PHYSICS 149: Lecture 14

PHYSICS 149: Lecture 14• Chapter 5: Circular Motion

5 4 Circular Orbits of Satellites and Planets– 5.4 Circular Orbits of Satellites and Planets

• Chapter 6: Conservation of EnergyChapter 6: Conservation of Energy– 6.1 The Law of Conservation of Energy– 6.2 Work Done by a Constant Forcey

Lecture 14 Purdue University, Physics 149 1

ILQ 1Two children ride on a merry-go-round. Bob is 2 m from the axis of rotation and Mary is 4 m from thefrom the axis of rotation and Mary is 4 m from the axis. Which is true:

A) Mary has larger speed and acceleration but the sameangular speed

B) Bob has larger speed, acceleration, and angular speedC) M h l d l i d l dC) Mary has larger speed, acceleration, and angular speed

( )22rωRecall: v = ω r

Lecture 14 Purdue University, Physics 149

( ) 2ca r

rω= =Recall: v ω r

2

ILQ 2A spider sits on a turntable that is rotating at a constant 33 rpm The acceleration a of the spider isconstant 33 rpm. The acceleration a of the spider is

A) nonzero and independent of the location of the spider onA) nonzero and independent of the location of the spider on the turntable.B) greater the closer the spider is to the central axis.C) greater the farther the spider is from the central axis.D) zero.


Kepler’s Laws

Elliptical Equal areas in T2 = R3porbits…

qequal time

T R


These were empirical laws4

Kepler’s First Law of Planetary Motion

• The orbit of each planet around the Sun is an ellipse with the Sun at one focus.ellipse with the Sun at one focus.


Kepler’s Second Law of Planetary Motion• As a planet moves around its orbit, a line

drawn from a planet to the Sun sweeps out l i l ti i t lequal areas in equal time intervals.

This means that a planet travels faster when it is nearer to the Sun and slower when it is farther from the Sun.


Kepler’s Third Law of Planetary Motion

• The square of the orbital period is proportional to the cube of the average distance from the planetthe cube of the average distance from the planet to the Sun.

T 2 ∝ r 3

This means that more distant planets orbit the S t l dSun at slower average speeds.


ILQThe figure below shows the orbit of a comet about the sun. The comet has the greatest velocity when t litraveling

A) from A to B.B) from B to C.C) from C to D.D) from D to E.


Circular Orbits of Planets and Satellites• Newton’s Law of Universal Gravitation:• In this case, the gravitational force (e.g., acting

1 22

Gm mFd

=

In this case, the gravitational force (e.g., acting on a planet due to Sun, or acting on a satellite due to Earth) is the centripetal force.– For example, if Earth’s orbit is a perfect circle,

Earth.)andSunbetweendisttheisr(where2

2∑ ==vMMGMF Earth

EarthSunr )(2∑ rr Earthr

Tr

rGMv Sun π2

==

Law)3rds(Kepler'4 32

2 rGM

TSun

π=

We can get similar results for Sun & a planet, Earth & Moon, Earth & a satellite, and so on.


Satellites and Planets2

2earth

r rGmM mvF ma

r r= = =∑

earthGMvr

=

- Speed is independent of mass of satellite- Satellites in lower orbits have greater speeds

Geostationary orbits:A circular orbit in Earth’s equatorialA circular orbit in Earth s equatorial plane whose period is equal to Earth’s rotational period.

35 786 km above ground


~35,786 km above ground

10

Satellites in Orbit• A satellite is held in orbit by the force of gravity of the

Earth. In this case, the centripetal force is simply gravity., p p y g y• The speed and orbital period of the satellite can be

determined by3

here r the the center to center distance

earthGMvr

=3

2earth

rTGM

π=

where r the the center-to-center distance– Speed is independent of mass of satellite– Satellites in lower orbits have greater speeds

• Geostationary orbits– A circular orbit in Earth’s equatorial plane whose period is equal to

Earth’s rotational period.


p– ~35,786 km above ground

11

Weightlessness• Inside the Space Shuttle, the astronauts experience

apparent weightlessness.• The force of gravity on the Earth pulls on the astronauts to

keep them in a circular orbit around the Earth, so we can’t accurately say there is “no gravity” there.y y g y

• Both the astronauts and the Space Shuttle are in uniform circular motion, and are continually accelerating towards the Earth The are both “falling” at the same ratethe Earth. The are both falling at the same rate.

