Physics

PHYS 205 Mechanics SectionLecture #1

Texts useful for Mechanics

Serway and Jewett. Physics for Scientists and Engineers. 7th Ed. [SJ]100-level text book.

C. Kittel et al., Mechanics (Berkeley Physics Course), McGraw-Hill.Old but good.

T.W.B. Kibble. Classical Mechanics.

Lecture 1 main concepts:

Inertial frames. Transformations between frames. Energy and momentum.

[Serway 1.2], [SJ 4,5], [Kittel 4]

A fundamental principle of mechanics is expressed by the following equivalent statements:

Basic laws of physics are unchanged in form in two reference frames connected by a Galileantransformation.

The laws of physics have the same form in all inertial reference frames.

Let us examine these statements for a Galilean transformation between two states S and S . Firstlylet us review Galilean transformations.

Galilean transformations

Consider two reference frames S with coordinates x, y, z and S with coordinates x, y, z. Thereference frame S moves at speed V with respect to S along the x-direction.

Consider at t = 0 that the origins of the twoframes are aligned.What is the coordinate transformation relatingthe two reference frames?

Now to examine the states about the laws of physics being the same in all inertial reference frameswe will consider Newtons law in both reference frames.

Firstly let us check the relationship betweenvx =

dxdt

and vx =dxdt

.

vx =dxdt

=

Is this the velocity relationship you wouldexpect between S and S ?

Now obtain the relationship between ax =dvdt

and ax =dvdt

.

In examining situations such as collisions it can be useful to work in alternative reference framessuch as the centre of mass or centre of momentum frame. Quantites such as the change in kineticenergy must be the same in either frame.

For example consider an inelastic condition of twoequal mass bodies with mass m. Consider the lab-oratory frame where initially object 1 has speed v1in the positive x direction and object 2 is at rest.After the collision the two objects stick together.What is the velocity of the combined objects?

What is the change in kinetic energy between the intial and final states?

Now consider the same situation in the centre ofmomentum (COM) frame. In this frame the totalmomentum is zero.What are the speeds of the two objects initiallyin this frame?

What is the final speed of the combined object in the COM frame?

What is the change in kinetic energy between the initial and final states?

Conservation of Linear Momentum

Main Concepts

Internal forces and centre of mass Collision problems

Inelastic

Elastic

Variable mass problems Conveyor belt

Rockets (gaining or losing mass)

Collections of particles

Total momentum:ptotal =

mivi

Internal forces balance:Fij = Fjiand dont affect ptotal.

PHYS202

Conservation ofLinear Momentum

Mike [email protected]

PHYS202 Linear Momentum, September 18, 2014 p.1/14

Concepts Internal forces and centre of mass Collision problems Inelastic Elastic

Variable mass problems Conveyor belt Dust Rockets

[SJ 9], [Kittel 6].


Collections of particlesTotal momentum:ptotal =

mivi

1 2F12

F21


Centre of mass (momentum) COM

rCOM =

rimimi

VCOM =

vimimi

1. If no external forces VCOM = constant.

2. mtotalaCOM =miai =

Fi = Fexternal


Completely inelastic collisionLaboratory frame

v1

m1 m2

v2 = 0 v

m1 +m2

m1v1 +m2 0 = (m1 +m2)vKi =

1

2m1v1

2

Kf =1

2(m1 +m2)v

2

=1

2(m1 +m2)

m21(m1 +m2)2

v12

=m21v1

2

2(m1 +m2) PHYS202 Linear Momentum, September 18, 2014 p.4/14


rCOM =

rimimi

VCOM =

vimimi

Completely inelastic collision

Laboratory frame

PHYS202

Conservation ofLinear Momentum

Mike [email protected]


Concepts Internal forces and centre of mass Collision problems Inelastic Elastic

Variable mass problems Conveyor belt Dust Rockets

[SJ 9], [Kittel 6].


