Physics 115 2013

Final Review

Calculus is about “rates of change”.

A TIME RATE is anything divided by time.

CHANGE is expressed by using the Greek letter, Delta, D.

For example: Average SPEED is simply the “RATE at which DISTANCE changes”.

The MEANING?

For example, if t = 2 seconds, using x(t) = kt3=(1)(2)3= 8 meters.

The derivative, however, tell us how our DISPLACEMENT (x) changes as a function of TIME (t). The rate at which Displacement changes is also called VELOCITY. Thus if we use our derivative we can find out how fast the object is traveling at t = 2 second. Since dx/dt = 3kt2=3(1)(2)2= 12 m/s

23

3)( ktdtktd

Derivative of a power function

Unit Vector Notation

direction z in the 1 r unit vecto-ˆdirectiony in the 1r unit vecto-ˆdirectionx in the 1 r unit vecto-ˆ

k

j

i

jiJ ˆ4ˆ2

The proper terminology is to use the “hat” instead of the arrow. So we have i-hat, j-hat, and k-hat which are used to describe any type of motion in 3D space.

How would you write vectors J and K in unit vector notation?

jiK ˆ5ˆ2

ExampleA boat moves with a velocity of 15 m/s, N in a river which flows with a velocity of

8.0 m/s, west. Calculate the boat's resultant velocity with respect to due north.

1.28)5333.0(

5333.0158

/17158

1

22

Tan

Tan

smRv

15 m/s, N

8.0 m/s, W

Rv

The Final Answer :

smjsmi

sm

WofNsm

/ˆ15/ˆ8

1.118/17

1.28@/17

Dot Products in PhysicsConsider this situation: A force F is applied to a moving object as it transverses over a frictionless surface for a displacement, d.

As F is applied to the object it will increase the object's speed!

But which part of F really causes the object to increase in speed?

It is |F|Cos θ ! Because it is parallel to the displacement d In fact if you apply the dot product, you get (|F|Cos θ)d, which happens to be defined as "WORK" (check your equation sheet!)

cos

cos

xFxFW

BABA

Work is a type of energy and energy DOES NOT have a direction, that is why WORK is a scalar or in this case a SCALAR PRODUCT (AKA DOT PRODUCT).

ExampleSuppose a person moves in a straight line from the lockers( at a position x = 1.0 m) toward

the physics lab(at a position x = 9.0 m). “To the right” is taken as positive, as shown below

The answer is positive so the person must have been traveling horizontally in the positive direction.

ExampleSuppose the person turns around!

The answer is negative so the person must have been traveling horizontally in the negative direction

What is the DISPLACEMENT for the entire trip?

What is the total DISTANCE for the entire trip?

mxxx initialfinal 00.10.1 D

m1688

Instantaneous VelocityInstantaneous velocity is a

measure of an object’s displacement per unit time at a particular point in time.

dtdxv

txv t

DD

D 0lim

Example: A body’s position is defined as:?)(,ˆ4ˆ7)( 3 tvj

tittx

dt

jt

itd

dtdxv

)ˆ4ˆ7( 3 smj

tittv /]ˆ4ˆ21[)( 2

2

Instantaneous AccelerationInstantaneous velocity is a measure

of an object’s velocity per unit time at a particular point in time.

dtdva

tva t

DD

D 0lim

?)(,/]ˆ4ˆ21[)( 22 tasmj

tittv

If the velocity of an object is defined as:

dt

jt

itd

dtdva

)ˆ4ˆ21( 22

ssmjt

it //]ˆ8ˆ42[ 3

What do the “signs”( + or -) mean?Quantity Positive NegativeDisplacement Your position has

changed toward the positive.

Your position has changed toward the negative.

Velocity You are traveling in the +x or +y direction.

You are traveling in the –x or –y direction.

Acceleration If moving in the positive direction, you are speeding up. If moving in the negative direction, you are slowing down.

If moving in the positive direction, you are slowing down. . If moving in the negative direction, you are speeding up.