• The apparent weight (how heavy you feel) is defined as the magnitude of the normal force acting on the body:

where g is the local gravitational field strength:

'W m g a= −

g =GM


where g is the local gravitational field strength: g

r2

12

Astronaut’s Weight• Why do astronauts appear

weightless?C i ht th t ti• Compare weight on the space station with weight on earth.

22 2

2

( ) 6400 0.84( ) 7000

E

spacestation E E

EE th E

GM mW r h r

GM mW r h+ ⎛ ⎞= = = =⎜ ⎟+ ⎝ ⎠

2( ) 7000

( )EEarth E

E

W r hr

+ ⎝ ⎠

• Weight on space station is about 80%.• Astronaut has the same acceleration of the space

station

Purdue University, Physics 149

' '( ) 0W m g a if g a then W= − = =Lecture 14 13

Example: Artificial Gravity

A little girl is having fun on a swing ofA little girl is having fun on a swing of length 8.0 m. If her weight is 180 N, fi d h t i ht t th l tfind her apparent weight at the lowest point, where her speed is 9.7 m/s.


Example: Artificial Gravityv=9.7 m/sR=8 mR=8 mW=180NW’=?

vW ?

2mvT Wr

− = T2 2

1

rmv vT W W

r rg⎛ ⎞

= + = + =⎜ ⎟⎝ ⎠

W

2 2 2

2

9.7 /180 1 4008 9 8 /

r rg

m sN N W

⎝ ⎠⎛ ⎞

′+ = =⎜ ⎟⎝ ⎠

Purdue University, Physics 149

28 9.8 /m m s⎜ ⎟⎝ ⎠

Lecture 14 15

The Law of Conservation of Energy

• The Law of Conservation of Energy• The Law of Conservation of Energy– The total energy in the universe is unchanged by any

physical process:physical process:total energy before = total energy after.

• Conservation law– A physical law that identifies a quantity that does not

change with time.


Energy is Conserved• Energy is “Conserved” meaning it can not be

created nor destroyedcreated nor destroyed – Can change form– Can be transferred

• Total Energy of an isolated system does not gy ychange with time

• This is a BIG deal!


Energy• Forms

Kinetic Energy Motion– Kinetic Energy Motion– Potential Energy Stored– HeatHeat– Mass (E=mc2)

• Energy is a scalar quantity (not a vector quantity).

• Units: Joules = 1 N⋅m = kg m2 / s2

• Force times distance


Meaning of Conservation of Energy

• Energy can be converted from one form to another, or transferred from one place to another.p

• When we produce or generate energy, we are NOTcreating any new energy; we’re just “converting” energy from one form into another formfrom one form into another form.

• For example, when we press a car’s brakes, the kinetic energy of the car is converted into heat energy due to the gy gyfriction between the brakes and road.– The total energy of the car-brakes-road-atmosphere system is the

same during the eventsame during the event.– The energy of the car “alone” (an isolated system) is not conserved.

Its kinetic energy is decreased.

D i “ k” i l t d t h it• Doing “work” on an isolated system may change its “energy,” but the total energy in the universe is unchanged.


Definition of “Work” in Physics

• Work is a scalar quantity (not a vector quantity).• Units: J (Joule), N⋅m, kg⋅m2/s2, etc. .

– Unit conversion: 1 J = 1 N⋅m = 1 kg⋅m2/s2Unit conversion: 1 J 1 N m 1 kg m /s• Work is denoted by W (not to be confused by weight W).


Work: Energy Transfer due to Force

• Force to lift trunk at constant speed– Case a T – mg = 0 T = mgCase a Ta mg 0 T mg– Case b 2Tb - mg =0 or T = ½ mg

• But in case b, trunk only moves ½ distance you pull ropey p p

• F × distance is the same in both!F × distance is the same in both!

TaTbTb

W F d (θ)Lecture 14 Purdue University, Physics 149

mg mgW = F d cos(θ)

21

ILQHilda holds a gardening book of weight 10 N at a height of 1 0 m above her patio for 50 s How muchheight of 1.0 m above her patio for 50 s. How much work does she do on the book during that 50 s?

