FE Review dynamics - Louisiana State University1 Fundamentals of Engineering Review for Dynamics...

1

Fundamentals of Engineering Review for Dynamics

Dr.’s Yannitell (retired)

&

Dr Waggenspack [email protected]

Louisiana State University

__________________________________________________________ Last Revised: 02/14

Review Notes available in PDF format @ http://www.eng.lsu.edu/students/current/feinfo.html

Dynamics Problem Decomposition Dynamics!

Kinetics!Kinematics!

m

!=5°

Pm

Lµ != 0.4k

A

!r

30°

300 mm

O

v!!, aA A

Geometric descriptions of motion & constraints

Loading relationships which dictate CHANGES in motion

2

Dynamic Studies Dynamics:!

(Kinematics & Kinetics)!

Particles!

1 mh

v02.8 m

!20 m

•  M - mass •  (x,y) - position in 2D •  2 Degrees of Freedom

(DOF) in the plane

•  M & I - both mass & rotational inertia

•  (x,y,θ) - position & orientation in 2D (3DOF)

Rigid Bodies!

4 cm

!

A = (x,y)

B

x

y

Getting Started => Particle Kinematics •  Rectilinear Motion

– Movement along a straight line in 1-2 or 3D •  1 Degree of Freedom (DOF)* - s(t)

X

Y

O

e

s

r

p

+s

-s

•  Curvilinear Motion – Movement of particle along an arbitrary path

through space

p

isi

s=0s(t)

X

Y

O

r i

3

Rectilinear Motion Overview (Calculus/Physics Review!):

•  Position - s(t)

(1) v = dsdt

= s•

•  Speed - v(t)

•  Acceleration - a(t)

(2) a = dvdt

= v•=d2sdt 2 = s

••

DIFFERENTIATE

INTEGRATE

+s

-s

ts

+v

-v

v

+a

-a a

•  Typical Functions ?? – Polynomials – Trigonometric – Logarithms – Exponentials

s min

v = 0

v min

a = 0

dsdt

dv dt

•  Basic Calculus! – Max’s & Min’s ? – Undo Differentiation?

Rectilinear Motion Summary: •  Position - s(t)

(1) v = dsdt

= s•

•  Speed - v(t)

•  Acceleration - a(t)

(2) a = dvdt

= v•=d2sdt 2 = s

••(2*) v dv = a ds

•  Alternate form ?

DIFFERENTIATE

INTEGRATE

a(t) => Solid Rocket Propulsion a(v) => aerodynamic drag

a(s) => Gravitational fields, springs, conservative forces etc.

+s

-s

ts

+v

-v

v

+a

-a a s min

v = 0

v min

a = 0

4

Given: s =2t2 - 8t + 3 Find: Displacement from t = 1 to t = 3 Distance traveled from t = 1 to t = 3

Since, s(3) - s(1) = 0 displacement = 0

distance traveled = | s(2) - s(1) | + |s(3) - s(2) | = | -5 - (-3) | + | (-3) - (-5) | = 2 + 2 = 4

-8

3

-3 -5 -3

4

s

v t

t

1 2 3

s = 2t2 - 8t + 3 v = s = 4t - 8 a = s = 4 ••

• s(2) = -5 v(2) = 0 a(2) = 4

s(1) = -3 & s(3) = -3

Reversal @ v|t=2 = 0

Rectilinear Kinematics: Accel. a function of velocity – a(v)

t f = t(v) =dva(v)v i

v f

! " ti =dv"kv28

v f

! + 0

Given: •  A freighter moving at 8 knots when engines are stopped •  Deceleration a= -kv2

•  Speed reduces to 4 knots after ten minutes

! t f =1kv 8

v f

=1k1v f"18

#

$ % %

&

' ( ( ! vf = v(tf ) = 8

8kt f + 1 (knots)

v = 8 knots a = - k v 2

s0

Solution: (A) With a, v & t parameters given/requested, use a=dv/dt form

Find: (A) Speed of the ship as a function of time v(t) (B) How far does the ship travel in the 10 minutes it takes to

reduce the speed by 1/2 ?

5

and the resulting expression for speed of the ship as a function of time v(t) is as follows

v f = v(t) =8

6t +1

(knots)

•  Substituting BCʼs helps resolve the unknown constant k!

t =10 (min)

60 (min/hr)=

16

hr , v = 4 knots

! k =

34 1

nm"

# $ $

%

& ' ' ! v(1/ 6) = 8

8k 1/ 6( ) +1= 4 (knots)

•  From here, there are two alternatives for resolving the second question!

v = 8 knots a = - k v 2

s0


and the resulting expression for position of the ship as a function of time s(t)

(B) !METHOD 1: Now, knowing the velocity as a function of time!

v(t) = 86t +1

=dsdt

ds0

sf

! =8

6t +1

dt0

t f

!

s f ! 0 = 43

ln 6t +1( ) 0

t f =43

ln 6t +1( ) ! ln(1)( )

s f = s(t) =4

3

ln 6t +1( )

can now be used to find the particular displacement/distance at t=1/6 hr !

s(1/ 6) = 4 3

ln 6(1/ 6) +1( ) =4

3

ln(2) (nautical miles)

v = 8 knots a = - v 2

s0

34

!the boatʼs position can be found by integration!


6

and as was seen before

(B) !METHOD 2: With a, v & s parameters given/requested, use ads=vdv form!

s(t = 1/ 6)! s(v = 4) = 4 3

ln(2) (nautical miles)

s f = s(v) =vdva(v)vi

v f

! + si " s(4) = vdv#3/ 4v2( )8

4

! + 0

s(4) = !4

3dvv8

4" = !43 ln v 8

4=!43 ln4! ln8( ) = 4

3ln 84

Q.E.D.

v = 8 knots a = - v 2

s0

34

!and the boatʼs displacement (position?) can again be found by integration!


