mechanics

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) is zero however the centripetal acceleration (pointing radially inward) isnot zero 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

q1q2r2

r 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 to 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.

Reduced mass

In the situations that we considered above, when we calculated the relationship between T and r,one mass was much larger than the other mass, and we assumed that the large mass was essentiallystationary while the second mass orbited around it. If the masses of the objects are similar thenthe second mass will also be accelerating, and in the case of a bound system, for example, the twomasses will orbit around their centre of mass. An example of this is a binary star system.

The centre of mass position will move at constant velocity and can be used to define an inertialframe.

In the diagram m1 and m2 interact via a centralforce colinear with the vector r, r1 and r2 are theposition vectors of m1 and m2 referred to someinertial frame origin.

Recall that the position of the centre of mass is given by

RCOM =m1r1 +m2r2m1 +m2

.

In this case we need to consider the coupled pair of Newtons law equations

m1d2r1dt2

= F (r12)r (eq.1) m2d2r2dt2

= F (r12)r (eq.2)

Note, the minus sign in the second equation, as the masses feel equal but oppositely directed forces.

It is possible to rearrange these equations in such a way that their solution is equivalent to solvinga single particle equation such as we solved above but instead of, for example m1 appearing as itdid on the left-hand side above, the reduced mass must be used:

d2r

dt2= F (r12)r (eq.3)

where =

m1m2m1 +m2

.

Subtract (eq. 2) from (eq. 1) to obtain (eq. 3) :

Show that if instead we had chosen to add (eq.) 1 and (eq.) 2 we would have merely obtained astatement of the conservation of linear momentum:


Rigid Bodies

Serway and Jewett Chapter 9

Main Concepts

1. Review of basics of rigid bodies.

2. Inertial tensor, wobbles, gyroscopes.

We wish to consider rotating, solid objects andcalculate quantites such as their kinetic energyand angular momentum. The normal relation-ships for these quantities still hold. However itis complicated to use them as different parts ofthe solid body have different speeds, as the speedof a particular part of the solid is dependent onthe radial distance of that part from the rotationaxis. It is more convenient to be able to use theangular speed which is constant throughout thesolid. In order to do this a quantity called themoment of inertia which takes into account howthe mass in an object is distributed with respectto the rotation axis.

The moment of inertia isI = imir

2i

for rotating discrete masses and

I =

V

r2dm

for rotating continuously distributed masses. In the first expression ri is the perpendicular distanceof the mass mi from the rotation axis and similarly in the second expression r is the perpendiculardistance from the rotation axis for the differential mass dm.

The moment of inertia of an object depends on the orientation and placement of therotation axis.

Examples:

Form an expression for the moment of inertia of the object in the figure for the two differentplacements and orientations of the rotation axes indicated in the figure.

Consider the rod in the adjacent figure.(a) Calculate the moment of inertia about the yaxis.(b) Calculate the moment of inertia about the y

axis.

Parallel axis theorem

Calculating the moment of inertia can be difficult but the parallel axis theorem often simplifies thecalculation. The parallel azis theorem is that

I = ICM +MD2

where I is the moment of inertia about the desired axis an dICM is the moment of inertia about anaxis which is parallel to the desired axis which passes through the centre of mass of the object, Mis the total mass of the object and D is the perpendicular distance between the desired axis andthe centre of mass.

We will now prove this theorem:

We want to calculate the moment of inertia through the point marked by the origin O. This isgiven by

I =

r2dm =

(x2 + y2)dm.

Now we can write the coordinates x and y in terms of the x and y coordinates of the centre of massusing the length definitions shown in the diagram, x = x + xCM and y = y + yCM.

Substitute these expressions for x and y into the expression for I and expand to obtain the parallelaxis theorem.

Perpendicular axis theorem

A second useful for finding the moment of inertia of a planar object is the perpendicular axistheorem. For any planar object such a a round plate or square tile, the theorem states that themoment of intertia abour an axis perpendicular to the plane of the object is equal to the sum ofthe moments of inertia of the object about any two axes in the plane that pass through the samepoint and are perpendicular to each other

Iz = Ix + Iy.

Consider the adjacent diagram, (but ignore theaxis going the the centre of mass - I didnt have acopy of the relevant figure). Pick a point on theobject in the diagram and draw a line from theorigin to the point and label its length r, its xcordinate x and its y coordinate y.The contribution of dm to the moment of inertiaIz with respect to the z axis is r

2dm . If we shouldconsider rotation about the x axis it would con-tribute y2dm and considering rotation about they axis it would contribute x2dm

Ix =

y2dm

Iy =

x2dm

Add Ix and Iy together to obtain the perpendic-ular axis theorem result.


