Lecture 4: Discrete model - Division of Mechanics · Lecture 4: Discrete model Klarbring...

Lecture notes Models of Mechanics

Anders Klarbring

Division of Mechanics, Linköping University, Sweden

Lecture 4: Discrete model

Klarbring (Mechanics, LiU) Lecture notes Linköping 2012 1 / 29

Contents

1 Discrete geometric model

2 Mass

3 Momentum

4 Force system

5 Euler’s laws explicit

6 Newton’s law

7 The theorem of action and reaction

8 Summary (so far)

9 Complete problems

10 Equilibrium and quasi-equilibrium



2 Mass

3 Momentum

4 Force system


6 Newton’s law


8 Summary (so far)

9 Complete problems



Discrete geometric model

We will now define, for the discrete geometry, mass. momentum andforces, which are functions appearing in the universal laws:

∂

∂tM(P) = 0 (1)

∂

∂tp(P) = f (P) (2)

∂

∂tho(P) = co(P). (3)



2 Mass

3 Momentum

4 Force system


6 Newton’s law


8 Summary (so far)

9 Complete problems



Discrete geometric model – Mass

Every X ∈ B is given a scalar mX > 0 and

M(P) =∑

X∈P

mX

Take P = {X} and use (1). This results is

∂mX

∂t= 0

Thus, the mass of each particle is constant.



2 Mass

3 Momentum

4 Force system


6 Newton’s law


8 Summary (so far)

9 Complete problems



Discrete geometric model – Momentum

The linear momentum is defined by:

p(P) =∑

X∈P

V (X , t)mX .

We calculate the time derivative

∂

∂tho(P) =

∑

X∈P

r(X , t) × A(X , t)mX , (4)

where the acceleration is defined by

A(X , t) =∂

∂tV (X , t).


Discrete geometric model – Momentum

The angular momentum is defined by:

ho(P) =∑

X∈P

r(X , t) × V (X , t)mX .

The derivative becomes

∂

∂tho(P) =

∑

X∈P

r(X , t) × A(X , t)mX . (5)



2 Mass

3 Momentum

4 Force system


6 Newton’s law


8 Summary (so far)

9 Complete problems



Discrete geometric model – Force system

Two types of forces:

(i) For every material point X ∈ B there is a geometric vectorf e(X ), called an external force vector.

(ii) For every ordered pair (Xk , Xℓ) of material points of B,there is a geometric vector f i(Xk , Xℓ), called the internalforce with which Xk acts on Xℓ.



2 Mass

3 Momentum

4 Force system


6 Newton’s law


8 Summary (so far)

9 Complete problems



Discrete geometric model – Euler’s laws explicit

Euler’s law of linear momentum∑

X∈P

f (X ,P) =∑

X∈P

A(X , t)mX . (6)

Euler’s law of angular momentum∑

X∈P

r(X , t) × f (X ,P) =∑

X∈P

r(X , t) × A(X , t)mX . (7)

wheref (X ,P) = f e(X ) +

∑

Xk∈P−1

f i(Xk , X )

is the part of the total force related to the material point X when part Pof the body B is considered.



2 Mass

3 Momentum

4 Force system


6 Newton’s law


8 Summary (so far)

9 Complete problems



Discrete geometric model – Newton’s law

Euler’s law for one material point, X , is usually called Newton’s(second) law, i.e.

f e(X ) +∑

Xk∈B\{X}

f i(Xk , X ) = A(X , t)mX . (8)



2 Mass

3 Momentum

4 Force system


6 Newton’s law


8 Summary (so far)

9 Complete problems



Discrete geometric model – action and reaction

The following results can be (mathematically) proved asconsequences of Euler’s laws. They say that the two forces that actsbetween two particles are oppositely directed and directed along theline that unites the particles.

f i(A, B) = −f i(B, A). (9)

r(A, t) × f i(B, A) + r(B, t) × f i(A, B) = 0. (10)



2 Mass

3 Momentum

4 Force system


6 Newton’s law


8 Summary (so far)

9 Complete problems



Discrete geometric model – Summary

Euler’s universal laws are made explicit by choosing particularforms for mass, momentum and force.

Conservation of mass is equivalent to constant particle massesmX .

Euler’s laws are equivalent to Newton’s law + the action andreaction formulas.



2 Mass

3 Momentum

4 Force system


6 Newton’s law


8 Summary (so far)

9 Complete problems



Discrete geometric model – Complete problems

What has to be added to Newton’s law and the action and reactionprinciple in order to have a complete problem?

Suggestions for complete problems:

The force problem We know the masses and the motion (frommeasurements) and want to determine the forces.

The motion problem We know the external forces and want todetermine the motion (and the internal forces).



We conclude that the force problem is (in general) not solvable unlesswe have extra information (specific laws) such as knowledge ofexternal forces (say, gravity) or know that particles are connected bysprings (which means that we know the internal forces).The motion problem has similar properties.



A fairly general specific law for internal forces is the following

f i(Xk , Xℓ) = −f i(Xℓ, Xk ) = f (d) n(φt(Xk ), φt (Xℓ)), (11)

where

n(φt(Xk ), φt(Xℓ)) =φt(Xk ) − φt(Xℓ)

|φt(Xk ) − φt(Xℓ)|,

is the unit vector pointing from φ(Xℓ) to φ(Xk ), and f (d) is a possiblynon-linear function of the distance between the pair of points, thedistance being given by

d = |φt(Xk ) − φt(Xℓ)|.



In the case of Newton’s gravitation, the scalar force is given by

f (d) = Gmkmℓ

d2 ,

In the first home assignment you are to test a modification of thisequation where you replace the exponent 2 of the denominator by ageneral constant α, say in the interval from -1 to 2.



A degenerated form of (11) is when f (d) → ∞.This results in the condition

|φt(Xk ) − φt(Xℓ)| is independent of time. (12)

That is the distance between the material points is constant. If this isprescribed for all pairs of material points we get the model of a rigidbody.



2 Mass

3 Momentum

4 Force system


6 Newton’s law


8 Summary (so far)

9 Complete problems



Equilibrium and quasi-equilibrium

DefinitionA body is said to be in equilibrium if it is in a state of constanttranslatation, meaning that there is a vector V const such that

V (X , t) = V const, (13)

for all X and t, i.e., the velocity is constant for all material points andtime.

DefinitionA body is said to be in quasi-equilibrium if it satisfies

co(B) = 0, f (B) = 0. (14)



From (13) we find that A(X , t) = 0, so by Euler’s laws we obtain (14).Thus

equilibrium =⇒ quasi-equilibrium

However, the opposite direction is not true, which is shown by simpleexamples.



For a rigid body we have:

Constant translation of the center of mass and constant rotationaround a principle axis implies quasi-equilibrium


Lecture 4: Discrete model - Division of Mechanics · Lecture 4: Discrete model Klarbring...

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