FE Exam: Dynamics review

D. A. LynSchool of Civil Engineering

3 March 2013

Preliminaries

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

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

addition and subtraction, 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, size unimportant) 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 (before and after)

– Work (distances involved) and energy (velocities involved)• working with scalars (usually easier) (before and after)

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 (in magnitude and direction)

acceleration, (initial conditions needed to determine integration constants)

( ) , ( ) = ( , , ), ( , , ), ( , , )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 a

20 0

1( ) ( 0) , ( ) ( 0) ( 0)

2t 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 q=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 0Given : 0, , 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;

L

v 0

x

y

tend=3.1 s and L=269.2 ft

Kinetics of a particle• Linear momentum, L=mu (mass, m, i.e., measure of inertia)• Newton’s 2nd law:• Forces determined from free body diagram (as in statics)– Types of forces: gravitational, frictional, spring, 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

q=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

• kinetics problem (force balance in s dir’n):

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

– Tangential (t) and normal (n) coordinates

ρ is the radius of curvature of particle path at the particle position (advantage: vn = 0, zero normal comp.)– Radial (r) and transverse () or polar coordinates

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

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

2( ) , ( ) /t n tt v t v v v e a e e

22 2

2

, 0, , [ , ]

0, ,

( ) , ( )

r n t

r

r v r rv

r r r v r r rr

a t r a t r

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=

(a is the angular acceleration)

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

acceleration, a, at the instant shown if at the position shown the sphere of mass, m=10 kg, has a tangential velocity of v0=4 m/s. R = 0.6 m, and 0=30.

• Choose a radial-transverse 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

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

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, Fx(x2-x1)

– Gravitational force, -W(y2-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 q=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• Motion constraints: sB=sA/2 (and DyA=-2DyBsinq), 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 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

travel before its speed is 4 m/s?

Rigid body (plane) dynamics• rigid body: distance between any two arbitrary points on

body remain same• types of motion

– pure translation (can be treated as a particle)– pure rotation about a fixed axis

convenient to characterize in terms of angular frequency, w, and angular acceleration, a

– general plane motion (combined translation and rotation about a fixed axis)

• rotation is distinguishing feature of rigid-body dynamics– moment equation: product of mass moment of inertia and

angular acceleration analogous to ma

• effective forces and moments and mass centers in force and moment equations

Rigid body (plane) kinematics• general plane motion of a

rigid body is sum of a translation of a point and rotation about that point.

/

/

/

/ /

( )

( )

2/ / /

( )

=

=

B A t

B A n

B A t

B A B A A B A

B A B A A B A B A

v

a

a

v v v v k r

a a a a k r r

• choose point A to write equation for the motion of point B• direction of vB/A, (aB/A)t, and (aB/A)n

crucial to solution• directions of unknown w and a

may be assumed• velocity problem: knowing vA and

direction of vB and geometry, find magnitude of vB and w.

• acceleration problem: knowing aA and direction of aB and geometry, find magnitude of aB and a

for plane problems, the ‘direction’ of and especially is known, usually key to solution.

/B Ak r/B Ar

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: assuming clockwise rotation about A (may not be correct)

/

2 2/ /

222

cos , sin / ( sin ) 3.85/s, 1.16 m/s

0 sin cos cossin cos sin sin

B A B A B A

B A A B

B A B A B A B A

A A ABx A

By B

V V L

V L V L V L V

a a L L

a a Va a L L

L L L

a a

v v k r

a a k r r

2

2 2cos sin 1 cos 11.7 m/ssin

AA

VL L a

L

V , aA , A

A

BV ?B

a ?B

x

y

aA

aB

2L

rB/A

VA

VB

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

− May be more convenient to deal with effective forces and effective moments

• 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

= and G G Gm F a L M H

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, a, 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=Ra)

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 2

0 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

+