Moment of Inertia (continued)

Example: A uniform solid cylinder has a radius R, mass M and length L.

(a) First it is set spinning about its central longaxis. Calculate the moment of inertia.

(b) Then it is set spinning about an axis which parallel to the lenght of the cylinder but througha point half way between its outer surface and centre. Calculate the moment of intertia.

Angular Velocity

In the Lecture # 3 Lecture Notes we recalled the relation v = r between the linear speed v andthe angular velocity for an object moving in circular motion where r is the radius of the circularmotion. We also introduced the notion that can be considered a vector quantity . Here weexamine again the relation between and v in particular when considering solid objects.

Consider the adjacent figure.

The distance R is the quantity referred to as rin the expression v = r while r is the positionvector which has magnitude r. We can see thatR = r sin . In the case that r is the positionvector, as in the diagram, then the relationshipbetween and v is

v = r

Kinetic energy and angular momentum of rotation solid objects

We want to examine quantities such as the kinetic energy and angular momentum of a rotatingsolid object such as the one in the figure. We can consider a mass element mi in the figure, find thekinetic energy and angular momentum of the mass element, and then sum all the mass elementswhich make up the body. It is necessary to consider individual mass elements as they have differentvi . However all of the elements have the same and throughthe use of the moment of inertia wecan obtain expressions for the kinetic energy and angular momentum which apply to the body asa whole (rather than individual mass elements). We show this as follows:

we consider a thin slab or a thin distribution ofmatter as in the adjacent figure (other objectscan always be divided into many thin slabs whichmeans the results we are about to obtain applyin general). The slab is rotating in its own planeabout the point marked O. C in the diagram isthe centre of mass and rC is the position vectorof the centre of mass. The vector is constant inits direction and magnitude. In the case of a thinslab r lies in the slab and is perpendicular to sothat the speed of each mass element is vi = ri.

The kinetic energy for the total slab is

K =1

2miv

2i =

1

2

(mir

2i

)2 =

1

2Iz

2

where we recognized the quantity (mir2i ) as the moment of inertia about the z axis Iz. Here we

have used the subscript z to denote that this is the moment of inertia for rotation about the zaxis. (Note (once again!) that the moment of inertia depends on the position and direction of therotation axis.)

In this case O is not the centre of mass for the object. We can use the parallel axis theorem torelate the moment of inertia of the object about the axis through O to that through the centre ofmass,

IOz = ICz +Mr2c

where rc is the separation betwen C and O, and in turn the kinetic energy in terms of ICz is

K = ICz2 +Mr2c

2

The angular momentum of the rotating slab is

J = ri pi = rimiviz = mir2iz = Izz

which again we can use the parallel axis theorem to write as

J = ICzz +Mr2cz

PHYS 205 Mechanics SectionLectures # 8 and #9

Rotations about fixed axes: time dependence of the motion

We wish to apply the angular equation of motion

dJ

dt=

to problems with rigid bodies rotating about fixed axes so as to learn about the time dependenceof the motion in response to given torques. Since the direction of the rotation axis is constrainedto be fixed in space and in relation to the body the inertial properties of the body will be constantin relation to this axis. For this reason we only need to look at the component of the anglularmomentum related to this axis. For this reason the equatoin of motion can be treated as a scalarequation dJaxial/dt = axial where Jaxial and axial refer to the components of and J parallel to therotation axis.

In this case it can be shown that the equation of motion is

Iaxiald

dt= axial

where Iaxial is the moment of intertia about the rotation axis.

To show that dJaxial/dt = axial is equivalent to Iaxiald/dt = axial we need to find the axialcomponent of the vector J . The axial component is the component in the direction of so we canfind it by taking the dot product of J with the unit vector /:

Jaxial = J

=1

[ri mi( ri)]

The last part of this expression can be simplified using the vector identity

A (BC) = (A C)B (A B)C

meaning thatri ( ri) = r2i (ri )ri

and[ri ( ri)] = r2i2 (ri )2 = r2i2 r2i2 cos2 = 2r2i sin2

so that

Jaxial =1

imi

2(ri sin )2 = imi(ri sin )

2

and we recognise imi(ri sin )2 as the moment of inertia I of the object about the rotation axis.

