Introduction to Dynamics

Introduction to Dynamics (N. Zabaras)

Review LectureProf. Nicholas Zabaras

Warwick Centre for Predictive Modelling

University of Warwick

Coventry CV4 7AL

United Kingdom

Email: [email protected]

URL: http://www.zabaras.com/

April 16, 2016

1

Includes highlights from the various lectures – Please do not assume that the exam material is related or restricted

to these notes.

mailto:[email protected]

http://www.zabaras.com/


Velocity as a Function of Time

c

dva

dt

cdv a dt

0o

v t

c

v

dv a dt

0 cv v a t


Position as a Function of Time

0 c

dxv v a t

dt

0

0

( )

o

x t

c

s

dx v a t dt

2

0 0

1

2cx x v t a t


Velocity as a Function of Position

0 0

v x

c

v s

v dv a dx

cv dv a dx

2 2

0 02 ( )cv v a x x

2 2

0 0

1 1( )

2 2cv v a x x


Tangential and Normal Components

tevv

• With the velocity vector expressed as

the particle acceleration may be written as

t tt t

de dedv dv dv d dsa e v e v

dt dt dt dt d ds dt

but

vdt

dsdsde

d

edn

t

After substituting,

22 va

dt

dvae

ve

dt

dva ntnt

• Tangential component of acceleration reflects

change of speed and normal component reflects

change of direction.

• Tangential component may be positive or

negative. Normal component always points

toward center of path curvature.

5


Radial and Transverse Components

• When particle position is given in polar coordinates,

it is convenient to express velocity and acceleration

with components parallel and perpendicular to OP.

rr e

d

ede

d

ed

dt

de

dt

d

d

ed

dt

ed rr

dt

de

dt

d

d

ed

dt

edr

erer

edt

dre

dt

dr

dt

edre

dt

drer

dt

dv

r

rr

rr

• The particle velocity vector is

• Similarly, the particle acceleration vector is

errerr

dt

ed

dt

dre

dt

dre

dt

d

dt

dr

dt

ed

dt

dre

dt

rd

edt

dre

dt

dr

dt

da

r

rr

r

22

2

2

2

2

rerr

6


Equations of Motion

• Newton’s second law provides

amF

• Solution for particle motion is facilitated by

resolving vector equation into scalar component

equations, e.g., for rectangular components,

zmFymFxmF

maFmaFmaF

kajaiamkFjFiF

zyx

zzyyxx

zyxzyx

• For tangential and normal components,

2vmF

dt

dvmF

maFmaF

nt

nntt

7


Linear Momentum of a Particle

• Replacing the acceleration by the derivative of

the velocity yields

particle theof momentumlinear

L

dt

Ldvm

dt

d

dt

vdmF

• Linear Momentum Conservation Principle:

If the resultant force on a particle is zero, the

linear momentum of the particle remains

constant in both magnitude and direction.

8


Sample Problem

The two blocks shown start from

rest. The horizontal plane and the

pulley are frictionless, and the

pulley is assumed to be of

negligible mass.

Determine the acceleration of each

block and the tension in the cord.

9


Sample Problem

• Write equations of motion for blocks and pulley.

:AAx amF

AaT kg1001

:BBy amF

B

B

BBB

aT

aT

amTgm

kg300-N2940

kg300sm81.9kg300

2

22

2

:0 CCy amF

02 12 TT

SOLUTION:

• Write the kinematic relationships for the dependent

motions and accelerations of the blocks.

ABAB aaxy21

21

x

y

O

10


Angular Momentum of a Particle• moment of momentum or the

angular momentum of the particle about O. VmrHO

• Derivative of angular momentum with respect to time,

O

O

M

Fr

amrVmVVmrVmrH

• It follows from Newton’s second law that the sum

of the moments about O of the forces acting on

the particle is equal to the rate of change of the

angular momentum of the particle about O.

zyx

O

mvmvmv

zyx

kji

H

• is perpendicular to plane containingOH

Vmr

and

2

sin

mr

vrm

rmVHO

11


Eqs of Motion in Radial & Transverse Components

rrmmaF

rrmmaF rr

2

2

• Consider particle at r and , in polar coordinates,

rrmF

rrrm

mrdt

dFr

mrHO

2

22

2

2

• This result may also be derived from

conservation of angular momentum,

12

rr r v e e


Conservation of Angular Momentum

• When only force acting on particle is directed

toward or away from a fixed point O, the particle

is said to be moving under a central force.

