introduction engineering mechanics statics dynamics...

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ME 3133 Dynamics •  Topics Covered

–  Introduction to Dynamics –  Rectilinear Kinematics (1D)

•  Position / Displacement / Distance Traveled •  Speed / Velocity •  Acceleration { a(t), a(s), a(v) } •  Additional considerations

– Continuous vs Piecewise Continuous –  Average vs Instantaneous –  Vector vs Scalar (soon)

•  Assignments: –  Read: Meriam & Kraige: Chp 1, Chp 2.1-2.2 –  Practice Problems: 2.26, 29, 54 [57] –  Moodle Quiz:

•  4-Digit Class ID (e.g. 5907) •  Permission to Post Grades •  Pledge of Academic Honesty •  Teaching technique likes/dislikes

–  Class representatives

01/14/15 Lecture #01

Introduction

Engineering Mechanics

Statics Dynamics Strengths

€

F ii=1

N

∑ = 0 & M jj=1

N

∑ = 0

€

F ii=1

N

∑ ≠ 0 & M jj=1

N

∑ ≠ 0

2

Dynamics Problem Decomposition

Dynamics

KineticsKinematics

The image cannot be displayed.

m

θ=5°

Pm

Lµ = 0.4k

Success in Dynamics:

•  Dynamics is about Coordinate Systems: –  Much of dynamics is essentially looking at relative motion of

coordinate systems => some fixed/inertial, some moving –  Begin thinking in terms of VECTORS!

•  Review Vector Operations & Properties (Appedix C in text) •  Think in terms of VECTOR Properties (has a physical significance)

–  Can simplify the strategy for solving many problems –  Coordinate system independent!

–  Answers can be expressed in terms of any coordinate system •  Prudent choice of coordinate systems can simplify problem

formulation & solution strategies (both kinematics & kinetics)

•  Dynamics is WORD PROBLEMS:

3

Introduction

Newton�s Laws

Work - Energy

Impulse - Momentum

Engineering Mechanics

Statics Dynamics

KineticsKinematics

Dynamic Studies

Dynamics:(Kinematics & Kinetics)

Particles

•  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

θ

A = (x,y)

B

x

y

4

Particle vs Rigid Body

θ

β

v

r

A

•  Choice often dictated by: –  Scale of which problem is viewed & –  Precision required in desired measurement

Particle Rigid Body

Getting Started => Particle Kinematics

•  Rectilinear Motion – Movement along a straight line in 1-2 or 3D

•  1 Degree of Freedom (DOF)* - s(t)

•  Curvilinear Motion – Movement of particle along an arbitrary path

through space

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Particle Kinematics

•  Rectilinear Motion – Movement along a straight line in 1-2 or 3D

•  1 Degree of Freedom (DOF)* - s(t)

Geometric Parametric in tf(x,y) = a x + by + c = 0

y = (a x + c)/b = 0x(t) = a1 t + a0y(t) = b1 t + b0s(t) = s1 t + s0

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

•  Typical Functions ?? –  Polynomial, Trigonometric, Logarithmic, Exponential

v a

6

Rectilinear Kinematic Relationships:

•  Two independent kinematic constraints w/ four scalar variables (s, v, a, t )

€

(1) v =dsdt

•  Another useful form

€

(2) a =dvdt

€

v =dsdt

⇒ dt =dsv

a =dvdt

⇒ dt =dva

⎫

⎬ ⎪

⎭ ⎪

€

dsv

=dva

or(2*) v dv = a ds

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

+v

-v

+s

-s

t

+a

-a

smax

v min

a = 0v = 0

•  Alternate form

DIFFERENTIATE

INTEGRATE

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

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

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Works for ROTATION (1D) too!

•  Angular Position - θ(t)

€

(1) ω =dθdt

= θ•

•  Angular speed - ω(t)

•  Ang. Acceleration - α(t)

€

(2) α =dωdt

=ω•

=d2θdt 2 =θ

••

€

(2*) ω dω =α dθ•  Alternate form

DIFFERENTIATE

INTEGRATE

B

θ, ω, α

d

O

Particle Kinematics Summary & Looking Ahead!

•  Rectilinear Kinematic Relationships – Scalar Parameters { s, v, a, t } or { θ, ω, α, t } – Position / Displacement / Distance Traveled – Speed / Velocity – Acceleration {a(t),a(s),a(v)} or {α(t),α(θ),α(ω)} – Additional considerations

•  Continuous vs Piecewise Continuous •  Average vs Instantaneous •  Vector vs Scalar

*  LOOKING AHEAD: The basic scalar relations developed today will apply to the more general curvilinear path discussions later

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Rectilinear Kinematics: Acceleration a function of velocity – a(v)

€

t f − ti( )0

t f=

dv−kv 28

v f∫

Given: •  A ship 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

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

€

⇒ v f = v(t f ) =8

8kt f +1 (knots)

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 ?

€

a(v) =dvdt⇒ dt =

dva(v)

⇒ dtti

t f∫ =dv−kv 2vi

v f∫


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 BC�s helps resolve the unknown constant k

€

t f =10 (min)

60 (min/hr)=

16

hr , v f = 4 knots

€

⇒ k =34

1nautical miles⎛

⎝ ⎜

⎞

⎠ ⎟

€

⇒ v(1/6) =8

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

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

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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=ds/dt

€

v(t) =8

6t +1

=dsdt

€

ds0

s f∫ =8

6t +1

dt0

t f∫

€

sf − 0 =43

ln 6t +1( ) 0

t f =43

ln 6t +1( )− ln(1)( )

€

sf = s(t) =4

3

ln 6t +1( )