Example (tutorial 3 question 1)

Rotations about fixed axes: behaviour of the angular momentum vector

In general the angular momentum vector J will not be parallel to the axis of rotation and in thiscase the direction of J will change as the object rotates. Since rotation about a fixed axis impliesa circular motion of all features of the object we can see that the vector bj will rotate about therotation axis. If J is changing in direction then dJ/dt can involve a term related to the change indirection as well as a term related to any change in magnitude of J .

To illustrate this we consider a rigd body whichconsists of two equal masses connected by a mass-less rod which is made to rotate about a fixed axisthrough the centre of mass and orientated at anangle to the rod. The adjacent figure shows thissystem at an instant in its rotation when the rodcoincides with the xy plane (but note that as itrotates the rod will not in general lie in the xyplane). The rod is length 2a and its angular ve-locity vector lies along the x-axis (this is alwaysthe case).

The angular momentum is by its general definition:

J = r1 m( r1) + r2 m( r2)where r1 and r2 are the position vectors of the two masses and we denote the particle which is inthe positive xy quadrant as particle 1 and the other as particle 2. Write down expressions for theposition vectors of the two masses:

r1 = x+ y

r2 = x+ y

= x

Evaluate J using

a b = (aybz azby)x+ (azbx axbz)y + (axby aybx)z

J = 2ma2 sin (sin x cos y)(Note that this expression for J applies for the instant of the rotation shown in the figure whenthe rod lies in the xy plane.)

Now since J rotates its time derivative dJ/dt is not zero.

At the instant of the rotation we are considering, what is the direction of dJ/dt? Think about inwhich direction J will next point and about the vector dJ .

We can compare to the direction of v = dr/dt for a particle undergoing circular motion. We notedearlier that

v = rand similarly we can see that

dJ

dt= J

which means that

=dJ

dt= J

Show that in the case we are considering at this instant:

= 2m2a2 sin cos z

General situation

We have just considered a specific example (two masses connected to a rod with the rotation axisat an angle to the rod) where J is changing. To deal with the general case we write

J =i

ri mi( ri)

in terms of the components:

ri = xixR + yiyR + zizR

i = xxR + yyR + zzR

where xR, yR, zR are unit vectors fixed relative to the rotating body.

Show that:

Jx =i

mi(y2i + z

2i )x

i

mixiyiy i

mixiziz

Jy = i

miyixix +i

mi(z2i + x

2i )y

i

miyiziz

Jz = i

mizixix i

miziyiy i

mi(x2i + y

2i )z

For neatness and convenience these equations are written

Jx = Ixxx + Ixyy + Ixzz

Jy = Iyxx + Iyyy + Iyzz

Jz = Izxx + Izyy + Izzz

where the Ixx etc are just defined by comparing to the corresponding terms in the first set ofequations. It can be seen that the diagonal elements Ixx, Iyy, Izz are just the moments of intertiaabout the respective axes, for example, Ixx is the moment of inertia aout the x-axis. The off-diagonalterms are known as the products of inertia and occur in symmetric pairs Ixy = Iyx.

It is possible (we wont prove but you can hopefully see for symmetric objects that its true) thatit is possible to pick axes in the body in such a way that the products of inertia vanish. Such axesare known as principal axes.

So assuming we have principal axes, so the products of inertia (Ixy etc) vanish and then we candrop the double subscripts off the moments of inertia (Ixx etc) and we have:

J = IxxxR + IyyyR + IzzzR

and we remind here that Ix means the moment of inertia of the object about the x-axis. Now thetime-dependence involves the change of angular velocities and change of orientation:

dJ

dt= Ix

dxdtxR + Iy

dydtyR + Iz

dzdtzR

+IxxdxRdt

+ IyydyRdt

+ IzzdzRdt

Again thinking about the direction of the time variation of the unit vectors as we thought aboutthe direction of dJ/dt we can see that

dxRdt

= xR, dyRdt

= yR,dzRdt

= zRso

dJ

dt=dJdt

+ Jwhere the first term is from the variation of angular velocities and the second from rotation of theprincipal axes.

Finally we have the equations of motion:

=dJ

dt=dJdt

+ J

which in components is:

x = Ixdxdt (Iy Iz)yz

y = Iydydt (Iz Ix)zx

z = Izdzdt (Ix Iy)xy

These equations are known as the Euler Equations.