Curvilinear Kinematics Summary: •  Position

r(t)= x(t)i + y(t)j+ z(t)k= r(t)er + zez

•  Velocity: v = d r / dt

v(t)= r•

(t)= x•

i+ y•

j + z•

k

= ve t = s•

e t = r•

er + r!•

e! + z•

ez

•  Acceleration: a = d v / dt

a(t)= v•

(t)= r••

(t) = x••

i + y••

j + z••

k = v•

e t + !"• 2

en = v•

e t +v2

!en

= r••

# r"• 2$

%

& & &

'

(

) ) ) er + r"

••

+ 2r•

"•$

% & &

'

( ) ) e" + z

••

ez

O

p

r(t)

v(t )

a(t )

7

Curvilinear Kinematics Summary: •  Position

r(t)= x(t)i + y(t)j+ z(t)k= r(t)er + zez

•  Velocity: v = d r / dt

v(t)= r•

(t)= x•

i+ y•

j + z•

k

= ve t = s•

e t = r•

er + r!•

e! + z•

ez

•  Acceleration: a = d v / dt

a(t)= v•

(t)= r••

(t) = x••

i + y••

j + z••

k = v•

e t + !"• 2

en = v•

e t +v2

!en

= r••

# r"• 2$

%

& & &

'

(

) ) ) er + r"

••

+ 2r•

"•$

% & &

'

( ) ) e" + z

••

ez

er

O

e!

ref. line

r(t) p2D

!

v(t )

i

j

ere!

et

en

!

"a

2D Curvilinear Motion: Coordinates & Conversions

•  Cartesian <-> Polar <-> Path r(t) = xi + y j = rer

er = cos !i + sin!j = x

ri+ y

rj

e! = k " er = cos ! j # sin!i

i = cos!er " sin!e! j = k ! i = cos"e" + sin"er

v( t) = r•(t) = x

•i+ y

•j = ve t = s

•e t

et =vv=x•

vi+ y

•

vj

en = k ! e t =x•

vj " y

•

vi

i = cos!et " sin!enj = k ! i = cos"en + sin"et

er

O

e!

ref. line

r(t) p2D

!

v(t )

i

j

ere!

et

en

!

"a

8

15

Curvilinear Motion: Cartesian Coordinates!

•  Projectile Motion

a( t) = 0i ! g j = 0, !g[ ]•  z component drops out!

– Scale w.r.t. earth such that gravity g is ~constant |g| = 32.2 ft/s2 = 9.81 m/s2

– Neglect any air resistance – Motion is PARABOLIC thus PLANAR! – Typically align

•  y-axis along gravity vector •  x-axis horizontal in direction of motion

a = g

O x

y

r=(x,y)

v

16

Curvilinear Motion: Projectile Motion!

–  Integrate rectilinear relations •  Two (2) scalar relations •  One VECTOR relationship a(t) = 0i! g j= 0,!g[ ] = ac

! dvvi

v f" = ac dtti

t f"

v f ! vi( ) = ac t f ! ti( ) ! drri

r f" = v(t)dtti

t f"! v f (t) = ac t f " ti( )+ vi ! r f =

ac2t f " ti( )

2+ vi t f " ti( )+ ri

vx f = 0 t f ! ti( )+ vxi = vxivyf = -g t f ! ti( )+ vyi

x f = vxi t f ! ti( ) + xiyf =

-g2tf ! ti( )

2+ vyi tf ! ti( ) + yi

O x

y v = ( v ,v )

a = g = (0,-g )r =(x ,y )i i i

i x yii

9

17

Curvilinear Kinematics: Projectile Motion example ref: Hibbler 12-104!

Given:!–  Figure shown w/ ground !–  t0=0, (x0,y0)=0, v0=v0 @ θ above horizon!

y = !kx 2

Find: In terms of v0 , θ & k –  (A) The location at impact (xI,yI) –  (B) Velocity & Speed @ impact, vI, vI –  (C) Elapsed time @ impact, tI

Solution: –  2D projectile motion –  Get expressions for vx(t),vy(t) then x(t), y(t) –  Substitute into ground constraint expression

•  Solve for time of impact –  With tI known, substitute & solve for (xI,yI)

y = - kx 2

O x

y a = g = (0,- g )

v = v @ θ 0 θ

(x ,y )I I

! v f (t) = ac t f " ti( )+ vi ! r f =ac2t f " ti( )

2+ vi t f " ti( )+ ri

18

Curvilinear Kinematics: Projectile Motion!

a( t) = 0i ! g j = 0, !g[ ]

! (B) v I = a tI " 0( ) + v0 = vx I ,vyI[ ] vx I

= v0 cos!

vy I= -gtI + v0 sin!

•  IC’s => t0=0, (x0,y0)=0, v0=v0 @ θ!

s•= v = vxI

2 + vy I2

= v0 cos!( )2 + -gtI + v0 sin!( )2

= v02 " 2gv0 sin!tI + gtI( )2

–  Speed y = !kx 2

O x

y a = g = (0,-g )

v = v @ "0

(x ,y )I I

!

10

19

Curvilinear Kinematics: Projectile Motion!

! (A) rI =a2tI " 0( )2

+ v0 tI " 0( ) + 0

xI = v0cos! tI

yI =-g2tI

2 + v0sin! tIy = !kx 2

O x

y a = g = (0,-g )

v = v @ "0

(x ,y )I I

-g2tI

2 + v0sin! tI = "k v0cos! tI( )2

! (C) tI =2v0sin"

g # 2k v0cos"( )2

, tI = 0

! (B) v I = a tI " 0( ) + v0 = vx I ,vyI[ ]

–  Substitute value for tI into position, velocity & speed relations for solution

y = !kx 2

20

Curvilinear Motion: Projectile Motion!

a( t) = 0i ! g j = 0, !g[ ]

– Other typical P.M. queries •  Max Height •  Max Range •  Time @ some place along trajectory •  Later w/ Path & Polar Coord

–  Velocity (speed,direction/tangent) –  Curvature, rate of speed change ….

! v f = a tf " ti( ) + vi ! r f =a2tf " ti( )2 + v i t f " ti( ) + r i

O x

y v = ( v ,v )

a = g = (0,-g )r =(x ,y )i i i

i x yii

–  Reconsider problems w/ different axes placement

11

21

Given: launch at 3600 m altitude vo = 180 m/s angle 30°

x = 0 y = -9.81

for h set y = 0 t = 9.17 h = y = 4013 m

x = 180 (cos 30) = 156 y = 180 (sin 30)-9.81 t = 90-9.81 t x = 156 t y = 90 t - 4.905 t2 +3600

for tT set y = 0 t = 31.18

.. ..

. .

