FE Exam: Dynamics review D. A. Lyn School of Civil Engineering 21 February 2012.

FE Exam: Dynamics review

D. A. Lyn

School of Civil Engineering

21 February 2012

Preliminaries

• Units (relevant quantities: g, displacement, velocity, acceleration, energy, momentum, etc.)

• Notation (dot, vector)• Vectors (components and directions/signs,

addition (graphical), dot and cross products, vector polygons)

• Coordinate systems (Cartesian and curvilinear, fixed and moving or relative, unit vectors)

• Statics (free body diagram)

Classification of dynamics and problems• Kinematics: description of motion without reference to forces

– Particle (no rotation about itself) and rigid-body– Coordinate systems (Cartesian, curvilinear, rotation)– Constraints on motion

• Kinetics: inclusion of forces (mass, or momentum or energy)– Types of forces: conservative (gravitational, spring, elastic

collisions) and non-conservative (friction, inelastic collisions)– Newton’s 2nd law: linear and angular momentum

• Use of free body diagram to deal with external forces– Particles and rigid body (system of particles)– Impulse (time involved) and momentum

• still working with vectors– Work (distances involved) and energy (velocities involved)

• working with scalars (usually easier)

Particle kinematics• General relations between displacement

(r), velocity (u), and acceleration (a)

• Given a formula for (or graph of) r as function of t, take derivatives to find u and a– Given a formula for u or a as function of t,

integrate to find r or uSpecial case: constant acceleration,

( ) , ( ) = ( , , ), ( , , ), ( , , )d d

t t x y z x y z x y zdt dt

r u

u r a u r r u a

0a a2

0 01

( ) ( 0) , ( ) ( 0) ( 0)2

t t t t t t t t

u u a r r u a

Sample problems• The position of a particle moving horizontally is described by

, with s in m and t in s. At t = 2 s, what

is its acceleration? Soln: Take derivatives of s with respect to t, and evaluate at

t=2s ( ) so a(t=2s) = 4 m/s2.

• Projectile problem: A projectile is launched with an initial speed of v0=100 ft/s at =30° to the horizontal, what is the horizontal distance, L, covered by the projectile when it lands again?

Soln: constant acceleration (only gravitational acceleration involved) problem, so apply formulae in two directions

2( ) 2 8 3s t t t

( ) ( ) 4 8, ( ) ( ) 4u t s t t a t s t

20 0

20 0

/ 2

/ 2

end x end x end

end y end y end

x x v t a t

y y v t a t

0 00, , 100cos30 , 100sin30x y x ya a g v v wish to find L=xend-x0, for yend-y0=0, so we solveL=v0xt end and 0=v0ytend-g(tend

2/2) for tend and L;tend=3.1 s and L=269.2 ft

L

v 0

x

y

Kinetics of a particle• Linear momentum, L=mu (appearance of mass, i.e., inertia)

• Newton’s 2nd law:• Forces determined from free body diagram (as in statics)

– Types of forces: gravitational, frictional, external• Angular momentum (about a point O) ,• Newton’s 2nd law:

• Impulse (used in impact and collision problems),– momentum conservation:

– mini-problem: A golf ball of mass 50-g is hit with a club. If the initial velocity of the ball is 20 m/s, what is the impulse imparted to the ball? If the contact duration was 0.05 s, what was the average force on the ball?

m F a L

0 m H r u

0 0M H2

1

1 2 t

t

dt Imp F2

1

2 1 1 2 t

t

dt L L Imp F

1 2 1 2 2

1 20

0 Imp (0.05 kg)(20 m/s) = 1 Ns

Imp 1 Ns 1 Ns / 0.05 s 20 Nt

avg avg

L L mv

F dt F t F

Problem: kinetics of a particle (truck)• A truck of weight W = 4000 lbf moves down a

=10° incline at an initial speed of u0 = 20 ft/s. A constant braking force of Fbrk=1200 lbf is experienced by the truck from a time, t = 0. What is the distance covered by the truck before it stops from the time that the braking force is applied?

