

3.15 /(50)4 (50) v m s Ansv ==

B-10

2 g: ([1] Sample Prob. e/1g)

The flatbed truck, which carries an 80-kg crate, starts from rest and attains a speed of 72 km/h in a distance of 75 m on a level road with constant acceleration. Calculate the work done by the friction force acting on the crate during this interval if the static and kinetic coefficients of friction between the crate and the truck bed are (a) 0.3 and 0.28, respectively, or (b) 0.25 and 0.20, respectively.

1) FBD ( 80 kg

2) (() G(G

( ) ( ) ( )22 22 20 02 72 / 3.6 2.0 2 75 67 /0 #v v a s s a a m s = + = = +

( 2(80 )(2.67 / ) 213 #F ma k s Ng m= = =

3) ( ( . ) G

(

Case a): 0.3, 0.28s k = = ,max 0.3(80)(9.8 235 21 )31) (f s N n sF N lN o ip= = >=

No slip: = G 213f NF F= =

( )( ) 1: [ ] 213 7 65fWork done by friction U Fs kJ AnsF s N m= = =

Case b): 0.25, 0.20s k = =

,max 0.25(80)(9. 196.81) )2 1 (2 3f s N N slipF N= = = <

Slip: = G

( )( )0.2 80 9.81 157.0f kF N N= = = G [ ] 2157.0 80 1.962 / #F m a a a m s== =

((( ( )1.962 752.67

55.2 .#s mm = =

( )( ): [ ] 157.0 55 6.2 8.6fWork doneby fric kJ Anstion U Fs F s N m= = =

: 0 ,

B-11

2 e: ([1] Sample Prob. e/1e)

The 50-kg block at A is mounted on rollers so that it moves along the fixed horizontal rail with negligible friction under the action of the constant 300-N force in the cable. The block is released from rest at A, with the spring to which it is attached extended an initial amount x1 = 0.233 m. The spring has a stiffness k = 80 N/m. Calculate the velocity of the block it reaches position B.

1) FBD ( 50 kg

2) Bv (( A-B

[ ]1 1 2 2T U T+ = 2 21 1 2 21 12 2v U vm m+ = (1)

( 1 2 1 2, 1 2,Spring CableU U U = +

( )22 2 21 2, 2 11 180 0.233 1.2 0.2332 2 80.0Spring x JU k x = = + =

( )2 21 2, 300 1.2 0.9 0.9 180Cable cableU Fs J = = =

1 2, 80.0SpringU J = 1 2, 180Cable JU = (1)

( ) 22210 80 180 2 2.00 /(50)v v m s Ans+ = + =

: 0 ,0

B-12

2 h: ([1] Sample Prob. e/1h)

The power winch A hoists the 360-kg log up the 30o incline at a constant speed of 1.2 m/s. If the power output of the winch is 4 kW, compute the coefficient of kinetic friction k between the log and incline. If the power is suddenly increased to 6 kW, what is the corresponding instantaneous acceleration a of the log?

a) . : k

1) FBD ( (log) 360 kg

2) (( ( ( )k G

[ ]x xmaF = ( )sin30 0k oT N mg m =

( )cos30 sin30 0ok omT g mg = (1)

T (1) (Power) G

4,00 3,330 #0 1.2P F V P T T T Nv = = = =v v

(G( (1) ( k

( )3,330 360(9.81)cos30 360(9.81)sin30 0 0.513ko ok Ans = =

b) a 22 (winch) 2M 6 kW

6,00 5,000 #0 1.2P F V P T NT Tv = = = =v v

G (( x

[ ] ( )cos30 sin30o ox x k xma T mg mg maF = =

( ) 25,000 0.513 360(9.81)cos30 360(9.81)sin30 360 4.63 /o o x xa a m s Ans = =

B-13

2 k: ([1] Sample Prob. e/1k)

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 v1 = 30 000 km/h. Determine the velocity v2 of the satellite as it reaches point B, a distance h2 = 1,200 km from the earth.

1) FBD ( m

2) 2v (( A-B

[ ]1 1 2 2T U T+ = 2 21 1 2 21 12 2v U vm m+ = (1)

( 21 21 2

1 1mgRr r

U

= , r ( (1)

2 2 21 2

1 2

1 1 1 12 2

v mgR vr r

m m =

2 2 2 2 2 21 2 2 1

1 2 2 1

1 1 1 1 1 122 2

v gR v v v gRR h R h R h R h = = + + + + +

( )( ) ( ) ( )2

32

22 3

3

30,000 1 12 9.81 6,371 103.6 6,371 1,200 10 6,371 500 10

v = +

+ +

2 27,663 / 27,590 /v m s v km h Ansro= =

B-14

:

2 _: ([1] Sample Prob. e/1l)

The 3-kg slider is released from rest at position 1 and slides with negligible friction in a vertical plane along the circular rod. The attached spring has a stiffness of 350 N/m and has an unstretched length of 0.6 m. Determine the velocity of the slider as it passes position 2.