The 3 Kinematic equations

There are 3 major kinematic equations than can be used to describe the motion in DETAIL. All are used when the acceleration is CONSTANT. )(2

21

22

2

oo

oo

o

xxavv

attvxx

atvv

Kinematics for the VERTICAL DirectionAll 3 kinematics can be used to analyze one dimensional

motion in either the X direction OR the y direction.

)(2)(22

12

1

2222

22

ooyyoox

oyooxo

oyyo

yygvvxxavv

gttvyyattvxx

gtvvatvv

ExamplesA stone is dropped at rest from the top of a cliff. It is

observed to hit the ground 5.78 s later. How high is the cliff?

What do I know?

What do I want?

voy= 0 m/s y = ?

g = -9.8 m/s2

yo=0 m

t = 5.78 s

Which variable is NOT given andNOT asked for?

Final Velocity!

2

21 gttvyy oyo

yy 2)78.5(9.4)78.5)(0(

-163.7 m

H =163.7m

ExamplesA pitcher throws a fastball with a velocity of 43.5 m/s. It is determined that

during the windup and delivery the ball covers a displacement of 2.5 meters. This is from the point behind the body when the ball is at rest to the point of release. Calculate the acceleration during his throwing motion.

What do I know?

What do I want?

vo= 0 m/s a = ?

x = 2.5 m

v = 43.5 m/s

Which variable is NOT given andNOT asked for?

TIME

)(222oo xxavv

aa )05.2(205.43 22

378.5 m/s/s

ExamplesHow long does it take a car at rest to cross a 35.0 m intersection after

the light turns green, if the acceleration of the car is a constant 2.00 m/s/s?

What do I know?

What do I want?

vo= 0 m/s t = ?

x = 35 m

a = 2.00 m/s/s

Which variable is NOT given andNOT asked for?

Final Velocity

t

t 2)2(21)0(035

2

21 attvxx oxo

5.92 s

ExamplesA car accelerates from 12.5 m/s to 25 m/s in 6.0 seconds.

What was the acceleration?

What do I know?

What do I want?

vo= 12.5 m/s a = ?

v = 25 m/s

t = 6s

Which variable is NOT given andNOT asked for?

DISPLACEMENT

atvv o

aa )6(5.1225

2.08 m/s/s

SummaryThere are 3 types of MOTION graphs• Displacement(position) vs. Time• Velocity vs. Time• Acceleration vs. Time

There are 2 basic graph models• Slope• Area

Summary

t (s) t (s) t (s)

x (m)v (m/s) a (m/s/s)

slope = v

slope = a

area = xarea = v

Comparing and Sketching graphsOne of the more difficult applications of graphs in physics is when given a certain type of graph and asked to draw a different type of graph

t (s)

x (m)

slope = v

t (s)

v (m/s)

List 2 adjectives to describe the SLOPE or VELOCITY1. 2.

The slope is CONSTANTThe slope is POSITIVE

How could you translate what the SLOPE is doing on the graph ABOVE to the Y axis on the graph to the right?

Example

t (s)

x (m)

t (s)

v (m/s)

1st line• •

2nd line•

3rd line• •

The slope is constant

The slope is constantThe slope is “+”

The slope is “-”The slope is “0”

Example – Graph Matching

t (s)

v (m/s)

t (s)

a (m/s/s)

t (s)

a (m/s/s)

t (s)

a (m/s/s)

What is the SLOPE(a) doing?

The slope is increasing

Force Diagrams

A pictorial representation of forces complete with labels.

W1,Fg1 or m1g

•Weight(mg) – Always drawn from the center, straight down•Force Normal(FN) – A surface force always drawn perpendicular to a surface.•Tension(T or FT) – force in ropes and always drawn AWAY from object.•Friction(Ff)- Always drawn opposing the motion.

m2g

T

TFN

Ff

Force Diagrams

mg

FNFf

New’s 1st Law and Equilibrium

Since the Fnet = 0, a system moving at a constant speed or at rest MUST be at EQUILIBRIUM.

