MA4248 Weeks 6-7.

Topics Work and Energy, Single Particle Constraints,Multiple Particle Constraints, The Principle of VirtualWork and d’Alembert’s Principle

1

Work and Energy: the English words originated, viaGermanic and Greek branches, respectively, from theProto Indo-European word Werg about 7 k years ago my homepage under courses/Ussc2001/Energy1.pdf

The work done on a particle that is displaced byin a constant force field

F

equals

FThis work has units of energy.

CONSTANT FORCE FIELDS Let us consider this situation in detail. Let

2

Then hence

Aq

Ap,rqp whererF)p(U

and if

2/)t(r)t(rm))t(p(U particle whose trajectory is

then construct the function

A denoteaffine space and choose a point

RA:U by

F)p(U)p(U

UgradF

)t(rq)t(p

is constantthen

since Newton’s second law implies that

)t(rF2/)t(r)t(rmdtd

F

is the net force on a

CONSTANT FORCE FIELDS and

3)t(rF)t(r))t(p(Ugrad

))t(p(U

t]t)t(r[))]t(p(Ugrad[

t))t(p(U))t)t(r)t(p(U

t))t(p(U))tt(p(U

dtd

0tlim

0tlim

0tlim

CONSERVATIVE FORCE FIELDS

4

The argument on page 20 in Week 1-3 Vufoils showthat this total energy is constant for any conservativeforce field. Let us consider the following converse

2/)t(r)t(rm))t(p(U

There exists a force field F and a function U that arefunctions on A (time independent) such that

is constant for

every particle of mass m that moves with net force F.

0)t(r))t(p(F)t(r))t(p(Ugrad Then

hence UgradF

is conservative.

WORK OVER A SMOOTH CURVE

5

Thomas, p. 1062. The work done by a force (field)

F

over a smooth curve parameterized by a smooth

vector valued function on the interval [a,b] is

bt

at

bt

atdt)t(rFrdFW

If UgradF

is conservative then

r

bt

at))a(p(U))b(p(UdtUW

PATH INDEPENDENCE AND COMPONENT TEST

6

F

Thomas, p. 1072 A vector field is conservative

W depends only on the endpoints if and only if

)a(p and of the curve.

)b(p

Thomas, p. 1074 kPjNiMF

is conservative if and only if

yM

xN

xP

zM

zN

yP ,,

SINGLE PARTICLE CONSTRAINT

7

In Tutorial 3, Prob. 5 you computed the trajectory of a ring sliding down a straight rod by assuming either1. that the total energy is conserved, or 2. that the force of constraint is orthogonal to the rodWhy do these assumptions yield the same trajectory?

Consider a particle having mass m that is constrainedto move along a curve parameterized by a function

of a variable s, called a generalized coordinate. Thismay be the case if the particle consists of a ring thatslides along a rigid wire. We will first assume that thecurve does not move so that it is independent of time

]b,a[s),s(hq)s(c

SINGLE PARTICLE CONSTRAINT

8

Therefore, the trajectory of the particle must equal

where t denotes time and s(t) is a function of time

cadtd FF))t(s(hm

2

2

Newton’s second law implies that

]b,a[)t(s)),t(s(hq)t(p

where aF

cF

is the applied force that would be there

if the physical constraint (wire) was removed, and is the force of constraint defined by this equation

SINGLE PARTICLE CONSTRAINT

9

Define ))t(s(h)t(r

over the curve

]t,t[r 21

cF

2

1

2

1

tt

tta

tt

ttc rdFrmrdF

The work performed by

)t,t(W 21cis

SINGLE PARTICLE CONSTRAINT

10

Then

2

1

2

1

tt

tt

tt

ttdtrrmrdrm

)t(E)t(Errm 1kin2kin

2

1

tt

tt 21

dtd

is the change of kinetic energy over the time interval

SINGLE PARTICLE CONSTRAINT

11

If the applied force is conservative UgradFa

0rdFc

2

1

2

1

tt

tt

tt

tta dtrUgradrdF

and

)t(U)t(UdtU 12

2

1

tt

tt

hence )t,t(W 21c is the change in total energy

It equals zero for all time intervals iff

SINGLE PARTICLE CONSTRAINT

12

In this case, it is very convenient to use arc lengthparameterization of the curve, then

and energy conservation implies that the trajectory isdetermined, up to the initial position, by the first orderdifferential equation

221

21 smrrmEkin

)t(UE)t(sm2

SINGLE PARTICLE CONSTRAINT

13

We now consider the case where the particle isconstrained to move along a curve that is moving with time as in Tutorial 5, Problems 4 and 5.

In this case the force of constraint may perform workon the particle, yet it is reasonable to assume that ateach value of time the force of constraint is orthogonalto the curve described at that time.

