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PHYS 1110 Lecture 2 Professor Stephen Thornton August 30, 2012

PHYS 1110 Lecture 2 Professor Stephen Thornton August 30, 2012

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Page 1: PHYS 1110 Lecture 2 Professor Stephen Thornton August 30, 2012

PHYS 1110

Lecture 2

Professor Stephen Thornton

August 30, 2012

Page 2: PHYS 1110 Lecture 2 Professor Stephen Thornton August 30, 2012

An IssueOther classes are also using iClickers nearby. Therefore, we will need to change frequencies.iClicker 1 (old one): Press and hold the On/Off power button on the remote until the blue Power light begins flashing. Then press BB. A green Vote Status light on your remote will indicate that you have successfully reset the remote frequency.

iClicker 2: Press and hold the On/Off power button on the remote until the BB on the LCD begins flashing.

Page 3: PHYS 1110 Lecture 2 Professor Stephen Thornton August 30, 2012

Reading Quiz:What is the total displacement from start to finish? A) - 2 m

B) +2 mC) +3 mD) +7 mE) +10 m

finish

start

Page 4: PHYS 1110 Lecture 2 Professor Stephen Thornton August 30, 2012

Reading Quiz:What is the total displacement from start to finish? A) - 2 m

B) +2 mC) +3 mD) +7 mE) +10 m

finish

start

Page 5: PHYS 1110 Lecture 2 Professor Stephen Thornton August 30, 2012

How can we change things?

• Be more energy efficient. Could reduce electricity need by 15% by 2020, 30% by 2030.

• Develop more renewable energy.

• Energy policy like tax credits, policy changes.

• Carbon capture and storage in order to use fossil fuels.

• Revolutionary nuclear reactors that are simpler and safer. They probably are already imagined.

Page 6: PHYS 1110 Lecture 2 Professor Stephen Thornton August 30, 2012

Hydraulic fracturing video: http://www.oerb.com/Default.aspx?tabid=242 - man talking

http://www.youtube.com/watch?v=kv3cQngRPmw – watered down, woman talking

Page 7: PHYS 1110 Lecture 2 Professor Stephen Thornton August 30, 2012

Solar and wind energy did not even show up in 2008, < 1% in US.

By mid-2012 wind energy had grown to 50 GW (4.5%).

Solar is still way behind (~0.2%), but growing by 30% a year. By some estimates it is as much as 30 GW or 3%.

Wind and solar energy represent the greatest potential increase of renewable energy.

There are 104 nuclear reactor power plants operating in the United States, 4 in Virginia, which generates 38% of its power.

Page 8: PHYS 1110 Lecture 2 Professor Stephen Thornton August 30, 2012

Growth of Fuel Inputs to World Power Generation

Page 9: PHYS 1110 Lecture 2 Professor Stephen Thornton August 30, 2012

Estimates of Levelized Cost of Electricity for New Baseload and Intermittent Sources for 2020. Dashed is actual 2007 price; shaded is range in 2007.

Page 10: PHYS 1110 Lecture 2 Professor Stephen Thornton August 30, 2012

And then there is the transportation energy problem.

The United States is committed to ethanol. It has been growing for 25 years and is now a political issue. US law requires us to use 10% ethanol in our gas – 15% in some places. Ethanol use is required to increase every year.

Biofuel generation has not worked, but there is progress.

Page 11: PHYS 1110 Lecture 2 Professor Stephen Thornton August 30, 2012

The electrical distribution system is a huge problem. At least 10% of our electricity is lost. It is a patchwork and archaic system.

Page 12: PHYS 1110 Lecture 2 Professor Stephen Thornton August 30, 2012
Page 13: PHYS 1110 Lecture 2 Professor Stephen Thornton August 30, 2012

Quiz:Which of the following energies had the greatest increase in the last few years in the US?

A) Concentrated solar powerB) WindC) BiomassD) GeothermalE) Hydropower

Page 14: PHYS 1110 Lecture 2 Professor Stephen Thornton August 30, 2012

Quiz:Which of the following had the greatest increase in the last few years in the US?

A) Concentrated solar powerB) WindC) BiomassD) GeothermalE) Hydropower

Page 15: PHYS 1110 Lecture 2 Professor Stephen Thornton August 30, 2012

One-Dimensional Coordinate System

Page 16: PHYS 1110 Lecture 2 Professor Stephen Thornton August 30, 2012

Distance = total length traveled

Example: You can run 2 m/s. How far can you run in 4 s?