TIPS for solving problems• Draw a FBD• Resolve anything into COMPONENTS• Write equations of equilibrium• Solve for unknowns

Example10-kg box is being pulled across the table to the right at a constant speed

with a force of 50N at an angle of 30 degrees above the horizontal.a) Calculate the Force of Friction

b) Calculate the Normal Force

mg

FN Fa

Ff30

NFFNFF

axf

aax

3.433.4330cos50cos

Fax

Fay

NF

FmgF

mgFFmgF

N

ayN

ayN

N

73

30sin50)8.9)(10(

!

Springs – Hooke’s Law

One of the simplest type of simple harmonic motion is called Hooke's Law. This is primarily in reference to SPRINGS.

kxorkxFkk

xF

s

s

N/m):nitConstant(U SpringalityProportion ofConstant

The negative sign only tells us that “F” is what is called a RESTORING FORCE, in that it works in the OPPOSITE direction of the displacement.

Hooke’s Law from a Graphical Point of View

x(m) Force(N)0 0

0.1 120.2 240.3 360.4 480.5 600.6 72

graph x vs.F a of Slope

kxFk

kxF

s

sSuppose we had the following data:

Force vs. Displacement y = 120x + 1E-14R2 = 1

0

10

20

30

40

50

60

70

80

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7

Displacement(Meters)

Forc

e(Ne

wto

ns)

k =120 N/m

ExampleA load of 50 N attached to a spring hanging vertically stretches the spring

5.0 cm. The spring is now placed horizontally on a table and stretched 11.0 cm. What force is required to stretch the spring this amount?

kkkxFs

)05.0(501000 N/m

s

s

s

FF

kxF)11.0)(1000(

110 N

Newton’s Second Law

The acceleration of an object is directly proportional to the NET FORCE and inversely proportional to the mass.

maFmFa

maFa

NETNET

NET

1 FFNET

Tips:•Draw an FBD•Resolve vectors into components•Write equations of motion by adding and subtracting vectors to find the NET FORCE. Always write larger force – smaller force.•Solve for any unknowns

Newton’s 2nd Law

A 10-kg box is being pulled across the table to the right by a rope with an applied force of 50N. Calculate the acceleration of the box if a 12 N frictional force acts upon it.

mg

FN Fa

Ff

2/8.3

101250

sma

a

maFFmaF

fa

Net

In which

direction, is this object accelerating?

The X direction!

So N.S.L. is worked out using the forces in the “x” direction only

Example

m1g

m2g

T

TFN

A mass, m1 = 3.00kg, is resting on a frictionless horizontal table is connected to a cable that passes over a pulley and then is fastened to a hanging mass, m2 = 11.0 kg as shown below. Find the acceleration of each mass and the tension in the cable.

amTamTgm

maFNet

1

22

2

21

2

122

122

212

/7.714

)8.9)(11()(

smmmgma

mmagmamamgmamamgm

Example (cont.)

amTamTgm

maFNet

1

22

NT 1.23)7.7)(3(

Where does the calculus fit in?

dtxdm

dtdvmamF

2

220.03)( tttv

There could be situations where you are given a displacement function or velocity function. The derivative will need to be taken once or twice in order to get the acceleration. Here is an example.

You are standing on a bathroom scale in an elevator in a tall building. Your mass is 72-kg. The elevator starts from rest and travels upward with a speed that varies with time according to:

When t = 4.0s , what is the reading on the bathroom scale (a.k.a. Force Normal)?

)4(40.03)4(

40.03)20.03( 2

a

tdt

ttddtdva

4.6 m/s/s

)6.4)(72()8.9)(72(N

NN

net

FmgmaFmamgF

maF

1036.8 N

TWO types of Friction• Static – Friction that keeps an object at rest and prevents it

from moving (no sliding)• Kinetic – Friction that acts during motion (two surfaces sliding)

Force of Friction• The Force of Friction is

directly related to the Normal Force.

Nkkf

Nssf

Nf

FF

FF

FF

friction oft coefficien alityproportion ofconstant

The coefficient of friction is a unitless constant that is specific to the material type and usually less than one.