3600 h

30° 180 m/s

T

.

a = - g j

Path Coord. Example ref: Meriam&Kraige 2-8

Given: –  A rocket at high altitude with –  a0=6i-9j (m/s2) –  v0=20 (km/hr) @ 15° below horizontal

Find: At instant given (A) The normal & tangential

accelerations (B) Rate at which speed is increasing (C) Radius of curvature of the path (D) Angular rotation rate of the radial from CG to center of curvature

cg

C

Solution: –  “High altitude” means negligible air resistance –  Interested only at this instant (NO Integration required) –  V given is TANGENT TO THE PATH

•  Use this to relate path to cartesian coordinates

v = 15 km/hr

O x

y15°

a = g = (6,-9)20

12


Solution (cont’d):

!

e t =vv

=vv

= cos15°i " sin15°j

!

en = ± k " e t( )= # cos15°j# sin15°i

(A) an & at =?

(C) ρ=?

at = a • e t = 6i ! 9j( ) • cos15°i! sin15°j( ) = 8.12 (m / s2 ) = at

!

an = a •en = 6i " 9j( ) • "cos15°j" sin15°i( ) = 7.14 (m /s2) = an

v•= at = 8.12 (m / s2 )(B) v

•= ??

an =v2

! " ! =

v2

an

=(20 *103km /hr )2

7.14(m / s2 )1hr

3600s* 103m

km#

$ % %

&

' ( (

2

= 4.32(106 )m

v = 15 km/hr

O x

y15°

a = g = (6,-9)

cg

C

eten 20 (2D shortcut!)


Solution (cont’d):

an = !"• 2

#"•=

an!=

7.14 (m / s 2 )4.32 *106(m)

= 12.9 *10$4 1s

(D) –  Look either at an or velocity !•= ??

v

= !"•

# "• =

v!

=20*103km / hr4.32 *106 (m)

1hr3600s

* 103mkm

$

% & &

'

( ) ) = 12.9 *10*4m

v = 15 km/hr

O x

y15°

a = g = (6,-9)

cg

C

eten 20

13

RELATIVE MOTION

rB = rA + rB/A vB = vA + vB/A aB = aA +aB/A

When A & B are two points on the same rigid body then: •  i.e. the relative motion is circular •  vB/A is perpendicular (⊥) to rB/A & •  | vB/A | = | ωAB AB |

rB

B

rB/A

A rA

rB/A

B

A vB/A

Special Case: Rigid Bodies

vB = vA + ωAB k × ABuB/A (2D)

ωAB

26

Relative Motion: ref ~DWY FE Review

Given:!•  A river flows south at 5 m/s!•  A boat can travel at 10 m/s relative

to the water.!Find: •  In what direction should the boat

head from A, in order to reach point B, directly across the river?

Solution: •  Set up Cartesian Axes @A •  Write Relative Velocity expression

VB = VR + VB/R DM DM DM

A B

v = 5 m/sR

x

y

Ov

v = -5 j m/sR

v = ? i B

v = 10 m/s @! ?B/R

!•  Draw Velocity Polygon diagram

–  VR magnitude & direction known –  VB direction known, not magnitude –  VB/R magnitude known, not direction

14

27

Relative Motion: ref ~DWY/WNW FE Review

A B

v = 5 m/sR

x

ySolution (cont’d):

VB = VR + VB/R DM DM DM

vB i = vR j + vB/R (cos θ i + sin θ j) vB i = -5 j + 10 (cos θ i + sin θ j) (m/s)

•  Equating components: i => vB = 10 cos θ (m/s) j => 0 = -5 + 10 sin θ (m/s)

•  2 equations <-> two unknowns v = ? i BOv

v = -5 j m/sR

v = 10 m/s @!B/R

!

sin θ = 5/10 = 1/2 (m/s) => θ = sin-1 (1/2) = 30°

•  Lagniappe vB = 10 cos 30° (m/s)

28

Relative Motion: ref ~DWY/WNW FE Review

Alternate Solution (cont’d):

A B

v = 5 m/sR

x

y

•  With velocity polygon drawn, use the Law of Sines

v = ? i BOv

v = -5 j m/sR

v = 10 m/s @!B/R

!

sin!vR

=sin90o

vB/R

! = sin"1 vR sin90o

vB/R

= sin"1 510

#

$ % %

&

' ( ( = 30o

θ

15

29

Relative Motion: ref ~Meriam & Kraige 2/13

Given:!•  Two cars A & B at the instant shown!

vA= 72 i km/hr !aA= 1.2 i m/s2!

vB= 54 et km/hr, constant speed!

Find: (A) vB/A =? (B) aB/A =?

Solution: •  Convert to consistent units

x

y

A

30°

B

150 m e n

e t

en = cos30°i ! sin30°je t = k ! en = cos30°j + sin 30°i

(km /hr)* 13.6

= (m / s) ! vA = 72(km /hr) = 20(m / s)vB = 54(km /hr) = 15(m / s)

•  Motion RELATIVE TO A of interest •  Two coordinate axes are used

–  Simplifies v & a definitions –  Illustrates “coordinate conversion” for

expressing answers “in terms of” a unified set.

30

Relative Motion: ref ~Meriam & Kraige 2/13 (A) !Relative Velocity!

x

y

A

30°

B

150 m e n

e t

vB/A = vB ! vA

=15e t ! 20 i (m / s)=15(sin30°i + cos30°j) ! 20 i (m / s)

vB/A = !12.5i +13.0j (m / s) = 18 (m / s)@! 46°

•  Velocity Polygon Approach (Graphical)

v = 20 m/sA

v = 15 m/sB

Ov

30°

46°

vA

vB! vB/A = vB " vAvB = vA + vB/A

VB/A = 18 m/s!

16

31

Relative Motion: ref ~Meriam & Kraige 2/13 x

y

A

30°

B

150 m e n

e t

)/( 2.1)30sin30(cos5.1

)/( 2.1 5.12

2

smsm

ijiie

aaa

!°!°=

!=

!=

n

ABB/A

°!=!= 82@)/(76.0)/( 75.0 1.0 22 smsmjiaB/A

•  Acceleration Polygon (Graphical)

(B) Relative Acceleration aA = 1.2 i (m / s2 )

aB = v• e t +

v2

! en (m / s 2) = 15 m / s( )2

150 men

= 1.5en (m / s2 )

a = 1.2 m/s2A

a = 1.5 m/sB2

Oa30°

a = 1.5 m/sB2

a = 1.2 m/s2A

aB = aA + aB/A! aB/A = aB " aA

-82°!aB/A = 0.76 m/s2!

