1Chapter 2Motion along a Straight

Line

Terms we will use:

Position, distance, displacement Speed, velocity (average and instantaneous)

Acceleration (average and instantaneous)

Coordinate Systems

A coordinate system is used to describelocation.

A coordinate system consists of: a fixed reference point called the origin

a set of axes a definition of the coordinate variables

coords

Example

The position of an object is its location in a coordinate system.

x

y

(4,2)

cartesian

The arrow indicates thepositive direction.

(-3,5)

Today well deal only with one-dimension, and usually call it x

x

(0,0)(0,0) = the origin

Distance and displacement

Displacement is defined as the change inposition of an object.

It can be positive, negative or zero. Displacement is a vector (see Chapter 3)

Distance is the total length of travel.

It is always positive.

It is measured by the odometer in your car.

xx1x2

2Average Speed and Velocity

Speed and velocity are not the same in physics!

SI units of speed and velocity are m/s.

(always positive, or zero)

(positive, negativeor zero)

velocity is a vector

xx1x2

ExampleWhat is the average speed of a person at the equator dueto the Earths rotation?

What is the average velocity of a person at the equator dueto the Earths rotation?

The Dragster Graph of the Position of the Dragsteras a Function of Time

3Position vs. Time Plots

The average velocity between two times is the slopeof the straight line connecting those two points.

average velocity ispositive

average velocity isnegative

Instantaneous VelocityThe velocity at one instant in time is known as the instantaneousvelocity and is found by taking the average velocity for smallerand smaller time intervals:

(e.g. the speedometer indicates themagnitude of instantaneous velocity)

x

t

in the limit this gives thetangent to the curve

Velocity as a Slope of Tangent Lineon a Graph of x vs. t

On a graph of position as a function of time, the slope of the tangent lineat any point is the velocity at that point.

Acceleration

Often, velocity is not constant, rather itchanges with time

The rate of change of velocity is known asacceleration

This is the average acceleration

Acceleration is a vectorThe SI unit of acceleration is m/s2

positive, negative or zero

4Acceleration

for example:

car is advertised to do zero to sixty in 6 secondsin this case 60 is 60mph = 27m/s

so the average acceleration is

Instantaneous Acceleration

If we wish to know the instantaneousacceleration, we once again let t 0:

Graph of Velocity vs. Timefor Accelerating Object Four Ways to Represent the Motion

of a Race Car Moving with ConstantAcceleration

in a few slides well see that we can describe these graphs by somesimple equations

5DecelerationDeceleration

refers to decreasing speed

is not the same as negative acceleration

occurs when velocity and acceleration have opposite signs

Example: A ball thrown up in the air. The velocity is upwardbut the acceleration is downward. The ball is slowing down asit moves upward. (Once the ball reaches its highest point andstarts to fall again, it is no longer decelerating, but isaccelerating)

Velocity vs. Time PlotsFor these curves, theaverage accelerationand theinstantaneousacceleration are thesame, because theacceleration isconstant.

in what region do we have deceleration?

can we sketch the position versus time?

Example: Velocity vs. Time Plot

time(s)

velocity (m/s)

0

4

2

-2

6

5

-4

10A

C

B

1. What is the velocity at time t = 3 sec?

2. What is the velocity at point A?

3. When is the acceleration positive?

4. When is the acceleration negative?

5. When is the acceleration zero?

6. When is the acceleration constant?

7. When is there deceleration?

8. What is the acceleration at point C?

9. What is the acceleration at time t = 6 sec?

10. During what 1 s interval is the magnitude of theaverage acceleration greatest?

Example be careful with signs

A car moves from a position of +4.0 m to a position of -1.0 m in2.0 seconds. The initial velocity of the car is -4.0 m/s and the finalvelocity is -1.0 m/s.

(a) What is the displacement of the car?(b) What is the average velocity of the car?(c) What is the average acceleration of the car?

Answer: (a) x = x2 x1 = -1.0 m - (+4.0 m) = -5.0 m

(b) vav = x/t = -5.0 m / 2.0 s = -2.5 m/s

(c)deceleration!

6Motion with Constant Acceleration

If acceleration is constant, there are four useful formulaerelating position x, velocity v, acceleration a and time t:

v0 = initial velocity

x0 = initial position

t0 = initial time assumedhere to be at 0 s.

If t0 0, replace t in these formulae with t t0

Understanding theequations

always check whatis being plotted !

Understanding theequations

straight line, slope a

parabola

7Example: Lets go back to our previous example ofthe car and assume that the acceleration isconstant. We found that the average accelerationwas = 1.5 m/s2

Lets calculate the acceleration.Recall

v = v0 + at -1 m/s = -4 m/s + a(2 s) 3 m/s = a(2 s)

a = 1.5 m/s2

better still - rearrangethe equation for a - thenwe can use it for anyvalues:

this was our definition of average accn (with t0 = 0) - if the accnis constant then it is the same as the average

Whats the use of these equations?

Much of physics involves knowing the values of some quantities andusing equations to find the values of unknowns

e.g. say we know that a car was moving with velocity 4 m/swhen it was 10m down a track, and we wanted to knowhow fast it was moving 50m down the track if we also knowits acceleration is constant at 3 m/s2

which equation do we use?

Whats the use of these equations?

Much of physics involves knowing the values of some quantities andusing equations to find the values of unknowns

e.g. now say we were traveling at constant acceleration ofunknown value and moved from rest at the origin to aposition 10 m away where we had velocity of 3 m/s

which equation do we use to find:

(a) the time taken?

(b) the acceleration?

Freely Falling Objects

Near the earths surface, the acceleration due to gravity (g)is roughly constant:

g = aEarths surface = 9.80 m/s2 toward the center of the earth

Free fall is the motion of an object subject only to theinfluence of gravity (not air resistance).

An object is in free fall as soon as it is released, whether itis dropped from rest, thrown downward, or thrownupward

Question: What about the mass of an object? Answer: The acceleration of gravity is the same for all

objects near the surface of the Earth, regardless of mass.

8Calculate y and v as a function of t.

Example 2.10 - Falling Euro in PisaCoin is dropped. Starts from rest and falls freely. Compute its positionand velocity after 1.0, 2.0, 3.0 seconds.

-29.4-44.13.0

-19.6-19.62.0

-9.8-4.91.0

v (m/s)y (m)t (s)

throwing a ball in the airvelocityvector

accelerationvector

throwing a ball in the air

?at the highest point

velocityvector

accelerationvector

v = 0a = 0

v = 0

A:

B:

C:

9throwing a ball in the air

v = 0

throwing a ball in the air

a

t

-9.80 m/s2

constant a = -g = -9.80 m/s2

throwing a ball in the air

v

t0

v = v0 + a t = v0 - g t

v0

throwing a ball in the air

y

t

10

Example 2.11 - A ball on the roofSuppose you throw a ball vertically upward from the flat roof of a tall building. The ball

leaves your hand at a point level with the roof railing, with an upward velocity of15.0 m/s. On its way back down it just misses the railing. Find

(a) the position and velocity of the ball 1.00 s and 4.00 s after it leaves your hand

(b) the velocity of the ball when it is 5.00 m above the railing

(c) the maximum height reached and the time at which it is reached

Ignore the effects of the air.

Example 2.11 - A ball on the roof

What do we know?

initial velocity (v0 = 15.0 m/s)

acceleration (constant)(a = -g = -9.80 m/s2)

initial height (y0 = 0)

Example 2.11 - A ball on the roofProblem Solving Strategy

Make a list of given quantities Make a sketch

Draw coordinate axes identify thepositive direction

Identify what is to be determined Pick the right equation(s)

Be consistent with units Check that the answer seems reasonable

1Reviewing what weve learntVectors

components

x

45

4.0 m

x

30

3.0 mA

B

y y

Ay

Ax

By

Bx

x

y

Ay

By

BxAx

A

B

C

Cx

Cy

+ =

A + B = C

adding vectors, by components

note B = C - A

Reviewing what weve learntVelocity, acceleration and equations of motion for constant

acceleration

slope of tangent of x vs. t graph

slope of tangent of v vs. t graph

Chapter 3Motion in Plane

Motion in the x direction is independent frommotion in the y direction. We use the sameequations as in Chapter 2, but for eachdimension separately.

There are not really many new equations in thischapter.

Motion in a plane

Often, motion in three-dimensions is really motion in a plane (two-dimensions).

e.g. if theres no wind, a field-goal kick can be considered to move in vertical plane

In this lecture well consider some generalities then two specific,common examples:

projectile motion (in a plane, with gravity vertically)

uniform circular motion

2Position Vector Specifies Location and Displacementof an Object in an x-y Coordinate System

A position vector points from the origin to the particle.

note: y vs. x

recall:

co-ordinates indicate positionrelative to fixed axes (here x, y)

position vector points from theorigin to the object

vector has components - for theposition they are the co-ordinates

Velocity in two-dimensions

Velocity is a vector:average velocity defined as thechange in position vector in time

note: y vs. x

Velocity in two-dimensions - components

Velocity is a vector:average velocity defined as thechange in position vector in time

can completely separate intocomponents - hence nothingnew in this chapter !

