Summary Dynamics(1)

8/13/2019 Summary Dynamics(1)

1/14

Particle Kinematics

r i j k ( ) ( ) ( ) ( )t x t y t z t r(t)

r(t+dt)

v(t) dt

i

j

path of particle

kO

( ) ( ) ( ) ( )

( ) ( ) ( )

x y z

x y z

t v t v t v t

d dx dy dz x y z

dt dt dt dt

dx dy dz v t v t v t

dt dt dt

v i j k

i j k i j k

Direction of velocity vector is parallel to path

Magnitude of velocity vector is distance traveled / time

Inertial framenon accelerating, non rotating reference frame

Particlepoint mass at some position in space

Position Vector

Velocity Vector

Acceleration Vector

2 2 2

2 2 2

( ) ( ) ( ) ( )

( ) ( ) ( )

yx zx y z x y z

yx zx y z

dvdv dvdt a t a t a t v v v

dt dt dt dt

dvdv dvd x d y d z a t a t a t

dt dt dt dt dt dt

a i j k i j k i j k


2/14

Particle Kinematics

Simple Harmonic Motion

x(t)

k,L0

m

i

j

02

2

sin(2 / ) cos(2 / ) sin(2 / )

2 2 4

X X t T V t T A t T

X V XV A

T T T

r i v i a i

Circular Motion at const speed

2

2 2

cos sin

sin cos

(cos sin )

R

R V

VR R

R

r i j

v i j t

a i j n n

Straight line motion with constant acceleration

2

0 0 0

1

2X V t at V at a

r i v i a i

t s R V R

t

R

i

j

Rcos

Rsin

tsin

cos

n


3/14

Summary

General circular motion

Arbitrary path

R

i

j

Rcos

Rsin

tsin

cos

n

2

22

cos sin

sin cos

( sin cos ) (cos sin )

R

R V

R R

dV VR R

dt R

r i j

v i j t

a i j i j

t n t n

2 2/ / /

/

d dt d dt d dt

s R V ds dt R

2

V

dV V

dt R

v t

a t n

t

R

t

n2 2

2 2

3/222

1

d y dydx d x

d dd d

Rdydx

d d

( ) ( )x y r i j


4/14

Summary

Polar Coords

r

i

j

e

er

22 2

2 2 2

r

r

dr dr

dt dt

d r d d dr d r r

dt dt dt dt dt

v e e

a e e


5/14


6/14

Calculating forces required to cause prescribed

motion of a particle

Idealize system Free body diagram

Kinematics

F=mafor each particle. (for rigid bodies or frames only)

Solve for unknown forces or accelerationsc M 0


7/14

Deriving Equations of Motion for particles

1. Idealize system

2. Introduce variables to describe motion(oftenx,y coords, but we will see other

examples)

3. Write down r, differentiate to get a

4. Draw FBD

5.

6. If necessary, eliminate reaction forces

7. Result will be differential equations for coordsdefined in (2), e.g.

8. Identify initial conditions, and solve ODE

mF a

2

02 sin

d x dxm kx kY t

dtdt


8/14

Motion of a projectile

i

j

kX0

V00 0 0

0 0 0

0

x y z

X Y Z

tdV V V

dt

r i j k

ri j k

2

0 0 0 0 0 0

0 0 0

1

2x y z

x y z

X V t Y V t Z V t gt

V V V gt

g

r i j k

v i j k

a k


9/14

Work and Energy relations

Rate of work done by a force(power developed by force)i

j

k

O

PF

vP F v

Total work done by a force

ij

k

O

PF(t)

r0

r1

1

0

t

W dt

F v

1

0

W d

r

rF r

2 2 2 21 1

2 2 x y zT m m v v v vKinetic energy

i

j

k

O

P

v

r0

r1Work-kinetic energy relation1

0

0W d T T

r

r

F r

Power-kinetic energy relationdT

Pdt


10/14

Potential energy

i

j

k

O

P

Fr0

r1

0

( ) constantV d r

r

r F r

Potential energy of a

conservative force (pair)

grad( )V F

Type of force Potential energy

Gravity acting on aparticle near earthssurface

V mgy

Fm

j

i y

Gravitational forceexerted on mass mby

massMat the origin

GMmV

r

r

F

r m

Force exerted by a

spring with stiffness kand unstretched length

0L

2

0

1

2V k r L F

i

j r

Force acting betweentwo charged particles

1 2

4

QQV

r

r

+Q1+Q2

i

j

1

F2

Force exerted by onemolecule of a noble gas(e.g. He, Ar, etc) on

another (Lennard Jonespotential). ais the

equilibrium spacingbetween molecules, and

Eis the energy of thebond.

12 6

2a a

Er r

r

i

j

1

F2


11/14

m1

m2

m3

m4

R21

R12

R13 R31

R23

R32

F2

ext

F3

ext

F1

ext

Energy relations for

conservative systems subjected to external forces

ijR

Internal Forces: (forces exerted by

one part of the system on another)

External Forces: (any other forces) extiF

0

0

( ) ( )

t t

ext

ext i

forces t

W t t dt

F vWork done by external forces

0 0

extW T V T V

System is conservative if all internal forces are

conservative forces (or constraint forces)

Energy relation for a conservative system 0 0t t t t

Kinetic and potential energy at time 0t 0 0T V

Kinetic and potential energy at timet

T V


12/14

Linear Impulse of a force

i

jk

O

F(t) v

m

1

0

( )

t

t

t dt F

Linear momentum of a particle mp v

Impulse-momentum relations d

dt

pF 1 0 p p

Impulse-momentum relations

m1

m2

m3

m4

R21

R12

R13 R31

R23

R32

F2

ext

F3

ext

F1

ext

0

0

( )

t t

ext

tot i

particles t

t dt

FTotal external impulse

tot i i

particles

m p vTotal linear momentum

( )ext totiparticles

dt

dt

p

FConservation lawtot tot

p

Impulse-momentum for a system of particles

Impulse-momentum for a single particle


13/14

Collisions

*

A B

A B

vx

B1vx

A1

vx

B0

vx

A0

1 1 0 0A B A B

A x B x A x B xm v m v m v m v

1 1 0 0B A B A

v v e v v

1 0 0 0

1 0 0 0

(1 )

(1 )

B B B AA

A B

A A B AB

A B

mv v e v v

m m

mv v e v v

m m

A

B

A B

B0vA0

v

vA1

B1

v

n

1 1 0 0 0 0(1 )B A B A B Ae v v v v v v n n

1 1 0 0B A B A

B A B Am m m m v v v v

1 0 0 0(1 )

B B B AA

B A

me

m m

v v v v n n

1 0 0 0(1 )A A B ABB A

me

m m

v v v v n n


14/14

Angular Impulse-Momentum Equations for a Particle

i

j

k

x

y

z

O

F(t)

r(t)

0

0

( ) ( )

t t

t

t t dt

r F

m h r p r v

d

dt

hMImpulse-Momentum relations

1 0 h h

Useful for central force problems

Angular Momentum

Angular Impulse

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Summary Dynamics(1)