32

Given: A balloon at an altitude of 60 m is rising at steady rate of 4.5 m/s. A car passes below at constant speed of 72 kph.

Find: Relative rate of separation 1 second later: aC = 0 aB = 0 (m/s2)

vC = 20i vB = 4.5j (m/s)

72 km/hr

60 m

4.5 m/s

Alternative Method (Vectors!):

!

r•B /C =V B /C •

rB /C

rB /C

= V B "VC( ) •rB " rC( )rB /C

= "20,4.5( ) •"20,64.5( )

67.5=

690.367.5

=10.2 (m/s)

( )222B/C 5.460)20(r tt ++=

5.4)5.460(2)20)(20(2r2r B/CB/C tt ++=•

Divide through by 2 rB/C & set t = 1 )/(22.1052.67/25.690r / smCB ==

•

rC = 20t i rB = (60 + 4.5t)j (m)

= 20 m/s

J

I

rB/C = rB - rC

j

i

rB/C = rB/C = rB/C •rB/C

17

33

Given: The ferris wheel rotates at θ= 2 r/s, θ = -1 r/s2 and the boy (B) walks to the right at a constant speed of 2 m/s.

Find: The velocity and acceleration of girl (G) on the ferris wheel relative to boy B

B

R= 4m θ

G

j i

eθ er

• •

Defining unit vectors i, j, er, & eθ shown velocities & accelerations are then:

vB = 2i (m/s) aB = 0

•

vG = 4 m (2 r/s) eθ = 8 eθ (m/s) aG = -4 m (2 r/s)2 er + 4 (-1 r/s2) eθ = -16 er - 4 eθ (m/s2)

Solution:

=> vG/B = vB - vG = ? aG/B = aB – aG = ?

θ •θ • •

er = cos θ i + sin θ j eθ = -sin θ i + cos θ j

or i = cos θ er - sin θ eθ j = sin θ er + cos θ eθ

vG/B = vG - vB = 8 eθ - 2i = -(8 sin θ + 2) i + 8 cos θ j OR = -2 cos θ er + (2 sin θ + 8) eθ (m/s)

aG/B = aG = -16 er -4 eθ (m/s2)

Must convert axes to combine vectors:

and since aB = 0 :

ωAB

VB

VA

B

vB = vA + vB/A

vB j = vA i + ωAB k × AB uB/A

vB j = vA i

- AB ωAB(sinθ i + cosθ j)

θ

A

Equating i & j components:

i → vA - AB ωAB sinθ = 0

j → vB = AB ωAB cosθ

vA AB ωAB sinθ vB AB ωAB cosθ =

Two points on a rigid body:

18

VA = AC ω AB [ i ]

VB = - BC ωAB [ j ] ωAB

C I.C.

VB

VA

B

θ

A

Using Instant Centers (IC):

AC = AB sinθ BC = AB cosθ

vA AB ωAB sinθ vB AB ωAB cosθ =

O

ωo

VB ⊥ OB A VA

B

VB

VA

VB/A ⊥ AB

Slider Crank Velocities Using Graphical & Instant Centers (IC): C

ωAB

VB = OB ωo = CB ωAB VA = CA ωAB Be sure to account for direction!

VA = (OB / CB) CA ωo

19

Given: ωc = 2 r/s αc = 6 r/s2

Find: vD, aD

AC

ωc αc

D

| 6”| 6"| 8” |

D

CA

B

F E2 E1

rA = 6" rB = 12" rC = 8"

ωC = 2 r/s αC = 6 r/s2

VE1 = rCωC = 8•2 = 16 in/s

VE2 = 16 = rBωB = 12 ωB

so: ωB = 4/3 = ωA [r/s]

VD = VF = ωArA = 4/3 (6) = 8 [in/s]

aE1 = αCrC + ωC2rC

= 6 (8) + 4(8) = 48 + 32

aE2 t = aE1t = 48 = αBrB = αB(12) αB = 4 = αA

aF t = αArA = 4(6) = 24 = aD

[in/s2 ]

[in/s2]

20

Given: ro = 3' ri = 2' vo = 10 f/s no slip

Find: vB

vo = 10 ft/s

vc = vo + ω r = 10 - 2 ω = 0

ω = 5 or -5 k

vB = vo + ω x rB/o = 10 i + -5k x -3j = -5 i [ft/s]

or vB = vc + ω x rB/c = 0 + 5k x -1j = -5 i [ft/s]

O 45°

A

C

B

vA = ?

Kinetics Summary •  Three general solution approaches for establishing the governing

equations of motion (EOM) => Which one to use? i) Newton’s Laws ii) Work- Energy & Conservation of Energy iii) Impulse - Momentum & Conservation of Momentum

MP! = ICG " + reff maCG

UA!B = matdssA

s B

" = mvdvvA

vB

" = 12m vB

2 ! vA2( ) = #TA!B

–  Typical forces •  Springs •  Friction •  Gravitation

F f = µs / k NF = k s - s0( )

F =mg

UNC = !T + !Vg + !Ve = !ETOT

I = FRdt! = dL! = "L

F! = m aCG

21

Particle Kinetics: Free Body Diagrams!•  Free Body Diagrams:!

–  Isolate the particle/system of interest (i.e. boundaries)!–  For noting action-reaction between particles/bodies it is

important to identify the common normal-tangent @ the point of contact (often one or the other is easily identified)!

mAm B

FAB

FBA

The image part with relationship ID rId6 was not found in the file.

–  Include ALL forces (& later => moments)!•  Field forces (gravity, electro-magnetic fields etc)!•  Viscous forces (aerodynamic drag, fluid flows, etc)!•  Contact forces (touching elements) -- Most common!

–  For motion over an interval --- draw in a general position!!

n!

t!

n!

t!

Kinetics: Σ F = maC Σ MC = IC α

F2 C = cg

etc.

M1

F1

C IC α

maC

FBD EFD

MP! = IC " + reff maC

22

Given: the 20# force is applied to the sliding door which weighs 100 #

6'

10 '

5'

20#

Find: the reactions at the frictionless roller supports

Find: the reactions at the frictionless roller supports

ΣF → = 20 = (100/g) a a = g/5

ΣMA = 10B - 5(100) + (20) = 3 (100/g)(g/5)

=> B = 54 #

A = 46 #

6' 5'

20#

A B10 '

100# 100 g a

ΣF = A + B - 100 = 0 ΣMCG = 5B - 5A - (20) 2 = 0

23

α

O!45°!