Answer: 2 m/s x 4 s = 8 m

Page 17: PHYS 1110 Lecture 2 Professor Stephen Thornton August 30, 2012

Displacement

Definition:

displacement = change in position

= final position – initial position

Δx = xf - xi

Page 18: PHYS 1110 Lecture 2 Professor Stephen Thornton August 30, 2012

One-Dimensional Motion Along the x-axis – do quiz

What is the total distance traveled (0-4s)?

What is the displacement? – Reading Quiz

start

finish

Page 19: PHYS 1110 Lecture 2 Professor Stephen Thornton August 30, 2012

What is the total distance traveled from start to finish?

A) - 4 mB) +4 mC) +3 mD) +8 mE( +10 m

start

finish

Page 20: PHYS 1110 Lecture 2 Professor Stephen Thornton August 30, 2012

What is the total distance traveled from start to finish?

A) - 4 mB) +4 mC) +3 mD) +8 mE( +10 m

start

finish

Page 21: PHYS 1110 Lecture 2 Professor Stephen Thornton August 30, 2012

Average Speed

distanceaverage speed =

elapsed time

Note that this is always a positive number.

Page 22: PHYS 1110 Lecture 2 Professor Stephen Thornton August 30, 2012

Average Velocity

Velocity is different than speed, because velocity is a vector.

displacementaverage velocity =

elapsed time

f iav

f i

x xxv

t t t

Page 23: PHYS 1110 Lecture 2 Professor Stephen Thornton August 30, 2012

One-Dimensional Motion Along the x Axis

What is average velocity (0 – 4 s)?

( 1m) - (1m) 2m0.5 m/s

4s-0s 4savv

start

finish

Page 24: PHYS 1110 Lecture 2 Professor Stephen Thornton August 30, 2012

Motion Along the x axis represented with an x-versus-t graph

Page 25: PHYS 1110 Lecture 2 Professor Stephen Thornton August 30, 2012

Average Velocity on an x-Versus-t Graph

trajectory

Page 26: PHYS 1110 Lecture 2 Professor Stephen Thornton August 30, 2012

Instantaneous Velocity

0

dlim

dt

x xv

t t

In this way we can find the velocity at any particular instant of time.

Page 27: PHYS 1110 Lecture 2 Professor Stephen Thornton August 30, 2012

What is the instantaneous velocity at t = 1 s?

Page 28: PHYS 1110 Lecture 2 Professor Stephen Thornton August 30, 2012

Graphical Interpretation of Average and Instantaneous Velocity

Page 29: PHYS 1110 Lecture 2 Professor Stephen Thornton August 30, 2012

AccelerationJust like velocity is given by the rate of change of position with respect to time, the acceleration is given by the rate of change of velocity with respect to time.

We are still dealing with one-dimensional motion, so vector direction is simple.

Page 30: PHYS 1110 Lecture 2 Professor Stephen Thornton August 30, 2012

Average Acceleration

f iav

f i

v vva

t t t

We must be very careful with units. What are they?

m/s2

Page 31: PHYS 1110 Lecture 2 Professor Stephen Thornton August 30, 2012

Instantaneous Acceleration

0

dlim

dt

v va

t t

0

dlim

dt

x xv

t t

Similarity between velocity and acceleration is clear.

Page 32: PHYS 1110 Lecture 2 Professor Stephen Thornton August 30, 2012

Graphical Interpretation of Average and Instantaneous Acceleration

If you don’t remember about tangents, please review!

Page 33: PHYS 1110 Lecture 2 Professor Stephen Thornton August 30, 2012

We use signs to denote the directions of both velocity and acceleration along a particular axis. x

When v is +, motion is to right. When v is -, motion is to left. When motion is to the right, and a is +, then object speeds up (accelerates) to the right. When motion is to the left, and a is +, then object is slowing down and will eventually turn to the right. (not shown here)

Page 34: PHYS 1110 Lecture 2 Professor Stephen Thornton August 30, 2012

Cars Accelerating or Decelerating

speeding up

speeding upslowing down

slowing down

Much easier to see what is happening, when we draw a picture.