Instantaneous Velocity

The velocity vector has a magnitude equal to the speed of theobject and points in the direction of motion.

be careful here - this is not a graph of x vs. t, so the slope is not the velocity.

all were getting here from the tangent is the direction

3 Instantaneous Velocity

The velocity vector has a magnitude equal to the speed of theobject and points in the direction of motion.

recall the trig. propertiesof vectors:

A model carRadio-controlled model car on a tennis court. The surface of the court represents

the x-y plane and you stand at the origin. At time t1 = 2.00 s the car has x, yco-ordinates (4.0 m, 2.0 m) and at t2 = 2.50 s it is at (7.0 m, 6.0 m). For thetime interval t1 to t2 find

(a) the components of the average velocity

(b) the magnitude and direction of the average velocity

DragonflyA dragonfly follows the path shown, moving from point A to

point B in 1.50 s. Find

(a) the x & y components of its position vector at A(b) the magnitude and direction of its position vector at A

(c) the x & y components of the average velocity betweenA & B

(d) the magnitude and direction of the average velocitybetween A & B

(e) indicate thedirection of theinstantaneousvelocity at A and B onthe diagram

Acceleration Vector

Just as in one-dimension, we can define an acceleration interms of the change in velocity

average

instantaneous

4Acceleration Vector

Important point:

If the velocity changes direction, there is an acceleration, even if there is no change in speed

this is what we considered inthe last chapter

this is new to two-dimensional motion

Acceleration Vector Components

Acceleration, just like all vectors, can be expressed in components

The Model Car Again

Were given more info. Suppose that at t1=2.00 s the car has components ofvelocity vx=1.0 m/s and vy=3.0 m/s and that at time t2=2.50 s the componentsare vx=4.0 m/s and vy=3.0 m/s. Find

(a) the components of the average acceleration

(b) the magnitude and direction of the average acceleration

5Projectile Motion

= motion in two-dimensions under gravity (well neglect airresistance)

Project a body with some initial velocity and watch it travele.g. firing a cannon

kicking a field-goal

throwing a baseball many other things

Only acceleration is in the vertical direction, is constant, andis due to gravity

Components of Projectile Motion

The motion separates into components.We separately solve the equations for the x-direction, then

the y-direction

y

x

y-position doesnt changeif the particle has an x-component of velocity

Components of Projectile Motion

y

x

notice that the velocitypoints in the direction ofthe tangent to the curve

Equations for Projectile Motion

y

x

6demo - cup catch

Shape of Projectile Motion

y

x

using our equations we can prove the trajectory has this parabolic shape

take the simplified case of x0 = 0, y0 = 0

Paintball GunA paintball is fired horizontally at a speed of 75.0 m/s from a point 1.50 m above

the ground.(a) for how many seconds is the ball in the air?(b) find the maximum horizontal displacement (or range)Ignore air resistance

for (a) we can solve the y-component problem: y0 = 1.50 m, v0y = 0, y = 0, t = ?

= 0.553 s

for (b) we must solve the x-component problem: x0 = 0, v0x = 75.0 m/s, x = ?,t = 0.553 s = 41.5 m

A home-run hitHit with an initial speed 37.0 m/s at an initial angle of 53.1o. Find(a) the balls position and magnitude and direction of velocity when t =2.00 s(b) the time when the ball is at its highest point and its height at that time(c) the horizontal range of the ball

well neglect the fact that the ball is probably hit about 1 m above the ground,and treat it as though it were at ground level

initial velocity components:

7A home-run hitHit with an initial speed 37.0 m/s at an initial angle of 53.1o. Find(a) the balls position and magnitude and direction of velocity when t =2.00 s(b) the time when the ball is at its highest point and its height at that time(c) the horizontal range of the ball

x = 44.4 m

y = 39.6 m

(a) @ t=2.00 s

(b) highest point, so not getting higher any more, i.e. vy = 0

t = 3.02 s

vx = 22.2 m/s

vy = 10.0 m/s

y = 44.7 m

A home-run hitHit with an initial speed 37.0 m/s at an initial angle of 53.1o. Find(a) the balls position and magnitude and direction of velocity when t =2.00 s(b) the time when the ball is at its highest point and its height at that time(c) the horizontal range of the ball

(c) horizontal range?

many ways to solve this

a simple way is to notice that the trajectory is symmetric about the midpoint

then t2 = 2 t1 = 6.04 sx = 134 m

Getting angry at the physics bookA physics book slides off a horizontal tabletop with a speed of 1.10 m/s. It strikes the floor in

0.350 s. Ignore air resistance. Find(a) the height of the tabletop above the floor(b) the horizontal distance from the edge of the table to the point where the book strikes

the floor(c) the horizontal and vertical components of the books velocity, and the magnitude and

direction of its velocity, just before the book hits the floor

person jumping off a cliffA daring swimmer dives off a cliff with a running horizontal leap, as shown below.

What must her minimum speed be just as she leaves the top of the cliff so thatshe will miss the ledge at the bottom?

8Uniform Circular MotionIf a particle moves along a circular path with constant

speed, it is said to be in uniform circular motion

Uniform Circular MotionInstantaneous acceleration vector is perpendicular to the

tangential velocity vector

It thus points toward the center of the circle -centripetal acceleration

a bit of geometry and algebra will give us auseful new formula:

Uniform Circular MotionInstantaneous acceleration vector is perpendicular to the tangential

velocity vector

It thus points toward the center of the circle -centripetal acceleration

Fast car, flat curveA car claims a lateral acceleration of 0.92 g. If this represents the maximum

centripetal acceleration that can be attained without skidding out of acircular path, and if the car is traveling at a constant 45 m/s (about 100mi/hr), what is the minimum radius of curve that the car can negotiate?

R = 230 m

9Carnival RideThe passengers in a carnival ride travel in a circle with radius 5.0 m.

They make one complete circle in a time T = 4.0 s. what istheir acceleration?

in 4.0 s they travel a distance

so their circular speed is

v = 7.9 m/s

= 12 m/s2

about 1.2g !

Stone on a stringYou swing a 2.2 kg stone in a circle of radius 75 cm. At what speed should you

swing it so its centripetal acceleration will be 9.8 m/s2 ?

1Chapter 6: Circular Motion andGravitation

We briefly discussed motion at constant speed in acircle

Now we know about forces we can look again atthis

Well also introduce Newtons theory of gravitationwhich often causes circular motion

Uniform Circular Motion - a reminderAn object moving at constant speed around a circular path has an accelerationpointing radially inward

no acceleration componentparallel to the velocity?

- that would change the speed!

we found a formula

Uniform Circular Motion - dynamicsIn Chapter 4 we learnt that accelerations are caused by forces - Newtons second law

there must a radial force causing

the acceleration

what this force is depends uponthe problem

e.g. ball on a string - force is thetension in the string

Uniform Circular Motion - dynamicsSo what happens if we remove the force? e.g. what if the string is cut?

vote:

1. particle stops where it is

2. particle moves at constantspeed along the radialdirection

3. particle moves at constantspeed in the tangentialdirection

2Uniform Circular Motion - dynamicsNewtons second links the force to the radial acceleration

REMEMBER: is not the force!

Example 6.1 Model airplane on a stringFly a propeller-driven model airplane on a 5.00m string in a horizontal circle. The airplane, which hasmass 0.500kg flies level and at constant speed and makes one revolution every 4.00 seconds. How hardmust you pull on the string to keep the plane flying in a circle?

radial accn

speed

Example 6.4 Rounding a banked curveAn engineer wants to bank a curve so that, at a certain speed, v, no friction is needed for a car to takethe curve. At what angle should it be banked if v = 25 m/s (56 mph)?

vertically - no acceleration

were using the horizontal component of the normal force to cause the radial acceleration

Example 6.5 Dynamics of a Ferris wheel rideA passenger of mass m, riding on a Ferris wheel moves in a vertical circle of radius R with constant

speed v.(a) assuming that the seat remains upright during the motion, derive expressions for the magnitude of

the upward force the seat exerts on the passenger at the top and bottom of the circle(b) What are these forces if m = 60.0 kg, the radius of the circle is R = 8.00 m and the wheel makes on

revolution in 10.0 s? How to they compare to the passengers actual weight?

at the top:

smaller than your weight

at the bottom:

larger than your weight

3Example 6.5 Dynamics of a Ferris wheel rideA passenger of mass m, riding on a Ferris wheel moves in a vertical circle of radius R with constant

speed v.(a) assuming that the seat remains upright during the motion, derive expressions for the magnitude of

the upward force the seat exerts on the passenger at the top and bottom of the circle(b) What are these forces if m = 60.0 kg, the radius of the circle is R = 8.00 m and the wheel makes on

revolution in 10.0 s? How to they compare to the passengers actual weight?