ExampleA 1500 N crate is being pushed

across a level floor at a constant speed by a force F of 600 N at an angle of 20° below the horizontal as shown in the figure.

a) What is the coefficient of kinetic friction between the crate and the floor?

mg

FNFa

20

Ff

Fay

Fax

331.021.170582.563

21.17051500)20(sin600

1500sin

82.563)20(cos600cos

k

k

N

aayN

aaxf

Nkf

NF

FmgFF

NFFF

FF

ExampleIf the 600 N force is instead pulling the block

at an angle of 20° above the horizontal as shown in the figure, what will be the acceleration of the crate. Assume that the coefficient of friction is the same as found in (a)

mg

FN

Ff

20

Fa

Fax

Fay

2/883.0

1.15357.4288.5631.153)20sin6001500(331.020cos600

)sin(coscos

sma

aa

maFmgFmaFF

maFFmaF

aa

Na

fax

Net

Inclines

cosmg

sinmg

mg

FNFf

Tips•Rotate Axis•Break weight into components•Write equations of motion or equilibrium•Solve

Example

gmamTamgmTmaFNET

1111

m2

m1

Masses m1 = 4.00 kg and m2 = 9.00 kg are connected by a light string that passes over a frictionless pulley. As shown in the diagram, m1 is held at rest on the floor and m2 rests on a fixed incline of angle 40 degrees. The masses are released from rest, and m2 slides 1.00 m down the incline in 4 seconds. Determine (a) The acceleration of each mass (b) The coefficient of kinetic friction and (c) the tension in the string.

m1g

m2g

FNT

T

Ff

40

40 amTFgm f 22 )(sin m2gcos40

m2gsin40

Example

cossin

cossincossin

sin

)(sin

sin

2

2112

22112

21122

2112

2112

22

gmamgmamgm

gmamgmamgmamgmamgmgm

amgmamFgm

amgmamFgm

amTFgm

k

k

k

Nk

f

f

2

2

2

/125.0

)4(2101

21

sma

a

attvx ox

gmamTamgmTmaFNET

1111

amTFgm f 22 )(sin

NT 7.39)8.9(4)125(.4

235.057.67

125.12.395.07.56

k

Horizontally Launched ProjectilesTo analyze a projectile in 2 dimensions we need 2

equations. One for the “x” direction and one for the “y” direction. And for this we use kinematic #2.

212oxx v t at

oxx v t

Remember, the velocity is CONSTANT horizontally, so that means the acceleration is ZERO!

212y gt

Remember that since the projectile is launched horizontally, the INITIAL VERTICAL VELOCITY is equal to ZERO.

Horizontally Launched ProjectilesExample: A plane traveling with a

horizontal velocity of 100 m/s is 500 m above the ground. At some point the pilot decides to drop some supplies to designated target below. (a) How long is the drop in the air? (b) How far away from point where it was launched will it land?

What do I know?

What I want to know?

vox=100 m/s t = ?y = 500 m x = ?voy= 0 m/s

g = -9.8 m/s/s

2 2

2

1 1500 ( 9.8)2 2102.04

y gt t

t t

10.1 seconds(100)(10.1)oxx v t 1010 m

Vertically Launched ProjectilesThere are several

things you must consider when doing these types of projectiles besides using components. If it begins and ends at ground level, the “y” displacement is ZERO: y = 0

Vertically Launched ProjectilesYou will still use kinematic #2, but YOU MUST use

COMPONENTS in the equation.

vo

vox

voy

oxx v t 212oyy v t gt

cossin

ox o

oy o

v vv v

ExampleA place kicker kicks a football with a velocity of 20.0 m/s and at an

angle of 53 degrees.(a) How long is the ball in the air?(b) How far away does it land?(c) How high does it travel?

v o=20.0 m

/s

53

cos20cos53 12.04 /

sin

20sin 53 15.97 /

ox o

ox

oy o

oy

v vv m sv v

v m s

ExampleA place kicker kicks a

football with a velocity of 20.0 m/s and at an angle of 53 degrees.

(a) How long is the ball in the air?

What I know What I want to know

vox=12.04 m/s t = ?voy=15.97 m/s x = ?y = 0 ymax=?g = - 9.8 m/s/s

2 2

2

1 0 (15.97) 4.9215.97 4.9 15.97 4.9

oyy v t gt t t

t t tt

3.26 s

Example

A place kicker kicks a football with a velocity of 20.0 m/s and at an angle of 53 degrees.