Page 35: PHYS 1110 Lecture 2 Professor Stephen Thornton August 30, 2012

Conceptual Quiz: The graph shows position as a function of time for two trains running on parallel tracks. Which of the following is true? A) At time tB, both trains have the same velocity.B) Both trains speed up all the time. C) Both trains have the same velocity at some time before tB.D) Somewhere on the graph, both trains have the same acceleration. time

Page 36: PHYS 1110 Lecture 2 Professor Stephen Thornton August 30, 2012

Conceptual Quiz: The graph shows position as a function of time for two trains running on parallel tracks. Which of the following is true? A) At time tB, both trains have the same velocity.B) Both trains speed up all the time. C) Both trains have the same velocity at some time before tB.D) Somewhere on the graph, both trains have the same acceleration. time

Page 37: PHYS 1110 Lecture 2 Professor Stephen Thornton August 30, 2012

Motion with Constant AccelerationIf the acceleration is constant, then we have

where v = v0 at t = 0. This result is easy to show from our definition of a.

0v v at

Page 38: PHYS 1110 Lecture 2 Professor Stephen Thornton August 30, 2012

The Average Velocity

constant accelerationiv

fv

Page 39: PHYS 1110 Lecture 2 Professor Stephen Thornton August 30, 2012

The Average Velocity

nonconstant acceleration

Page 40: PHYS 1110 Lecture 2 Professor Stephen Thornton August 30, 2012

Constant accelerationAnother important result:

Note that I have used vi and vf , which is more general than using v0 and v, because we may want to find the average between some initial and final position other than v0 and a general v.

Let’s determine some important equations.

av 0

1 1( )

2 2 i fv v v v v

Page 41: PHYS 1110 Lecture 2 Professor Stephen Thornton August 30, 2012

0av

0av

av 0

0 av

0

or

solve for : ****

f i

f i

x xx x xv

t t t t

x xv

tv t x x

x x x v t

Insert our previous result for vav

0 0

1( )

2x x v v t

Only for constant acceleration!!!!!

Page 42: PHYS 1110 Lecture 2 Professor Stephen Thornton August 30, 2012

0 0

1( )

2x x v v t

This is a very important equation. It relates the position x to the velocity v as a function of time t.

But we also know the relationship between velocity v and acceleration a. It was

0 constantv v a t

Page 43: PHYS 1110 Lecture 2 Professor Stephen Thornton August 30, 2012

0 0

0 0 0 constant

20 0 constant

1( )

21

( )2

1

2

x x v v t

x x v v a t t

x x v t a t

Page 44: PHYS 1110 Lecture 2 Professor Stephen Thornton August 30, 2012

0

0constant

constant

0 0

00 0

constant

2 20

0constant

solve for , is constant

remember

1We had ( )

2Substitute in for from above

1( )

2

2

v v at t a

v vt a a

a

x x v v t

t

v vx x v v

a

v vx x

a

Page 45: PHYS 1110 Lecture 2 Professor Stephen Thornton August 30, 2012

0

av 0

20 0

2 20 0

1( )

21

2

2 ( )

v v at

v v v

x x v t at

v v a x x

= +

= +

= + +

= + -

Four important equations

Page 46: PHYS 1110 Lecture 2 Professor Stephen Thornton August 30, 2012

0 0

av 0

20 0 0 0

2 20 0

( )

1( )

21

( ) ( )2

2 ( )

v v a t t

v v v

x x v t t a t t

v v a x x

= + -

= +

= + - + -

= + -

Four important equations with initial time t0

Page 47: PHYS 1110 Lecture 2 Professor Stephen Thornton August 30, 2012

Freely falling objects

Most important example of constant acceleration; a = ± g = 9.81 m/s2

Do demo: Paper and racquetball

Nickel and feather

Galileo – father of physics

Page 48: PHYS 1110 Lecture 2 Professor Stephen Thornton August 30, 2012

We let x be downward. Look at our previous equation:

x

20 0 constant

1

2x x v t a t

Let’s release an object at x0 = 0 at t = 0. We then also have v0 = 0. a = g

21

2x gt

Note that g is always positive. Here x is down.

Page 49: PHYS 1110 Lecture 2 Professor Stephen Thornton August 30, 2012

Our previous equations for v become

0 0v v at gt gt 2 2 2

00

2

02 2

which can be rewritten as

2 or 2

v v vx x

a g

v gx v gx

These are useful equations. Drop from rest.