(b) arad = v2/R = 3.16 m/s2

nT = 60.0 * ( 9.80 - 3.16 ) N = 398 N

nB = 60.0 * ( 9.80 + 3.16 ) N = 778 N

Conceptual QuestionA box slides down a circular, frictionless ramp. The forces on the box at point A are:(a) the upward push of the ramp, which is greater than the inertial force ma of the box(b) the upward push of the ramp, which equals the downward pull of the earth(c) the upward push of the ramp, which is greater than the downward pull of the earth

Newtons Law of GravitationBy considering in detail the motion of the planets around the sun, Newton (after

much work by others) concluded the following:

every particle of matter in the universe attracts every other particle witha force that is directly proportional to the product of their masses andinversely proportional to the square of the distance between them

note the third law at work here

Newtons Law of GravitationThis is OK for particles, what about objects with size?

for spherically symmetric objects the law holds as if all the mass where at thecenter of the sphere

Hence the gravitational force on you, due to the earth is

4Newtons Law of GravitationWhat about this constant G, that tells us how strong gravity is?

We can do a high-precision experiment to try to measure it:say that we have two known masses, and we place them a known distance apart,then if we can measure the force between them, we can find G.

Cavendish torsion balance:

Gravity is very weak, but adds upSo G is very small indeed and hence gravity is very weak. Two 1kg masses, 1m apart

would exert a pull on each other of 7x10-11 N= 7 / 10,000,000,000 N !

But, gravity adds up - masses are always positive, there is, as far as we know, noanti-gravity

So we can get big gravitational pulls when objects are very massive, like planets orstars,e.g. the earth has a mass 6x1024 kg, and I have a mass 80kg. I am standingRE=6x106 m from the center of the earth, so that force on me from the earthis about 7x10-11 x 6x1024 x 8x101 / (6x106)2 N = 9x102 N,

easily enough to stop me jumping off into space - so gravity isnt always weak

Example 6.7 Acceleration of Cavendish spheresSuppose one large (m = 0.500kg) and one small (m = 10.0g) sphere are placed 5cm

apart (distance between their centers). There are no other objects within avery large distance. What is the instantaneous acceleration of each sphere asviewed by a stationary observer?

WeightA while ago we suggested that weight was the force on you due to the gravitational

attraction from the earth.Now we have Newtons law of gravitation we can see whats going on:

but we said before that w = m g , where g is the acceleration due to gravity

as Galileo found, the acceleration due togravity is constant for all bodies, independentof their mass

5Example 6.9 Gravity on MarsMars has a radius 3.38x106m and a mass 6.42x1023kg. Weve built a lander whose

weight on earth is 39,200 N. Calculate the weight of the Mars lander and theacceleration due to Marss gravity

(a) at the surface of Mars(b) 6.00x106m above the surface of Mars

mass of lander?

(a) weight of lander at Martial surface?

= force on the lander due to Marss gravity

acceleration due to Marss gravity then follows fromNewtons 2nd law (F = ma)

Satellite Motionlets try out a thought experiment - imagine a motorcyclist drives off the edge of a

cliff

we studied his free-fall motion,

the faster his initial horizontal speed,the further away he lands

now imagine he lands far enough awaythat we have to consider the curvatureof the earth

eventually he could go so fast that henever hit the earth

he becomes a satellite

Satellite Motionwell just consider the case where satellite motion is circular - this is approximately

true for most man-made satellites and for several of the planets in the solarsystem

We have a force acting on a satellite due to the gravity of the planet it orbits. Thisis the only force and it is directed toward the planets center

by Newtons 2nd laworbital time?

Conceptual ProblemsA satellite in circular orbit travels at constant speed because(a) the net force acting on it is zero(b) the pull of gravity is balanced by centrifugal force(c) there is no component of force along its direction of motion

The moon is pulled directly toward the earth by the earths gravity, yet it does notfall into the earth because

(a) the net force on it is zero(b) the pull of the earth is balanced by the centrifugal force(c) the moon pulls back on the earth with an equal and opposite force(d) but it is constantly falling!

6Example 6.10 A weather satelliteSuppose we want to place a weather satellite into a circular orbit 300km above the

earths surface. What speed, period, and radial acceleration must it have?

radius of orbit = 6380km + 300km = 6680km = 6.68x106m

v = 7730 m/s

T = 5430 s = 90.5 min

arad = 8.95 m/s

mE = 5.98x1024 kg

Determining the masses of planetsNot an easy thing to do. If we can see moons orbiting the planet we can use the

law of gravitation to estimate the mass of the planet. We need only to measurethe moons orbital period and radius of orbit.

(Apparent) WeightlessnessAn astronaut in a spacecraft orbiting the earth appears to be weightless. How can

this be, the earths gravity still acts upon them?

Easier to understand this way. Imagine you are in an elevator without a safety breakwhen the cable is cut. Both you and the elevator are in free-fall withacceleration g.

Imagine you are standing on a bathroom scale in the elevator - it reads zero, youare weightless. It must since you are accelerating downwards at g and hence thenet force must just be gravity, therefore no normal force, therefore no readingon the scale.

Orbital motion is also, essentially, free-fall and hence we observe the samephenomenon.

QuizzesIf the Earth had twice its present mass, its orbital period around the sun (at the

present radius) would be(a) years (b) 1 year (c) year (d) 2 years

If the sun had twice its present mass, the Earths orbital period around the sun (atthe present radius) would be

(a) years (b) 1/2 year (c) year (d) 2 years

In the 1960s, during the cold war, the USSR put a rocket into a circular orbit aboutthe earth. A US senator expressed concern that a nuclear bomb could bedropped on the US as it passed over. What would happen if a bomb werereleased from the rocket

(a) it would drop on the US, directly below(b) it would follow a curved path and land elsewhere(c) it would remain in the same orbit as the rocket(d) it would move in a straight line into space

1Chapter 7: Work & Energy

The law of conservation of energy is a verypowerful one that can make solving complicatedproblems simple.

We need to introduce the energy of a system andhow one may introduce energy into a system bydoing work

making a start on Chpt 7 : Work & Energya very important and general law of physics is that the energy in a closed system is

conserved. That is while it can be transformed into different forms, it cannotbe created or destroyed.

what types of energy do we need to know aboutkinetic energy - the energy a body has due to its motion

potential energy - energy stored in a system that has the potential to be released

example would be gravitational potentialenergy being changed into kinetic energy

conservation of energy

vertical projectile motion can be analyzed interms of the conserved energy of the particle

kinetic

potential

kinetic

energy dissipationkinetic and potential energies are mechanical

there is also non-mechanical energy

e.g. there is some air resistance on the rock, slightly slowing it down and hence reducingits energy

this energy is not lost but is transferred when it slightly heats the air - becomesinternal energy of the air

As another example think of friction, where the friction force slows objects down,reducing their kinetic energy.

2Worktheres a very specific definition of what physicists mean by work.

the simplest case is when a constant force is used to move an object with thisdisplacement parallel to the force

then the work done by the force, W = F s

more generally, if the force vector makes an angle with the displacementvector, then the work done by the force is W =

WorkSI units of work are clearly N m (Newton - meters). 1 N m is also known as 1 J,or one Joule (pron. jewel)

Work is a scalar quantity - it does not possess a direction.

Well see soon that work is the transfer of energy

Example 7.2 - Sliding down a rampA package with mass, m, is unloaded from a truckwith an inclined ramp, as shown. The ramp hasrollers that eliminate friction, and the truck unloadsfrom a height, h. The ramp is inclined at an angle .Find an algebraic expression in terms of thesequantities for the work done on the package duringits trip down the ramp.

normal force perpendicular todisplacement - does no work

weight has a component parallel to thedisplacement - does some work

look at the geometry of the slope

so

only the vertical distance matters

Example 7.3 - Work done by several forcesTractor pulls a sled of wood a distance of 20.0 m along level frozen ground. The total weight of sledand load is 14700 N. The tractor exerts a constant force FT with magnitude 5000 N at an angle 36.9o

above the horizontal. A constant 3500 N frictional force opposes the motion. Find the work done onthe sled by each force individually and the total work done on the sled by all the forces.

normal force and weight force are perpendicular tothe motion, so they do no work on the sled

tractor does work WT = FT cos s = 80.0 x 103 J

friction forces is in the opposite directionto the motion, so = 180o,

cos 180o = -1

Wf = - f s = -70.0 x 103 J

negative? means the sled is doing work -losing energy through friction (to theground)

3Example 7.3 - Work done by several forces

WT = 80.0 x 103 J

Wf = -70.0 x 103 J

Wtot = WT+Wf = 10.0 x103 J

OR, net force in direction of motion,

Fx,tot = FT cos - f = 4000 N - 3500 N = 500 N

Wtot = Fx,tot s = 10.0 x 103 J

Work & Kinetic EnergyThe work we do on an object is energy transferred to that object. This can takethe form of kinetic energy.

e.g. object moving along x under the action of a constant force Fx. The particlewill have constant acceleration, ax = Fx / m . We had formulae for motion withconstant acceleration, e.g.

so the work done increases the kinetic energy from

to

Kinetic EnergyThe energy possessed by a body due to its motion. Depends only on the mass ofthe body and its speed. Is a positive, scalar quantity.