• kinematics problem:

2sin , sin 4.1 ft/sbrkbrk net s s

FF W F ma W mg a g

W

0

2 20 0

( ) ( 0) / 4.9 s

( ) ( 0) ( / 2) / 2 48.8 ft

end s end end s

end end s end s

u t t u t a t t u a

s t t s t u t a t u a

Notes: forces involved – kinetics problem, rectilinear (straight-line) motion: determine net force on truck in direction of motion, apply Newton’s 2nd law to evaluate distance coveredFrom free body diagram, sum of forces in direction of motion,

u 0

WF b rk

W s in

m a ss

Curvilinear coordinates and motion• Plane motion (motion on a surface, i.e., in only two

dimensions)– Tangential (t) and normal (n) coordinates

where is the radius of curvature of particle path– Radial (r) and transverse () or polar coordinates

– Special case: pure circular motion at an angular frequency, t

2( ) , ( ) ( ) ( 2 )r rt r r t r r r r v e e a e e

2

( ) , ( )t n tv

t v t v

v e a e e

22 2

, 0, , ,

0, ,

t r nr v r rv

r r r v r r rr

e e e e

xx 1

y

y 1

r

e re n

e t

e p a r tic le

p a th

p a r t ic lea t tim e t

r

v =r

a rr=

2

ar

r=

.( is the angular acceleration)

Particle kinetics problem• Find the tension, T, in the string and the

angular acceleration, , if at the position shown the a sphere of mass, m=10 kg, has a tangential velocity of v0=4 m/s.

• Choose a polar coordinate system, perform free body analysis to determine sum of forces, and set equal to ma.

202

0 0 0

20 0

dir'n: - cos / cos 352 N

dir'n: sin sin / 8.2/s

rv

r T W ma mv R T m gR

W ma mR W mR

W

T

r

R0

v0

m

Energy and work• Work of a force,F, resulting in a change in position from state 1

to state 2:

– Constant force in rectilinear motion, Fxx2-x1)

– Gravitational force, -Wy2-y1), y>0 upwards

– Spring force, -k(x22-x1

2)/2, (x2<x1, returning to undeformed state)

• Kinetic energy,

• Relation between work and kinetic energy:

• for conservative forces (such as gravitational and spring forces, but not frictional forces), a potential energy function, V, can be defined such that – Gravitational force: V = Wy, spring force, V=kx2/2

• For conservative forces, an equation for conservation of energy can be expressed as or

2

1 21

U d F r

2 / 2T mv1 2 2 1U T T

1 2 1 2U V V

1 2 2 1V V T T 1 1 2 2T V T V

A problem solved using energy principles• A 2-kg block (A) rests on a frictionless

plane inclined at an angle =30°. It is attached by an inextensible cable to a 3-kg block (B) and to a fixed support. Assume pulleys are frictionless and weightless. If initially both blocks are stationary, how far will the 2-kg block travel before its speed is 4 m/s?

• Motion constraints: sB=sA/2 (and yA=-2yBsin), and vB=vA/2• Frictionless system conservative gravitational forces only, only

distances and speeds explicitly involved apply energy equation

22 2 21 2Initially, 0; at end, / 2 / 2 / 2 1 / /A A B A B A

A BT T mv mv m v m m v v

1 2 2 1 2 1 21 2

22

22

11 1 1

2 2sin

1 2.24 m 0 (2 1 / 2 sin

A A B B A A B B A A B B

A A B B B B BA A A A

A A A A A

A B BA

B A A A

V V T T T W y W y W y W y W y W y U

m v m v W y WW y W y

m v W y W

v m vy

g W W m v

Block A rises)

/ sin 2 4.48 mA As y y

3 kg

2 kg

= 3 0

S ta te 1

B

A

3 kg3 kg

2 k g

2 kg

= 3 0 = 3 0

S ta te 1 S ta te 2

s yB = B

y A

v

B B

AA s

A

Constrained motion, reference frames, relative motion

• Constrained-motion problems – choice of reference frames: relative motion (in a plane)

• Choice of reference frames – motion relative to a point A in a moving reference frame

– For plane motion, note direction of components, e.g., rB/A is perpendicular to rB/A, etc.

– For points on the same rigid body,

/ / /= , = , = B A B A B A B A B A B A r r r u u u a a a / / / / / / / /, 2B A B A B A B A B A B A B A B A u Ω r r a Ω r Ω Ω×r r Ω r

/ /0B A B A r r

Problem: Kinematics of rigid body example

• The end A of rod AB of length L = 0.6 m moves at velocity VA = 2 m/s and acceleration, aA = 0.2 m/s2, both to the left, at the instant shown, when = 60°. What is the velocity, VB ,and acceleration, aB , of end B at the same instant?