1) : ( 1

2) 2v

(Conservation of energy equation) G

1 22 2 2 21 1 1 2 2 2

1 1 1 1 (1)2 2 2 2

mv mgh kx U mv mgh kx+ + + = + +

(1) ( 21

12

mv 1mgh+ 1 221

12

kx U + + 2 22 2 2

1 12 2

mv mgh kx= + +

(1) 2 2 21 2 2 2

1 1 1 (2)2 2 2

kx mv mgh kx= + +

(

( ) ( )22 2 2 2

21 1 12 2 2

350 0.6 3 3(9.81)( 0.6) 350 0.6 0.6 0.6v + += +

22

12

69.849 3v=

2 6.824 /v m s Ans=

: 0 h 0

h+ 00 h 0

, ( W X )

B-15

2 g: ([1] Sample Prob. e/1m)

The 10-kg slider moves with negligible friction up the inclined guide. The attached spring has a stiffness of 60 N/m and is stretched 0.6 m in position A, where the slider is released from rest. The 250-N force is constant and the pulley offers negligible resistant to the motion of the cord. Calculate the velocity vC of the slider as it passes point C.

1) : ( A

2) 2v Cv

(Conservation of energy equation) G

1 22 2 2 21 1 1 2 2 2

1 1 1 1 (1)2 2 2 2

mv mgh kx U mv mgh kx+ + + = + +

(1) (

21

12

mv 1mgh+ 1 22 2 21 2 2 2

1 1 12 2 2

kx U mv mgh kx+ + = + +

(1)

1 22 2 21 2 2 2

2 2 21 2 2 2

1 1 12 2 2

1 1 1 (2)2 2 2

kx U mv mgh kx

kx Fs mv mgh kx

+ = + +

+ = + +

(

( ) ( ) ( ) ( )2 22 2 221 1 12 2 260 0.6 250 1.2 0.9 0.9 10 10(9.81) 1.2sin30 60 0.61.2ov+ + ++ = + + 221

24.74 10v=

2 0.974 /v m s Ans=

C-1

Section C: (Impulse and Momentun)

() (Impulse and Momentum) . (Force Mass and

Acceleration, Work and Energy, Impulse and momentum) ... Kinetics .

.. . ()

I .I

[1] II .

3C.1 (Linear Impulse and Linear Momentum)

... 3C.1

m .. (path) Fv

... v r=v v&

.. .

.. a v=v v&

Fv

I..

( )mvdt

dF ma mv

F G

= = =

=

vv v v&

vv &

(3C.1)

. G mv=v v (Linear

momentum)

. (3C.1)

..

x-y-z . (3C.1)

x x

y y

z z

F G

F G

F G

=

=

=

vv &

vv &

vv &

(3C.2)

9 3C.1

. . 1 2t t

..-

(3C.1)

2 12 2

1 1

2

1

G

G

t

t

t

t

dt

dt

F dG G G

F G

= =

=

v v vv

vv (3C.3)

C-2

21

t

tdtFv

(linear

impulse) .

. (3C.3)

1 22

1

2

11 2

t

t

t

t

dt

dt

G F G

mv F mv

+

+

=

=

v vv

vv v (3C.3a)

. (3C.3 a)

1t

1 2t t

2t . . 3C.2

9 3C.2

. (3C.3a)

vector . I

x-y-z

( ) ( )

( ) ( )

( ) ( )

2

1

2

1

2

1

1 2

1 2

1 2

tx x xt

ty y yt

tz z zt

dt

dt

dt

mv F mv

mv F mv

mv F mv

+

+

+

=

=

=

(3C.3b)

2 am bm

.. avv bv

v

I..

t . . Fv F v .

.I

. .. 3C.3

. t

(C.a)

am : 0t

a adtF m v = v v

bm : 0t

b bdtF m v = v v

9 3C.3

I

0 a a ab b bG Gm v m v = = + v vv v

( ) 0a bG G+ =v v

I

I a bG G G= +v v v

( ) ( ) ( ) ( )

1 2

1 21 2

0

a a a ab b b b

or

or

G G G

m v m v m v m v

=

=

=

+ +

v v v

v v v v (3C.4)

. (3C.4)

(the principle of conservation of

linear momentum)

C-3

3C.2 (Angular Impulse and Angular Momentum)

... 3C.4 P

m .. (path) .

r x i y j z k= + +vv vv

O . x-y-z

9 3C.4

m vv

O (Angular momentum: oHv

)

O ( )r mvv v

o

o

x y z

H r mv

i j kH m x y z

v v v

=

=

v v v

vv v

v (3C.5)

. x-y-z

( )( )

( )

x z y

y x z

z y x

v y v z

v z v x

v x v y

H m

H m

H m

=

=

=

SI N-m-s

Fv

O

oM r F r mv = = v vv v v&

(3C.5)

oH r mv r mv v mv r mv= + = + v v v v v v v v v& & & &

. 0v mv =v v I oH r mv= v v v& & .