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

€

s(1/6) =4

3

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

3

ln(2) (nautical miles)

the boat�s position can be found by integration

€

⇒ dssi

s f∫ =8

6t +1

dtti

t f∫


and as was seen before

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

€

sf (t =1/6)⇒ sf (v = 4) =4

3

ln(2) (nautical miles)

€

a(v)ds = vdv⇒ dssi

s f∫ =vdva(v)vi

v f∫

€

sf =−43

dvv8

4∫ = −4

3lnv 8

4=−43

ln4 − ln8( )=43

ln 84

and the boat�s displacement (position?) can again be found by integration

€

sf − si( )0

s f=

vdv−3/4v2( )8

4∫

10

Additional Considerations:

•  Continuous vs Piecewise Continuous –  Two (2) or more distinct motion intervals –  Example: automobile acceleration profile

•  Foot on accelerator •  Coasting •  Foot on the brake

–  Match physical parameters (boundary conditions- position, speed) between time interval boundaries

t

+a

-a

Foot on

Accelerator

Foot on

Brake

Coasting

Rectilinear Kinematics: Acceleration - a(t) => Example

Given: –  A particle moves along the x axis –  t=0, x=0, vx=50 m/s , ac=0 constant –  t=4, vx=50 m/s , ac=-10 m/s2 constant

Find: –  (A) x|t=8 & 12 & v|t=8 & 12 –  (B) x max –  (C) stot traveled in 12 seconds

Solution: –  Motion has two constant acceleration segments –  Individually resolve kinematics of each segment –  Match kinematic parameters at common boundary (t=4) –  Results across intervals used in the total solution

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€

x(t) =1

2

ac t f − ti( )2

+ vi t f − ti( )− xi = 50t f

€

v(t) = ac t f − ti( )+ vi = 50m /s ⇐ constant

–  At the end of this interval (e.g. start of next interval)

€

v(4) ≡ v(0) = 50m /s

€

x(4) = 50m /s * 4s( )= 200m

€

v(t) = ac t f − ti( )+ vi = −10 t − 4( )+ 50

(2) For the subsequent interval t=[4,--]–  t=4, position & speed from above, ac=-10 m/s2 constant

€

x(t) =1

2

ac t f − ti( )2

+ vi t f − ti( )− xi = −5 t f − 4( )2

+ 50 t f − 4( )− 200

•  Now, cleaning up the clutter & looking at it graphically

Check for consistent Units!

(1) Over the interval t=[0,4] –  t=0, x=0, vx=50 m/s , ac=0 constant


Interval t=[0,4]

€

v(t) = −10 t − 4( )+ 50 (m /s)€

x(t) = 50t (m)

Interval t=[4,--]€

a(t) = 0 m /s2

€

x(t) = −5 t − 4( )2+ 50 t − 4( )− 200 (m)

!  Now on to answer the specific questions @ t=8, 12 seconds

€

v(t) = 50 m /s

€

a(t) = −10 m /s2The image cannot be displayed. Your computer may not have enough memory to open the image, or the image may have been corrupted. Restart your computer, and then open the file again. If the red x still appears, you may have to delete the image and then insert it again.

t (sec)

+v

-v

02 6 8

10 12

4

-50

+50(m/s)

The image cannot be displayed. Your computer may not have enough memory to open the image, or the image may have been corrupted. Restart your computer, and then open the file again. If the red x still appears, you may have to delete the image and then insert it again.

t (sec)

+x

02 6 8 10 124

400 (m)

200

12


(A) x|t=8,12 & v|t=8,12

€

v(t) = −10 t − 4( )+ 50

€

x(t) =−5 t f − 4( )2

+ 50 t f − 4( )−200

€

⇒ v(8) = −10(4)+ 50 =10m /s

⇒ v(12) = −10(8)+ 50 = −30m /s

€

⇒ x(8) = −5 4( )2 + 50 4( )−200=320m

⇒ x(12) = −5 8( )2 + 50 8( )−200=280m


t (sec)

+x

02 6 8 10 124

400 (m)

200x(4)=200

x(8)=320

x(12)=280


(B) x max =?

€

v(t) = −10 t − 4( )+ 50=0⇒ t = 9 seconds

€

x(9) = −5 9− 4( )2 + 50 9− 4( )−200=325m

@ v=0 OR at interval boundaries!


t (sec)

+x

02 6 8 10 124

400 (m)

200x(4)=200

x(8)=320

x(12)=280

x (9)=325max


t (sec)

+v

-v

02 6 8

10 12

4

-50

+50(m/s)

v(9)=0!

13


(Area enclosed v-t curve)

€

stot = (4)50

+129− 4( ) 50( )

+1212 − 9( ) −30( )

= 200+125+ 45 = 370m


t (sec)

+v

-v

02 6 8

10 12

4

-50

+50(m/s)

v(9)=0!

(C) Distance Traveled


t (sec)

+x

02 6 8 10 124

400 (m)

200x(4)=200

x(8)=320

x(12)=280

x (9)=325max

(Alternative- x(t) graph)

€

stot = (325− 0)− (280− 325) = 325 + 45 = 370m

ME 3133 Dynamics •  Topics Covered

–  Introduction to Dynamics –  Rectilinear Kinematics (1D)

•  Position / Displacement / Distance Traveled •  Speed / Velocity •  Acceleration { a(t), a(s), a(v) } •  Additional considerations

– Continuous vs Piecewise Continuous –  Average vs Instantaneous –  Vector vs Scalar (soon)

•  Assignments: –  Read: Meriam & Kraige: Chp 1, Chp 2.1-2.2 –  Practice Problems: 2.26, 29, 54 [57] –  Moodle Quiz:

•  4-Digit Class ID (e.g. 5907) •  Permission to Post Grades •  Pledge of Academic Honesty •  Teaching technique likes/dislikes

–  Class representatives

01/14/15 Lecture #01

introduction engineering mechanics statics dynamics...

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