SI units are Joules (J)

note that work done on the body can reduce the kinetic energy

if W is negative then vf < vi - this is the case where the force opposes themotion (e.g. v in positive x-direction, F in negative x-direction)

Example 7.4 - Using work and energy to compute speed

We found the total work done on the sled by all the forces is 10,000 J, so the kinetic energy of thesled must increase by 10,000 J. The mass of the sled is 1500 kg. Suppose the sleds initial speed vi is2.00 m/s, what is its final speed?

vf = 4.16 m/s

we can rearrange so the final speed is the subject:

4Example 7.5 - Forces on a hammerhead

In a pile driver, a hammerhead of mass 200kg is lifted 3.00m above the top of a vertical I-beam beingdriven into the ground. The hammer is dropped, driving the I-beam 7.40cm into the ground.The vertical rails guiding the hammerhead exert a constant 60.0N frictional force on it.

Use the work-energy relation to find

(a) the speed of the hammerhead just as it hits the I-beam, and

(b) the average force the hammerhead exerts on the I-beam

Example 7.5 - Forces on a hammerheadIn a pile driver, a hammerhead of mass 200kg is lifted 3.00m

above the top of a vertical I-beam being driven intothe ground. The hammer is dropped, driving the I-beam 7.40cm into the ground. The vertical railsguiding the hammerhead exert a constant 60.0Nfrictional force on it.

Use the work-energy relation to find

(a) the speed of the hammerhead just as it hits the I-beam, and

(b) the average force the hammerhead exerts on the I-beam

total work done by the hammerhead during the drop(with gravity and against friction)

Wdrop = Ftotal s = (w - f) s = 1900 N * 3.00 m = 5700 J

the hammerhead starts from rest, vi = 0 so

vf = 7.55 m/s

as the hammerhead pushes the I-beam 7.40cminto the ground it slows from 7.55 m/s down to0 m/s . Hence it loses kinetic energy of 5700 Jdoing work on the I-beam (via the normal force),gravity and the guide-rail frictionFtotal = (w - f - n) = 1900 N - nW = 5700 J = Ftotal s = (1900 N - n) * 0.00740m n = 79,000 Nthis is the force the I-beam exerts on thehammerhead, so by Newtons third, this is theforce the hammerhead exerts on the I-beam

Work done by a varying forceWeve dealt with the case of a constant force moving an object through adistance, what about the case where the force varies?

e.g. remember that the force required to stretch a spring obeys Hookes lawF = k x , so the more you stretch it, the more force is required to stretch itfurther.

there is a graphical technique that allows us to compute the work done by avarying force

Work done by a varying forceconsider an object moving along a straight line thathas a force acting on it that varies as it travels along

we can plot a graph of the force at each point alongthe path

we could split this up into very small segments ofpath - during each segment the force isapproximately constant, so the work done is

which is the area of the rectangle

hence the total work done is the sum of therectangle areas, or better yet, the area underthe curve

5Applying this to the stretched springWe apply a force equal to the Hookes law force at one end of a tethered spring toextend it

the area under the curve is the area of a triangle ofbase X and height kX

makes sense, the average force is and thetotal displacement is X

Potential EnergyWe just did work on a spring by stretching it - that is we supplied energy to it.Where did this energy go? Not into kinetic energy since the spring isnt moving.

We consider the spring to have potential energy. The stretched spring isstoring the energy. We could release it by say, putting a block on the end of thespring and releasing the block. Then the potential energy of the spring would betransferred into kinetic energy of the block.

An important form of potential energy is gravitational potential energy,which is energy associated with the height of an object.

Gravitational Potential EnergyConsider moving an object of mass m, from aheight yi to a height yf .

We can compute the work done againstgravity in increasing the height

Wgrav = w (yf - yi) = m g (yf - yi)

if the object fell from yf to yi this would be thework done by gravity on the object

Why not call m g y the gravitationalpotential energy as then the work done simplychanges the amount of gravitational potentialenergy the object possesses.

Conservation of energyIf the only force acting on an object is gravity, then the sum of the kinetic andpotential energy is conserved:

this is very useful

N.B. it doesnt matter where we put the origin (y = 0) - only differences in position everappear ( yf - yi & )

6Example 7.7 - Height of a baseball from energy conservation

You throw a 0.150kg baseball straight up, giving it an initial upward velocity of magnitude 20.0 m/s .Use conservation of energy to find how high it goes (ignoring air resistance).

initial total energy

at the highest point, v = 0,

so total energy

would also follow from

Elastic potential energyrecall that to stretch a spring by an extension X, we had to do work

which became the energy stored in the spring

- this is elastic potential energy

Example 7.8 - Potential energy on an air-track

A glider of mass 0.200kg sits on a frictionless, horizontal air-track, connected to a spring of negligible mass with forceconstant k = 5.00N/m. You pull on the glider, stretching the spring 0.100m, and then release it with no initial velocity.The glider begins to move back towards its equilibrium position (x=0). What is its speed when x=0.0800m?

initial kinetic energy = 0

initial potential energy = 1/2 k x2 = 0.0250 J

at x=0.080 m,

kinetic energy = 1/2 m v2 = ?

potential energy = 1/2 k x2 = 0.0160 J

0.0250 J = 0.0160 J + 1/2 m v2

Example 7.9 Maximum height of a home-run hitA few weeks ago we derived the maximum height of a projectile launched with initial speed at anangle with the horizontal. Derive this using energy conservation.

just kinetic and gravitationalpotential energy to deal with

initially: kinetic energy =

grav. potential energy =

at the highest point, kinetic energy =

grav. potential energy =

conservation of total energy:

pythagoras on the velocity vector

y-component of velocity is zero at the highest point,& there is no acceleration in x so vx = v0x

Example 7.10 - Speed on a vertical circleSkateboard down a quarter-pipe of radius R. Total mass of board and skater is m. Start from rest andmoves with no friction. Derive an algebraic relation for the speed at the bottom of the ramp.

7Non-conservative forcesEnergy is conserved in a closed system. We also may have to deal with the casewhere work is being done on the system by an external force. In this case energyis still conserved provided we take account of the incoming or outgoing energythrough work.

An example of this is when we have non-conservative forces doing work, e.g.friction which transfers energy out of the system

Example 7.12 - Work and energy on an air-track

Glider on an air-track with spring again. Suppose the glider is initially at rest at x=0, with the springunstretched. Then you apply a constant force of magnitude 0.610 N to the glider. What is the glidersspeed at x=0.100m

your applied force is external tothe glider-spring system

Example 7.13 - Skate with frictionSuppose the skate-ramp isnt frictionless so that the skaters speed at the bottom is 7.00m/s for a piperadius 3.0m and a skater mass 25.0kg. What work was done by friction on the skater?

energy being lost from the skater system through friction

1Chapter 8: Momentum

A simple way to understand what happens in acollision is to use a quantity called momentum.

Well find that the total momentum of a system isconserved if no external forces act on the system

Collisions and Momentuman 18-wheeler moving at 30mph collides with a Honda Civic moving at 30mph -

who comes off worse? Why?

We intuitively know the Civic is in trouble and the reason is that it has much lessmass.

Which is more dangerous - a bullet moving at 10mph or one moving at 50mph?

It is clear that both mass and velocity are important in these circumstances, so wedefine a quantity that includes both,

momentum,

(18 wheeler has more momentum,faster bullet has more momentum)

Momentum and Newtons 2nd Lawrecall the mathematical formulation of Newtons 2nd law

and recall that the acceleration vector was defined as the instantaneous change inthe velocity vector

if the mass is constant with time then

so forces actually cause changes in momentum

and this even works when the mass changes with time(an example would be rocket propulsion)

Total momentumif we have a system of particles, the total momentum of the system is simply the

vector sum of the momenta of the particles

Example 8.1 - Preliminary analysis of a collision

A compact car with a mass of 1000kg is traveling north at speed 15m/s when it collides with a truck ofmass 2000kg traveling east at 10m/s. Treating each vehicle as a particle, find the total momentum justbefore the collision.

2Conservation of momentumLast time we learnt of an example of a conservation law. We found that provided

no external forces acted on a system to do work, the total energy in thesystem was conserved.

With momentum we have another conservation law:Provided no external forces act on a system, the total momentum of a systemis conserved

Internal forces cannot change the total momentum by Newtons 2nd and 3rd laws -say particle A applies a force to particle B, changing Bs momentum (2nd law)then by the 3rd law, B applies an equal and opposite force to A changing Asmomentum by an equal and opposite amount. Then

is unchanged and we say that the totalmomentum is conserved

If an external force was applied to the AB system, then the total momentum wouldchange according to

Example 8.2 - Astronaut RescueAn astronaut finds herself floating in space 100m from her ship. She and the ship are motionless

relatively. The astronauts mass is 100kg and she holds a 1kg wrench. She has 20mins of airremaining. She decides to use conservation of momentum to propel herself back to the ship bythrowing the wrench. Where should she throw it? What will her magnitude of velocity be if shethrows the wrench at 10m/s. Will she make it back to the ship before her air runs out?

before throwing the wrench

after

Example 8.4 - Head-on collisionTwo gliders on an frictionless air-track. Glider A has mass 0.50kg and glider B a mass of 0.30kg. Both

gliders move with an initial speed of 2.0m/s toward each other. After they collide, glider B movesaway with a final velocity in the x-direction of +2.0m/s. What is the final velocity of A?

after the collision

Example 8.5 - Collision in a horizontal planeChunks of ice sliding on a frozen pond. Chunk A, of mass 5.0kg moves with initial velocity 2.0m/s parallel

to the x-axis. It collides with chunk B of mass 3.0kg that is initially at rest. After the collision, thevelocity of A is found to be 1.0m/s at an angle 30o to the x-axis. What is the final velocity of B?

split the problem into components - momentumis conserved separately in x and y

3Inelastic collisionsThe preceding examples have been elastic collisions, where the total kinetic energy

before the collision is that same as after.