Note: the curve at a specific time does not describe theactual trajectory of the particle, this very important factis illustrated in Fig. 2.04 on page 31 of the textbook.

SINGLE PARTICLE CONSTRAINT

14

Therefore, the trajectory of the particle must equal

since we now have a time varying family of curves.

cadtd FF))t(s(hm

2

2

Newton’s second law implies that

)]t(b),t(a[)t(s),t),t(s(hq)t(p

and the orthogonality condition implies that

0s)t(FhFs

)t),t(s(hcc

There are 4 unknowns and 4 equations

SINGLE PARTICLE CONSTRAINT

15

We now consider a particle that is constrained to movealong a (possibly moving) surface. Then the trajectoryis determined by the principle that constraint force attime t is orthogonal to the constraint surface at time t

cadtd FF))t(s(hm

2

2

Parameterize the surface at time t is by a (possibly timevarying) function of generalized coordinates

so that at time t )t),t(q),t(q(hq)t(p 21

21 q,q h

Newton implies

SINGLE PARTICLE CONSTRAINT

16

Note that we now have 5 unknowns, the 2 generalizedcoordinates and 3 components of the constraint force.Newton gives us three equations. We need 2 more. They are provided by the orthogonality principle:

22

21c1

1

21c

c

qq

)t,q,q(hFq

q

)t,q,q(hF

rF0

2,1i,0q

)t,q,q(hF

i

21c

SINGLE PARTICLE CONSTRAINT

17

If the applied force is conservative and if the surface isindependent of time then energy is conserved. Energy conservation is not sufficient to determine the motionsince it provides only 1 additional equation.

If the surface is moving then the forces of constraintmay (and usually do) perform work since

dtt

)t,2q,1q(hrFrdFW cc

dtt

)t,q,q(h 21cF

since 0rFW c

SINGLE PARTICLE CONSTRAINT

18

In the previous discussion of a single particle, we usedgeneralized coordinates. However, we could have

usedrectangular coordinates. If we did then we would have6 unknown variables – 3 coordinates x, y, z for theposition of the particle and 3 coordinates of the forceof constraint. These 6 variables are determined (by thesolution of differential equations) by the constraint equations (1 for a surface and 2 for a curve), Newton’ssecond law (3 equations), and the principle that theforce of constraint is orthogonal to the constraint set (2 for a surface, 1 for a curve)

MULTIPLE PARTICLE CONSTRAINTS

19

For a system with N particles, Newton’s 2nd law gives

N,...,1i,FFFrm coni

appi

netiii

3N equations in 6N variables. We need 3N more!

M=3N-f holonomic constraints, given by equationsM,...,1,0)t;r,...,r,r(G N21

give a total of 6N-f equations, we need f more!

These f equations will be provided by the Principle Of Virtual Work. For 1 particle, this principle says that the force of constraint is orthogonal to the surface(M=1) or curve (M=2) that the particle moves on.

MULTIPLE PARTICLE CONSTRAINTS

20

For multiple particle constraints virtual displacements

are any displacements that satisfy

The principle of virtual work says that the total work done bythe forces of constraint over these displacements equals zero

N1 r,...,r

M,...,1,0rNi

1iG

ir i

0rNi

1iFW i

coni

GENERALIZED COORDINATES

21

The set of virtual displacements form a vector spacehaving dimension f = 3N-M, these are the number of degrees of freedom of the system

If we introduce generalized coordinates then

N,...,1i),q,...,q(rr f1ii f1 q,...,q

N,...,1i,qf

1 q

irri

and we can find a basis for this vector space

GENERALIZED COORDINATES

22

Then the principle of work can be expressed by

This holds for all choices of iff

0qq

rNi

1iF

f

1W

iconi

f,...,1,q

f,...,1,0q

rNi

1iF

iconi

These are the f additional equations that we require.

D’ALEMBERT’S PRINCIPLE

23

Then the principle of work can be expressed by

0q

q

rNi

1i irimFf

1

Ni

1i irimFW

iappi

appi ir

The work done by the applied forces, plus the work done bythe inertial forces , in a virtual displacement is zero

ii rm

EXAMPLES

24

Two particles connected by a light rigid rod (pp.29-30)

The forces of constraint are proportional to

0)rr(FrFrFW 2112211

The constraint

1F

2F

21 FF

a|rr| 21

implies that

0)()( 2121 rrrr

and Newton’s second law implies that

21 rr

hence

EXAMPLES

25

Inclined plane (pp.34)

If the block undergoes a virtual displacement

Inclined plane p. 34

ssinmg down the plane then the applied force, gravity, does work

m

sg

generalized coordinateS is the distance downthe inclined plane

s

and the inertial force (oriented up the plane)

ssm which yields the well-known resultdoes work

sings