Collections of particlesTotal momentum:ptotal =

mivi

1 2F12

F21



rCOM =

rimimi

VCOM =

vimimi

1. If no external forces VCOM = constant.

2. mtotalaCOM =miai =

Fi = Fexternal


Completely inelastic collisionLaboratory frame

v1

m1 m2

v2 = 0 v

m1 +m2

m1v1 +m2 0 = (m1 +m2)vKi =

1

2m1v1

2

Kf =1

2(m1 +m2)v

2

=1

2(m1 +m2)

m21(m1 +m2)2

v12

=m21v1

2

2(m1 +m2) PHYS202 Linear Momentum, September 18, 2014 p.4/14

What is the initial momentum?

What is the final momentum?

Obtain an expression for v?

What is the initial kinetic energy?

What is the final kinetic energy?


Continuation of Completely inelastic collision in Laboratory frame

Considering that the difference between the final and initial kinetic energy is the change in internalenergy we can write

Einternal = Ki KfWhat is Einternal?

CoM frame

Obtain an expression for VCoM .

Consider a transformation from the laboratory reference frame to the CoM refeence frame.

Obtain expressions for v1 and v2 (in terms of v1, m1 and m2).

Check that Einternal = Ki Kf is the same as in the laboratroy frame.

Completely Elastic Collision

Kinetic energy and momentum are both conserved in an elastic collision. We will consider onlythe one-dimensional case when all motion is constrained to one dimension and some special three-dimensional cases.

Laboratory frame

Completely inelastic (2)KfKi

=m1

m1 +m2

Kinternal = Ki Kf= Ki

(1 Kf

Ki

)

= Ki

(1 m1

m1 +m2

)

=1

2v1

2(

m1m2m1 +m2

)


Inelastic COMv1 v

2

m1 m2

v = 0

m1 +m2

plab = m1v1 +m2 0VCOM = v =

m1v1m1 +m2

sov1 = v1 VCOM = v1

(1 m1

m1 +m2

)

v2 = 0 VCOM = v1(

m1m1 +m2

)v = v VCOM = 0Kinternal = Ki Kf = ?


Completely Elastic CollisionThe general (non-1D) case is really interesting, butwell confine ourselves to 1D and special cases fornow.

Laboratory frame

v1

m1 m2

v2 = 0 w1

m1

w2

m2

Momentum: m1v1 +m2v2 = m1w1 +m2w2

Energy:1

2m1v1

2 +1

2m2v2

2 =1

2m1w1

2 +1

2m2w2

2

Soluble, but tedious!


Elastic COMVCOM =

m1v1m1 +m2

v1

m1

v2

m2

w1

m1

w2

m2

|w1| = |v1| |w2| = |v2| Easy!3D version: In COM frame these equalities stillapply, but the objects can recoil in any direction.

Possibilities:m1 < m2 m1 = m2 m1 > m2


The one-dimensional situation is depicted in the figure. Write down the equations which resultfrom applying conservation of momentum and conservation of kinetic energy.

These equations are soluble but there is quite a lot of algebra involved. The CoM frame is simpler...

CoM frame

Completely inelastic (2)KfKi

=m1

m1 +m2

Kinternal = Ki Kf= Ki

(1 Kf

Ki

)

= Ki

(1 m1

m1 +m2

)

=1

2v1

2(

m1m2m1 +m2

)


Inelastic COMv1 v

2

m1 m2

v = 0

m1 +m2

plab = m1v1 +m2 0VCOM = v =

m1v1m1 +m2

sov1 = v1 VCOM = v1

(1 m1

m1 +m2

)

v2 = 0 VCOM = v1(

m1m1 +m2

)v = v VCOM = 0Kinternal = Ki Kf = ?


Completely Elastic CollisionThe general (non-1D) case is really interesting, butwell confine ourselves to 1D and special cases fornow.