We can also have collisions where the kinetic energyafter the collision is less than it was initially - we callthese inelastic

Inelastic collisions

momentum is conserved, no external forces: Pfx=Pix

say, vBx=0

Example 8.6 - Inelastic collision on an air trackVelcro covered gliders on an air track. A has mass 0.50kg and B has mass 0.30kg, both move with initial

speeds 2.0m/s. Find the final velocity of the joined glider and compare the initial and final kineticenergies.

equating the x-components of momentum

Example 8.7 - The ballistic pendulumA system for measuring the speed of a bullet. Bullet of mass mB is fired into a block of wood of mass mW

suspended like a pendulum. The bullet makes an inelastic collision with the block, becomingembedded in it. The block swings up to a maximum height h. Given measured values of h, mB, mWwhat is the speed of the bullet?

conservation of momentum

immediately after the collision, the block haskinetic energy

when the block is at the highest point of its swing, ithas zero velocity - all the K has been turned intogravitational potential energy

4Elastic collisionsHave conservation of both kinetic energy and momentum.

In the special case of a moving object (A) hitting a stationary object (B), we canderive the following

consider special cases:(a) mA >> mB(b) mB >> mA(c) mB = mA

can generally derive

for elastic collision- this is not an extra conservation law - it follows from conservation of bothkinetic energy and momentum

Example 8.10 - Moderator in a nuclear reactorHigh-speed neutrons are produced in a nuclear reactor. Before it can trigger another fission the neutron

needs to be slowed down, this is done by collisions with nuclei of a moderator material.Suppose a neutron of mass 1u traveling at 2.6x107 m/s makes an elastic head-on collision with a carbon

nucleus (mass 12u) at rest. What are the velocities after the collision?

conservation of x-component ofmomentum

relative velocity relation

1u = 1.66 x 10-27 kg

Center of MassSuppose we have several particles at various positions, then the position of the

center of mass is defined to be

e.g.

solid objects also havea center of mass

1kg 1kg

CM

1kg 4kg

CM

1kg 1kg

6kg

CM

velocity of center of mass

total mass

Center of Mass

the effect of an external force then is to accelerate the center of mass

1Chapter 9: Rotational Motion

Weve considered the physics of the motion ofparticles, and the motion of the center of mass ofextended objects.

But extended objects can also rotate about an axis.

Rotation and anglesThink about a rigid body that rotates about a fixed axis.

Firstly we need to sort out a better unit for angles

Weve been using degrees [0:360], but there is a betterunit we can use called radians

1 radian is defined as the angle subtended at the centerof a circle by an arc with length equal to the radiusof the circle

all round the circle,

1 rad = 57.3o

Angular velocityDefine an angular velocity as the rate of change of an angle

units are radians/second

who has the larger angular velocity, the boyor the girl?

boy

girl

Angular accelerationDefine an angular acceleration as the rate of change of the angular velocity

units are radians/second2

2Example 9.1 - Rotation of a compact diskA CD rotates at high speed. The disk has radius 6.0cm and while data is being read by the laser it spins

at 7200rev/min. What is the CDs angular velocity in radians per second. How much time isrequired for it to rotate through 90o? If it starts from rest and reaches full speed in 4.0s, what is itsaverage angular acceleration?

DirectionsRotation can be clockwise or counterclockwise - our sign definition is

Constant angular accelerationEquations describing this are analogous to those for constant linear acceleration

applying these formulae to rotational problems is just the same as we did previously

Example 9.2 - Rotation of a bicycle wheelAngular velocity of the rear wheel of an exercise bike is 4.00rad/s at t=0 and it has constant angular

acceleration of 2.00rad/s2. One of the spokes lies parallel with the x-axis at t=0, what angle doesthis spoke make with the x-axis at t=3.00s. What is the wheels angular velocity at this time?

one full rotation is rad

3Kinetic energy of rotation and moment of inertiaConsider a rigid body rotating to be made up of a large number of particles: A,B,C, then its kinetic

energy is

The linear velocity of a particle is related to the angular velocity of the rigid body and the distance to thefixed axis

all the particles have the same

we can define a property of the rigid body (for a particular axis) called the moment of inertia

so that

Example 9.6 - An abstract sculptureThree massive disks are connected by light rods. Find the moment of inertia about an axis passing

through disks B and C. Find the moment of inertia about an axis through disk A, perpendicular todisk As surface. If the object rotates about the axis through A with angular velocity 4.0rad/s, findits kinetic energy.

axis BCperpendicular distance to A, rA = 0.40m

perpendicular distance to B, rB = 0

perpendicular distance to C, rC = 0

axis Aperpendicular distance to A, rA = 0

perpendicular distance to B, rB = 0.50m

perpendicular distance to C, rC = 0.40m

Example 9.7 - Cable from a winchA cable is wrapped around a winch drum - a solid cylinder of mass 50kg and diameter 0.12m that rotates

about a horizontal axis. The free end of the cable is pulled with a constant force of magnitude 9.0Nfor a distance of 2.0m. If the cylinder is initially at rest, find its final angular velocity and the finalspeed of the cable.

need a fact: the moment of inertia of a solidcylinder about its central axis is

assuming no energy is lost as friction, all the work donein pulling the cable becomes kinetic energy of the winch

work done

kinetic energy of a rotating cylinder

cable moves transverse to the circle at the radius

Rotation about a moving axisEvery possible motion of a rigid body can be expressed as a combination of the motion of the centre of

mass and rotation about an axis passing through it.

The kinetic energy can be split into a part for the motion of the center of mass and a part for therotation

=

4a primitive yo-yoa primitive yo-yo is made from a solid disk of radius R and

mass M. It is released from rest by a stationarysupporting hand and the string unwinds withoutslipping.

(a) Find an expression for the speed of the center of massof the disk after it has dropped a distance h.

(b) What would the speed have been if the disk had beendropped without the string

(c) Why are they different?

motion = rotation + center of mass

and are related:

in one complete rotation, the amount of stringunwound is , which must be the same distancetraveled by the center of mass

torqueWeve considered rigid bodies undergoing angular acceleration. What causes

this? In the case of linear acceleration it was a force, as dictated byNewtons 2nd law.

For angular acceleration we introduce the concept of torque

torquesign convention for torques:counterclockwise torques are positiveclockwise torques are negative

torquecomputing the torque:

only the perpendicular piece of the force/moment arm contributes

5a weekend plumberthe weight of a 900N man is applied to the end of an extended wrench 0.80m

long which makes an angle 19o to the horizontal. Find the magnitude anddirection of the torque exerted out the center of the pipe fitting.

compute the component of force perpendicular to the lever

the torque is the lever length times this

torque is positive since it causes a counterclockwise rotation

torque and angular accelerationusing only Newtons 2nd law we can derive a relationship between torque and

angular acceleration

torque and angular accelerationusing only Newtons 2nd law we can derive a relationship between torque and

angular acceleration

rotational analogue ofNewtons 2nd law

unwinding a winchcable is wrapped around a uniform solid cylinder of diameter 0.12m and mass

50kg that can rotate freely about its axis. The cable is pulled by a force ofmagnitude 9.0N. Assuming that the cable unwinds without stretching orslipping find the acceleration of the cable.

6a primitive yo-yoa primitive yo-yo is made from a solid disk of radius R and

mass M. It is released from rest by a stationarysupporting hand and the string unwinds withoutslipping.

(a) Find an expression for the speed of the center of massof the disk after it has dropped a distance h.

(b) What would the speed have been if the disk had beendropped without the string

(c) Why are they different?

motion = rotation + center of mass

and are related:

in one complete rotation, the amount of stringunwound is , which must be the same distancetraveled by the center of mass

forces on the primitive yo-yofind the acceleration of the yo-yo and the tension in the string

Newton II for the center of mass:

torque/ ang. accl. eqn:

work in rotational motionangular acceleration causes a change in kinetic energythis energy comes from work done by the torque

hence for a constant torque:

work in rotational motionthe flywheel of a motor has a mass of 300.0kg and a moment of inertia of 580 kg m2 . The

motor develops a constant torque of 2000.0 Nm and the flywheel starts fromrest.

(a) what is the angular acceleration of the flywheel?(b) what is its angular velocity after it makes 4.00 revolutions(c) how much work is done by the motor during the first 4.00 revolutions?