Fv

O

I

o oM H =v v

& (3C.6)

. (3C.6)

m O

O

x-y-z . (3C.6)

ox ox

oy oy

oz oz

M H

M H

M H

=

=

=

v v&

v v&

v v&

(3C.6a)

.. 1 2t t

.

.

. (3C.6)

2 2

1 1

2

1

2 1

H

H

to o o ot

t

o ot

dt

dt

M dH H H

M H

= =

=

v v v v

v v (3C.7)

. 1 11oH r mv= v v v 2 22oH r mv=

v v v

21

tot

dtMv

(angular impulse)

. 1 2t t . .

C-4

.

. (3C.7) 2

11 2

too ot

dtH M H+ =v v v

(3C.7a)

. (3C.7a)

1t

1 2t t

2t

21

0t

otdtM =

v

1 2t t . 3C.7a

1 2o oH H=v v

(3C.8)

. (3C.8)

(the principle of conservation of

angular momentum)

1) x, y (Free Body Diagram: FBD)

2) ()

3) m

3.1) ..

- 21

1 2

t

tdtmv F mv+ =vv v n

- 1 2G G= v v

( ) ( ) ( ) ( )1 21 2a a a ab b b bm v m v m v m v=+ +v v v v

m I

3.2) .. 2

11 2

too ot

dtH M H+ =v v v

. ( )o r mvH = v vv ( )o r FM =

vvv

: Impulse and Momentum vector ()

C-5

: (Linear Impulse and Linear Momentum)

9 1: ([1] Sample Prob. I/19) A tennis player strikes the tennis ball with her racket when the ball is at the uppermost point of its trajectory as shown. The horizontal velocity of the ball just before impact with the racket is v1 = 15 m/s and just after impact its velocity is v2 = 21 m/s directed at the 15o angle as shown. If the 60-g ball is in contact with the racket for 0.02 s, determine the magnitude of the average force R

v exerted by the racket on the ball.

Also determine the angle made by Rv

with the horizontal.

1) FBD . 60 g

2) Rx Ry

( ) ( )2 21 1

1 21 2: cos15t t o

x x x xt tdt dtmv F mv mv R mv+ ++ = =

1 2 cos15o

xmv R t mv + =

( ) ( ) ( )15 0.020.06 0.060 21cos15105.9 #x

ox

R

R

N

+

=

=

( ) ( ) ( )2 2 21 1 1

21 2: (0) sin15

t t t oy y y yt t t

dt mg dt dtmv F mv m R mv+ ++ = =

( )( ) ( ) ( ) ( ) ( ) ( )

2 sin15

0.06 9.81 0.02 0.02 0.06 21sin15

16.89 #

o

oy

y

y

mg t R t

R N

m v

R

+

+

=

=

=

I Rv

2 2 2 2

1 1 16.89105.9

107.2

9

105.9 16.89

tan ta .07n

x

o

y

y

x

N Ans

n

R R

s

R R

ARR

= + = +

=

=

= =

v v

C-6

9 2: ([1] Sample Prob. I/20) A 0.2-kg particle moves in the vertical y-z plane (z up, y horizontal) under the action of its weight and a force F

v which varies with time. The linear momentum of the particle in newton-seconds is given by the expression ( ) ( )2 33 2

2 33 4G j kt t= +

v vv , where t is time in seconds. Determine F

v and its magnitude for the instant when t = 2 s.

Fv

( )dt

dF G G = =v vv &

( ) ( )2 3

2

3 2

2 33 4

3 2

dtF mg

m

k j k

kg

d t t

F t tj k

+ =

+

=

v vv

v

v

v vv

. t = 2 sec

( ) ( ) ( )

( )22

23(2) 2 2 0.2 9.81

6.04

6.04

6

8.51 .6

j k k

j k

F

F

sF N An

= +

=

= + =

v vv

vv

v

v

v

9 3: ([1] Sample Prob. I/21) A particle with a mass of 0.5 kg has a velocity of 10 m/s in the x-direction at time t = 0. Forces 1F

v and 2F

v act on the particle, and

their magnitude change with time according to the graphical schedule shown. Determine the velocity 2v

v of the particle at the end

of the 3-s interval. The motion occurs in the horizontal x-y plane.

FBD . 0.5 kg

2 2, 2,x yi

01-engineering dynamics - (v 2558)

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