#1!

#2!

100 mm!

Particle Kinetics: Path Coord Example ref ~Meriam & Kraige 3/74!Given:!•  The slider (m=2 kg) fits loosely in the smooth slot of

the disk which lies in a horizontal plane and rotates about a vertical axis through point O.!

•  The slider is free to move only slightly along the slot in either direction before one (but not both) of the two wires #1 or #2 becomes taut. !

•  The disk starts from rest at time t = 0 and has a constant clockwise angular acceleration of α=0.5 r/s2.!

Find:!(A)  Determine the TENSION (T2) in wire #2 at t =1 second!(B)  Determine the REACTION FORCE (N) between the

slot and the block, again at t =1 second.!(C)  Determine the TIME (t ) at which the tension in wire #2

goes slack and wire #1 becomes taut. Solution:!•  Asks for FORCES (T,N) so we must first establish kinematics (accelerations!)!•  “Move only slightly” means it is effectively fixed relative to the slot/disk, thus!•  The slider travels a circle about O & path (en-et) axes !! ! ! ! !or polar (er-eθ) axes are convenient!

er!

eθ!

er!

eθ!

en!

et!

en!et!

Particle Kinetics: Path Coord Example ref ~Meriam & Kraige 3/74!Solution (continued):!•  Construct FBD!•  Use disk kinematics (α=0.5 r/s2 CW constant) to

determine sliderʼs total acceleration!

α

O!45°!

#1!

#2!

100 mm!er!

eθ!

en!

et!

en ,-er!

et ,eθ

45°!

T2 or -T1 !

N!

!

" = 0.100m = r # constant

!

" #•

= #••

= r•

= r••

= 0

!

d"0

"

# = $dt0

t# = 0.5dt

0

t#

" = 0.5t

•  Not instantaneous - integrate angular acceleration!

!

a s= " re# $ r% 2 er

= v•

e t +v 2

&en = " re t + % 2 ren

24

Particle Kinetics: Path Coord Example ref ~Meriam & Kraige 3/74!Solution (continued):!•  Newtonʼs Law can be applied along ANY two

independent directions to resolve unknown reactions!–  Sum force components along (n-t, r-θ)

α

O!45°!

#1!

#2!

100 mm!er!

eθ!

en!

et!

en ,-er!

et ,eθ

45°!

T2 or -T1 !

N!

!

T2 cos45+N sin 45 = m"rT2 sin 45#N cos45 = # m$ 2r

!

T = m " r cos45#$ 2r sin 45( ) =mr 2

2" #$ 2( )

N = m " r cos45+$ 2r sin 45( ) =mr 2

2" +$ 2( )

–  OR to simplify algebra of unknowns, choose the directions along the unknown reactions and sum both forces and acceleration components!

–  ASIDE: This IS the geometric equivalent to simultaneously solving the first set of constraints to yield expressions for the unknowns!

–  Noting the similarity of the expressions (± : + for N, - for T)

!

N,T2 =mr 22

" ±# 2( )

Particle Kinetics: Path Coord Example ref ~Meriam & Kraige 3/74!Solution (continued):!•  Substituting the known expressions for α & ω(t) α

O!45°!

#1!

#2!

100 mm!er!

eθ!

en!

et!

en ,-er!

et ,eθ

45°!

T2 or -T1 !

N!

(A) !So for t=1, the TENSION T2 is !

N,T2 =mr 2

2" ±# 2( )

=2kg*0.1m* 2

2 0.5 ± 0.5 t( )2{ }(r /s2)

N,T2 =2

20 1 ± 0.5 t 2{ }(N)

!

T2 =2

20 1 " 0.5(1)2{ } (N )= 2

40(N ) = 0.035(N )

(B) !At t=1, the NORMAL REACTION N is

!

N =2

20 1 + 0.5(1)2{ } (N ) =

3 240

(N ) = 0.106(N )

(C) !The time when TENSION T2 goes to zero is!

!

T2 =2

20 1 " 0.5t2{ } (N ) = 0 #1 " 0.5t2 = 0 # t= 2 # t= 1.414 (s)

25

en ,-er!

et ,eθ

45°!

T2 or -T1 !N!

Particle Kinetics: Path Coord Example ref ~Meriam & Kraige 3/74!Langiappe:!•  The acceleration vector starts off completely in the lateral (θ or t) direction here

(ω=0). Since cables/wires/ropes cannot PUSH, only T2 can be engaged in balancing the (r or n) component of the side wall reaction N!

•  The tangential acceleration component remains constant!•  As the disk speeds up (ω >0), the normal component increases!•  When the total acceleration vector aligns with the normal reaction force

between the block & slot, the cord/wire tensions are both zero momentarily, and as T2 goes slack, T1will become taut.!

increasing ω

T1=T2=0!!

a t= 0 = a t or a"

Impulse / Momentum

If mutual forces cancel, impulses cancel ( not true of work )

coef. Of restitution e =

Angular Momentum:

rel. norm. sep. vel. rel. norm. app. vel.

∫ F dt = ∫ madt

= ∫ mdv

= mv2 - mv1

t2

t1

∫ Mc dt = ∫ Ic α dt

= ∫ Ic dω

= ∫ Ic ω2 - Ic ω1

° °

°

° °

26

Example: Conservation of Momentum Given: •  An artillery gun (mG) fires a shell

(mP) with a speed vp Find:

(A) The recoil speed (vR) of the gun

Solution: •  FBD of system components, just as

shell leaves the gun •  Rectilinear motion (i.e. only horizontal

motion of interest here) •  Propellant firing is internal to the

system –  System momentum is conserved in the

horizontal direction

vR

vP

FBD! mGg! mPg!

Fpropellant!

N!

!

"Lsys = 0

x!

!

"Lx#system = mG (vG # 0) + mP (vP # 0) = 0

!

vR = " vG = mP

mG

vp

Example: Conservation of Momentum Given: •  More often, a “muzzle velocity” (vP/G)or

speed of the shell relative to the gun barrel is specified

Find: (A) The recoil speed (vR) of the gun Solution: •  FBD (same), Rectilinear motion & Propellant

firing is internal

FBD!

vR

vP/G

mGg! mPg!

Fpropellant!

N!

!

"Lsys = 0

x!

!

mGvG + mP (vG + vP /G ) = 0

!

vR = " vG = mP

mG + mP

#

$ %

&

' ( vP /G

!

vP = vG + vP /G

!