7angular momentumis there a quantity like momentum that is conserved in rotational motion?

angular momentum is defined by

so if there is no net torque we have

so angular momentum is conserved

conservation of angular momentumprof stands on a turntable holding his arms extended horizontally with a 5.0kg dumbbell in

each hand. He is set rotating making one revolution in 2.0s. His moment of inertiawithout the dumbbells is 3.0 kg m2 with arms outstretched and 2.2 kg m2 with armstucked in. The dumbbells start 1.0 m from the axis and are pulled in to 0.20m. Findthe professors angular velocity with the dumbbells pulled in. Also compute thechange in kinetic energy.

1Chapters 11 & 12 Periodic motion & waves

Periodic Motion

e.g. a pendulum - keeps swinging from side to side with a fixed period -hence we can use it to keep time (pendulum clock)

it has an equilibrium position, but if we give it an initial push away fromthere, conservation of energy prevents it from coming to rest even thoughthere is a force pulling it back

simplest example of periodic motion is called simple harmonic motion this turns out to be one of the most important idealizations in physics, it just

keeps coming up e.g. quantum field theories that describe the smallest known particles in the

universe are really just fsncy coupled harmonic oscillators

Simple Harmonic Motion

a physical system that is described well by simple harmonic motion is aglider on an air-track tethered by a spring

free-body diagram of block when x0 (spring is extended)

acceleration is back towardequilibrium position (x=0)

acceleration is back towardequilibrium position (x=0)

Simple Harmonic Motion

a physical system that is described well by simple harmonic motion is aglider on an air-track tethered by a spring

free-body diagram of block when x=0 (spring is in equilibrium)

but the block cant stop because it has velocity (Newtons 1st law)

no acceleration

2Simple Harmonic Motion

a physical system that is described well by simple harmonic motion is aglider on an air-track tethered by a spring

amplitude = max |x| = A

period = T = time to travel one cycle andreturn to original position

frequency = f = 1/ T = number of cycles in agiven unit of time

(units Hz = s-1)

Energy in Simple Harmonic Motion

no external or non-conservative forces here, so mechanical energy shouldbe conserved

when the block is at the maximumdisplacement, x=A, v=0,

Energy in Simple Harmonic Motion

energy sloshes between kinetic and potential

Example 11.5 - Simple harmonic motion on an air track

A spring is mounted horizontally on an air track, with the left end heldstationary. We attach a spring balance to the free end of the spring, pulltoward the right, and measure the elongation. A force of 6.0 N causesan elongation of 0.030 m. We remove the spring balance and attach a0.50 kg object to the end, pulling it a distance 0.040 m. After releasingthe object it oscillates in Simple Harmonic Motion.

Find(a) the force constant of the spring(b) the maximum and minimum velocities of attained by the object(c) the maximum and minimum acceleration(d) the velocity and acceleration when the object has moved 0.020 m(e) the kinetic, potential and total energies at 0.020 m

3Equations of simple harmonic motion

how does the position, velocity and acceleration of the block vary as afunction of time?

with calculus these equations are easy to find, they are:

what is ?

check on energy conservation:

equations of motion

how do these look?

Equations of simple harmonic motion

notice that when , all quantities return to the same value

- thus one cycle has been completed

looks a bit like circular motion - is there a connection?

YES, and we can use it to derive the above equations

The circle of reference projection of circular motion onto a flat screen

4equations of motion

check we can correctly use them: at t=0, x=A=20 cm & v=0 the period is found to be 10 s

what is the position, velocity andacceleration at t = 2 s ?

what is the position, velocity andacceleration at t = 8 s ?

what is the position, velocity andacceleration at t = 102 s ?

at what time(s) is the position x=10cm ?

what is the maximum velocity?

The simple pendulum

a point mass suspended by a weightless string.

45o 90o force proportional to displacement - suggestsSHM

making a simple timepiece

lets say we want to build a pendulum clock that will count seconds the bob should pass the equilibrium position once every second

what is the period of the motion? how long should our pendulum be?

if I make the pendulum length four times as long, how does the angularfrequency change? what about the period?

Mechanical waves

a disturbance that propagates through a medium

transverse, longitudinal or a combination of both

waves transport energy, but not matter

5periodic mechanical waves

a disturbance that propagates through a medium, repeatingitself over time

a special case is where the particles undergo SHM - then thewave is said to be sinusoidal

sinusoidal mechanical waves

a special case is where theparticles undergo SHM - then thewave is said to be sinusoidal

amplitude defined as before as themaximum deviation of any particlefrom its equilibrium position

wavelength, = spatial distanceover which the wave-patternrepeats

wave propagates with a speedgiven by

longitudinal waves

e.g. a tube filled with air beingcompressed and expanded by aplunger

get regions of increased pressure(compressions) and regions ofdecreased pressure (rarefactions)

transverse waves on a rope

how fast will these waves propagate?

depends on the tension in the rope and the mass per unit length

example 12.2 - Mineral samples in a mine shaft

Nylon rope fixed at the top end and held taut by a 20.0 kg boxof rocks at the bottom. The rope has a mass of 2.00 kg. Guyshakes the bottom of the rope from side to side. What will thespeed of the transverse waves be? If the shake is transversesimple harmonic of frequency 20 Hz, what wavelength waveswill we get?

6reflections and superposition

what happens if a wave reaches a boundary, e.g. the fixed end of a rope?

note - not a periodic wave here

principal of superposition

consider a related process, where two waves traveling on the same ropemeet each other

superposition indicates that we add(vectorally) the displacements of eachwave to make a total displacement

standing waves

if we fix both ends of a string, what sort of waves do we get?

call these standing waves since they are not transferring energy alongthe string

standing waves by superposition

wave traveling to the left (red) combines with an identical wave travelingto the right (blue)

7standing waves

if we fix both ends of a string, what sort of waves do we get?

clearly we have to have a node at each end, which limits the allowedwavelengths

allowed set of frequencies (harmonic series) :

example 12.3 - a giant bass viol

string of length 5.00m between fixed points, mass per unit length40.0g/m and a fundamental frequency of 20.0 Hz.

what is the tension in the string? what is the frequency and the wavelength of the second harmonic? what is the frequency and wavelength of the second overtone?

Interference

standing waves are one example of an interference between two wavesthat overlap in a region of space

t

t

constructiveinterference

destructiveinterference

Example 12.6 - Interference from two speakers

two speakers are driven by the same amp and emit sinusoidal waves inphase. If the speed of sound is 350 m/s for what frequencies doesmaximum constructive interference occur at point P? What frequenciesgive destructive interference?

8Beats

interference of waves with differing frequencies - tends to cause apulsing effect called beats

beat frequency is the difference of the two wave frequencies: peak intensity at some time - next peak intensity when wave 1 has gone

through one more whole cycle than wave 2

Sound & Hearing

define the intensity of a wave as the time-averaged rate that energy istransferred by a wave, per unit area.

rate of energy flow called power and measured in Watts, W=J/s consider sound spreading out from a point, then at a distance r, the

power is spread over a spherical shell of area

human hearing is effective over a huge range of intensities, from faintwhispers of intensity about 10-12 W/m2 to the pain threshold at 100 W/m2- 14 orders of magnitude

an appropriate scale is logarithmic - define intensity level by

Sound & Hearing

human hearing is effective over a huge range of intensities, from faintwhispers of intensity about 10-12 W/m2 to the pain threshold at 100 W/m2- 14 orders of magnitude

an appropriate scale is logarithmic - define intensity level by

10-120hearing threshold

10-1110rustle of leaves

10-1020whisper

10-750quiet car

3 x 10-665conversation

10-570busy street

3 x 10-395riveter

1120pain threshold

100140rock gig

intensity, W/m2intensity level, dBnoise

the Doppler effect

when a car passes at high speed you might notice that the pitch of theengine or the horn seems to drop as it passes

this is caused by the relative velocity of the car and the observer(you)and is called the Doppler effect

listener receives waves at relative speed

wavelength is a fixed length in space, so is the same for the listener,hence

stationary source

listener hears a higher frequency

9the Doppler effect - with moving source

now suppose the source is moving

now the wavelength is changed by the source movement during one period , an entire wavelength is emitted, but the source

has moved a distance , so to a stationary observer the distance

between wavecrests is

hence the moving listener hears a frequency

plus - left

minus - right

Examples 12.10-12 - Police car chasing a speeder

police car siren emits a sinusoidal wave with frequency fS = 300 Hz. The speed ofsound is 340 m/s.(a) find the wavelength of the waves if the siren is at rest in the air(b) if the siren is moving with a velocity vS = 30.0 m/s. Find the wavelength infront of and behind the siren

(a) when the source is at rest

(b) in front of the siren the wavelength is reduced

(c) behind the siren the wavelength is increased

Examples 12.10-12 - Police car chasing a speeder

a bystander (at rest) observes the chase - what frequency does he hear when thepolice car has passed him?

what if the bystander gets in his car and drives away at 30 m/s ?