Laboratory frame

v1

m1 m2

v2 = 0 w1

m1

w2

m2

Momentum: m1v1 +m2v2 = m1w1 +m2w2

Energy:1

2m1v1

2 +1

2m2v2

2 =1

2m1w1

2 +1

2m2w2

2

Soluble, but tedious!


Elastic COMVCOM =

m1v1m1 +m2

v1

m1

v2

m2

w1

m1

w2

m2

|w1| = |v1| |w2| = |v2| Easy!3D version: In COM frame these equalities stillapply, but the objects can recoil in any direction.

Possibilities:m1 < m2 m1 = m2 m1 > m2


What is VCoM?

What can we deduce about the relationships of the magnitudes of v1 and w1 and the magnitudes

of v2 and w2?

3D Elastic collision

In three dimensions the analysis is more complicated. The special case when m1 = m2 and v2 = 0is considered in the tutorial where it is shown that in this case the angle between the outgoingtrajectories w1 and w2 is 90

.

Variable Mass Problems

We can still use Newtons second law

F =dp

dt=dmv

dt

but we need to take into account that the mass m is changing with time.

Use the product rule to evaluate dmvdt

in the case when m is changing in time.

We will consider two examples: mass being added to a conveyor belt and a spacecraft accumulatingdust; and these and other problems will also be examned in the tutorial and assignment.

Conveyor belt

A conveyor belt is moving at a constant speed v. A mass n is being deposited onto the belt persecond.Compare the work done per second with the kinetic energy change per second to determine whetherthis is an elastic or inelastic process.

Space craft

A spacecraft of mass m is travelling at speed v relative to interplanetary dust. The dust is stickingto the spacecraft, so its mass is increasing at a rate proportional to its speed.

What is the acceleration of the spacecraft?


Angular momentum and Central forces

Concepts:

1. Angular Momentum, Torque, Conservation of Angular Momentum.

2. Central forces such as gravity and electrostatic forces.

3. Orbits, Keplers Laws.

[SJ 11-13, 42], [Kittel 6, 7, 8]

Angular quantities

An object (or point in an extended object) which rotates through an angle at a distance r fromthe rotation axis moves distance s along a circular arc where s is given by

s = r

where is in radians.

The angular speed of the rotating object (or point in an extended object) is denoted by and isrelated to the linear speed by

v = r

where the units of are radians/second.

The angular acceleration of the rotating object (or point in an extended object) is denoted by and is related to the linear acceleration by

a = r

where the units of are radians/second2.

For an object undergoing constant-speed circular motion the angular acceleration (related to thetangential acceleration) iszero however the centripetal acceleration (pointing radially inward) is notzero as the velocity direction is changing. The centripetal acceleration is given by

acentripetal =v2

r.

and are vectors

The expressions above refer to the magnitudes ofthe vector quantities and . These quantitesare vectors where the direction of the vector isparallel with the rotation axis and the direction isgiven by a right hand rule with the fingers curledin the direction of the rotation.

Angular Momentum and Torque

Torque: = r FAngular momentum: L = r p = mr v

Recap cross product

Given any two vectors A and B the cross productAB is defined as a third vector C which has amagnitude AB sin where is the angle betweenA and B. The direction of C is perpendicular tothe plane formed by A and B and a right handrule is use to determine which way C points.

Show that dLdt

=

This means that if = 0, dLdt

= 0 and angular momentum is conserved.


Conservative forces and the potential energy function SJ 7.8

The work done on the member of a system by a conservative force between the members of asystem does not depend on the path taken by the moving member. For such a system a potentialenergy function U can be defined such that work done within the system equals the decrease inthe potential energy of the system. Consider a case when the motion of moving particle is alongthe x-xis. The work done by the force F is

W = x1x2

Fxdx = U.

For an infintesimal displaced dx we can express the infinitesimal change of the potential energy dUas

dU = Fxdxwhere in the infinitesimal limit the integral is the area of the rectangle Fxdx.

Therefore the conservative force is related to the potential function through the relationship

Fx = dUdx.