Page 50: PHYS 1110 Lecture 2 Professor Stephen Thornton August 30, 2012

0 0

av 0

20 0 0 0

2 20 0

( )

1( )

21

( ) ( )2

2 ( )

v v g t t

v v v

y y v t t g t t

v v g y y

= - -

= +

= + - - -

= - -

Now use acceleration of gravity, with . Note y is up.

y

a g=-

Page 51: PHYS 1110 Lecture 2 Professor Stephen Thornton August 30, 2012

Free fall from restx = 14.7 m9

x=4.91 mx = 24.5 m

x = 34.4 m 21

2

v gt

x gt

x = 4.9 m

Page 52: PHYS 1110 Lecture 2 Professor Stephen Thornton August 30, 2012

So what would happen if we dropped a rope that had masses at equal intervals?

Do demo. Free fall

What would happen if we dropped a rope that had masses spaced out as t2?

Do demo. Free fall

Page 53: PHYS 1110 Lecture 2 Professor Stephen Thornton August 30, 2012

What happens if we throw a ball up?

We throw a ball up at x = 0 with speed v0.

What is its speed when it returns? v0

How long does it take to return? 2v0/g

How can we determine these numbers?

The equations we have determined must tell us these answers!

x = 0

Page 54: PHYS 1110 Lecture 2 Professor Stephen Thornton August 30, 2012

Conceptual Quiz: Throw ball up.Initial speed = v0. Round trip time is 2v0/g. What is minimum speed?

A) 0

B) -v0

C) v0

D) -2v0

E) 2v0

Page 55: PHYS 1110 Lecture 2 Professor Stephen Thornton August 30, 2012

Conceptual Quiz: Throw ball up. Initial speed = v0. Round trip time is 2v0/g. What is minimum speed?

A) 0

B) -v0

C) v0

D) -2v0

E) 2v0

Page 56: PHYS 1110 Lecture 2 Professor Stephen Thornton August 30, 2012

Conceptual Quiz: Throw ball up.Initial speed = v0. Round trip time is 2v0/g. What is time when minimum speed is reached?

A) 0

B) v0/g

C) 2v0/g

D) Can not be determined

Page 57: PHYS 1110 Lecture 2 Professor Stephen Thornton August 30, 2012

Conceptual Quiz: Throw ball up.Initial speed = v0. Round trip time is 2v0/g. What is time when minimum speed is reached?

A) 0

B) v0/g

C) 2v0/g

D) Can not be determined

Page 58: PHYS 1110 Lecture 2 Professor Stephen Thornton August 30, 2012

Review of Vectors

A scalar is a number with units. It can be positive, negative, or zero.

A vector has both a magnitude and direction.

We will put an arrow over a quantity that is a vector. Sometimes a vector is in boldface.

Page 59: PHYS 1110 Lecture 2 Professor Stephen Thornton August 30, 2012

Directions to the library

3 blocks west, 3 blocks north.

Start

Page 60: PHYS 1110 Lecture 2 Professor Stephen Thornton August 30, 2012

2 2 1

cos sin

tan

x y

yx y

x

A A A A

AA A A

A

Page 61: PHYS 1110 Lecture 2 Professor Stephen Thornton August 30, 2012

The Sum of Two Vectors

We can add vectors.

C = A + B

Page 62: PHYS 1110 Lecture 2 Professor Stephen Thornton August 30, 2012

Component Addition of Vectors

Page 63: PHYS 1110 Lecture 2 Professor Stephen Thornton August 30, 2012

Unit Vectors

More common to use ˆ ˆi and j or just i, j or ,

ˆ ˆthan x, y.i j

Page 64: PHYS 1110 Lecture 2 Professor Stephen Thornton August 30, 2012

Multiplying a Vector by a Scalar

We can multiply a vector by a scalar.

Page 65: PHYS 1110 Lecture 2 Professor Stephen Thornton August 30, 2012

Vector Component Use

Unit vectors make vector addition and subtraction reasonably easy.

ˆ ˆ ˆ ˆA 3i + 4j B 2i -2j

ˆ ˆ ˆ ˆ ˆ ˆA B (3i+2i) (4j 2j) = 5i 2 j

ˆ ˆ ˆ ˆ ˆ ˆA - B (3i+4j) (2i 2j) i 6 j

Page 66: PHYS 1110 Lecture 2 Professor Stephen Thornton August 30, 2012

Good review of vector use:

http://www.physics.uoguelph.ca/tutorials/vectors/vectors.html