1Chapters 14 - Heat,Thermodynamics etc

temperature

temperature intuitively related to hotness orcoldness

quantitative definition comes via the thermometer two examples of thermometers

thermal equilibrium

measure temperature by placing thermometer incontact with the object

the two interact, the object cooling a little and thethermometer heating up - at some point they reachand stay at the same temperature they are said to be in thermal equilibrium

the zeroth law of thermodynamics

two systems that are in thermal equilibrium with a thirdsystem are also in equilibrium with each other

verified experimentally their temperature is the quantity A, B & C have in common

2a useful thermometer

a thermometer becomes a quantitative device if we assign anumerical scale to the changes it undergoes

common scale choice is Celsius water freezes @ 0oC

water boils @ 100oC

another common choice is Fahrenheit water freezes @ 32oF

water boils @ 212oF

scales are related

the zero of temperature?

it looks like were free to choose the zero of our temperature scale

is there a minimum temperature that we ought to use?

absolute zero = -273.15oC

define Kelvin scale - same interval as Celsius, but zerod at absolute zero

heat

so a bodys hotness is described by its temperature - but how did it gethot?

consider putting a cold metal spoon into a hot cup of coffee - the spoonwill heat up until it is in thermal equilibrium with the coffee (they bothhave the same temperature)

heat transfer from the coffee to the spoon has taken place.

James Joule showed experimentally that there is an equivalence betweenwork and heat transfer

heat is a form of energy

heat and temperature

heat flow to an object is likely to increase its temperature - how are theyrelated?

define specific heat capacity, c

if Q is positive, heat is flowing to the object and the temperature increases if Q is negative, heat is flowing from the object and the temperature

decreases

specific heat capacity of water is c = 4.190 x 103 J kg -1 K-1

e.g. how much heat must be transferred to raise 0.25 L of water by 80oC?

3heat and temperatureif Q is the amount of heat required to raise the temperature of an object from

10oC to 20oC, how much heat is required to raise the temperature from20oC to 25oC ?

A. Q/2

B. Q

C. 2Q

D. 25Q/20

phase changes

heat flow does not always cause a change in temperature - it can cause achange of phase, for example from solid to liquid or from liquid to gas

e.g. add heat to ice at 0oC and its temperature doesnt increase, instead inturns into water at 0oC

the amount of heat required to cause such a phase change is called alatent heat

e.g. the heat required to turn 1kg of ice at 0oC into 1kg of water at 0oC iscalled the latent heat of fusion and is

for an arbitrary mass

process is reversible so we get this heat energy out when the water freezes

also a latent heat (of vaporization) to turn water into water vapor (a gas)

phase changes

add heat continuously to a sample of ice

ice

Suppose over 30 seconds I add 10,000 J of heat to a 1 kg sample of ice,initially at a temperature of 0oC. Which of the following is true of thesample after Ive added the heat? (the system is in thermalequilibrium)

A. it is entirely ice at a temperature of 5oC

B. it is a mixture of ice and water at a temperature of 0oC

C. it is entirely water at a temperature of 0oC

D. it is entirely water at a temperature of 5oC

4example 14.2 - chilling soda

how much ice (@ -20oC) should we add to 0.25 kg of cola (@ +20oC) tochill it to 0oC with all the ice melted?

heat lost from the cola is gained by the ice

heat gained by the ice does two things raises the temperature from -20oC to 0oC

melts the ice

total heat flow is zero

mechanisms of heat flow how can heat move from one system to another? three basic mechanisms are conduction, convection and radiation

conduction occurs within objects that are in physical contact

heat is really just the energy of motion of huge numbers of atomic particles,heat transfer by conduction is when neighbouring atomic particles transferenergy through collisions

note that still air is actually not very good at conducting heat

mechanisms of heat flow

how can heat move from one system to another?

three basic mechanisms are conduction, convection and radiation convection occurs when bulk amounts of a substance move, carrying heat

with them

mechanisms of heat flow

how can heat move from one system to another?

three basic mechanisms are conduction, convection and radiation

radiation is the transfer of heat in the form of electromagnetic waves radiation allows heat transfer even through a vacuum where there

are no atoms and hence no concept of heat

all hot objects emit electromagnetic radiation - some of it in thevisible region (e.g. white light from a filament lightbulb), but also somein the frequency region not visible to our eyes

this is how the Earth manages to be heated by the sun, despite thevacuum separating the Earth from the Sun

5the Dewar (thermos) flask designed to prevent heat transfer between

the flask contents and the outside world what method of heat transfer does the screw

top mostly prevent?A. conductionB. convectionC. radiation

if you hold the flask in your hand whatmethod of heat transfer from the contents toyour hand does the vacuum region mostlyprevent?

A. conductionB. convectionC. radiation

what method of heat transfer does thesilvering mostly prevent?

A. conductionB. convectionC. radiation

the atomic levelall materials, solid, liquid or gas, are made up of atoms - the essential differences

between materials being which atoms are present and how they are arranged.

how many atoms are we talking about in a lump of stuff?

take for example 12g of carbon - this would be the lead of a few pencils:

this contains 6.022 x 1023 atoms of carboni.e. roughly 600,000,000,000,000,000,000,000 atoms!

This number is called Avogadros number and it has a more general meaning as

the number of molecules in one mole of substance(one mole being the amount of substance that has as many molecules as 12g of carbon)

this is the scale of macroscopic amounts of stuff and is clearly a very large number

Avogadros number & homeopathyA LESSON IN JUNK-SCIENCE

Homeopathy is a medical treatment that claims to work by taking an active compoundand repeatedly diluting it (usually with water) - the claim is that this dilution makesthe cure more potent.

The originator of homeopathy recommended what he termed a 30C dilution (diluted toone part in 100, 30 times), this corresponds to a dilution by a factor of 1060

lets say were too lazy to do all that diluting and only did a 15C dilutionroughly speaking wed have 1030 water molecules for every molecule of active compound1030 water molecules is about 106 moles of water which would have a mass of about 104

kg and a volume of about 10 m3

to get a single molecule of active ingredient you would need to swallow 10,000 liters ofwater!

a typical homeopathic dose is not of this scale and it is more than likely to youswallowing just water

for comparison the EPA allows 50 ppb (parts per billion) of Arsenic in drinking waterso there is something like ten thousand billion billion times as much Arsenic in your remedy than there isactive ingredient

heat at the atomic levelis the random motion of molecules, all carrying kinetic energy

this random motion can be observed by adding some visible tracers into a liquid and iscalled Brownian motion

6Joules experiment

demonstrates that the same temperature change can be causedby

doing work on a system

directly heating the system

so there must be a connection between heat & work

the first law of thermodynamics

combines the effects of heat transfer and work

thermodynamic system

fluid-based thermodynamic system

the working substance in a thermodynamic system is often a fluid hence work can be done via a volume change.

very simple system do work by pushing or pulling on the plunger

supply heat by putting a hotplate under the cylinder

heat transfer depends upon path consider expanding 2.0L of fluid to 5.0L keeping the temperature constant

W > 0

Q > 0

W = 0

Q = 0same final state reached through two different processes (paths)

7the first law of thermodynamics

there is a property of the state of a thermodynamic system that doesntcare about the path used to get to that state

the internal energy can be thought of as the sum of all the kinetic andpotential energies of the constituent particles of a system

the first law of thermodynamics relates the heat transfer to a system andthe work done by it to the internal energy of the system

in words - when heat Q is added to a system, some of this added energyremains within the system changing the internal energy by an amount

.The remainder leaves the system as it does work W on itssurroundings.

gas in an insulated rigid container

weights are added to the frictionless piston, compressing the gas, asthis is done

A. the internal energy of the gas stays the samebecause the container is insulated

B. the internal energy of the gas increasesbecause heat is added to the system

C. the internal energy of the gas increasesbecause work is done on the gas

D. the internal energy of the gas decreasesbecause there is less gas

example 15.8 - thermodynamics of boiling water

one gram of water becomes 1671cm3 of steam when boiled at a constant pressureof (1atm = 1.013x105Pa). The latent heat of vaporization is Lv = 2.256x10

6J/kg.

Compute the work done by the water when it vaporizes and its increase ininternal energy

at constant pressure W = p(V2-V1) = (1.013x105Pa)(1671x10-6m3 - 1.0x10-6m3)

W = 169 J

heat added Q = m Lv = 2256 J

a heat engine

a device which absorbs heat from a hot reservoir, does some work, anddiscards heat to a cold reservoir.

assumed to be cyclic - the enginereturns to its original state

net heat absorbed per cycle

first law then says we can get outuseful work

8a very familiar heat engine - the internal combustion engine

the Otto cycle is anidealized model ofthe internalcombustion engine

fuel consumption in a truck

engine takes in 2500J of heat by burning gasoline and delivers 500J ofmechanical work per cycle.

(a) what is the thermodynamic efficiency?

(b) how much heat is discarded per cycle?

refrigerator as a reversed heat engine

use incoming work to remove heat

practical refrigerator

9the second law of thermodynamics

it is impossible for any system to undergo a process in which it absorbs heat from areservoir at a single temperature and converts the heat completely into mechanical workwith the system ending in the state in which it began.

it is impossible for any process to have asits sole result the transfer of heat from acooler to a hotter object

the most efficient heat engine

The Carnot cycle is the most efficient heat engine possible - i.e. obeying the secondlaw of thermodynamics

the second law of thermodynamics

what is the origin of these impossibilities? they dont come from the firstlaw (conservation of energy)

the issue is one of disorder at the microscopic level - in a hot body themolecules are flying around with a high degree of randomness, while in acolder body they are less random

once a closed system is randomised it will not un-randomise itself

an alternative statement of the second law of thermodynamics is theentropy (degree of disorder) in a closed system either remains constantor increases.