In three dimensions F is related to U by

F = U

where is the gradient vector. In cartesian coordinates is expressed as

=(

x,

y,

z

)

and in spherical polar coordinates as

=(

r,1

r

,

1

r cos

).

The total energy of an object is given by the sum of the kinetic energy and the potential energy

E = Ek + U.

Example: Simple harmonic oscillator

What is the potential energy function for a simpleharmonic oscillator (in one dimension)?

Sketch this potential.

Use the relation Fx = dUdx to determine the force.

Indicate on your sketch where the equilibriumpoint is. Note that the force is zero at the equi-librium point and equilibrium is when dU

dx= 0.

Central forces

Forces in which the force is directed along the line connecting two objects are called central forces.Examples are the gravitational force (force directed along the line connecting two masses) and theelectrostatic force (force directed along the line connecting two charges).

For the gravitational and electrostatic forces U(r) 1r

and as the only variation is radial is inthe radial direction with compoment /r in that direction.

What is U for U 1r?

Gravity: U = Gm1m2r

F = Gm1m2r2

r The minus sign indicates that the forceis directed in the opposite direction tothe position vector

Electrostatics: U = 14pi0

q1q2r

F = 14pi0

q1q2r2r The direction of the force vector de-

pends on the signs of the relativecharges

Consequences of a central force

Keplers First Law

Keplers first law states that All planets move in elliptical orbits with the Sun at one focus. Thisis actually a particular case when the force is an inverse-square law force (F 1/r2). In general itcan be shown that in the case of an inverse-square law particles will follow a conic section

1

r=

1

se(1 e cos )

where e is known as the eccentricity and s determines the scale of the figure.

There are four types of possible curve and they are determined by the value of e which is relatedto the total energy (E = Ek + U) of the particle

Hyperbola e > 1 E > 0Parabola e = 1 E = 0Ellipse 0 < e < 1 E < 0Circle e = 0 E < 0

Keplers Second Law

Keplers second law which is stated as Orbit sweeps out an equal area per unit time is actuallyan expression that angular momentum is conserved.

Show that for a central force where F is directed along r angular momentum is conserved.Hint: Consider the torque in this case.

Show that if angular momentum is conserved an orbit sweeps out an equal area per unit time.

(Note it is possible to do this by considering themagnitude of the area which is swept out but itcan also be done vectorially.

The vector which is associated with an area pointsperpendicularly out from the area and its magni-tude is the magnitude of the area.)

The adjacent diagram shows A the area sweptout when the position vector changes by r.

Forces

Gravity: U = Gm1m2r

, F = Gm1m2r2

r

Electrostatics: U =1

40

q1q2r

, F =1

40

q1q2r2r

Nuclear forces are short range: U er

r,

What depends on:1. Force being central?

2. Potential being 1r?

PHYS202 Angular Momentum, September 18, 2014 p.5/14

Angular MomentumConservation

= r FBut for a central force F r so = 0.

r

F

v


Keplers Second Law[SJ 13]

Orbit sweeps out equal area per unit time.

r

r+r

r

A

AreaA =

1

2rr.

dA

dt=

1

2r v = 1

2m(r p) = 1

2mJ. Conserved!


Keplers First LawElliptical with sun at one focus.

If force is not 1r2

the ellipse precesses.

PHYS202 Angular Momentum, September 18, 2014 p.8/14Keplers Third Law

Keplers third law is The square of the orbital period of any planet is proportional fo the cube ofthe semi-major axis of the elliptical orbit.

We will consider the case of circular orbits. The relationship between T and r depends on the formof the potential or force.

Show that in the case of a circular orbit by a mass m1 about a large mass m2 with magnitude ofthe force given by

F =Gm1m2r2

that T 2 r3.

Show that in the case of circular orbit of a particle in a 3-dimensional harmonic oscillator potential

U =1

2kr2

that the period is independent of the radius.

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