A) 10 J B) No work is doneB) No work is done C) 100 J D) 500 JD) 500 J


Example• You pull a 30 N chest 5 meters across the floor at a

constant speed by applying a constant force of 50 N at an p y pp y gangle of 20 degrees. How much work have you (tension) done?

50N

– WTension = F Δr cos θ= (50 N)⋅(5 m)⋅(cos 20º)

=50N

(50 N) (5 m) (cos 20 )= 235 J

– What are the works done by friction, normal force, and gravity?


Work by Constant Force• Only component of force parallel to direction of

motion does work!F– W = F Δr cos θ F

θF

θ (θ)Δr WF > 0: 0< θ < 90 : cos(θ) > 0F

W 0 θ 90 (θ) 0Δr

WF = 0: θ =90 : cos(θ) =0

FW 0 90 θ 270 (θ) 0

ΔrWF < 0: 90< θ < 270 : cos(θ) < 0

Δr W > 0 0< θ < 90 (θ) > 0


F WF > 0: 0< θ < 90 : cos(θ) > 0

24

Examples

W 0 W 0 W > 0W = 0 W = 0 W > 0

W > 0 A t PW = 0W = 0 W < 0: P to A

W > 0: A to PConsider the sign of cosθ !Lecture 14 Purdue University, Physics 149 25

ILQDoes the Earth do work on the Moon? Assume that the Moon’s orbit is a perfect circlethe Moon s orbit is a perfect circle.

A) Yes it does positive workA) Yes, it does positive work.B) Yes, it does negative work.C) No it does no work at allC) No, it does no work at all.


ILQ: Ball TossYou toss a ball in the air. What is the work done by gravity as the ballWhat is the work done by gravity as the ball goes up?

A) Positive B) Negative C) Zero

What is the work done by gravity as the ball goes down?

A) Positive B) Negative C) Zero


ILQA box with mass m1 is being pulled up a roughincline by a rope-pulley-weight system How manyincline by a rope-pulley-weight system. How many forces are doing work on the box?

A) One forceA) One forceB) Two forcesC) Three forcesC) Three forcesD) Four forcesE) No forces are doing workE) No forces are doing work


ILQA box with mass m1 is being pulled up a roughincline by a rope-pulley-weight system. The works y p p y g ydone by tension, friction, and gravity are:

A) Positive, negative, & 0B) Negative, positive, & 0C) &C) Positive, negative, & negativeD) Negative, positive, & positiveE) N f d i kE) No forces are doing work


Work by Constant Force• You pull a 30 N chest 5 meters across the floor at a constant speed

by applying a constant force of 50 N at an angle of 20 degrees. How h k h (t i ) d ? H h k h th f i timuch work have you (tension) done? How much work has the friction

done?– WTension = F Δr cos θ

50N= (50 N)⋅(5 m)⋅(cos 20º)= 235 J

=50N

– constant speed net force = ΣFx = 0ΣFx = T⋅cos θ – fk = 50⋅cos 20º – fk = 0

f = 47 Nfk = 47 N

– Wfriction = fk Δr cos θf Δr

= (47 N)⋅(5 m)⋅(cos 180º)= –235 J


180

Where did the Energy go?• Example: You pull a 30 N chest 5 meters across the floor at a

constant speed, by applying a force of 50 N at an angle of 30 degrees. • How much work did gravity do?

90

ΔrW = F Δr cos θ = 30 × 5 cos(90)

• How much work did friction do?mg

90 = 30 × 5 cos(90) = 0

x-direction: ΣF = maT cos(30) – f = 0f = T cos(30)f T cos(30)

f ΔrW = F Δr cos θ

= 50 cos(30) × 5 cos(180)


180

( ) ( ) = −217 Joules

31

Total Work

• When several forces act on an object the “total”• When several forces act on an object, the total work is the sum of the work done by each force individually:individually:


Dissipative Forces• A dissipative force is opposite to the direction of

motionmotion.

• It always tends to reduce the speed of the object• It always tends to reduce the speed of the object, and always does negative work.

• The work they do shows up as thermal energy (heat)(heat).

• Example includes friction and air resistance• Example includes friction and air resistance.


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PHYSICS 149: Lecture 14

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