"Lx#system = mG (vG # 0) + mP (vP # 0) = 0

27

Example: Conservation of Momentum!Given:!•  Numerous examples with similar

circumstances, rephrasing the wording!–  Kid(s) on a boat in still water, one

jumps off!–  Car lands on a barge & skids to rest

relative to barge!–  Rail cars collide & stay attached!

Find:!(A)  The resulting speeds of each element!(B)  A time it takes to “skid to rest” !

Solution:!•  Similar conservation of momentum relations !

!

"Lsys = 0

A B C

50 km/hr

A B

•  Impact Problems: –  Reformulation of one type of Impulse-Momentum

Particle Kinetics: Impulse-Momentum

–  Impulsive Forces characterized by •  LARGE MAGNITUDE •  SHORT TIME DURATION •  Ex: explosions, ball-bat, club-golf ball

–  Neglect other conventional forces of lesser effect for the short time interval

•  Springs •  Gravity •  Many Reaction forces (BUT NOT ALL!)

–  Good opportunity to look at the SYSTEM of particles in simplifying the problem (reactions are internal!)

L = mv|F |

t (sec) Δt

n

t

F

F

mewagg

Text Box

28

Particle Kinetics: Impulse-Momentum/ Impact •  Impact

–  Locate Common Normal/Tangent •  Line of contact/impact - the NORMAL!

–  Forces of interaction •  Equal, Opposite, Co-linear

–  Very complex internal phenomena, captured by Coefficient of Restitution

(good derivation in B&J text --- READ IT!)

– Central & Oblique Impacts •  Velocities are NOT co-linear with the line

of impact (i.e. the common normal)

n

t

F

F

n

t

vB

A

B

vA

vA*

Oblique* Impact

Central Impact

vB*

e =

Vrelative Separation( )Vrelative Approach( ) Common Normal

Particle Kinetics: Impulse-Momentum/ Impact •  Solving Impact Problems !

n

t

vA

vB

vA*

vB*

A

B

(1) Tangential Direction: individual particles have no net external impulsive forces in!

mAvAt =mAvAt* & mBvBt = mBvBt

*

(2) System of particles: No net impulsive forces normal direction!

!Ln SYS= 0" mAvAn +mBvBn = mAvAn

* +mBvBn*

(3) Coefficient of Restitution: Rel. Velocities along Common NORMAL!

e = vRelative Separation

vRelative Approach Normal

=vBn* ! vAn

*vAn ! vBn

!

(Perfectly Plastic) 0 " e "1 (Perfectly Elastic)

29

Particle Kinetics: Impulse-Momentum/ Impact!•  Solving constraint relations !!

n

t

vA

vB

vA*

vB*

A

B

*BtBt

*AtAt vvvv == &

( ) (1) *AnAnBn

*Bn vvvv !+=

B

A

mm

( ) **(2) AnBnAnBn vvvv +!= e

•  From which the unknown rebound (normal) component of velocities become!

( ) 1 (1) BnAn*An vvv e

mmm

mmemm

BA

B

BA

BA +!!"

#$$%

&

++!!

"

#$$%

&

+'

=

( ) 1 (2) BnAn*Bn vvv !!

"

#$$%

&

+'

++!!"

#$$%

&

+=

BA

AB

BA

A

mmemme

mmm

Particle Kinetics: Impulse-Momentum/ Impact!•  What if mA >>> mB ?!

n

t

vA

vB

vA*

vB*

A

B

*BtBt

*AtAt vvvv == &

( ) (1) *AnAnBn

*Bn vvvv !+=

B

A

mm

( ) **(2) AnBnAnBn vvvv +!= e

•  From which the unknown rebound (normal) component of velocities become!

( ) 1 (1) BnAn*An vvv ee ++!=

(2) Bn*Bn vv =

0

30

Given: two balls of equal mass with the velocities shown collide, coefficient of rest = 0.8

Find: velocities after impact

. .

45° 4

35 m/s 10 m/s

x - mom.: u1 + u2 = 5/√2 - 8

Rest: u2 - u1 = 0.8 [(5/√2) + 8 ]

u2 = 2.76 u1 = -7.23 5/√2

5/√2 6

8

5/√2

u1

6

u2

x

y

Work / Energy :

T1 + V1 + WNC = T2 + V2

where T = m vc2 = Ic ω2 = Ic ω2

Vgrav = -mgh

Vspring = k x2 where x = l - lo= stretch

1 2

1 2

1 2

1 2

31

61

v

A

!=15°µ!!=0.3k

B

10 m

Work-Energy Example ref ~Meriam & Kraige 3/11 Given: •  A crate of mass m slides down an incline •  m =50 kg, θ=15°, µk=0.3, •  Reaches A with speed 4 m/s

Find: (A) Speed of crate vB as it reaches a point B

10 m down the incline from A

Solution: •  Rectilinear motion, align axes accordingly -

i.e. || & ⊥ to incline •  FBD of crate in general position (working

over a motion interval here)

Fy! = mg cos" #N = 0 $ N = mgcos"

•  No movement ⊥ to incline so

y

x

The

θ

θ

m g

N F µ

62

Work-Energy Example ref ~Meriam & Kraige 3/11 Solution (cont’d): •  Work done is due to the resultant forces in

direction of displacement (i.e. down incline) & includes Friction & component of Weight

•  Principle of Work-Energy then says

U A-B = (mg sin! "Nµk )#xAB

= (mg sin! "mg cos!µk )#xAB

UA-B = !TA-B = TB " TA ! TB = UA -B + TA12mvB

2 = mg(sin! " cos!µk )#xAB +12mvA

2

vB = 2g(sin! " cos!µk )#xAB + vA2

!

vB = 2*9.81(m / s2 )*(sin15°" cos15°*0.3)*10m+ (4m / s)2

vB = 3.15 m / s

v

A

!=15°µ!!=0.3k

B

10 my

x

The

θ

θ

m g

N F µ

32

63

Work-Energy: Example ref ~Meriam & Kraige 3/13 Given: •  Block (m =50 kg) mounted on rollers •  Massless spring w/ k=80 N/m •  Released from rest at A where spring has

initial stretch of 0.233 m •  Cord w/ constant tension P=300 N attaches

to block & routed over frictionless/massless (ideal) pulley @ C

Find: (A) Speed of block vB as it reaches a point B

directly under the pulley.