Consider the example of the two masses at the end of the rod that we looked at earlier and findthe torque using Eulers equations.


Gyroscope or spinning top

The diagram opposite shows a simple spinning topconsisting of a circular disk of mass M and radiusa. The tip of the stem is at O and the centre ofmass of the disk is at C a distance l from the tip.There are two coordinate frames shown. XY Z isan inertial reference frame and xyz are rotatingprinciple axes. The principal axes move with thestem of the top but do not spin with the diskabout the stem.The axis Oz is along the stem, Ox is always inthe horizontal XY plane, and Oy inclines belowthis plane by an angle which is the same as theangle by which Oz inclines away from OZ.

The projection of the centre of mass upon the XY plane falls at C and OC is shown at an angle from the X axis in the horizontal plane.

So the orientation of the stem is given by and and the motion of the stem is given by thevariation of these angles. The disk spins about the stem at the rate S rad/s as viewed from thexyz frame but the total angular velocity will generally involve any variations in the and so thetotal angular velocity vector will be the sum

= x+ Z + Sz

We can express this entirely in terms of the principal axes if we write Z in terms of y and z.

Z =

Show that this means that

= x sin y + ( cos + S)z

What are Ix, Iy and Iz?

UseJ = Ixxx+ Iyyy + Izzz

to show that

J =

(1

4Ma2 +M`2

)(x sin y

)+

1

2Ma2

( cos + S

)z

In general a gyroscope has very complex behaviour. We will examine a specific simplified case whenthere is steady precession = 0 and assume that S . In this case

J =1

2Ma2Sz

and the z axis rotates around the Z axis:

Z = Z.

Only the second term ofdJ

dt=dJdt

+ J

contributes, because and S are constant in magnitude (recall the first term in the dJ/dt expressionis referring the change in magnitude of the angular velocity, with time and the second term to thechange in direction of the J vector).

dJ

dt= Z J

= Z 12Ma2Sz

Show thatdJ

dt= 1

2Ma2S sin x.

The torque is due to gravity. Draw a diagram to show that

= Mg` sin x

recalling that = r F.

Using = dJ/dt show that

=Mg`

12Ma2S

=Mg`

IzS

for this special case of a gyroscope with = 0 and S .


Noninertial Coordinate Systems

The S frame has coordinates x0(t), speed v0(t)and acceleration a0(t) relative to frame S.

x = x + x0dxdt

= dx

dt+ dx0

dt

v = v + v0d2xdt2

= d2xdt2

+ d2x0dt

a = a + a0

We can apply Newtons law in the S frame

ma = F

using the expression for a above we have

ma +ma0 = F

F = FInertial force

ma0Fictitious force

So we can use Newtons law in S as long as we modify it by adding a noninertial or fictitiousforce.

Tutorial problem example

A boy racer travelling at 80 km/hr notices a police road block and applies his brakes. A bystandernotices that the cord that attaches a pair of fluffy dice to the boy racers rear-view mirror is at anangle of 40 from vertical. Use the concept of non-inertial forces to calculate the acceleration ofthe car. Explain any assumptions you have to make.

Rotating frames

Rotating frames are very relevant given that theEarth is a rotating frame.SI is an inertial frame with coordinates for pointP as xI , yI , zIFrame SR is rotating at constant about z. PointP has coordinates xR, yR, zR

xI = xR cost yR sintyI = xR sint+ yR cost

zI = zRNow we want to find expressions for the rotating v and a

xI = xR cost xR sint yR sint yR costyI = xR sint+ xR cost+ yR cost yR sintzI = zR

xI = xR cost 2xR sint 2xR cost yR sint 2yR cost + 2yR sint

yI = xR sint + 2xR cost 2xR sint+ yR cost 2yR sint 2yR cost

zI = zR

Rotation Coriolis Centrifugal

Example

Imagine an object projected radially outward.The upper figure shows the path of the particleas viewed from an inertial reference frame, andthe lower panel shows the path of the particle (forvarious values of the ratio of the rotation speedand the speed of the particle) as viewed from arotating reference frame.

yI = yR = 0, yR = 0 sint = 0, cost = 1

xI = xR + 0 2xR inward centripetal acceleration

FRx = FIx +m2xR

outward centrifugal force

yI = yR +2xR upward acceleration

+0

FRy = FIy 2mxR downward Coriolis force

mechanics

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