(b) How far away does it land?

What I know What I want to know

vox=12.04 m/s t = 3.26 svoy=15.97 m/s x = ?y = 0 ymax=?g = - 9.8 m/s/s

(12.04)(3.26)oxx v t 39.24 m

ExampleA place kicker kicks a football

with a velocity of 20.0 m/s and at an angle of 53 degrees.

(c) How high does it travel?

CUT YOUR TIME IN HALF!

What I know What I want to know

vox=12.04 m/s t = 3.26 svoy=15.97 m/s x = 39.24 my = 0 ymax=?g = - 9.8 m/s/s

2

2

12

(15.97)(1.63) 4.9(1.63)

oyy v t gt

yy

13.01 m

A special case…What if the projectile was launched from the ground at an angle and did

not land at the same level height from where it started? In other words, what if you have a situation where the “y-displacement” DOES NOT equal zero?

Let's look at the second kinematic closely!

Assuming it is shot from the ground. We see we have one squared term variable, one regular term variable, and a constant number with no variable. What is this?

A QUADRATIC EQUATION!

Circular Motion and New’s 2nd Law

Recall that according to Newton’s Second Law, the acceleration is directly proportional to the Force. If this is true: ForcelCentripetaF

rmvFF

rvamaF

c

cNET

cNET

2

2

Since the acceleration and the force are directly related, the force must ALSO point towards the center. This is called CENTRIPETAL FORCE.

NOTE: The centripetal force is a NET FORCE. It could be represented by one or more forces. So NEVER draw it in an F.B.D.

Examples

rgv

rmvmg

rmvF

FF

N

cf

2

2

2

What is the minimum coefficient of static friction necessary to allow a penny to rotate along a 33 1/3 rpm record (diameter= 0.300 m), whenthe penny is placed at the outer edge of the record?

mg

FN

Ff

Top view

Side view 187.0)8.9)(15.0(

)524.0(

/524.080.1

)15.0(22

sec80.1555.0

sec1sec555.0

sec60min1*

min3.33

22

rgv

smTrv

Trevrev

revrev

c

ExamplesThe maximum tension that a 0.50 m

string can tolerate is 14 N. A 0.25-kg ball attached to this string is being whirled in a vertical circle. What is the maximum speed the ball can have (a) the top of the circle, (b)at the bottom of the circle?

mgT

smvmmgTrv

mvmgTrrmvmgT

rmvmaFF ccNET

/74.525.0

))8.9)(25.0(14(5.0)(

)( 22

2

ExamplesAt the bottom?

smvmmgTrv

mvmgTrrmvmgT

rmvmaFF ccNET

/81.425.0

))8.9)(25.0(14(5.0)(

)( 22

2

mg

T

N.L.o.G – Putting it all together

221

2227

221

1067.6

Constant nalGravitatio UniversalGalityproportion ofconstant

rmmGF

kgNmxG

GrmmF

g

g

earth eLEAVING th areyou when thisUse

earth on the areyou when thisUse

221

rmmGF

mgF

g

g

Kepler’s 3rd Law – The Law of Periods"The square of the period of any planet is proportional to

the cube of the semi major axis of its orbit."

Gravitational forces are centripetal, thus we can set them equal to each other!

Since we are moving in a circle we can substitute the appropriate velocity formula!

Using algebra, you can see that everything in the parenthesis is CONSTANT. Thus theproportionality holds true!

The expression in the RED circle derived by setting the centripetal force equal to the gravitational force is called ORBITAL SPEED.

ExampleA 2-kg sliding puck whose initial velocity magnitude is v1 = 10 m/s strikes a wall at a 30 degree angle and bounces off. If it leaves the wall with a velocity magnitude of v2 = 10 m/s, and if the collision takes a total of 0.02 seconds to complete, what was the average force applied to the puck by thewall?

There is something you need to consider:Momentum is a VECTOR!!!

Let’s look at this problem using a X-Y axis for reference

Example cont’

If we did the same thing for the Y direction we would discover that the Force Net is equal to ZERO!