1Physics 112NSpring Semester 2008

Prof. Jo Dudekjdudek@odu.eduOCNPS Rm 325

Physics 112N this semester well continue our tour through

physics considering electromagnetic phenomena optics aspects of modern physics

as before well need some math skills and well alsomake use of physics learnt last semester

youre likely to have experienced static electricalcharges - they accumulate if you rub a balloon onyour sweater, they also accumulate on clouds andare the origin of lightning discharges

it turns out that charge is quantized, it comes inpackets which cant be subdivided

it is conserved - the total charge in a closed systemwill not change with time

Chapter 17 - Electric Charge and Electric Field1. Which of the following is NOT TRUE? The electric force

A. decreases with the inverse square of the distance between the two chargedparticles

B. between two protons separated by a distance d is larger than between twoelectrons separated by d

C. may be either attractive or repulsive

2. A material which permits electric charge to move through it is calledA. an insulatorB. a conductorC. a capacitorD. an inductor

Reading Quiz

2simple experiments involving electriccharge

like charges repel, opposite charges attract DEMO 56

the charges we encounter are properties of the particles thatmake up matter - namely electrons and protons

origin of charges

some materials permits electric charge tomove from one region to another - we callthese conductors

materials that do not permit suchmovement are titled insulators

conductors and insulators induction and polarization

DEMO 58

3A positively charged object is placed close to a conducting objectattached to an insulating glass pedestal (a). After the oppositeside of the conductor is grounded for a short time interval (b), theconductor becomes negatively charged (c). Based on thisinformation, we can conclude that within the conductor

A. both positive and negative charges move freely.B. only negative charges move freely.C. only positive charges move freely.D. We cant really conclude anything.

the sum of charges in a closed system is constant the magnitude of the charge of a proton or an

electron is the indivisible unit of charge* any charge seen in experiment will be an integer

multiple of this basic charge unit

units of charge are Coulombs

conservation and quantisation of charge

*the prevailing nuclear theory suggests quarks with fractional charges exist, butcannot be separated far enough to be seen

Three pithballs (insulators) are suspended from thinthreads. Various objects are then rubbed againstother objects (nylon against silk, glass againstpolyester, etc.) and each of the pithballs is chargedby touching them with one of these objects. It isfound that pithballs 1 and 2 repel each other andthat pithballs 2 and 3 repel each other. From thiswe can conclude that

A. 1 and 3 carry charges of opposite sign.B. 1 and 3 carry charges of equal sign.C. all three carry the charges of the same sign.D. one of the objects carries no charge.

four lightweight plastic spheres are suspended from insulatingthreads

the net charge on sphere d is +Q the other three spheres have net charges of either +Q, -Q or 0

plastic spheres

a d b c cd b d

A. a=0, b=+Q, c=-QB. a=+Q, b=0, c=-QC. a=-Q, b=+Q, c=0D. a=-Q, b=0, c=+Q

4 Coulomb studied theinteraction forces betweencharges in a torsion balanceexperiment

Coulombs law

like gravity, Coulombs law gives a force thatbehaves as the inverse square of distance

1/r2

a hydrogen atom contains one proton and one electron. The Bohr model picturesthe electron as moving around the proton in a circular orbit of radius r = 5.29 x10-11 m. What is the ratio of the electric force between the electron and the protonto the magnitude of the gravitational force between them?

example 17.2 - gravity in the hydrogen atomA hydrogen atom is composed of a nucleuscontaining a single proton, about which a singleelectron orbits. The electric force between the twoparticles is 2.3 x 1039 greaterthan the gravitational force! If we can adjust thedistance between the two particles, can we find aseparation at which the electric and gravitationalforces are equal?

A. Yes, we must move the particles farther apart.B. Yes, we must move the particles closer together.C. No, at any distance

5Two uniformly charged spheres are firmly fastened to andelectrically insulated from frictionless pucks on an air table. Thecharge on sphere 2 is three times the charge on sphere 1.Which force diagram correctly shows the magnitude anddirection of the electrostatic forces:

A B

C D

two point charges lie on a line. q1 = 2.0 C is located on the positive x-axis at x = 0.30 m,and an identical charge q2 is at x= - 0.15 m. Find the magnitude and direction of the totalforce that these two charges exert on a third charge q3 = 4.0 C which lies at the origin.

vector addition of forces

a point charge q1 = 2.0 C is located on the positive y-axis at y = 0.30 m, and an identical charge q2 is atthe origin. Find the magnitude and direction of the total force that these two charges exert on a thirdcharge q3 = 4.0 C that is on the positive x-axis at x = 0.40 m

example 17.4 - vector addition of forces

61. Which of the following is NOT TRUE?A. The electric field obeys the principle of superpositionB. The tangent to an electric field line gives the direction of the field at that

pointC. The density of electric field lines is directly proportional to the strength of

the fieldD. Negative charges are sources of electric field lines and positive charges are

sinks

2. Which of the following is NOT TRUE?A. The electric field from a point charge at a distance r is the electrostatic

force on a test charge at r divided by the test chargeB. The electric field from a point charge falls as the inverse square of distance

from the point chargeC. The units of electric field strength are NewtonsD. The electric field lines from a point charge point radially to or from the

charge

Reading Quiz electric forces act, like gravity, at a distance a powerful way to think of the cause of the force on

a charge B due to a distant charge A is to use theconcept of a force field from charge A that iseverywhere in space

then the force on B arises because B feels theelectric field

electric field is defined as the force on B, with the charge of Bdivided out (so the properties of B dont appear)

Electric Field

strictly speaking we call B a test charge and make its chargeso small that it does not produce a significant field of its own

if we have many sources of charge, each will produce a field(having a vector value at each point in space) which can beadded vectorally to produce a net electric field

Electric FieldTwo test charges are brought separately into the vicinity of acharge +Q. First, test charge +q is brought to a point a distance rfrom +Q. Then this charge is removed and testcharge q is brought to the same point. The electrostatic field atthe position of the test charge is greater for:

A. +qB. qC. It is the same for both.

7Two test charges are brought separately into the vicinity of acharge +Q. First, test charge +q is brought to a point a distance rfrom +Q. Then this charge is removed and testcharge 2q is brought to the same point. The magnitude ofelectric force felt by the test charge is largest for

A. +qB. 2qC. It is the same for both.

-2q

a vacuum chamber contains a uniform electric field directeddownward. If a proton is shot horizontally into this region, itfeels a force directedA. downward ?B. upward ?C. to the right ?D. to the left ?

a moving proton

p+

E

a vacuum chamber contains a uniform electric field directeddownward. If an electron is shot horizontally into this region,its acceleration isA. downward and constant ?B. upward and constant ?C. upward and changing ?D. downward and changing ?

a moving electron

e-

E

when the terminals of a 100V battery are connected to twolarge parallel horizontal plates 10cm apart, the resultingcharges on the plates produce an electric field in the regionbetween the plates that is very nearly uniform and hasmagnitude E = 1.0 x 104 N/C. Suppose the lower plate haspositive charge so that the electric field points upward.

if an electron is released from the upper plate, what is itsspeed just before it reaches the lower plate? how much timeis required for this motion?

explain why we can neglect acceleration due to gravity

example 17.5 - accelerating an electron

8 recall that Coulombs law gave for the force

hence the electric field from particle 1 is obtained by dividingout the charge of particle 2, considered to be a test charge

electric field from a point charge an electric field has a value at each point in space - this is not

an easy thing to display - electric field lines are a way toattempt this

a field line is drawn at a tangent to the field vector if lines are drawn close together this indicates a large

magnitude of electric field we assign a direction to the lines which is the direction of the

field vectors (hence it runs from positive to negative charges)

electric field lines

some examples

field lines only start and end on positive charges they are otherwise continuous

electric field lines some exampleselectric field lines

9Consider the four field patterns shown. Assuming there areno charges in the regions shown, which of the patternsrepresents a possible electrostatic field:

A B

C D

a small object S with a charge of magnitude q creates anelectric field. At a point P located 1.0m west of S, the field hasa value of 9 N/C directed to the west. At a point 1.0m east ofS, the field isA. 10 N/C, directed west ?B. 10 N/C, directed east ?C. 20 N/C, directed west ?D. 20 N/C, directed east ?

what is the value of q ? (k = 9.0x109 Nm2/C2)A. 10 CB. 1x10-9 C = 1 nCC. 1 CD. 1.0x109 C

magnitude and direction

Consider the field patterns shown. Identify theunknown charge

Q?+q

A. Q = -2qB. Q = -qC. Q = +qD. Q = +2q

Consider the field patterns shown. Identify theunknown charges

Q2Q1A. Q1 = -q , Q2 = -qB. Q1 = -q , Q2 = +qC. Q1 = +q , Q2 = +qD. Q1 = +q , Q2 = -q

10

Consider the field patterns shown. Identify theunknown cha