Solution: •  Again, rectilinear motion, align axes

accordingly •  FBD of block in general position (working


•  Look at alternative - include the rope in as part of the SYSTEM - reduce FDB to an ACTIVE Force Diagram!

k P=300N x A 1.2 m 0.9 m A B C

P

Fs

mg

R

N

P

Fs

mg

N

!

k The image part with relationship ID rId10 was not found in the file.

P=300N

x A

1.2 m

0.9 m

A B

C lA

ps

y

x

64

Work-Energy: Example ref ~Meriam & Kraige 3/13 Solution (cont’d):

k P=300N x A 1.2 m 0.9 m A B C

y

x

P

FsActive Force

Diagram

ACTIVE Force Diagram!

•  Eliminate Normal Forces ⊥ to displacement @ their point of contact {THEY DO NO WORK!}

P

Fs

mg

R

N–  Weight (mg) & Roller reactions (N) –  Pulley force on rope (R)

•  Active forces DO work on the system –  Spring Force (Fs) => opposes motion

Fs = !kx

–  Assuming block can actually reach B

UABFs = FsdxxA

xB

! = "kxdxxA

xB

! =12kx 2

xA

xB

= " 12k(xB

2 " xA2 )

UABFs = !1280(N / m){(1.2 + 0.233)2 ! 0.2332}(m 2) = !80Joules


P=300N

x A

1.2 m

0.9 m

A B

C lA

ps

33

65

Work-Energy: Example ref ~Meriam & Kraige 3/13 Solution (cont’d):

k P=300N x A 1.2 m 0.9 m A B C

y

x

P

FsActive Force

Diagram

P

Fs

mg

R

N

Lcord = sP + l = constant!sP = "!l = lA " lB

= 1.22 + 0.92 " 0.9 # 0.61m

–  Cord Tension (P) => constant –  Displacement of P

•  Calculate Work done on system

UABP= P!s = 300(n) * 0.61(m)

=180Joules

lA

= lB

•  Work-Energy UTOT = !TA-B = TB " TA

!80 +180(Joules) = 12m(vB

2 ! 0)

! vB =100(Joules) * 2

50Kg= 2.0m / s

ps


P=300N

x A

1.2 m

0.9 m

A B

C lA

ps

66

Conservation-Energy Example ref ~Meriam & Kraige 3/17 Given: •  m =3 kg slider on circular track shown •  Starting from A with vA=0 •  lo=0.6 m (unstretched), k=350 N/m •  µ=0 (i.e friction is negligible)

Find: (A) Velocity of slider as it passes B

Solution: •  FBD of crate in general position (working


•  Identify –  Conservative Forces

NFs

mgFBD

mg (Weight/Gravity) & Fs (Spring) –  Non-working Constraint Forces

N (Track reaction force)

The image part with relationship ID rId7 was not found in the file.The image part with relationship ID rId8 was not found in the file.The image part with relationship ID rId9 was not found in the file.The image part with relationship ID rId10 was not found in the file.

A R=0.6m B v B 0.6m k =350 N/m

AR=0.6m

B vB

0.6m

k=350 N/m

y

34

67

A R=0.6m B v B 0.6m k =350 N/m

Conservation-Energy Example ref ~Meriam & Kraige 3/17 Solution: •  ALL Forces are either Conservative or

Non-working constraints, therefore Cons. Of Energy applies!

•  Pulling together all components & isolating vB NFs

mgFBD

!ETOT = !T + !Vg + !Ve = 0

!TAB =12m(vB

2 " vA2 ) = 1

2m(vB

2 " 0)

!VABg = mg(yB " yA ) = mg(0 " R)

!VABe =12k (lB " l0 )

2 " (lA " l0 )2{ }

vB = 2gR + km

R2 ! ( 2R ! R)2{ }•  Incorporating numerical values of all terms

!

vB = 2 * 9.81(m /s2)(0.6m) +350N /m

3kg(0.6m)2 " ( 2 *0.6m " 0.6m)2{ } = 6.82 m /s

AR=0.6m

B vB

0.6m

k=350 N/m

y

Work/ Energy T1 + V1 + WNC = T2 = V2

Given: k, m, Θ block released from rest with spring compressed δ

T1 = 0 V1= 1/2 k δ2

WNC = -Fd

where N = mg cosΘ, F = µkN while sliding

T2=0 V2 = 1/2k(d - δ)2 - mgd sin Θ

Find: distance travelled to stop, d

k

m

Θ

35

Given: k, m, R, Θ, I released from rest, no slip, no initial deflection in spring

Find: speed after 1/2 revolution

Θ

m,R

T1 = 0 V1 = 0 WNC = 0

T2 = 1/2 m v22 + 1/2 I ω2

2, V2 = 1/2 k (π R)2 - mg π R sin Θ need also v2 = R ω2

(no slip condition) 0 = 1/2 m v2

2 + 1/2 I (v2/ R)2 + 1/2 k (π R)2 - mg π R sin Θ

πR k

Kinetics Summary •  Three general solution approaches for establishing the governing

equations of motion (EOM) => Which one to use? i) Newton’s Laws ii) Work- Energy & Conservation of Energy iii) Impulse - Momentum & Conservation of Momentum

MP! = ICG " + reff maCG

UA!B = matdssA

s B

" = mvdvvA

vB

" = 12m vB

2 ! vA2( ) = #TA!B

–  Typical forces •  Springs •  Friction •  Gravitation

F f = µs / k NF = k s - s0( )

F =mg

UNC = !T + !Vg + !Ve = !ETOT

I = FRdt! = dL! = "L

F! = m aCG

36

Given: Box placed on conveyor with zero initial velocity.