The temptation is to treat momentum as a SCALAR...DO NOT DO THIS! SIGNS COUNT!

Momentum is conserved!The Law of Conservation of Momentum: “In the absence

of an unbalanced external force, the total momentum before the collision is equal to the total momentum after the collision.”

smkgpsmkgpsmkgp

smkgp

smkgp

smkgmvp

total

car

truck

totalo

caro

otrucko

/*3300/*18005.4*400

/*15003*500

/*3300

/*800)2)(400(

/*2500)5)(500(

)(

)(

)(

Several Types of collisionsSometimes objects stick together or blow apart. In this

case, momentum is ALWAYS conserved.

2211)(

022011

2211022011

vmvmvmvmvmvmvmvmvmvm

pp

totalototal

totaltotal

afterbefore

When 2 objects collide and DON’T stick

When 2 objects collide and stick together

When 1 object breaks into 2 objects

Elastic Collision = Kinetic Energy is ConservedInelastic Collision = Kinetic Energy is NOT Conserved

WorkThe VERTICAL component of the force DOES NOT cause the block to move the right. The energy imparted to the box is evident by its motion to the right. Therefore ONLY the HORIZONTAL COMPONENT of the force actually does WORK.

When the FORCE and DISPLACEMENT are in the SAME DIRECTION you get a POSITIVE WORK VALUE. The ANGLE between the force and displacement is ZERO degrees. What happens when you put this in for the COSINE?

When the FORCE and DISPLACEMENT are in the OPPOSITE direction, yet still on the same axis, you get a NEGATIVE WORK VALUE. This negative doesn't mean the direction!!!! IT simply means that the force and displacement oppose each other. The ANGLE between the force and displacement in this case is 180 degrees. What happens when you put this in for the COSINE?When the FORCE and DISPLACEMENT are PERPENDICULAR, you get NO WORK!!! The ANGLE between the force and displacement in this case is 90 degrees. What happens when you put this in for the COSINE?

ExamplecosrFW

rFW

A box of mass m = 2.0 kg is moving over a frictional floor ( uk = 0.3) has a force whosemagnitude is F = 25 N applied to it at an angle of 30 degrees, as shown to the left. The box is observed to move 16 meters in the horizontal direction before falling off the table.

a) How much work does F do before taking the plunge?

JWNmW

W

rFWrFW

4.3464.346

30cos1625

cos

Example cont’What if we had done this in UNIT VECTOR notation?

JWNmW

W

rFrFWjiF

yyxx

4.3464.346

)05.12()1665.21(

)()(

ˆ5.12ˆ65.21

Example cont’

JW

FW

rFWrFW

N

0

90cos16

cos

Fn

How much work does the FORCE NORMAL do and Why?

There is NO WORK since “F” and “r” are perpendicular.

Ff

How much does the internal energy of the system increase?

34.08 J

Elastic Potential Energy

The graph of F vs.x for a spring that is IDEAL in nature will always produce a line with a positive linear slope. Thus the area under the line will always be represented as a triangle.

NOTE: Keep in mind that this can be applied to WORK or can be conserved with any other type of energy.

Elastic potential energy

20

2

00

21|

2|

)(

)()(

kxUWxkW

dxxkdxkxW

dxkxdxxFW

springxx

x

x

x

x

Elastic “potential” energy is a fitting term as springs STORE energy when there are elongated or compressed.

Energy is CONSERVED!

afterbefore

oo

oo

oo

EnergyEnergyUKUKUUKKUUKK

UKW

DD)(

ExampleA 2.0 m pendulum is released from rest when the support

string is at an angle of 25 degrees with the vertical. What is the speed of the bob at the bottom of the string?

L Lcos

h

h = L – Lcosh = 2-2cosh = 0.187 m

EB = EA

UO = Kmgho = 1/2mv2

gho = 1/2v2

1.83 = v2

1.35 m/s = v

Ballistic Pendulum

• Event includes a collision during which momentum conservation is the best model and post-collision motion during which energy conservation is the best model.

• Examples: last three problems on HW, Workshop 14,