• Since the line of action of the central force

passes through O, and 0 OO HM

constant OHVmr

• Position vector and motion of particle are in

a plane perpendicular to .OH

• Magnitude of angular momentum,

000 sin

constantsin

Vmr

VrmHO

massunit

momentumangular

constant

2

2

hrm

H

mrH

O

O

or

13


Sample Problem

The 0.5 lb pellet is pushed against

the spring and released from rest

at A. Neglecting friction,

determine the smallest deflection

of the spring for which the pellet

will travel around the loop and

remain in contact with the loop at

all times.

14


Sample Problem

SOLUTION:

• Setting the force exerted by the loop to zero, solve

for the minimum velocity at D.

:nn maF

222

2

sft4.64sft32.2ft 2

rgv

rvmmgmaW

D

Dn

• Apply the principle of conservation of energy

between points A and D.

0

18ftlb360

1

22

212

21

1

T

xxkxVVV ge

lbft5.0sft4.64sft2.32

lb5.0

2

1

lbft2ft4lb5.00

22

2

2

21

2

2

D

ge

mvT

WyVVV

25.0180 2

2211

x

VTVT

in. 47.4ft 3727.0 x

15


Principle of Impulse and Momentum

21

12

2

1

2

1

LdtFL

LLdtF

LF

t

t

t

t

21

12

2

1

2

1

HdtMH

HHdtM

HM

t

tO

t

tO

OO

• The momenta of the particles at time t1 and the impulse of the

forces from t1 to t2 form a system of vectors equipollent to the

system of momenta of the particles at time t2 .

16


2

10 1 0 0 2( ) ( )

t

tdt H M H

2

10 0 0 0 2 0 1( ) ( )

t

t

ddt

dt M H M H H

2 2

1 10Angular impulse

t t

t tdt dt M r F

2

1 21

t

tr mv M dt r F dt r m v

Angular Impulse and Momentum Principles

17

0 1 0 2( ) ( )H H

1 1 2 2r m r m

Conservation of

Angular

Momentum

when angular

impulse is zero

r

mv


W = 0.8 Ib

Smooth table

r1=1.7 ft

v1= 4 ft/s

vc=6ft/s (constant)

v2= ? at (r2)= 0.6ft

Work done = ?

1 2

1 1 2 2'B B

H H

r m r m

2

2

0.8 0.81.75 ( ) 4 0.6 ( ) '

32.2 32.2

' 11.67 /ft s

2 2

2 (11.67) (6) 13.1 /ft s

Work Done (energy balance for the ball)

1 1 2 2T U T 2 21 0.8 1 0.8

( )(4) ( )(13.1)2 32.2 2 32.2

FU

1.94 ft.IbFU

Example

18

The cord force F (not constant) on the

ball passes through the z axis, and

the weight and NB are parallel to it.

Hence the moments, or angular

impulses created by these forces,

are all zero about this axis.

Therefore, angular momentum is

conserved about the z axis.


two components of

acceleration:

one component of

velocity:

Fixed-Axis Rotation: Summary


Absolute and Relative Velocity in Plane Motion

• Any plane motion can be replaced by a translation of

an arbitrary reference point A and a simultaneous

rotation about A.

ABAB vvv

rvrkv ABABAB

ABAB rkvv

20


Instantaneous Center of Rotation in Plane Motion• The instantaneous center of rotation lies at the

intersection of the perpendiculars to the velocity

vectors through A and B .

cosl

v

AC

v AA

tan

cossin

A

AB

v

l

vlBCv

• The velocities of all particles on the rod are as if they

were rotated about C.

• The particle at the center of rotation has zero velocity.

• The particle coinciding with the center of rotation

changes with time and the acceleration of the particle

at the instantaneous center of rotation is not zero.

• The acceleration of the particles in the slab cannot be

determined as if the slab were simply rotating about C.

• The trace of the locus of the center of rotation on the

body is the body centrode and in space is the space

centrode.

21


Rolling wheels

22

vB r

Rolls without slipping

Cv (2 )r

C


IC

C

ICCC r /

D

Instantaneous center of zero velocity

ICDD r /

Rolling wheels

23


Absolute and Relative Acceleration in Plane Motion

• Absolute acceleration of a particle of the

slab,

ABAB aaa

• Relative acceleration associated with rotation about A

includes tangential and normal components,ABa

ABnAB

ABtAB

ra

rka

2

2

ra

ra

nAB

tAB

24


Absolute and Relative Acceleration in Plane Motion

• Given

determine

, and AA va

. and

Ba

tABnABA

ABAB

aaa

aaa

• Vector result depends on sense of and the

relative magnitudes of nABA aa and

Aa

• Must also know angular velocity .