µk

VBelt

Find: Time during which slip occurs

Find: Time during which slip occurs

VB= 10 ft/s

µ = 0.333

ΣF↑ = N - mg = 0

ΣF→ = µN = ma µmg = ma a= µg

v = µgt

slip stops where v = vB

so t = vB/ µg

mg ma

N F = µN

during slip

once slip stops F = 0 VBox

VB

t

37

Given: ball of radius r released from rest on incline, no slip, Θ = 30°

Find: acceleration

r

Θ

ΣF = N - mg cos 30 = 0

ΣF = mg sin 30 - F = ma

ΣMG = rF = 2/5 mr2α

→

if no slip, a = rα so mg sin 30 - 2/5 mrα = mrα

so α = (5/7)(g/r)(1/2); a = (5g/7)(1/2) = 5g/14

v = a t = 5gt 14

also: N = mg → Fmax = µsN = √3 2

√3 µ mg 2

F = mrα = 2 mg 5 7

F < Fmax requires < 1 7

√3 µ 2

or µ > 2 7 √3

N

F

30°

mg G c G

ma c

Iα

38

Given: given v , r

Find: Φ; tension

ma = m r ω2

ma = m v2/r

rv

Φ

Φ

ma

mg

T

ΣF ↑ = T cos Φ - mg = 0 ΣF → = T sin Φ = m v2/r

or ΣF = mg sin Φ = ma cos Φ

so tan Φ = = ag

v2 gr

Given: m1 released from rest, strikes m2 Find: max spring compression m1

m2 k

State 1 as shown State 2

vp

d

39

State 3

State 4

vʹ′p vm

δ

Vibrations

x

k c m

kx

cx .

mx ..

Σ F = - kx - cx = mx . ..

Or mx + cx + kx = 0 .. .

x + cx / m + kx / m = 0 .. .

general x + 2 ζ ω x + ω2 x = 0 ..

where w = natural freq. ζ = damping factor

ζ < 1 underdamped

ζ > 1 overdamped

ζ = 1 critically damped

40

!F

Solution:!•  Rectilinear motion! Constrained vertically so

align axes accordingly!

Particle Kinetics: Cartesian Example ref ~Meriam & Kraige 3/5!Given:!•  A collar of mass m slides vertically on a shaft with

kinetic coefficient of friction, µk.!•  Applied force F is constant but its direction varies as!! ! !θ = k t k=> constant

•  Collar starts from rest @ θ=0°!Find:!(A)! Magnitude of F which results in collar

coming to rest at θ=90°. !

•  FBD of collar in general position (working over a motion interval here)!

Fx! = Fsin" # N = 0 !

Fy! = #F cos" +mg +Nµk =may

•  Newtonʼs Laws!! N = Fsin"

& Fm

(sin"µk # cos") + g = ay

x

y

θ!F!

µkN!

mg!

N

!F

θ!F!

µkN!

mg!

x

y

N

Particle Kinetics: Cartesian Example ref ~Meriam & Kraige 3/5!

!

dvy0

vyf" = {Fm(sin(kt)µk # cos(kt)) + g}dt

0

t"

!

vyf =Fmk

("cos(kt)µk " sin(kt)) + gt# $ %

& ' ( 0

t

•  Donʼt know F but DO know itʼs constant!!•  Angle-time relation (θ = k t) cleans up!

–  Kinematic relationship variables & !–  Proper (Pos,Vel )BCʼs for integration!

•  Starts: vyi=0, θ=0 => t=0!•  Ends: vyf=0, θ=π/2 => t=π/(2k)!

–  Start w/ general upper limits!

!

Fm(sin"µk # cos") + g = ay =

dvydt

Solution (Contʼd):

!

vyf =Fmk

{[1" cos(kt)]µk " sin(kt)}+ gt

41

!F

θ!F!

µkN!

mg!

x

y

N


!F =" mg

2(1 # µk )

Solution (continued):!•  Now using the final BC for t=π/2k, vyf=0

Langiappe:!•  What are:!

–  The collarʼs vertical displacement as a function of time?!–  The collarʼs total distance traveled?

!

vyf =Fmk


!

0 =Fmk{[1" cos(# /2)]µk " sin(# /2)}+

g#2k

!

0 =Fmk{µk "1}+

g#2k

!F

θ!F!

µkN!

mg!

x

y

N


Langiappe (continued):!•  Returning to the expression for vy=f(t)

!

vyf =dydt

=Fmk


!

dy0

y f" =Fmk{[1# cos(kt)]µk # sin(kt)}+ gt

$ % &

' ( ) dt

0

t"

!

y f =Fmk

{[t " sin(kt)]µk + cos(kt)}+12gt 2

#

$ % &

' ( 0

t

!

y f =Fmk{[t " sin(kt)]µk + cos(kt) -1}+

12gt 2

!

y f =Fmk{[ "2k

# sin("2)]µk + cos("

2) -1}+

12g "2$

% &

'

( ) 2

!

y f =Fmk{[ "2k

#1]µk #1}+12g "2$

% &

'

( ) 2

42

!

Particle Kinetics: Path Coord Example ref ~Meriam & Kraige 3/7!

Given:!•  A box of mass m (particle) is released

from rest @ top of a smooth circular track.!

Find:!(A) !Normal force N as a function of

position θ along the circular track.!!

Solution:!•  FBD of box in general position - working over

motion interval A-B, then instantaneous @ B!!–  Track smooth ==> µ=0, no friction!!

Fn! = N "mgsin# = manFt! = mgcos# = mat

•  Attach normal-tangential coordinate axes!

•  Newtonʼs Laws for general θ!

A

R

r

B

!

(B) !The angular velocity (ω) of the pulley B such that the boxes donʼt slide onto the conveyor belt !

N

n

t

mg

!

N!

ω!

•  Must use kinematics to resolve unknowns!!


Solution (contʼd):!•  Kinematics: Circular track-> path coord!

Fn! = N "mgsin# = man = mv2

R

Ft! = mgcos# = mat = m v•

= m s••

•  Using tangential direction to resolve velocity by integrating alternate form => vdv=atds!

•  Now using the normal direction!

A

R

r

B

!

N

n

t

mg

!at = s••= gcos! s = R! " ds = Rd!

12v2 = gRsin! 0

!= gRsin!

vdv0

v

! = gRcos "d"0

"

!

v2 = 2gRsin!

! N = m v2

R+mgsin"

= 2mgsin" +mgsin"

(A) N = 3mgsin!

N!

ω!

s = R θ!

43


Solution (contʼd):!•  Belt Kinematics:!

–  From previous page, the box speed as it reaches B!

–  Belt must move at the same speed to avoid slip, i.e. relative velocity between belt/box is zero!

A

R

r

B

!

N

!

s=R!

v = 2gRsin!!=

"2= 2gR

(B) ! =2Rgr

r!

B Bv

vbelt

vB = vbelt2gR =! r

FE Review dynamics - Louisiana State University1 Fundamentals of Engineering Review for Dynamics...

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Transcript of FE Review dynamics - Louisiana State University1 Fundamentals of Engineering Review for Dynamics...