Pure kinematics problem:

/

2/ /

222

2

cos , sin cot 1.16 m/s

0 sin cos cossin cos sin sin

cos sin

B A B A B A

B A B A

B A B A B A B A

A A ABx A

By B

V V L

V L V L V V

a a L L

a a Va a L L

L L L

a a L L

u u Ω r

a a Ω r Ω Ω×r

2

21 cos 11.7 m/ssin

AA

Va

L

V , aA , A

A

BV ?B

a ?B

x

y

aA

aB

L

2L

rB/A

VA

VB

L

rB/A

Kinetics of a system of particles (or rigid body)

• For a system of particles (or a rigid body), analysis is performed in terms of the mass center, G, located at radial vector, rG, and total mass m

• Equations of motions:

where aG is the acceleration of the mass center, and HG is the angular momentum about the mass center− For a system with no external forces or moment acting, then linear

momentum, L, and angular momentum, H, is conserved, i.e., remains constant

• For a system of particles (or a rigid body), and where the mass moment of inertia I is defined by (Standard formulae for I = mk2, where k is the radius of gyration, for standard bodies are listed in tables; be careful about which axis I is defined, whether centroidal axis or not, remember parallel axis theorem)

or G i i Gm m m dm r r r r

eff eff= and G G Gm F a L F M H M

G G GI I H ω α 2 2 or i iI r m I r dm

G GIH ω

Problem: two-particle system• A particle A of mass m and and a particle B, of

mass 2m are connected by rigid massless rod of length R. If mass B is suddenly given a vertical velocity v perpendicular to the connecting rod, determine the location of the mass center, the velocity of the mass center, the angular momentum, and the angular velocity of the system soon after the motion begins.

/ /

2 2/ /

2 3 2

32 2

3 2 3 3

2 1 2 0 2

3 3 3

A B G A A B B G A B G A B A

G A B G A A B B G A B G B

G A G A A B G B B

G G G A A G B B G

m m m m m m

m m m m m m v

m m m R m R v mvR

I I m r m r

r r r r r r r r r r

L u u u u u u u u j

H r u r u k k k

H ω ω k2 2

2

020

2 1 22

3 3 3

2 2

3 3

m R m R mR

vmv R mR

R

k k

k k

v

x

y j

r A r B

r rB A-

A B

Gm 2 m

(2 /3 )v j

A BGm 2 m

Problem: rigid-body kinetics• What is the angular acceleration, , of the

60-kg (cylindrical) pulley of radius R = 0.2 m and the tension in the cable if a 30-kg block is attached to the end of the cable?

• Analysis of block− Kinematic constraint (ablock=R)

OR

m = 3 0 -k g

m p u lle y= 6 0 -k g

( )

y y yF ma T W ma m R

T m g R

• Analysis of pulley

2 20 0 0 pulley pulley where / 2 / 2M I I m R TR m R

2pulley pulley

2 1Solve for and : 147 N, 24.5

1 2 / s

mg TT T

m m m R

R

T T

W = m g

a Ry=

y

+

Dynamics Outline and Problem - Solutions

as Provided by Kaplan

Copyright Kaplan AEC Education, 2008

Dynamics Outline Overview

DYNAMICS, p. 205

KINEMATICS OF A PARTICLE, p. 206• Relating Distance, Velocity and the Tangential

Component of Acceleration• Constant Tangential Acceleration• Rectilinear Motion• Rectangular Cartesian Coordinates• Circular Cylindrical Coordinates• Circular Path


Dynamics Outline Overview Continued

RIGID BODY KINEMATICS, p. 203

• The Constraint of Rigidity

• The Angular Velocity Vector

• Instantaneous Center of Zero Velocity

• Accelerations in Rigid Bodies



NEWTON’S LAWS OF MOTION, p. 210

• Applications to a Particle

• Systems of Particles

• Linear Momentum and Center of Mass

• Impulse and Momentum

• Moments of Force and Momentum



WORK AND KINETIC ENERGY, p. 219

• A Single Particle

• Work of a Constant Force

• Distance-Dependent Central Force



KINETICS OF RIGID BODIES, p. 225

• Moment Relationships for Planar Motion

• Work and Kinetic Energy


Kinematics of Particles—1D Motion


Solution


Kinematics of Particles—1D Motion


Solution


Solution (continued)


2D Motion—Rectangular Cartesian Coordinates


Solution


2D Motion—Plane Polar Coordinates


Solution


Instantaneous Center of Zero Velocity


Solution


Evaluation of Accelerations


Solution


Newton’s 2nd Law


Solution


Newton’s 2nd Law


Solution


Work & Kinetic Energy


Solution


Moments of Force & Momentum


Solution

FE Exam: Dynamics review D. A. Lyn School of Civil Engineering 21 February 2012.

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Transcript of FE Exam: Dynamics review D. A. Lyn School of Civil Engineering 21 February 2012.