25


Equations of Motion for a Rigid Body• Consider a rigid body acted

upon by several external forces.

• Assume that the body is made

of a large number of particles.

• For the motion of the mass

center G of the body with

respect to the Newtonian frame

Oxyz, amF

• For the motion of the body with

respect to the centroidal frame

Gx’y’z’,

GG HM

• System of external forces is

equipollent to the system

consisting of . and GHam

26


Angular Momentum of a Rigid Body in Plane Motion

• Consider a rigid slab

in plane motion.

• Angular momentum of the slab may be

computed by

I

mr

mrr

mvrH

ii

n

iiii

n

iiiiG

Δ

Δ

Δ

2

1

1

• After differentiation,

IIHG

• Results are also valid for plane motion of

bodies which are symmetrical with respect

to the reference plane.

• Results are not valid for asymmetrical

bodies or three-dimensional motion.

27


( ) ( ) ( )P P G y G xM x m a y m a I kΜ

Kinetic Moment

28

( )

( )

x G x

y G y

G G

F m a

F m a

M I α

( )P PM kΜ


2( )

( )

n G n G

t G t G

G G

F m a mω r

F m a mr α

M I α

2 ( )

G G

G G

O

I α r m r α

I mr α

I α

( ) ( )O O G G G tM I α r m aΜ k

29

Rotation About a Fixed Axis


21

2GT mv 2 21 1

2 2G GT mv I ω

2 21 1

2 2

Translational Rotational

Kinetic energy Kinetic energy

G GT mv I ω

ωrv GG 2 2 21 1

2 2G GT m r ω I ω

2 21( )

2G GI m r ω

21

2OI ω

Translation Rotation General

Kinetic Energy

30


1 1 2 2T U T

2 2 2 2 2 2

1 1 2 1 2 1 2 2

1 1 1 1 1. ( ) ( ) ( )

2 2 2 2 2O Om I F S W y k s s M m I

S rr

Translation

n

Rotation

Principle of Work and Energy

31


0

( ) ( )

G

G

A G

L mv

H

H d mv

G

G G

O G G G

L mv

H I ω

H I ω r mv

( )

G

G G

A G G

L mv

H I ω

H I ω d mv

Translation Rotation About a Fixed Axis General Plane Motion

2

2( )

G G

G G

O

I ω mr ω

I m r ω

I ω

Angular Momentum

32


( )

G G

A G G

H I ω

H I ω d mv

1 2G G G GG

I mv r M dt F r dt I mv r

General Plane Motion

33


2

10 1 0 0 2( ) ( )

t

tdt H M H

2

1 21

t

tr mv M dt r F dt r m v

2

11 2

t

G G Gt

I ω M dt I ω

1 2G GI mv r M dt F r dt I mv r

Angular Impulse and Momentum Principles

34


Sample Problem

Uniform sphere of mass m and

radius r is projected along a

rough horizontal surface with a

linear velocity and no angular

velocity. The coefficient of

kinetic friction is

Determine a) the time t2 at which

the sphere will start rolling

without sliding and b) the linear

and angular velocities of the

sphere at time t2.

.k

1v

SOLUTION:

• Apply principle of impulse and

momentum to find variation of linear

and angular velocities with time.

• Relate the linear and angular

velocities when the sphere stops

sliding by noting that the velocity of

the point of contact is zero at that

instant.

• Substitute for the linear and angular

velocities and solve for the time at

which sliding stops.

• Evaluate the linear and angular

velocities at that instant.

35


Sample Problem

SOLUTION:

• Apply principle of impulse and

momentum to find variation of linear

and angular velocities with time.

0WtNt

y components:

x components:

21

21

vmmgtvm

vmFtvm

k

gtvv k 12

mgWN

moments about G:

22

52

2

mrtrmg

IFtr

k

tr

gk2

52

Sys Momenta1 + Sys Ext Imp1-2 = Sys Momenta2

• Relate linear and angular velocities

when sphere stops sliding by noting

that velocity of point of contact is zero

at that instant.

tr

grgtv

rv

kk

2

51

22

• Substitute for the linear and angular

velocities and solve for the time at

which sliding stops.

g

vt

k1

7

2

36

Introduction to Dynamics

Documents

Transcript of Introduction to Dynamics